Searching for a Physics Tutor? Contact Kumar Sir for CBSE, NEET, JEE Main, JEE Advanced, IB, IGCSE and A-Level Physics.

Phone: +91-9958461445 | Email: kumarsirphysics@gmail.com | Website: kumarphysicsclasses.com

WhatsApp
F2
CLASS 11 PHYSICS • LAWS OF MOTION

Newtons Second Law and Momentum

Master momentum, force, impulse, F = ma, variable force graphs and conservation of linear momentum.

CBSENEETJEE MainJEE AdvancedIBIGCSEA-Level

Section 1

Momentum

Momentum is the quantity of motion possessed by a body. It is defined as mass multiplied by velocity.

p = mv

Momentum is a vector because velocity is a vector. Its direction is exactly the same as velocity. SI unit is kg m s-1 and dimensions are [M L T-1].

DefinitionVectorSI unitDimensionsDirection
p = mvMomentum direction = velocity direction

Section 2

Force

Force is an interaction that changes or tends to change momentum. External force comes from outside the chosen system, while internal force acts between parts of the system.

  • Contact force: normal, friction, tension, applied push.
  • Non-contact force: gravitational, electric, magnetic.
External forceAcceleration

Sections 3 to 5

Rate of Change of Momentum, Newton's Second Law and F = ma

Rate of Change

Newton's second law defines force as the rate of change of momentum.

F = dp/dt

Constant Mass

If mass is constant, p = mv, so dp/dt = m dv/dt = ma.

F = ma

Physical Meaning

Greater force produces greater acceleration. For the same force, a larger mass gives smaller acceleration.

Shopping trolleyCricket ballFootballCar accelerationUnit: newton

Sections 6 and 7

Impulse and Impulse-Momentum Theorem

Impulse is force acting for a short time interval. For constant force, J = FΔt. For variable force, J = ∫F dt, equal to area under the force-time graph.

J = FΔt
J = ∫F dt

The impulse-momentum theorem states that impulse equals change in momentum.

J = Δp = p2 − p1

Applications include cricket catch, karate strike, airbag and jumping on sand.

Force FTime tImpulse = area under F-t curveJ = ∫F dt

New Section

Conservation of Linear Momentum

Conservation of linear momentum means the total vector momentum of an isolated system remains constant if the net external force on the system is zero.

Internal forces cannot change total momentum of the system because they occur in equal and opposite pairs. Momentum conservation is vectorial, so directions and signs are essential.

pinitial = pfinal
m1u1 + m2u2 = m1v1 + m2v2
Isolated systemNet external force zeroInternal forces cancelVector nature
Before collisionAfter collisionΣp before = Σp after

Two bodies before collision

Two bodies before collis

This diagram represents momentum before and after interaction. Choose a positive direction and apply vector conservation.

Two bodies after collision

Two bodies after collisi

This diagram represents momentum before and after interaction. Choose a positive direction and apply vector conservation.

Recoil of gun

Recoil of gun

This diagram represents momentum before and after interaction. Choose a positive direction and apply vector conservation.

Explosion into two parts

Explosion into two parts

This diagram represents momentum before and after interaction. Choose a positive direction and apply vector conservation.

Two carts pushing each other

Two carts pushing each o

This diagram represents momentum before and after interaction. Choose a positive direction and apply vector conservation.

Bullet-block system

Bullet-block system

This diagram represents momentum before and after interaction. Choose a positive direction and apply vector conservation.

Applications

Applications of Conservation of Momentum

Recoil of gun

Recoil of gun

Concept: Bullet and gun form an isolated system for a short time. Gun recoils opposite bullet momentum.

Formula: mbvb + mgvg = 0

Solved example: A light bullet at high speed gives the heavy gun a small backward speed.

Rocket propulsion

Rocket propulsion

Concept: Gases ejected backward carry momentum; rocket gains forward momentum.

Formula: Δprocket = -Δpgas

Solved example: If gases move backward faster, rocket acceleration increases.

Collision between two bodies

Collision between two bo

Concept: Total momentum before collision equals total momentum after collision when external force is zero.

Formula: pinitial = pfinal

Solved example: Two carts on an air track exchange momentum during collision.

Explosion problems

Explosion problems

Concept: Internal chemical forces separate parts but total momentum remains constant.

Formula: m1v1 + m2v2 = 0

Solved example: If one fragment moves right, the other moves left.

Bullet and block

Bullet and block

Concept: During short impact, external forces are negligible, so momentum is conserved.

Formula: m u = (M+m)v

Solved example: A bullet embedding in wood produces common velocity.

Two skaters pushing each other

Two skaters pushing each

Concept: Push forces are internal and equal-opposite; total momentum stays zero.

Formula: m1v1 = -m2v2

Solved example: Lighter skater moves faster.

Railway wagon collision

Railway wagon collision

Concept: Coupled wagons move with common velocity after inelastic collision.

Formula: v = (m1u1+m2u2)/(m1+m2)

Solved example: A moving wagon couples with a stationary wagon.

Air track experiments

Air track experiments

Concept: Air track reduces friction so momentum conservation is clearly observed.

Formula: Σp before = Σp after

Solved example: Gliders collide with negligible external horizontal force.

Advanced numericals

Momentum Conservation Numericals

1. One-dimensional collision

Question: A 2 kg body at 6 m/s collides with a 4 kg body at rest. They move separately with the 2 kg body at 2 m/s. Find speed of 4 kg body.

Given: m1=2,u1=6,m2=4,u2=0,v1=2

Formula: m1u1 + m2u2 = m1v1 + m2v2

Calculation: 12 = 4 + 4v2, so v2 = 2 m/s

Final Answer: 2 m/s

Exam Tip: Take one direction positive and keep signs consistent.

2. Two bodies stick together

Question: A 3 kg body at 4 m/s sticks to a 1 kg body at rest. Find common velocity.

Given: m1=3,u1=4,m2=1,u2=0

Formula: v = (m1u1+m2u2)/(m1+m2)

Calculation: v = 12/4 = 3 m/s

Final Answer: 3 m/s

Exam Tip: Sticking collision is perfectly inelastic.

3. Recoil of gun

Question: A 0.02 kg bullet leaves a 4 kg gun at 400 m/s. Find recoil speed.

Given: mb=0.02, vb=400, mg=4

Formula: 0 = mbvb + mgvg

Calculation: vg = -(0.02 x 400)/4 = -2 m/s

Final Answer: 2 m/s backward

Exam Tip: Initial total momentum is zero.

4. Bullet embedded in block

Question: A 0.01 kg bullet at 300 m/s embeds in a 1.49 kg block. Find common speed.

Given: m=0.01, u=300, M=1.49

Formula: v = mu/(m+M)

Calculation: v = 3/1.5 = 2 m/s

Final Answer: 2 m/s

Exam Tip: Momentum is conserved during short collision.

5. Explosion into two parts

Question: A body at rest explodes into 2 kg and 3 kg parts. The 2 kg part moves at 6 m/s. Find other speed.

Given: m1=2,v1=6,m2=3

Formula: 0 = m1v1 + m2v2

Calculation: v2 = -12/3 = -4 m/s

Final Answer: 4 m/s opposite

Exam Tip: Explosion has zero initial momentum if body was at rest.

6. Rocket propulsion basic

Question: A rocket throws gas backward at high speed. Explain forward motion.

Given: System: rocket + gases

Formula: pinitial = pfinal

Calculation: Gas gains backward momentum, rocket gains forward momentum

Final Answer: Rocket moves forward

Exam Tip: External force is small over short interval.

7. Two carts on smooth track

Question: Two carts push apart: 1 kg cart moves left at 3 m/s. Find speed of 1.5 kg cart.

Given: m1=1,v1=-3,m2=1.5

Formula: 0=m1v1+m2v2

Calculation: v2=3/1.5=2 m/s

Final Answer: 2 m/s right

Exam Tip: Smooth track means negligible external horizontal force.

8. Vector momentum

Question: A 2 kg object has velocity 3i+4j m/s. Find momentum magnitude.

Given: m=2,v=3i+4j

Formula: p=mv

Calculation: p=6i+8j, |p|=10

Final Answer: 10 kg m/s

Exam Tip: Use vector components.

9. JEE Main collision

Question: A 5 kg wagon at 2 m/s couples with a 10 kg wagon at 0.5 m/s in same direction. Find speed.

Given: m1=5,u1=2,m2=10,u2=0.5

Formula: v=(m1u1+m2u2)/(m1+m2)

Calculation: v=(10+5)/15=1 m/s

Final Answer: 1 m/s

Exam Tip: Same direction momenta add.

10. JEE Advanced multi-step

Question: A bullet embeds in block, then block compresses a spring. Find spring compression idea.

Given: Use bullet-block momentum, then energy after collision

Formula: m u = (M+m)v, then 1/2(M+m)v²=1/2kx²

Calculation: First find common v, then x = v sqrt((M+m)/k)

Final Answer: Formula result

Exam Tip: Momentum for collision, energy after collision.

Section 8

Variable Force

For variable force, instantaneous force is F = dp/dt and impulse is the area under the force-time graph.

F = dp/dt
J = ∫F dt
Graph-based QuestionShow Answer

Question: A triangular F-t graph has base 4 s and height 10 N. Find impulse.

Answer: Area = 1/2 x 4 x 10 = 20 N s. This equals change in momentum.

Force FTime tImpulse = area under F-t curveJ = ∫F dt

Section 9

Important Applications

Rocket launch

Rocket launch

Physics concept: Momentum changes when a force acts over time.

Momentum explanation: The object or person experiences a change in p = mv.

Impulse explanation: J = FΔt = Δp, so increasing time reduces average force for same Δp.

Bullet and gun recoil

Bullet and gun recoil

Physics concept: Momentum changes when a force acts over time.

Momentum explanation: The object or person experiences a change in p = mv.

Impulse explanation: J = FΔt = Δp, so increasing time reduces average force for same Δp.

Cricket bat

Cricket bat

Physics concept: Momentum changes when a force acts over time.

Momentum explanation: The object or person experiences a change in p = mv.

Impulse explanation: J = FΔt = Δp, so increasing time reduces average force for same Δp.

Airbag

Airbag

Physics concept: Momentum changes when a force acts over time.

Momentum explanation: The object or person experiences a change in p = mv.

Impulse explanation: J = FΔt = Δp, so increasing time reduces average force for same Δp.

Seat belt

Seat belt

Physics concept: Momentum changes when a force acts over time.

Momentum explanation: The object or person experiences a change in p = mv.

Impulse explanation: J = FΔt = Δp, so increasing time reduces average force for same Δp.

Hammer and nail

Hammer and nail

Physics concept: Momentum changes when a force acts over time.

Momentum explanation: The object or person experiences a change in p = mv.

Impulse explanation: J = FΔt = Δp, so increasing time reduces average force for same Δp.

Long jump pit

Long jump pit

Physics concept: Momentum changes when a force acts over time.

Momentum explanation: The object or person experiences a change in p = mv.

Impulse explanation: J = FΔt = Δp, so increasing time reduces average force for same Δp.

Boxing gloves

Boxing gloves

Physics concept: Momentum changes when a force acts over time.

Momentum explanation: The object or person experiences a change in p = mv.

Impulse explanation: J = FΔt = Δp, so increasing time reduces average force for same Δp.

Jet propulsion

Jet propulsion

Physics concept: Momentum changes when a force acts over time.

Momentum explanation: The object or person experiences a change in p = mv.

Impulse explanation: J = FΔt = Δp, so increasing time reduces average force for same Δp.

Water hose recoil

Water hose recoil

Physics concept: Momentum changes when a force acts over time.

Momentum explanation: The object or person experiences a change in p = mv.

Impulse explanation: J = FΔt = Δp, so increasing time reduces average force for same Δp.

Section 10

Free Body Diagrams

Block on table

Block on table

Draw all real forces first: weight, normal, tension, friction and applied force. Then apply Newton's second law along chosen axes.

Block on incline

Block on incline

Draw all real forces first: weight, normal, tension, friction and applied force. Then apply Newton's second law along chosen axes.

Hanging block

Hanging block

Draw all real forces first: weight, normal, tension, friction and applied force. Then apply Newton's second law along chosen axes.

Two-block system

Two-block system

Draw all real forces first: weight, normal, tension, friction and applied force. Then apply Newton's second law along chosen axes.

Elevator

Elevator

Draw all real forces first: weight, normal, tension, friction and applied force. Then apply Newton's second law along chosen axes.

Horizontal force problems

Horizontal force problem

Draw all real forces first: weight, normal, tension, friction and applied force. Then apply Newton's second law along chosen axes.

Section 11

High-Quality Numericals

CBSE 1: Basic momentum

Question: A 2 kg ball moves at 5 m/s. Find momentum.

Given: m=2 kg, v=5 m/s

Formula: p = mv

Calculation: p = 2 x 5 = 10 kg m/s

Final Answer: 10 kg m/s

Exam Tip: Momentum direction is same as velocity.

NEET 1: Impulse

Question: A force of 50 N acts for 0.2 s. Find impulse.

Given: F=50 N, t=0.2 s

Formula: J = FΔt

Calculation: J = 50 x 0.2 = 10 N s

Final Answer: 10 N s

Exam Tip: Impulse and momentum have same dimensions.

JEE Main 1: Force from momentum

Question: Momentum changes from 4 to 16 kg m/s in 3 s. Find average force.

Given: Δp=12 kg m/s, Δt=3 s

Formula: Favg = Δp/Δt

Calculation: F = 12/3 = 4 N

Final Answer: 4 N

Exam Tip: Average force comes from slope of p-t graph.

JEE Advanced 1: Vector impulse

Question: A particle receives impulse 3i + 4j N s. Find magnitude.

Given: J = 3i + 4j

Formula: |J| = sqrt(3²+4²)

Calculation: |J| = 5 N s

Final Answer: 5 N s

Exam Tip: Momentum is vector, so add components.

IB 1: Airbag

Question: Why does an airbag reduce force?

Given: Same Δp, larger Δt

Formula: Favg = Δp/Δt

Calculation: Increasing Δt decreases Favg

Final Answer: Smaller average force

Exam Tip: Safety devices increase stopping time.

IGCSE 1: Cricket catch

Question: A player moves hands backward while catching. Explain.

Given: Same momentum change

Formula: Favg = Δp/Δt

Calculation: Longer time reduces average force

Final Answer: Safer catch

Exam Tip: Use impulse-momentum theorem.

A-Level 1: Variable force

Question: Area under F-t graph is 18 N s. Find Δp.

Given: J=18 N s

Formula: J = Δp

Calculation: Δp = 18 kg m/s

Final Answer: 18 kg m/s

Exam Tip: Area under graph is impulse.

Section 12

NEET Question Bank

NEET Exam-style QuestionShow Answer

Question: NEET 1: A question on momentum. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 2: A question on impulse. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 3: A question on Newton second law. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 4: A question on variable force. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 5: A question on force-time graph. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 6: A question on vector momentum. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 7: A question on F=ma. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 8: A question on rate of change of momentum. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 9: A question on collision. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 10: A question on recoil. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 11: A question on momentum. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 12: A question on impulse. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 13: A question on Newton second law. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 14: A question on variable force. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 15: A question on force-time graph. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 16: A question on vector momentum. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 17: A question on F=ma. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 18: A question on rate of change of momentum. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 19: A question on collision. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 20: A question on recoil. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 21: A question on momentum. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 22: A question on impulse. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 23: A question on Newton second law. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 24: A question on variable force. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 25: A question on force-time graph. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 26: A question on vector momentum. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 27: A question on F=ma. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 28: A question on rate of change of momentum. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 29: A question on collision. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 30: A question on recoil. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 31: A question on momentum. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 32: A question on impulse. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 33: A question on Newton second law. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 34: A question on variable force. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 35: A question on force-time graph. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 36: A question on vector momentum. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 37: A question on F=ma. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 38: A question on rate of change of momentum. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 39: A question on collision. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 40: A question on recoil. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 41: A question on momentum. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 42: A question on impulse. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 43: A question on Newton second law. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 44: A question on variable force. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 45: A question on force-time graph. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 46: A question on vector momentum. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 47: A question on F=ma. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 48: A question on rate of change of momentum. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 49: A question on collision. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

NEET Exam-style QuestionShow Answer

Question: NEET 50: A question on recoil. Which option is most correct? Options: (A) Momentum is scalar. (B) Impulse equals change in momentum. (C) Momentum is independent of velocity. (D) Force-time area is displacement.

Answer: Correct option: B. Impulse J equals change in momentum Delta p. Momentum is a vector p = mv and the area under force-time graph gives impulse.

Section 13

JEE Main Question Bank

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 1: A 1 kg body initially moving at 2 m/s receives a variable/contact impulse equivalent to a constant force 10 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 10 x 0.10 = 1.00 N s. Initial momentum = mu = 2 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 2: A 2 kg body initially moving at 3 m/s receives a variable/contact impulse equivalent to a constant force 15 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 15 x 0.15 = 2.25 N s. Initial momentum = mu = 6 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 3: A 3 kg body initially moving at 4 m/s receives a variable/contact impulse equivalent to a constant force 20 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 20 x 0.20 = 4.00 N s. Initial momentum = mu = 12 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 4: A 4 kg body initially moving at 5 m/s receives a variable/contact impulse equivalent to a constant force 25 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 25 x 0.25 = 6.25 N s. Initial momentum = mu = 20 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 5: A 5 kg body initially moving at 6 m/s receives a variable/contact impulse equivalent to a constant force 30 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 30 x 0.30 = 9.00 N s. Initial momentum = mu = 30 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 6: A 1 kg body initially moving at 7 m/s receives a variable/contact impulse equivalent to a constant force 35 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 35 x 0.10 = 3.50 N s. Initial momentum = mu = 7 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 7: A 2 kg body initially moving at 8 m/s receives a variable/contact impulse equivalent to a constant force 40 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 40 x 0.15 = 6.00 N s. Initial momentum = mu = 16 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 8: A 3 kg body initially moving at 9 m/s receives a variable/contact impulse equivalent to a constant force 10 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 10 x 0.20 = 2.00 N s. Initial momentum = mu = 27 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 9: A 4 kg body initially moving at 2 m/s receives a variable/contact impulse equivalent to a constant force 15 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 15 x 0.25 = 3.75 N s. Initial momentum = mu = 8 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 10: A 5 kg body initially moving at 3 m/s receives a variable/contact impulse equivalent to a constant force 20 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 20 x 0.30 = 6.00 N s. Initial momentum = mu = 15 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 11: A 1 kg body initially moving at 4 m/s receives a variable/contact impulse equivalent to a constant force 25 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 25 x 0.10 = 2.50 N s. Initial momentum = mu = 4 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 12: A 2 kg body initially moving at 5 m/s receives a variable/contact impulse equivalent to a constant force 30 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 30 x 0.15 = 4.50 N s. Initial momentum = mu = 10 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 13: A 3 kg body initially moving at 6 m/s receives a variable/contact impulse equivalent to a constant force 35 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 35 x 0.20 = 7.00 N s. Initial momentum = mu = 18 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 14: A 4 kg body initially moving at 7 m/s receives a variable/contact impulse equivalent to a constant force 40 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 40 x 0.25 = 10.00 N s. Initial momentum = mu = 28 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 15: A 5 kg body initially moving at 8 m/s receives a variable/contact impulse equivalent to a constant force 10 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 10 x 0.30 = 3.00 N s. Initial momentum = mu = 40 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 16: A 1 kg body initially moving at 9 m/s receives a variable/contact impulse equivalent to a constant force 15 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 15 x 0.10 = 1.50 N s. Initial momentum = mu = 9 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 17: A 2 kg body initially moving at 2 m/s receives a variable/contact impulse equivalent to a constant force 20 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 20 x 0.15 = 3.00 N s. Initial momentum = mu = 4 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 18: A 3 kg body initially moving at 3 m/s receives a variable/contact impulse equivalent to a constant force 25 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 25 x 0.20 = 5.00 N s. Initial momentum = mu = 9 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 19: A 4 kg body initially moving at 4 m/s receives a variable/contact impulse equivalent to a constant force 30 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 30 x 0.25 = 7.50 N s. Initial momentum = mu = 16 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 20: A 5 kg body initially moving at 5 m/s receives a variable/contact impulse equivalent to a constant force 35 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 35 x 0.30 = 10.50 N s. Initial momentum = mu = 25 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 21: A 1 kg body initially moving at 6 m/s receives a variable/contact impulse equivalent to a constant force 40 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 40 x 0.10 = 4.00 N s. Initial momentum = mu = 6 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 22: A 2 kg body initially moving at 7 m/s receives a variable/contact impulse equivalent to a constant force 10 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 10 x 0.15 = 1.50 N s. Initial momentum = mu = 14 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 23: A 3 kg body initially moving at 8 m/s receives a variable/contact impulse equivalent to a constant force 15 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 15 x 0.20 = 3.00 N s. Initial momentum = mu = 24 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 24: A 4 kg body initially moving at 9 m/s receives a variable/contact impulse equivalent to a constant force 20 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 20 x 0.25 = 5.00 N s. Initial momentum = mu = 36 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 25: A 5 kg body initially moving at 2 m/s receives a variable/contact impulse equivalent to a constant force 25 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 25 x 0.30 = 7.50 N s. Initial momentum = mu = 10 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 26: A 1 kg body initially moving at 3 m/s receives a variable/contact impulse equivalent to a constant force 30 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 30 x 0.10 = 3.00 N s. Initial momentum = mu = 3 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 27: A 2 kg body initially moving at 4 m/s receives a variable/contact impulse equivalent to a constant force 35 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 35 x 0.15 = 5.25 N s. Initial momentum = mu = 8 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 28: A 3 kg body initially moving at 5 m/s receives a variable/contact impulse equivalent to a constant force 40 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 40 x 0.20 = 8.00 N s. Initial momentum = mu = 15 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 29: A 4 kg body initially moving at 6 m/s receives a variable/contact impulse equivalent to a constant force 10 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 10 x 0.25 = 2.50 N s. Initial momentum = mu = 24 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 30: A 5 kg body initially moving at 7 m/s receives a variable/contact impulse equivalent to a constant force 15 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 15 x 0.30 = 4.50 N s. Initial momentum = mu = 35 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 31: A 1 kg body initially moving at 8 m/s receives a variable/contact impulse equivalent to a constant force 20 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 20 x 0.10 = 2.00 N s. Initial momentum = mu = 8 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 32: A 2 kg body initially moving at 9 m/s receives a variable/contact impulse equivalent to a constant force 25 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 25 x 0.15 = 3.75 N s. Initial momentum = mu = 18 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 33: A 3 kg body initially moving at 2 m/s receives a variable/contact impulse equivalent to a constant force 30 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 30 x 0.20 = 6.00 N s. Initial momentum = mu = 6 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 34: A 4 kg body initially moving at 3 m/s receives a variable/contact impulse equivalent to a constant force 35 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 35 x 0.25 = 8.75 N s. Initial momentum = mu = 12 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 35: A 5 kg body initially moving at 4 m/s receives a variable/contact impulse equivalent to a constant force 40 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 40 x 0.30 = 12.00 N s. Initial momentum = mu = 20 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 36: A 1 kg body initially moving at 5 m/s receives a variable/contact impulse equivalent to a constant force 10 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 10 x 0.10 = 1.00 N s. Initial momentum = mu = 5 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 37: A 2 kg body initially moving at 6 m/s receives a variable/contact impulse equivalent to a constant force 15 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 15 x 0.15 = 2.25 N s. Initial momentum = mu = 12 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 38: A 3 kg body initially moving at 7 m/s receives a variable/contact impulse equivalent to a constant force 20 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 20 x 0.20 = 4.00 N s. Initial momentum = mu = 21 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 39: A 4 kg body initially moving at 8 m/s receives a variable/contact impulse equivalent to a constant force 25 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 25 x 0.25 = 6.25 N s. Initial momentum = mu = 32 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 40: A 5 kg body initially moving at 9 m/s receives a variable/contact impulse equivalent to a constant force 30 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 30 x 0.30 = 9.00 N s. Initial momentum = mu = 45 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 41: A 1 kg body initially moving at 2 m/s receives a variable/contact impulse equivalent to a constant force 35 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 35 x 0.10 = 3.50 N s. Initial momentum = mu = 2 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 42: A 2 kg body initially moving at 3 m/s receives a variable/contact impulse equivalent to a constant force 40 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 40 x 0.15 = 6.00 N s. Initial momentum = mu = 6 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 43: A 3 kg body initially moving at 4 m/s receives a variable/contact impulse equivalent to a constant force 10 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 10 x 0.20 = 2.00 N s. Initial momentum = mu = 12 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 44: A 4 kg body initially moving at 5 m/s receives a variable/contact impulse equivalent to a constant force 15 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 15 x 0.25 = 3.75 N s. Initial momentum = mu = 20 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 45: A 5 kg body initially moving at 6 m/s receives a variable/contact impulse equivalent to a constant force 20 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 20 x 0.30 = 6.00 N s. Initial momentum = mu = 30 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 46: A 1 kg body initially moving at 7 m/s receives a variable/contact impulse equivalent to a constant force 25 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 25 x 0.10 = 2.50 N s. Initial momentum = mu = 7 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 47: A 2 kg body initially moving at 8 m/s receives a variable/contact impulse equivalent to a constant force 30 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 30 x 0.15 = 4.50 N s. Initial momentum = mu = 16 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 48: A 3 kg body initially moving at 9 m/s receives a variable/contact impulse equivalent to a constant force 35 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 35 x 0.20 = 7.00 N s. Initial momentum = mu = 27 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 49: A 4 kg body initially moving at 2 m/s receives a variable/contact impulse equivalent to a constant force 40 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 40 x 0.25 = 10.00 N s. Initial momentum = mu = 8 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Main Exam-style QuestionShow Answer

Question: JEE Main 50: A 5 kg body initially moving at 3 m/s receives a variable/contact impulse equivalent to a constant force 10 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 10 x 0.30 = 3.00 N s. Initial momentum = mu = 15 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

Section 14

JEE Advanced Question Bank

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 1: A 1 kg body initially moving at 2 m/s receives a variable/contact impulse equivalent to a constant force 10 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 10 x 0.10 = 1.00 N s. Initial momentum = mu = 2 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 2: A 2 kg body initially moving at 3 m/s receives a variable/contact impulse equivalent to a constant force 15 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 15 x 0.15 = 2.25 N s. Initial momentum = mu = 6 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 3: A 3 kg body initially moving at 4 m/s receives a variable/contact impulse equivalent to a constant force 20 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 20 x 0.20 = 4.00 N s. Initial momentum = mu = 12 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 4: A 4 kg body initially moving at 5 m/s receives a variable/contact impulse equivalent to a constant force 25 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 25 x 0.25 = 6.25 N s. Initial momentum = mu = 20 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 5: A 5 kg body initially moving at 6 m/s receives a variable/contact impulse equivalent to a constant force 30 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 30 x 0.30 = 9.00 N s. Initial momentum = mu = 30 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 6: A 1 kg body initially moving at 7 m/s receives a variable/contact impulse equivalent to a constant force 35 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 35 x 0.10 = 3.50 N s. Initial momentum = mu = 7 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 7: A 2 kg body initially moving at 8 m/s receives a variable/contact impulse equivalent to a constant force 40 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 40 x 0.15 = 6.00 N s. Initial momentum = mu = 16 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 8: A 3 kg body initially moving at 9 m/s receives a variable/contact impulse equivalent to a constant force 10 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 10 x 0.20 = 2.00 N s. Initial momentum = mu = 27 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 9: A 4 kg body initially moving at 2 m/s receives a variable/contact impulse equivalent to a constant force 15 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 15 x 0.25 = 3.75 N s. Initial momentum = mu = 8 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 10: A 5 kg body initially moving at 3 m/s receives a variable/contact impulse equivalent to a constant force 20 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 20 x 0.30 = 6.00 N s. Initial momentum = mu = 15 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 11: A 1 kg body initially moving at 4 m/s receives a variable/contact impulse equivalent to a constant force 25 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 25 x 0.10 = 2.50 N s. Initial momentum = mu = 4 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 12: A 2 kg body initially moving at 5 m/s receives a variable/contact impulse equivalent to a constant force 30 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 30 x 0.15 = 4.50 N s. Initial momentum = mu = 10 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 13: A 3 kg body initially moving at 6 m/s receives a variable/contact impulse equivalent to a constant force 35 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 35 x 0.20 = 7.00 N s. Initial momentum = mu = 18 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 14: A 4 kg body initially moving at 7 m/s receives a variable/contact impulse equivalent to a constant force 40 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 40 x 0.25 = 10.00 N s. Initial momentum = mu = 28 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 15: A 5 kg body initially moving at 8 m/s receives a variable/contact impulse equivalent to a constant force 10 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 10 x 0.30 = 3.00 N s. Initial momentum = mu = 40 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 16: A 1 kg body initially moving at 9 m/s receives a variable/contact impulse equivalent to a constant force 15 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 15 x 0.10 = 1.50 N s. Initial momentum = mu = 9 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 17: A 2 kg body initially moving at 2 m/s receives a variable/contact impulse equivalent to a constant force 20 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 20 x 0.15 = 3.00 N s. Initial momentum = mu = 4 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 18: A 3 kg body initially moving at 3 m/s receives a variable/contact impulse equivalent to a constant force 25 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 25 x 0.20 = 5.00 N s. Initial momentum = mu = 9 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 19: A 4 kg body initially moving at 4 m/s receives a variable/contact impulse equivalent to a constant force 30 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 30 x 0.25 = 7.50 N s. Initial momentum = mu = 16 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 20: A 5 kg body initially moving at 5 m/s receives a variable/contact impulse equivalent to a constant force 35 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 35 x 0.30 = 10.50 N s. Initial momentum = mu = 25 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 21: A 1 kg body initially moving at 6 m/s receives a variable/contact impulse equivalent to a constant force 40 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 40 x 0.10 = 4.00 N s. Initial momentum = mu = 6 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 22: A 2 kg body initially moving at 7 m/s receives a variable/contact impulse equivalent to a constant force 10 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 10 x 0.15 = 1.50 N s. Initial momentum = mu = 14 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 23: A 3 kg body initially moving at 8 m/s receives a variable/contact impulse equivalent to a constant force 15 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 15 x 0.20 = 3.00 N s. Initial momentum = mu = 24 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 24: A 4 kg body initially moving at 9 m/s receives a variable/contact impulse equivalent to a constant force 20 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 20 x 0.25 = 5.00 N s. Initial momentum = mu = 36 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 25: A 5 kg body initially moving at 2 m/s receives a variable/contact impulse equivalent to a constant force 25 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 25 x 0.30 = 7.50 N s. Initial momentum = mu = 10 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 26: A 1 kg body initially moving at 3 m/s receives a variable/contact impulse equivalent to a constant force 30 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 30 x 0.10 = 3.00 N s. Initial momentum = mu = 3 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 27: A 2 kg body initially moving at 4 m/s receives a variable/contact impulse equivalent to a constant force 35 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 35 x 0.15 = 5.25 N s. Initial momentum = mu = 8 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 28: A 3 kg body initially moving at 5 m/s receives a variable/contact impulse equivalent to a constant force 40 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 40 x 0.20 = 8.00 N s. Initial momentum = mu = 15 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 29: A 4 kg body initially moving at 6 m/s receives a variable/contact impulse equivalent to a constant force 10 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 10 x 0.25 = 2.50 N s. Initial momentum = mu = 24 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 30: A 5 kg body initially moving at 7 m/s receives a variable/contact impulse equivalent to a constant force 15 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 15 x 0.30 = 4.50 N s. Initial momentum = mu = 35 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 31: A 1 kg body initially moving at 8 m/s receives a variable/contact impulse equivalent to a constant force 20 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 20 x 0.10 = 2.00 N s. Initial momentum = mu = 8 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 32: A 2 kg body initially moving at 9 m/s receives a variable/contact impulse equivalent to a constant force 25 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 25 x 0.15 = 3.75 N s. Initial momentum = mu = 18 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 33: A 3 kg body initially moving at 2 m/s receives a variable/contact impulse equivalent to a constant force 30 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 30 x 0.20 = 6.00 N s. Initial momentum = mu = 6 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 34: A 4 kg body initially moving at 3 m/s receives a variable/contact impulse equivalent to a constant force 35 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 35 x 0.25 = 8.75 N s. Initial momentum = mu = 12 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 35: A 5 kg body initially moving at 4 m/s receives a variable/contact impulse equivalent to a constant force 40 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 40 x 0.30 = 12.00 N s. Initial momentum = mu = 20 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 36: A 1 kg body initially moving at 5 m/s receives a variable/contact impulse equivalent to a constant force 10 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 10 x 0.10 = 1.00 N s. Initial momentum = mu = 5 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 37: A 2 kg body initially moving at 6 m/s receives a variable/contact impulse equivalent to a constant force 15 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 15 x 0.15 = 2.25 N s. Initial momentum = mu = 12 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 38: A 3 kg body initially moving at 7 m/s receives a variable/contact impulse equivalent to a constant force 20 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 20 x 0.20 = 4.00 N s. Initial momentum = mu = 21 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 39: A 4 kg body initially moving at 8 m/s receives a variable/contact impulse equivalent to a constant force 25 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 25 x 0.25 = 6.25 N s. Initial momentum = mu = 32 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 40: A 5 kg body initially moving at 9 m/s receives a variable/contact impulse equivalent to a constant force 30 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 30 x 0.30 = 9.00 N s. Initial momentum = mu = 45 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 41: A 1 kg body initially moving at 2 m/s receives a variable/contact impulse equivalent to a constant force 35 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 35 x 0.10 = 3.50 N s. Initial momentum = mu = 2 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 42: A 2 kg body initially moving at 3 m/s receives a variable/contact impulse equivalent to a constant force 40 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 40 x 0.15 = 6.00 N s. Initial momentum = mu = 6 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 43: A 3 kg body initially moving at 4 m/s receives a variable/contact impulse equivalent to a constant force 10 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 10 x 0.20 = 2.00 N s. Initial momentum = mu = 12 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 44: A 4 kg body initially moving at 5 m/s receives a variable/contact impulse equivalent to a constant force 15 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 15 x 0.25 = 3.75 N s. Initial momentum = mu = 20 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 45: A 5 kg body initially moving at 6 m/s receives a variable/contact impulse equivalent to a constant force 20 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 20 x 0.30 = 6.00 N s. Initial momentum = mu = 30 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 46: A 1 kg body initially moving at 7 m/s receives a variable/contact impulse equivalent to a constant force 25 N for 0.10 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 25 x 0.10 = 2.50 N s. Initial momentum = mu = 7 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 47: A 2 kg body initially moving at 8 m/s receives a variable/contact impulse equivalent to a constant force 30 N for 0.15 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 30 x 0.15 = 4.50 N s. Initial momentum = mu = 16 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 48: A 3 kg body initially moving at 9 m/s receives a variable/contact impulse equivalent to a constant force 35 N for 0.20 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 35 x 0.20 = 7.00 N s. Initial momentum = mu = 27 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 49: A 4 kg body initially moving at 2 m/s receives a variable/contact impulse equivalent to a constant force 40 N for 0.25 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 40 x 0.25 = 10.00 N s. Initial momentum = mu = 8 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

JEE Advanced Exam-style QuestionShow Answer

Question: JEE Advanced 50: A 5 kg body initially moving at 3 m/s receives a variable/contact impulse equivalent to a constant force 10 N for 0.30 s. Find the final momentum and discuss direction.

Answer: Impulse J = Ft = 10 x 0.30 = 3.00 N s. Initial momentum = mu = 15 kg m/s. Final momentum = initial momentum plus impulse in the force direction. If force acts opposite to motion, subtract the impulse vectorially.

Section 15

IB / IGCSE / A-Level Questions

IB Questions

IB QuestionShow Answer

Question: IB 1: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 2: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 3: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 4: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 5: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 6: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 7: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 8: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 9: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 10: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 11: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 12: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 13: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 14: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 15: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 16: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 17: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 18: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 19: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 20: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 21: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 22: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 23: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 24: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IB QuestionShow Answer

Question: IB 25: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE Questions

IGCSE QuestionShow Answer

Question: IGCSE 1: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 2: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 3: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 4: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 5: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 6: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 7: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 8: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 9: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 10: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 11: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 12: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 13: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 14: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 15: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 16: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 17: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 18: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 19: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 20: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 21: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 22: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 23: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 24: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

IGCSE QuestionShow Answer

Question: IGCSE 25: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level Questions

A-Level QuestionShow Answer

Question: A-Level 1: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 2: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 3: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 4: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 5: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 6: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 7: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 8: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 9: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 10: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 11: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 12: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 13: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 14: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 15: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 16: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 17: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 18: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 19: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 20: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 21: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 22: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 23: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 24: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

A-Level QuestionShow Answer

Question: A-Level 25: Explain one observation involving momentum, impulse, force-time graph or collision safety.

Answer: Use p = mv and J = Delta p. Increasing contact time reduces average force for the same change in momentum, which is why airbags, seat belts and soft landing pits reduce injury.

Section 16

Assertion Reason

Assertion-ReasonShow Answer

Question: Assertion 1: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 2: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 3: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 4: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 5: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 6: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 7: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 8: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 9: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 10: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 11: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 12: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 13: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 14: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 15: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 16: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 17: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 18: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 19: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 20: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 21: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 22: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 23: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 24: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 25: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 26: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 27: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 28: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 29: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Assertion-ReasonShow Answer

Question: Assertion 30: Impulse and change in momentum are equal in vector form. Reason: Force-time area gives impulse. Choose the correct assertion-reason relation.

Answer: Both statements are true. The reason supports the theorem because J = integral F dt = Delta p. Direction must be handled vectorially.

Section 17

Case Study Questions

Airbag system

Airbag system is a direct application of momentum, impulse or conservation principles.

Case StudyShow Answer

Question: Case 1: Explain airbag system using momentum and impulse.

Answer: Use J = Delta p and/or conservation of momentum. Identify the system, check external force, then compare initial and final momentum.

Seat belt system

Seat belt system is a direct application of momentum, impulse or conservation principles.

Case StudyShow Answer

Question: Case 2: Explain seat belt system using momentum and impulse.

Answer: Use J = Delta p and/or conservation of momentum. Identify the system, check external force, then compare initial and final momentum.

Rocket launch

Rocket launch is a direct application of momentum, impulse or conservation principles.

Case StudyShow Answer

Question: Case 3: Explain rocket launch using momentum and impulse.

Answer: Use J = Delta p and/or conservation of momentum. Identify the system, check external force, then compare initial and final momentum.

Cricket ball collision

Cricket ball collision is a direct application of momentum, impulse or conservation principles.

Case StudyShow Answer

Question: Case 4: Explain cricket ball collision using momentum and impulse.

Answer: Use J = Delta p and/or conservation of momentum. Identify the system, check external force, then compare initial and final momentum.

Recoil of gun

Recoil of gun is a direct application of momentum, impulse or conservation principles.

Case StudyShow Answer

Question: Case 5: Explain recoil of gun using momentum and impulse.

Answer: Use J = Delta p and/or conservation of momentum. Identify the system, check external force, then compare initial and final momentum.

Section 18

Common Student Mistakes

Confusing force and momentum

Fix it by writing definitions, drawing directions, choosing signs and checking whether net external force is zero before applying conservation.

Confusing impulse and force

Fix it by writing definitions, drawing directions, choosing signs and checking whether net external force is zero before applying conservation.

Forgetting vector nature

Fix it by writing definitions, drawing directions, choosing signs and checking whether net external force is zero before applying conservation.

Wrong sign conventions

Fix it by writing definitions, drawing directions, choosing signs and checking whether net external force is zero before applying conservation.

Wrong force-time graph interpretation

Fix it by writing definitions, drawing directions, choosing signs and checking whether net external force is zero before applying conservation.

Using conservation with external force present

Fix it by writing definitions, drawing directions, choosing signs and checking whether net external force is zero before applying conservation.

Scroll to Top