Impulse and Conservation of Momentum – Law of conservation of linear momentum and its applications: NEET MCQs with Answers

15 free questions on Impulse and Conservation of Momentum – Law of conservation of linear momentum and its applications 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: Law of conservation of linear momentum and its applications.

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Question 1

Previous-year question
A particle of mass 4M kg at rest splits into two particles of mass M and 3M. The ratio of the kinetic energies of mass M and 3M would be: [NEET 2022]
  1. 1 : 4
  2. 1 : 1
  3. 1 : 3
  4. 3 : 1
Show answer
Answer: (D) 3 : 1
Explanation: Both fragments carry momenta of equal magnitude p because the parent was at rest. Writing kinetic energy as E = p²/2m shows that energy is inversely proportional to mass, so E_M : E_3M = (1/M) : (1/3M) = 3 : 1. The lighter fragment therefore takes three-quarters of the released energy — the same reason a bullet, not the gun, carries most of the energy.

Question 2

Previous-year question
A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly off in mutually perpendicular directions with a speed of 21 m/s. The velocity of the heaviest fragment, in m/s, will be [AIPMT 1989]
  1. 3√2
  2. 7√2
  3. √2
  4. 5√2
Show answer
Answer: (B) 7√2
Explanation: With masses in the ratio 1 : 1 : 3 and a total of 5 kg, the fragments are 1 kg, 1 kg and 3 kg. Momentum of each of the two equal fragments: p₁ = p₂ = 1 × 21 = 21 kg m s⁻¹ Since these are mutually perpendicular, their resultant is p = √(p₁² + p₂²) = √(21² + 21²) = 21√2 kg m s⁻¹ The body was initially at rest, so the heaviest fragment must carry an equal and opposite momentum: 3v = 21√2 ⇒ v = 7√2 m/s.

Question 3

Previous-year question
A particle of mass m is moving with a uniform velocity v₁. It is given an impulse such that its velocity becomes v₂. The impulse is equal to [AIPMT 1990]
  1. m[|v₂| − |v₁|]
  2. ½m(v₂² − v₁²)
  3. m(v₁ + v₂)
  4. m(v₂ − v₁)
Show answer
Answer: (D) m(v₂ − v₁)
Explanation: Impulse equals the change in linear momentum: I = mv₂ − mv₁ = m(v₂ − v₁). This is a vector difference of the velocities, not a difference of their magnitudes, and an expression like ½m(v₂² − v₁²) has the dimensions of energy.

Question 4

Previous-year question
A man fires a bullet of mass 200 g at a speed of 5 m/s. The gun is of one kg mass. By what velocity does the gun rebound backwards? [AIPMT 1996]
  1. 1 m/s
  2. 0.01 m/s
  3. 0.1 m/s
  4. 10 m/s
Show answer
Answer: (A) 1 m/s
Explanation: The gun and bullet are initially at rest, so the total momentum is zero and must remain zero. m_gun v_gun = m_bullet v_bullet (1)v = (0.2)(5) v = 1 m/s. The gun recoils at 1 m/s in the direction opposite to the bullet. Its much larger mass is what keeps the recoil speed small compared with the bullet's.

Question 5

Previous-year question
A bullet is fired from a gun. The force on the bullet is given by F = 600 − 2 × 10⁵ t, where F is in newton and t in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet? [AIPMT 1998]
  1. 8 N-s
  2. zero
  3. 0.9 N-s
  4. 1.8 N-s
Show answer
Answer: (C) 0.9 N-s
Explanation: First find the time for which the force acts, i.e. when F = 0: 600 − 2 × 10⁵ t = 0 ⇒ t = 600/(2 × 10⁵) = 3 × 10⁻³ s Impulse is the time integral of the force: I = ∫₀^(3×10⁻³) (600 − 2 × 10⁵ t) dt = [600t − 10⁵ t²]₀^(3×10⁻³) = 600(3 × 10⁻³) − 10⁵(3 × 10⁻³)² = 1.8 − 10⁵(9 × 10⁻⁶) = 1.8 − 0.9 = 0.9 N-s. Equivalently, the force falls linearly from 600 N to 0, so its mean value is 300 N, and 300 × 3 × 10⁻³ = 0.9 N-s. ✔

Question 6

Previous-year question
A particle of mass 1 kg is thrown vertically upwards with a speed of 100 m/s. After 5 s it explodes into two parts. One part of mass 400 g comes back with a speed of 25 m/s. What is the speed of the other part just after the explosion? [AIPMT 2000]
  1. 100 m/s upwards
  2. 600 m/s upwards
  3. 100 m/s downwards
  4. 300 m/s upwards
Show answer
Answer: (A) 100 m/s upwards
Explanation: Just before the explosion the velocity is v = 100 − 10 × 5 = 50 m/s upwards. Momentum conservation with the 0.4 kg piece moving down at 25 m/s gives 1 × 50 = 0.4(−25) + 0.6v₂, so 0.6v₂ = 60 and v₂ = 100 m/s upwards.

Question 7

Previous-year question
If a cricketer catches a ball of mass 150 g moving with a velocity of 20 m/s, then he experiences a force of (time taken to complete the catch is 0.1 s) [AIPMT 2001]
  1. 300 N
  2. 30 N
  3. 3 N
  4. 0.3 N
Show answer
Answer: (B) 30 N
Explanation: Force is the rate of change of momentum: F = m(v − u)/Δt = 0.15 × (0 − 20)/0.1 = −30 N The magnitude of the force experienced is 30 N; the negative sign simply shows that it opposes the ball's motion, i.e. the ball does work against the cricketer's hands.
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Question 8

Previous-year question
An explosion blows a rock into three parts. Two parts go off at right angles to each other: a 1 kg first part moving with a velocity of 12 m s⁻¹ and a 2 kg second part moving with a velocity of 8 m s⁻¹. If the third part flies off with a velocity of 4 m s⁻¹, its mass would be [AIPMT 2009]
  1. 5 kg
  2. 7 kg
  3. 17 kg
  4. 3 kg
Show answer
Answer: (A) 5 kg
Explanation: Since the rock was initially at rest, the vector sum of the momenta of the three fragments must be zero. Momentum of the first part = 1 × 12 = 12 kg m s⁻¹ Momentum of the second part = 2 × 8 = 16 kg m s⁻¹ These are perpendicular, so their resultant is √(12² + 16²) = √(144 + 256) = 20 kg m s⁻¹ The third part must have momentum of the same magnitude, directed oppositely: 4 × M = 20 ⇒ M = 5 kg.

Question 9

Previous-year question
A man of 50 kg mass is standing in a gravity-free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed of 2 m s⁻¹. When the stone reaches the floor, the distance of the man above the floor will be [AIPMT 2010]
  1. 9.9 m
  2. 10.1 m
  3. 10 m
  4. 20 m
Show answer
Answer: (B) 10.1 m
Explanation: In gravity-free space the total momentum is conserved and stays zero. When the man throws the stone downwards, he recoils upwards: 50 v = 0.5 × 2 ⇒ v = 0.02 m s⁻¹ upwards The stone travels the 10 m to the floor at a constant 2 m s⁻¹ (no gravity to accelerate it): t = 10/2 = 5 s In this time the man drifts upwards through d = 0.02 × 5 = 0.1 m Hence his distance above the floor becomes 10 + 0.1 = 10.1 m.

Question 10

Previous-year question
A body of mass M hits a rigid wall normally with velocity v and bounces back with the same velocity. The impulse experienced by the body is [AIPMT 2011]
  1. Mv
  2. 1.5 Mv
  3. 2 Mv
  4. zero
Show answer
Answer: (C) 2 Mv
Explanation: Impulse equals the change in linear momentum. Taking the initial direction as positive: Initial momentum = +Mv Final momentum = −Mv (the body rebounds) Impulse = final − initial = (−Mv) − (Mv) = −2Mv The magnitude of the impulse experienced by the body is therefore 2Mv.

Question 11

Previous-year question
An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part, of mass 1 kg, moves with a speed of 12 m s⁻¹ and the second part, of mass 2 kg, moves with a speed of 8 m s⁻¹. If the third part flies off with a speed of 4 m s⁻¹, then its mass is [NEET 2013]
  1. 3 kg
  2. 5 kg
  3. 7 kg
  4. 17 kg
Show answer
Answer: (B) 5 kg
Explanation: The rock was initially at rest, so the total momentum after the explosion must also be zero. Momentum of the first part: p₁ = 1 × 12 = 12 kg m s⁻¹ Momentum of the second part: p₂ = 2 × 8 = 16 kg m s⁻¹ Since these two are at right angles, their resultant is p = √(p₁² + p₂²) = √(144 + 256) = √400 = 20 kg m s⁻¹ The third part must carry an equal and opposite momentum of 20 kg m s⁻¹: m₃ = p₃/v₃ = 20/4 = 5 kg.

Question 12

Previous-year question
A bullet of mass 10 g moving horizontally with a velocity of 400 m/s strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise through a vertical distance of 10 cm. The speed of the bullet after it emerges horizontally from the block will be [NEET 2016]
  1. 100 m/s
  2. 80 m/s
  3. 120 m/s
  4. 160 m/s
Show answer
Answer: (C) 120 m/s
Explanation: The block rises 0.1 m, so its speed just after impact is v = √(2gh) = √(2 × 10 × 0.1) = √2 m/s. Momentum conservation gives 0.01 × 400 = 2√2 + 0.01v′, so 0.01v′ = 4 − 2.83 = 1.17 and v′ ≈ 120 m/s.

Question 13

Previous-year question
A rigid ball of mass m strikes a rigid wall at an angle of 60° with the normal to the wall and is reflected without any loss of speed. The magnitude of the impulse imparted by the wall on the ball will be [NEET 2016] △ The original question carries a diagram; the arrangement it shows is described in full in the wording above, so no figure is needed to solve it.
Figure for question 13
  1. mv
  2. 2mv
  3. mv/2
  4. mv/3
Show answer
Answer: (A) mv
Explanation: Only the velocity component along the normal reverses. With the ball striking at 60° to the normal, that component is v cos 60° = v/2, so the impulse is J = 2m(v/2) = mv.

Question 14

Previous-year question
An object flying in air with velocity (20î + 25ĵ − 12k̂) suddenly breaks into two pieces whose masses are in the ratio 1 : 5. The smaller mass flies off with a velocity (100î + 35ĵ + 8k̂). The velocity of the larger piece will be [Odisha NEET 2019]
  1. 4î + 23ĵ − 16k̂
  2. −100î − 35ĵ − 8k̂
  3. 20î + 15ĵ − 80k̂
  4. −20î − 15ĵ − 80k̂
Show answer
Answer: (A) 4î + 23ĵ − 16k̂
Explanation: Let the masses be m and 5m, so the total mass is 6m. By conservation of linear momentum: 6m v = m v₁ + 5m v₂ 6(20î + 25ĵ − 12k̂) = (100î + 35ĵ + 8k̂) + 5v₂ 120î + 150ĵ − 72k̂ − 100î − 35ĵ − 8k̂ = 5v₂ 20î + 115ĵ − 80k̂ = 5v₂ v₂ = 4î + 23ĵ − 16k̂.

Question 15

Every second, N small balls, each of mass m, strike a wall normally with speed u and bounce straight back with the same speed. What average force do they exert on the wall?
  1. Nmu
  2. Nmu/2
  3. 2Nmu
  4. 4Nmu
Show answer
Answer: (C) 2Nmu
Explanation: Each ball's momentum changes from mu towards the wall to mu away from it, a change of 2mu. N balls strike per second, so the momentum change per second is 2Nmu. By the third law this is the average force on the wall.

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