Collision – Elastic and inelastic collisions in one and two dimensions: NEET MCQs with Answers

15 free questions on Collision – Elastic and inelastic collisions in one and two dimensions for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Elastic and inelastic collisions in one and two dimensions.

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Question 1

Previous-year question
A bullet of mass m hits a block of mass M elastically. The transfer of energy is maximum when: [NEET 2023]
  1. M = m
  2. M = 2m
  3. M
  4. M >> m
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Answer: (A) M = m
Explanation: The fraction of energy transferred in a head-on elastic collision is 4mM/(m + M)². Treating this as a function of the mass ratio, it reaches its maximum value of 1 when M = m, and falls away for any other ratio. Physically, equal masses exchange velocities completely, so the bullet stops and the block takes all the energy — the principle behind using light nuclei such as hydrogen to moderate neutrons.

Question 2

Previous-year question
A ball of mass m moving with some velocity strikes another ball of mass 2m at rest. If the collision is head-on and the final kinetic energy of the system is half the initial kinetic energy of the system, then the value of the coefficient of restitution e is: [NEET 2021]
  1. 1/2
  2. 1/√2
  3. 1/3
  4. 1/√3
Show answer
Answer: (A) 1/2
Explanation: Using the general one-dimensional results with restitution, v₁ = u(1 − 2e)/3 and v₂ = u(1 + e)/3. The final kinetic energy is ½mv₁² + ½(2m)v₂² = (mu²/9)[½(1 − 2e)² + (1 + e)²]. Setting this equal to half of ½mu² gives ½(1 − 2e)² + (1 + e)² = 9/4, which simplifies to 1.5 + 3e² = 2.25. Hence e² = 1/4 and e = 1/2.

Question 3

Previous-year question
A moving block of mass m collides with a stationary block of mass 4m. The lighter block comes to rest after the collision. If the initial velocity of the lighter block is v, the coefficient of restitution e is [NEET 2018]
  1. 0.5
  2. 0.25
  3. 0.4
  4. 0.8
Show answer
Answer: (B) 0.25
Explanation: Momentum conservation gives the speed of the heavier block as v₂ = mv/4m = v/4. Hence e = (velocity of separation)/(velocity of approach) = (v/4 – 0)/v = 0.25.

Question 4

Previous-year question
A body of mass 4m lying in the x-y plane at rest suddenly explodes into three pieces. Two pieces, each of mass m, move perpendicular to each other with equal speeds v. The total kinetic energy generated due to the explosion is [AIPMT 2014]
  1. mv²
  2. (3/2)mv²
  3. 2mv²
  4. 4mv²
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Answer: (B) (3/2)mv²
Explanation: The third piece of mass 2m must carry momentum √2 mv, so its speed is v/√2. Total KE = 2(½mv²) + ½(2m)(v²/2) = mv² + ½mv² = (3/2)mv².

Question 5

Previous-year question
The coefficient of restitution e for a perfectly elastic collision is [AIPMT 1988]
  1. 1
  2. zero
  3. infinite
  4. −1
Show answer
Answer: (A) 1
Explanation: e is defined as the ratio of the relative velocity of separation after the collision to the relative velocity of approach before it. In a perfectly elastic collision kinetic energy is conserved, which requires the relative speed of separation to equal the relative speed of approach. Hence e = 1. (e = 0 corresponds to a perfectly inelastic collision.)

Question 6

Previous-year question
A body of mass m moving with velocity 3 km/h collides with a body of mass 2m at rest. The two coalesce, and the combined mass then begins to move with a velocity of [AIPMT 1996]
  1. 1 km/h
  2. 2 km/h
  3. 3 km/h
  4. 4 km/h
Show answer
Answer: (A) 1 km/h
Explanation: Momentum conservation for a perfectly inelastic collision: m(3) + 2m(0) = (m + 2m)v 3m = 3mv ⟹ v = 1 km/h.

Question 7

Previous-year question
A metal ball of mass 2 kg moving with a velocity of 36 km/h collides with a stationary ball of mass 3 kg. If the two balls move together after the collision, the loss in kinetic energy due to the collision is [AIPMT 1997]
  1. 140 J
  2. 100 J
  3. 60 J
  4. 40 J
Show answer
Answer: (C) 60 J
Explanation: 36 km/h = 36 × 5/18 = 10 m/s. Momentum conservation: (2 × 10) + 0 = (2 + 3)v ⟹ v = 20/5 = 4 m/s. KE before = ½ × 2 × (10)² = 100 J. KE after = ½ × 5 × (4)² = 40 J. Loss = 100 − 40 = 60 J.
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Question 8

Previous-year question
Two equal masses m₁ and m₂ moving along the same straight line with velocities +3 m/s and −5 m/s respectively collide elastically. Their velocities after the collision will respectively be [AIPMT 1998]
  1. +4 m/s for both
  2. −3 m/s and +5 m/s
  3. −4 m/s and +4 m/s
  4. −5 m/s and +3 m/s
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Answer: (D) −5 m/s and +3 m/s
Explanation: In a one-dimensional elastic collision between equal masses, the velocities are exchanged. Hence m₁ moves off with −5 m/s and m₂ with +3 m/s. (Check: momentum before = 3 − 5 = −2; after = −5 + 3 = −2 ✓, and the KE is also unchanged.)

Question 9

Previous-year question
A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of the 18 kg fragment is 6 m/s. The kinetic energy of the other fragment is [AIPMT 2005]
  1. 256 J
  2. 486 J
  3. 524 J
  4. 324 J
Show answer
Answer: (B) 486 J
Explanation: Momentum conservation (initial momentum zero): 18 × 6 = 12 × u₂ ⟹ u₂ = 108/12 = 9 m/s. KE of the 12 kg piece = ½ × 12 × (9)² = 6 × 81 = 486 J.

Question 10

Previous-year question
A ball moving with velocity 2 m/s collides head-on with another stationary ball of double its mass. If the coefficient of restitution is 0.5, their velocities (in m/s) after the collision will be [AIPMT 2010]
  1. 0, 1
  2. 1, 1
  3. 1, 0.5
  4. 0, 2
Show answer
Answer: (A) 0, 1
Explanation: Using the standard results with m₁ = m, m₂ = 2m, u₁ = 2 m/s, u₂ = 0 and e = 0.5 : v₁ = [(m₁ − em₂)/(m₁ + m₂)]u₁ = [(m − 0.5 × 2m)/3m](2) = 0 v₂ = [(1 + e)m₁/(m₁ + m₂)]u₁ = [(1.5m)/3m](2) = 1 m/s. So the first ball stops and the heavier ball moves off at 1 m/s.

Question 11

Previous-year question
A mass m moving horizontally along the X-axis with velocity v collides with and sticks to a mass 3m moving vertically upward along the Y-axis with velocity 2v. The final velocity of the combination is [AIPMT 2011]
  1. (3/2)v î + (1/4)v ĵ
  2. (1/4)v î + (3/2)v ĵ
  3. (1/3)v î + (2/3)v ĵ
  4. (2/3)v î + (1/3)v ĵ
Show answer
Answer: (B) (1/4)v î + (3/2)v ĵ
Explanation: Total initial momentum = mv î + (3m)(2v) ĵ = mv î + 6mv ĵ. Combined mass = 4m, so V = (mv î + 6mv ĵ)/4m = (v/4) î + (3v/2) ĵ.

Question 12

Previous-year question
Two spheres A and B of masses m₁ and m₂ respectively collide. A is initially at rest and B is moving with velocity v along the X-axis. After the collision B has velocity v/2 in a direction perpendicular to its original motion. Sphere A then moves in the direction [AIPMT 2012]
  1. same as that of B
  2. opposite to that of B
  3. θ = tan⁻¹(1/2) to the X-axis
  4. θ = tan⁻¹(−1/2) to the X-axis
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Answer: (D) θ = tan⁻¹(−1/2) to the X-axis
Explanation: Along X: m₂v = m₁v_x ⟹ v_x = m₂v/m₁. Along Y: 0 = m₂(v/2) + m₁v_y ⟹ v_y = −m₂v/(2m₁). tan θ = v_y/v_x = [−m₂v/(2m₁)] / [m₂v/m₁] = −1/2 ⟹ θ = tan⁻¹(−1/2) with respect to the X-axis.

Question 13

Previous-year question
On a frictionless surface, a block of mass M moving at speed v collides elastically with another block of the same mass M initially at rest. After the collision the first block moves at an angle θ to its initial direction with a speed v/3. The second block's speed after the collision is [AIPMT 2015]
  1. 2√2 v/3
  2. 3v/4
  3. 3v/√2
  4. √3 v/2
Show answer
Answer: (A) 2√2 v/3
Explanation: For an elastic collision, kinetic energy is conserved: ½Mv² = ½M(v/3)² + ½Mv₂² v² = v²/9 + v₂² v₂² = 8v²/9 v₂ = (2√2/3)v.

Question 14

Previous-year question
Two particles A and B move with constant velocities v₁ and v₂. At the initial moment their position vectors are r₁ and r₂ respectively. The condition for particles A and B to collide is [AIPMT 2015]
  1. r₁ × v₁ = r₂ × v₂
  2. r₁ − r₂ = v₁ − v₂
  3. (r₁ − r₂)/|r₁ − r₂| = (v₂ − v₁)/|v₂ − v₁|
  4. r₁ · v₁ = r₂ · v₂
Show answer
Answer: (C) (r₁ − r₂)/|r₁ − r₂| = (v₂ − v₁)/|v₂ − v₁|
Explanation: The particles meet at some instant t when their position vectors coincide: r₁ + v₁t = r₂ + v₂t ⟹ r₁ − r₂ = (v₂ − v₁)t. Taking magnitudes, |r₁ − r₂| = |v₂ − v₁|t. Dividing the vector equation by this, (r₁ − r₂)/|r₁ − r₂| = (v₂ − v₁)/|v₂ − v₁|, i.e. the initial separation vector and the relative velocity must point along the same unit vector.

Question 15

Previous-year question
Two particles of masses m₁ and m₂ approach each other with initial velocities u₁ and u₂. During the collision one of the particles is raised to a higher energy level by absorbing energy ε. If the final velocities of the particles are v₁ and v₂, then we must have [AIPMT 2015]
  1. m₁²u₁ + m₂²u₂ − ε = m₁²v₁ + m₂²v₂
  2. ½m₁u₁² + ½m₂u₂² = ½m₁v₁² + ½m₂v₂² − ε
  3. ½m₁u₁² + ½m₂u₂² − ε = ½m₁v₁² + ½m₂v₂²
  4. ½m₁²u₁² + ½m₂²u₂² + ε = ½m₁²v₁² + ½m₂²v₂²
Show answer
Answer: (C) ½m₁u₁² + ½m₂u₂² − ε = ½m₁v₁² + ½m₂v₂²
Explanation: Energy ε is taken out of the mechanical (kinetic) energy pool and stored as internal excitation energy. Therefore the kinetic energy after the collision must be smaller by exactly ε : (Initial KE) − ε = (Final KE) ½m₁u₁² + ½m₂u₂² − ε = ½m₁v₁² + ½m₂v₂².

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