collisions and energy applications - physicstutor.kumarphysicsclasses.com

Collision Basics

A collision is a short-duration interaction with large impulsive forces. External impulse is usually negligible, so momentum is conserved.

Core Ideas

Interaction time is very small.
Impulsive forces are large internal forces.
Total momentum is conserved if external impulse is negligible.
Kinetic energy is conserved only in elastic collision.

Approaching Bodies

Separating Bodies

Elastic Collision

In an elastic collision, momentum and kinetic energy are both conserved, and e = 1.

1D Elastic Collision Formulae

v₁ = [(m₁-m₂)u₁ + 2m₂u₂] / (m₁+m₂)v₂ = [2m₁u₁ + (m₂-m₁)u₂] / (m₁+m₂)

Special Cases

Equal masses exchange velocities.
If target initially at rest and masses equal, first stops and second moves.
Heavy wall collision reverses velocity for elastic bounce.
Light body hitting heavy body rebounds nearly with same speed.

Inelastic Collision

Momentum is conserved but kinetic energy is not conserved; some KE becomes heat, sound and deformation.

Perfectly Inelastic

Bodies stick together after collision, so e = 0.

v = (m₁u₁ + m₂u₂) / (m₁ + m₂)

Sticking Together

Coefficient of Restitution

Coefficient of restitution measures bounciness of collision.

Formula

e = relative speed of separation / relative speed of approache = (v₂ - v₁) / (u₁ - u₂)

Classification

e	Type	Meaning
1	Elastic	KE conserved
0 < e < 1	Inelastic	Some KE lost
0	Perfectly inelastic	Bodies stick together

Energy Loss

Energy loss in inelastic collision depends on reduced mass, relative speed and coefficient of restitution.

Formula

Loss of KE = 1/2 [m₁m₂/(m₁+m₂)] (1-e²)(u₁-u₂)²Energy loss = 1/2 μ(1-e²)u_rel²

Terms

μ = m₁m₂/(m₁+m₂) is reduced mass.
u_rel = u₁ - u₂ is relative speed of approach.
If e = 1, loss is zero.
If e = 0, loss is maximum.

One-Dimensional Collision

Head-on collisions are solved using momentum equation and restitution equation with signs.

Method

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂v₂ - v₁ = e(u₁ - u₂)

Choose one direction positive and keep all velocities signed.

Solved Example

m1=2 kg moving at 6 m/s hits m2=4 kg at rest. e=0.5. Find velocities.

Momentum: 12=2v1+4v2. Restitution: v2-v1=0.5(6)=3. Solving gives v1=0, v2=3 m/s. Exam tip: use signed velocities.

Two-Dimensional Collision

In glancing or oblique collision, resolve momentum separately along x and y directions.

Momentum Components

m₁u_1x + m₂u_2x = m₁v_1x + m₂v_2xm₁u_1y + m₂u_2y = m₁v_1y + m₂v_2y

Glancing Collision

Realistic Collision Cases

Practical cases with concept, momentum equation, energy discussion and exam tip.

Car Collision

Momentum conserved approximately during impact; KE lost in deformation. Tip: use impulse for force-time questions.

Ball and Wall

Wall mass is huge; ball rebounds with speed ev. Equation: v=-eu.

Cricket Bat

Impulse changes ball momentum; sweet spot reduces vibration. Energy partly lost in sound/heat.

Bullet Block

Perfectly inelastic: mv=(M+m)V. KE not conserved during embedding.

Railway Wagon

Coupling often perfectly inelastic; common final speed by momentum.

Explosion

Momentum conserved, KE increases due to internal energy release.

Newton's Cradle

Nearly elastic; momentum and KE transfer through balls.

Floor Bounce

Use e = rebound speed / impact speed for fixed floor.

Rough Surface

Momentum during collision may conserve; after collision friction changes motion.

Spring Block

Collision energy may store in spring; use momentum then energy when appropriate.

Practical Applications

Application cards for safety, games and engineering.

Airbags

Increase stopping time, reducing average force.

Seat Belts

Provide controlled impulse to reduce injury.

Helmets

Extend impact time and absorb energy.

Sweet Spot

Minimizes vibration in bat collision.

Car Safety

Crumple zones dissipate KE.

Ball Games

Restitution controls bounce.

Newton's Cradle

Demonstrates elastic collision.

Pile Driving

Impact transfers momentum to pile.

Hammer and Nail

Short impulse gives large force.

Railway Coupling

Inelastic collision joins wagons.

High-Quality Numericals

Solved bank covering elastic, inelastic, e, energy loss, bullet block, ball-wall, floor and 2D collisions.

CBSE: Two equal masses collide elastically head-on. u1=4 m/s, u2=0. Find final speeds.

Diagram: one moving ball hits identical ball at rest. Given equal masses, elastic. Formula: equal masses exchange velocities. Final answer: v1=0, v2=4 m/s. Tip: special case saves time. Common mistake: conserving momentum only.

NEET: 2 kg mass at 6 m/s sticks to 4 kg mass at rest. Find common speed.

Perfectly inelastic. v=(2x6+4x0)/(6)=2 m/s. KE not conserved. Tip: use momentum only for collision.

JEE Main: Ball hits fixed wall with speed 10 m/s, e=0.8. Rebound speed?

For fixed wall, rebound speed=e u=8 m/s opposite direction. Final velocity=-8 m/s if toward wall is positive.

JEE Advanced: m1=1, m2=3, u1=8, u2=0, e=0.5. Find v1,v2.

Momentum: 8=v1+3v2. Restitution: v2-v1=4. Solving: v2=3, v1=-1 m/s. Tip: signs matter.

IB: Bullet 0.02 kg at 200 m/s embeds in 1.98 kg block. Find common speed.

v=mu/(M+m)=0.02x200/2=2 m/s. Common mistake: using KE conservation.

IGCSE: A 1 kg trolley at 3 m/s joins 2 kg trolley at rest. Speed?

v=3/(3)=1 m/s. Momentum conserved.

A-Level: Energy loss for μ=2 kg, e=0.5, relative approach 6 m/s.

Loss=1/2 μ(1-e²)urel²=1x(0.75)x36=27 J.

NEET Question Bank

50 high-quality NEET-style MCQs. Authentic years are not invented.

1. NEET Exam-style Question: In any isolated collision, conserved quantity is: A KE always B momentum C speed D force

Answer: B. Momentum is conserved when external impulse is negligible.

2. Elastic collision has e equal to: A 0 B 0.5 C 1 D 2

Answer: C.

3. Perfectly inelastic collision has e: A 1 B 0 C 2 D -1

Answer: B.

4. In inelastic collision, KE is: A always conserved B not conserved C infinite D zero always

Answer: B.

5. Equal masses in 1D elastic collision exchange: A masses B velocities C charges D radii

Answer: B.

6. e formula is: A approach/separation B separation/approach C momentum/KE D mass ratio

Answer: B.

7. Bullet block collision is usually: A elastic B perfectly inelastic C no collision D superelastic

Answer: B.

8. Ball rebounds from fixed wall with e=0.6 and impact speed 10. Rebound speed: A 4 B 6 C 10 D 16

Answer: B.

9. If e=1, energy loss formula gives: A maximum B zero C negative D infinite

Answer: B.

10. If e=0, bodies after collision: A always separate fastest B stick in 1D direct impact C vanish D exchange velocities

Answer: B for perfectly inelastic collision.

11. SI unit of coefficient of restitution: A m/s B kg C no unit D joule

Answer: C.

12. Reduced mass μ equals: A m1+m2 B m1m2/(m1+m2) C m1/m2 D zero

Answer: B.

13. Energy loss depends on relative speed as: A u B u² C 1/u D independent

Answer: B.

14. Collision time is generally: A very long B very short C infinite D one hour

Answer: B.

15. Impulse equals change in: A mass B momentum C charge D radius

Answer: B.

16. Newton's cradle is approximately: A elastic B perfectly inelastic C explosive D viscous

Answer: A.

17. Airbags reduce force by increasing: A mass B stopping time C speed D charge

Answer: B.

18. In 2D collision momentum is conserved along: A x only B y only C x and y separately D neither

Answer: C.

19. Perfectly inelastic common velocity formula uses conservation of: A momentum B KE C power D temperature

Answer: A.

20. KE loss appears as: A heat/sound/deformation B charge only C mass only D no form

Answer: A.

21. Two equal balls, one moving at v, elastic head-on. Moving ball after collision: A v B 0 C 2v D -v

Answer: B.

22. Target equal ball gets speed: A 0 B v C -v D 2v

Answer: B.

23. Heavy wall elastic collision changes velocity from u to: A u B 0 C -u D 2u

Answer: C.

24. If relative approach is zero, collision tendency is: A none B maximum C infinite D always elastic

Answer: A.

25. Momentum is vector, so in 2D use: A scalar only B components C only magnitudes D only energy

Answer: B.

26. Explosion increases KE due to: A internal energy B external impulse C zero momentum D wall

Answer: A.

27. A 2 kg mass at 3 m/s sticks to 1 kg at rest. v: A 1 B 2 C 3 D 6

Answer: B. v=6/3=2.

28. Same collision initial KE: A 3 B 6 C 9 D 18 J

Answer: C. 1/2*2*9=9 J.

29. Same collision final KE: A 2 B 4 C 6 D 9 J

Answer: C. 1/2*3*4=6 J.

30. Same collision KE loss: A 0 B 1 C 3 D 9 J

Answer: C.

31. If e=0.8, collision is: A elastic B perfectly inelastic C inelastic D impossible

Answer: C.

32. e greater than 1 indicates: A superelastic/explosive energy release B normal inelastic C no collision D wall only

Answer: A in special cases.

33. Momentum conservation may fail if significant external: A impulse B internal force C deformation D sound

Answer: A.

34. Cricket bat impact changes ball momentum by: A impulse B mass only C heat only D gravity only

Answer: A.

35. Seat belts reduce injury by: A increasing stopping time B increasing speed C reducing mass D removing impulse

Answer: A.

36. In elastic collision, relative speed of separation equals: A approach speed B zero C half D double

Answer: A.

37. In perfectly inelastic collision, separation speed is: A zero B approach C infinite D negative

Answer: A.

38. Energy loss formula contains factor: A 1+e² B 1-e² C e² only D 1/e

Answer: B.

39. For equal masses m, reduced mass is: A m B m/2 C 2m D zero

Answer: B.

40. For m and very heavy wall, reduced mass approximately: A m B wall mass C zero D 2m

Answer: A.

41. Pile driving uses: A collision impulse B only static force C no momentum D only heat

Answer: A.

42. Hammer and nail works due to: A impact force B no impulse C zero work D only mass

Answer: A.

43. Car crumple zones mainly dissipate: A kinetic energy B charge C mass D time

Answer: A.

44. Ball bouncing from floor, e = rebound/impact speed if floor is: A fixed B moving fast C absent D frictionless only

Answer: A.

45. Head-on collision means motion along: A one line B two axes C circle D random only

Answer: A.

46. Glancing collision is generally: A 2D B 1D only C no momentum D no impulse

Answer: A.

47. Momentum unit is: A kg m/s B joule C watt D newton/m

Answer: A.

48. KE unit is: A kg m/s B joule C watt D tesla

Answer: B.

49. Momentum conserved but KE increases in: A explosion B ordinary inelastic C perfectly inelastic D rest

Answer: A.

50. During collision, internal forces are equal and opposite by: A Newton's third law B Ohm's law C Hooke only D Snell's law

Answer: A.

JEE Main Question Bank

50 difficult JEE Main style questions. No fake years included.

1. JEE Main Exam-style Question: m1=1,u1=6; m2=2,u2=0 stick. Find v.

v=6/3=2 m/s.

2. Same collision KE loss.

Initial KE=18 J, final KE=1/2*3*4=6 J, loss=12 J.

3. Ball-wall e=0.5, u=12. Rebound velocity if toward wall positive.

v=-eu=-6 m/s.

4. m1=2,m2=2 elastic; u1=5,u2=0. Final?

Equal masses exchange: v1=0, v2=5.

5. m1=1,m2=3,u1=8,u2=0,e=1. Find v1,v2.

Elastic formula: v1=-4 m/s, v2=4 m/s.

6. m1=1,m2=3,u1=8,u2=0,e=0.5. Find v1,v2.

Momentum:8=v1+3v2; restitution v2-v1=4. v2=3, v1=-1.

7. Reduced mass of 2 kg and 6 kg.

μ=12/8=1.5 kg.

8. Energy loss for μ=1.5, e=0.5, urel=4.

Loss=1/2*1.5*(1-0.25)*16=9 J.

9. Two bodies approach with u1=5,u2=-3. Approach speed?

u1-u2=8 m/s if positive x along u1.

10. If e=0.25, separation speed for approach 8?

2 m/s.

11. Bullet 0.01 kg at 400 m/s embeds in 1.99 kg block. v?

v=4/2=2 m/s.

12. KE lost in previous.

Initial=800 J, final=1/2*2*4=4 J, loss=796 J.

13. Explosion: mass 3 kg at rest breaks into 1 kg at 6 m/s and 2 kg. Speed of 2 kg?

Momentum zero: 1*6+2v=0, v=-3 m/s.

14. 2D: masses stick, total initial momentum (6,8) kg m/s, total mass 2 kg. Velocity?

V=(3,4) m/s, speed 5 m/s.

15. Ball hits floor speed 10 m/s, rebounds 6 m/s. e?

e=6/10=0.6.

16. Drop from height h, rebound height h/4. e?

e=sqrt(h2/h1)=1/2.

17. Drop height 5 m, e=0.8. Rebound height?

h'=e²h=0.64*5=3.2 m.

18. Direct collision equation using e.

v2-v1=e(u1-u2).

19. Momentum equation for 1D collision.

m1u1+m2u2=m1v1+m2v2.

20. A 4 kg wagon at 3 m/s couples with 2 kg at rest. v?

v=12/6=2 m/s.

21. KE loss in previous.

Initial=18 J, final=12 J, loss=6 J.

22. Elastic collision with wall changes momentum by magnitude?

Δp=2mu for speed u reversal.

23. Inelastic wall bounce e, momentum change magnitude?

Δp=m(1+e)u.

24. Average force if impulse J acts in time Δt.

Favg=J/Δt.

25. Ball mass 0.2 kg reverses from 10 to -8 m/s in 0.02 s. Average force?

Δp=0.2(-8-10)=-3.6 Ns, F=-180 N.

26. Elastic equal masses approaching with +3 and -2. Final velocities?

Exchange: v1=-2, v2=+3.

27. Perfectly inelastic equal masses with +3 and -1. Common v?

v=(3-1)/2=1 m/s.

28. KE loss in previous if each mass m.

Initial=1/2m(9+1)=5m, final=1/2(2m)(1)=m, loss=4m.

29. e from velocities u1=6,u2=0,v1=1,v2=4.

e=(4-1)/(6-0)=1/2.

30. If e=1 and u2=0, m1=m2, v2?

v2=u1.

31. If e=0 and u2=0, common velocity formula?

v=m1u1/(m1+m2).

32. Collision on rough surface: during very short collision, friction impulse often?

Negligible compared with collision impulse, unless stated.

33. After collision, rough surface effect?

Friction changes kinetic energy and momentum of individual body-system with external impulse.

34. 2D inelastic: total p=(0,10), mass=5. Common velocity?

(0,2) m/s.

35. Two identical balls elastic, one at rest, after glancing collision velocities angle?

They move at right angles if collision is elastic and target initially at rest.

36. Energy loss if e=1/2 compared to maximum loss e=0.

Loss factor is 1-e²=3/4 of maximum.

37. For e=0.8, loss factor?

1-0.64=0.36.

38. If urel doubles, energy loss factor?

Four times.

39. If reduced mass doubles, energy loss?

Doubles.

40. Railway coupling is modeled as what collision?

Perfectly inelastic.

41. Newton cradle ideal model?

Nearly elastic collision of equal masses.

42. Car crash safety increases stopping distance/time, reducing?

Average force.

43. If impulse fixed and time doubled, average force?

Halves.

44. Momentum conservation in explosion of isolated system?

Total momentum remains constant.

45. KE in explosion generally?

Increases due to internal energy conversion.

46. Direct collision with e=0.75 and approach speed 20. Separation speed?

15 m/s.

47. If separation speed is 6 and approach speed 8, e?

0.75.

48. If a ball rebounds to same height from floor, e?

49. If ball does not rebound, e?

50. Which two equations solve most 1D collision questions?

Momentum conservation and coefficient of restitution equation.

JEE Advanced Question Bank

50 difficult JEE Advanced questions with compact complete solutions.

1. JEE Advanced Exam-style Question: Derive energy loss in terms of e.

In center of mass frame, relative kinetic energy before is 1/2 μu_rel² and after is 1/2 μ(eu_rel)². Loss=1/2 μ(1-e²)u_rel².

2. m1,m2 direct collision. Express reduced mass.

μ=m1m2/(m1+m2).

3. e=0.6, μ=3, urel=10. Loss?

Loss=1/2*3*(1-0.36)*100=96 J.

4. Bullet block with spring: bullet m speed u embeds in block M attached to spring k. Max compression?

After collision V=mu/(M+m). Then 1/2(M+m)V²=1/2kx², so x=mu/√(k(M+m)).

5. Bullet passes through block: bullet m u to v, block M speed V.

Momentum: mu=mv+MV, so V=m(u-v)/M. KE lost from initial minus final.

6. Equal masses elastic 2D with one initially at rest. Prove final velocities perpendicular.

Momentum vector u=v1+v2 and energy u²=v1²+v2². Squaring momentum gives u²=v1²+v2²+2v1.v2, so v1.v2=0.

7. Ball floor: coefficient e, initial drop height h. Rebound height?

Impact speed √2gh, rebound speed e√2gh. h'=v²/2g=e²h.

8. Total distance after repeated bounces from height h with coefficient e.

Distance=h+2he²+2he⁴+...=h(1+e²)/(1-e²).

9. Time after repeated bounces from height h.

Initial fall √(2h/g); rebound flight times form geometric series 2e√(2h/g)(1+e+...). Total=√(2h/g)[1+2e/(1-e)].

10. Superelastic collision has e>1. What happens to KE?

Kinetic energy increases due to release of internal energy.

11. In CM frame, what reverses in elastic collision?

Relative velocity reverses direction with same magnitude.

12. In CM frame for coefficient e, relative speed after?

e times relative speed before, opposite along line of impact.

13. Collision with moving wall velocity U. Formula for ball final velocity with e?

Use wall frame: relative approach u-U. After relative velocity = -e(u-U). Thus v=U-e(u-U)=(1+e)U-eu.

14. Ball dropped on upward moving platform speed U.

Apply moving wall formula with signs; rebound speed increases because platform supplies energy.

15. Oblique collision of smooth spheres: impulse acts along?

Along common normal/line of centers; tangential velocity components remain unchanged.

16. Smooth sphere oblique collision: restitution equation applies to which components?

Normal components along line of impact.

17. Tangential components in smooth collision?

Unchanged because no tangential impulse.

18. If rough collision, tangential impulse may change what?

Tangential velocities and rotational motion.

19. Two fragments masses m and 2m after explosion have momentum equal opposite. KE ratio?

For same p, K=p²/2m. Ratio K_m:K_2m=2:1.

20. Explosion of mass M at rest into m and M-m. Momentum relation?

m v1 + (M-m)v2=0.

21. Energy released in explosion equals?

Increase in total kinetic energy if no external work.

22. Two bodies stick, derive KE loss.

Loss=1/2 μ urel² for e=0.

23. For m1=m2=m, e=0, urel=u, loss?

μ=m/2, loss=1/4 m u².

24. For m1=m2=m, e=1, loss?

Zero.

25. A body hits heavy block elastically at rest. If M>>m, final light velocity?

Approximately -u, rebounds.

26. A heavy body hits light body at rest elastically. Heavy final velocity?

Nearly unchanged; light body moves about 2u.

27. Collision plus spring: two masses stick then compress spring on smooth surface.

Find common velocity from momentum, then use 1/2(m1+m2)V²=1/2kx².

28. Collision plus vertical rise: stuck masses climb height h.

Momentum gives V; then 1/2(Mtot)V²=Mtot gh.

29. Ballistic pendulum formula.

After embedding V=(m/(M+m))u; rise h satisfies 1/2(M+m)V²=(M+m)gh, so u=(M+m)/m √(2gh).

30. Collision with floor: impulse magnitude if speed before u and after eu upward.

J=m(u+eu)=m(1+e)u upward.

31. Average floor force including weight during contact time Δt?

N_avg Δt - mgΔt = m(1+e)u, so N_avg=mg+m(1+e)u/Δt.

32. Direct collision equations in matrix form solve?

Use momentum plus v2-v1=e(u1-u2); solve simultaneous linear equations.

33. If e known and one final velocity known, how find other?

Use restitution equation first, then momentum if needed.

34. 2D perfectly inelastic collision common velocity vector.

V=(m1u1+m2u2)/(m1+m2) vector form.

35. Energy loss in 2D sticking collision.

Loss=initial KE - 1/2(m1+m2)|V|².

36. Smooth oblique wall collision with e: normal component changes how?

v_n=-e u_n, tangential component unchanged.

37. Oblique wall collision speed after.

v²=u_t²+e²u_n².

38. Energy loss in oblique wall collision.

Loss=1/2m(1-e²)u_n².

39. Coefficient e from bounce angles and speeds off smooth wall.

Use normal components: e=|v_n|/|u_n|; tangential components unchanged.

40. When is KE conserved in collision?

Only elastic collision, e=1 for direct collision without internal energy changes.

41. When is momentum not conserved for chosen bodies?

When external impulse during collision is significant.

42. Can one body gain KE in inelastic collision?

Yes, individual KE may increase, but total KE decreases for ordinary inelastic collision.

43. Can total KE remain same while directions change?

Yes in elastic collisions, especially 2D.

44. Relative velocity method advantage?

Restitution directly relates separation and approach speeds, simplifying simultaneous equations.

45. Center of mass velocity formula.

Vcm=(m1u1+m2u2)/(m1+m2), unchanged in isolated collision.

46. In CM frame total momentum is?

Zero.

47. Energy loss occurs only in which part of motion?

Relative motion part; CM kinetic energy remains unchanged.

48. Loss formula in terms of CM relative KE.

Loss=(1-e²) times initial relative KE in CM frame.

49. If one body initially at rest and collision perfectly inelastic, fraction of KE lost?

Fraction lost=m2/(m1+m2), for projectile m1 hitting target m2 at rest.

50. Same case fraction of KE retained?

Fraction retained=m1/(m1+m2).

IB / IGCSE / A-Level Questions

Separate international question sets with answers and explanations.

IB Questions - 25

IB 1. Define collision.

A short interaction between bodies involving large forces and momentum change.

IB 2. State momentum conservation condition.

Total momentum is conserved when external impulse is negligible.

IB 3. Define elastic collision.

Momentum and kinetic energy conserved.

IB 4. Define inelastic collision.

Momentum conserved but kinetic energy not conserved.

IB 5. Define e.

Relative speed of separation divided by relative speed of approach.

IB 6. e for elastic collision.

IB 7. e for perfectly inelastic collision.

IB 8. What happens to lost KE?

It becomes heat, sound and deformation energy.

IB 9. Two equal masses elastic, one at rest.

They exchange velocities.

IB 10. Bullet block collision type?

Perfectly inelastic.

IB 11. Momentum equation in 1D.

m1u1+m2u2=m1v1+m2v2.

IB 12. Restitution equation.

v2-v1=e(u1-u2).

IB 13. Why airbags help?

They increase stopping time and reduce average force.

IB 14. Is momentum vector?

Yes.

IB 15. 2D collision momentum components?

Conserved separately along x and y.

IB 16. Ball hits fixed wall e=0.7 speed 10.

Rebound speed 7 m/s.

IB 17. Drop height h, rebound h/4. e?

1/2.

IB 18. Explosion momentum?

Conserved for isolated system.

IB 19. Explosion KE?

Can increase due to internal energy.

IB 20. Newton's cradle demonstrates?

Nearly elastic collision.

IB 21. Reduced mass formula.

m1m2/(m1+m2).

IB 22. Energy loss if e=1.

Zero.

IB 23. Energy loss if e=0.

Maximum for given masses and relative speed.

IB 24. Seat belt purpose.

Increase stopping time and control impulse.

IB 25. Momentum unit.

kg m/s.

IGCSE Questions - 25

IGCSE 1. What is momentum?

Mass times velocity.

IGCSE 2. Momentum formula.

p=mv.

IGCSE 3. Momentum unit.

kg m/s.

IGCSE 4. Collision conservation law.

Momentum conservation.

IGCSE 5. 2 kg at 3 m/s momentum.

6 kg m/s.

IGCSE 6. Two trolleys stick: use what conservation?

Momentum.

IGCSE 7. Elastic means KE?

Conserved.

IGCSE 8. Inelastic means KE?

Not conserved.

IGCSE 9. Lost KE becomes?

Heat, sound, deformation.

IGCSE 10. Airbags reduce?

Impact force.

IGCSE 11. Helmets increase stopping?

Time/distance.

IGCSE 12. A 1 kg object at 4 m/s hits and sticks to 1 kg at rest. Speed?

2 m/s.

IGCSE 13. Equal masses elastic exchange?

Velocities.

IGCSE 14. e=1 means?

Elastic.

IGCSE 15. e=0 means?

Perfectly inelastic.

IGCSE 16. Ball bounces lower because energy is?

Lost to heat/sound/deformation.

IGCSE 17. Hammer nail collision has large?

Impact force.

IGCSE 18. Railway coupling collision type?

Inelastic.

IGCSE 19. Momentum is scalar or vector?

Vector.

IGCSE 20. KE is scalar or vector?

Scalar.

IGCSE 21. External force absent, total momentum?

Constant.

IGCSE 22. Impulse equals?

Change in momentum.

IGCSE 23. Shorter stopping time means force?

Larger average force.

IGCSE 24. Longer stopping time means force?

Smaller average force.

IGCSE 25. Car crumple zone absorbs?

Kinetic energy.

A-Level Questions - 25

A-Level 1. Define coefficient of restitution.

Ratio of relative speed of separation to relative speed of approach.

A-Level 2. Direct impact restitution equation.

v2-v1=e(u1-u2).

A-Level 3. Energy loss formula.

1/2 μ(1-e²)u_rel².

A-Level 4. Reduced mass.

μ=m1m2/(m1+m2).

A-Level 5. Ball-wall oblique: tangential component?

Unchanged for smooth wall.

A-Level 6. Ball-wall normal component after?

Reverses and multiplies by e.

A-Level 7. Perfectly inelastic common velocity vector.

V=(m1u1+m2u2)/(m1+m2).

A-Level 8. Ballistic pendulum first step.

Use momentum during collision.

A-Level 9. Ballistic pendulum second step.

Use mechanical energy after collision.

A-Level 10. Smooth oblique sphere collision impulse direction.

Along line of centers.

A-Level 11. Equal masses elastic 2D angle result.

Final velocities are perpendicular.

A-Level 12. Explosion momentum.

Conserved if isolated.

A-Level 13. Explosion KE.

Usually increases.

A-Level 14. Moving wall collision formula.

v=(1+e)U-eu with signed velocities for wall speed U.

A-Level 15. Drop rebound height relation.

h2=e²h1.

A-Level 16. Repeated bounce total distance.

h(1+e²)/(1-e²).

A-Level 17. Impulse from floor collision.

J=m(1+e)u for fixed floor.

A-Level 18. Center of mass velocity.

Vcm=(m1u1+m2u2)/(m1+m2).

A-Level 19. CM velocity during isolated collision.

Constant.

A-Level 20. Energy loss occurs in which KE part?

Relative KE in CM frame.

A-Level 21. If e=0.6, loss fraction of relative KE.

1-e²=0.64.

A-Level 22. If e=0.8, retained relative KE fraction.

e²=0.64.

A-Level 23. Momentum in 2D collision.

Conserve vector momentum component-wise.

A-Level 24. Why not use KE conservation in bullet block?

It is perfectly inelastic; KE is lost in deformation/heat.

A-Level 25. What is superelastic collision?

A collision where KE increases due to internal energy release, e can exceed 1.

Assertion Reason

30 assertion-reason questions.

1. Assertion: Momentum is conserved in isolated collision. Reason: External impulse is zero.

Both true; reason explains assertion.

2. Assertion: KE is always conserved in collisions. Reason: Momentum is conserved.

Assertion false, reason true for isolated collisions.

3. Assertion: Elastic collision has e=1. Reason: Relative separation equals approach.

Both true.

4. Assertion: Perfectly inelastic collision has e=0. Reason: Bodies have same final velocity.

Both true.

5. Assertion: Inelastic collision loses KE. Reason: Energy becomes heat, sound and deformation.

Both true for ordinary inelastic collisions.

6. Assertion: e is dimensionless. Reason: It is ratio of speeds.

Both true.

7. Assertion: Equal masses exchange velocities in elastic head-on collision. Reason: Both momentum and KE are conserved.

Both true.

8. Assertion: Bullet block collision conserves KE. Reason: Bullet sticks in block.

Assertion false, reason indicates perfectly inelastic collision.

9. Assertion: Energy loss formula contains 1-e². Reason: Relative speed after is e times before.

Both true.

10. Assertion: Airbags reduce average force. Reason: They increase stopping time.

Both true.

11. Assertion: Momentum is vector. Reason: 2D collision requires components.

Both true.

12. Assertion: In glancing collision, momentum is conserved only in x. Reason: y component disappears.

Both false; conserve both components if isolated.

13. Assertion: Wall collision can reverse velocity. Reason: wall exerts impulse.

Both true.

14. Assertion: Newton's cradle is perfectly inelastic. Reason: balls stick.

Both false; it is nearly elastic and balls do not stick.

15. Assertion: Collision duration is short. Reason: impulsive forces are large.

Both true in collision model.

16. Assertion: Internal forces cancel in total momentum equation. Reason: Newton's third law pairs are equal and opposite.

Both true.

17. Assertion: Explosion can increase KE. Reason: internal energy converts into kinetic energy.

Both true.

18. Assertion: Momentum conservation needs no external force. Reason: external impulse changes momentum.

Both true.

19. Assertion: e=0.5 means separation speed is half approach speed. Reason: e is their ratio.

Both true.

20. Assertion: For e=1, energy loss is zero. Reason: 1-e²=0.

Both true.

21. Assertion: For e=0, energy loss is maximum. Reason: separation speed is zero.

Both true for fixed masses and approach speed.

22. Assertion: Seat belts remove impulse. Reason: momentum change becomes zero.

Both false; they increase time and reduce force, impulse remains momentum change.

23. Assertion: Car crumple zone dissipates KE. Reason: deformation absorbs energy.

Both true.

24. Assertion: Smooth oblique wall collision changes tangential velocity. Reason: tangential impulse exists.

Both false for smooth wall.

25. Assertion: Normal component changes in wall collision. Reason: impulse acts normal to wall.

Both true.

26. Assertion: Reduced mass appears in energy loss. Reason: loss is due to relative motion.

Both true.

27. Assertion: Momentum can be conserved while KE is not. Reason: momentum and KE are different quantities.

Both true.

28. Assertion: 2D collision problems require vector equations. Reason: momentum is vector.

Both true.

29. Assertion: Heavy wall can be treated as infinite mass. Reason: wall velocity change is negligible.

Both true.

30. Assertion: Collision with spring block may need momentum then energy. Reason: collision and compression are separate stages.

Both true.

Case Study Questions

Case studies with answers and explanations.

Case 1: Car crash and airbags. Why does airbag reduce injury?

The momentum change is almost fixed, but airbag increases stopping time. Since Favg=Δp/Δt, average force decreases. Energy is dissipated in deformation and airbag compression.

Case 2: Cricket bat collision. Ball reverses direction after impact.

Impulse from bat changes ball momentum. Sweet spot reduces vibration. Coefficient of restitution controls rebound speed; some energy is lost in sound and deformation.

Case 3: Bullet block. Bullet embeds in block then system rises.

During collision use momentum: mu=(M+m)V. During rise use energy: 1/2(M+m)V²=(M+m)gh. Do not conserve KE during embedding.

Case 4: Ball bouncing from floor.

For fixed floor, e=rebound speed/impact speed. Rebound height h2=e²h1. Energy loss is due to deformation, heat and sound.

Case 5: Newton's cradle.

Nearly elastic collisions transfer momentum and KE through identical balls. Ideal equal masses exchange velocities, so one ball emerges with similar speed.

Case 6: Railway wagon coupling.

Wagons stick, so perfectly inelastic model applies. Use total momentum to find common speed; KE loss appears as heat, sound and deformation in coupling.

Common Student Mistakes

Avoid these errors in collisions and energy applications.

Assuming KE Always Conserved

KE is conserved only in elastic collisions. Momentum is the safer first equation.

Wrong Sign Convention

Choose one positive direction and keep all velocities signed.

Approach vs Separation

e uses relative separation divided by relative approach, not individual speeds.

Forgetting e Equation

Use restitution with momentum for inelastic 1D collision.

Using Energy In Inelastic

Do not conserve KE in sticking, bullet-block or wagon coupling collision.

Ignoring 2D Vectors

Momentum is conserved component-wise in x and y.

Forgetting Energy Loss Formula

Loss = 1/2 μ(1-e²)u_rel² for direct collision.