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Momentum, impulse and conservation of linear momentum: NEET MCQs with answers

75 questions on this topic of Laws of Motion (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. A body of mass 4 kg is initially at rest. A force acting on it varies with time according to the relation F = (12 - 3t) N from t = 0 until the force drops to zero. The final speed acquired by the body is:

  1. 9 m/s
  2. 6 m/s
  3. 12 m/s
  4. 3 m/s
Show answer and explanation

Answer: (B) 6 m/s

The force drops to zero when 12 - 3t = 0, which gives t = 4 s. Impulse delivered to the body is the integral of F dt from t = 0 to 4 s: J = ∫(12 - 3t) dt = [12t - 1.5t²] from 0 to 4 = 48 - 24 = 24 N·s. From impulse-momentum theorem, J = m(v - u) = 4(v - 0) = 24, giving v = 6 m/s.

medium

2. A rocket of mass 1200 kg is ejecting gases at a constant speed of 600 m/s relative to the rocket. To give the rocket an initial upward acceleration of 20 m/s², the rate of combustion of fuel must be (take g = 10 m/s²):

  1. 40 kg/s
  2. 20 kg/s
  3. 60 kg/s
  4. 80 kg/s
Show answer and explanation

Answer: (C) 60 kg/s

The net upward equation of motion for the rocket is F_thrust - mg = ma, which gives F_thrust = m(g + a). Here F_thrust = u_rel × (dm/dt). Therefore, 600 × (dm/dt) = 1200 × (10 + 20) = 1200 × 30 = 36000 N. This gives dm/dt = 36000 / 600 = 60 kg/s.

hard

3. A helicopter of mass 2000 kg hovers stationary in air. It pushes down a column of air of cross-sectional area 20 m² with a uniform downward velocity v. Taking the density of air as 1.25 kg/m³ and g = 10 m/s², the value of v is:

  1. 20 m/s
  2. 40 m/s
  3. 28.3 m/s
  4. 14.1 m/s
Show answer and explanation

Answer: (C) 28.3 m/s

For hovering, the upward thrust on the helicopter balances its weight: Thrust = ρ A v² = Mg. Therefore, v² = Mg / (ρ A) = (2000 × 10) / (1.25 × 20) = 20000 / 25 = 800. Hence, v = √800 = 20√2 ≈ 28.3 m/s.

medium

4. Given below are two statements: Statement I: In any collision between two bodies in an isolated system, the total linear momentum is always conserved, regardless of whether kinetic energy is conserved or not. Statement II: Internal forces between colliding particles can alter the total momentum of the isolated system over the duration of the impact. In light of the above statements, choose the most appropriate answer:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct because by Newton's third law, internal forces cancel in pairs, and with no external force, linear momentum is conserved in all collisions. Statement II is incorrect because internal forces occur in equal and opposite action-reaction pairs and have zero sum, hence they can never change the total momentum of an isolated system.

easy

5. A ball of mass 100 g strikes a wall normally at 19 m/s and rebounds with the same speed. The magnitude of the impulse on the ball is:

  1. 1.9 N s
  2. 0 N s
  3. 7.6 N s
  4. 3.8 N s
Show answer and explanation

Answer: (D) 3.8 N s

Impulse = change in momentum = m(v − (−v)) = 2mv = 2 × 0.1 × 19 = 3.8 N s.

hard

6. A projectile of mass 3m is fired with velocity u at an angle of 45° to the horizontal. At the highest point of its trajectory, it explodes into two fragments of masses m and 2m. If the fragment of mass m immediately falls vertically downward with zero horizontal speed, the horizontal distance from the point of projection where the other fragment lands is:

  1. 3u² / (2g)
  2. 2u² / g
  3. 5u² / (4g)
  4. 7u² / (4g)
Show answer and explanation

Answer: (C) 5u² / (4g)

The horizontal range of the unexploded projectile is R = u² sin(2×45°)/g = u²/g. The explosion occurs at the highest point, where x = R/2 and horizontal speed of the system was u cos 45°. By momentum conservation along the horizontal: 3m(u cos 45°) = m(0) + 2m(v_x), giving v_x = 1.5(u cos 45°). Since the remaining fragment takes the same time t = T/2 to reach the ground, the additional horizontal distance it covers is Δx = v_x(T/2) = 1.5(u cos 45°)(T/2) = 1.5(R/2) = 0.75 R. Total landing distance from origin = 0.5 R + 0.75 R = 1.25 R = 5u² / (4g).

easy

7. A ball of mass 250 g strikes a wall normally at 12 m/s and rebounds with the same speed. The magnitude of the impulse on the ball is:

  1. 12 N s
  2. 0 N s
  3. 3 N s
  4. 6 N s
Show answer and explanation

Answer: (D) 6 N s

Impulse = change in momentum = m(v − (−v)) = 2mv = 2 × 0.25 × 12 = 6 N s.

easy

8. A ball of mass 300 g strikes a wall normally at 7 m/s and rebounds with the same speed. The magnitude of the impulse on the ball is:

  1. 2.1 N s
  2. 0 N s
  3. 8.4 N s
  4. 4.2 N s
Show answer and explanation

Answer: (D) 4.2 N s

Impulse = change in momentum = m(v − (−v)) = 2mv = 2 × 0.3 × 7 = 4.2 N s.

medium

9. A hunter of mass 60 kg stands on a frictionless ice surface and fires a rifle of mass 4 kg horizontally. The bullet of mass 40 g leaves the muzzle at 600 m/s relative to the ground. The recoil velocity of the hunter-rifle system is:

  1. 0.040 m/s
  2. 0.400 m/s
  3. 0.375 m/s
  4. 0.250 m/s
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Answer: (C) 0.375 m/s

By conservation of linear momentum, the total momentum before firing is zero. Thus: (M_hunter + M_rifle) V + m_bullet v_bullet = 0. We have (60 + 4) V + (0.040 kg)(600 m/s) = 0, giving 64 V + 24 = 0. Hence, V = -24 / 64 = -0.375 m/s. The magnitude is 0.375 m/s.

easy

10. A ball of mass 100 g strikes a wall normally at 10 m/s and rebounds with the same speed. The magnitude of the impulse on the ball is:

  1. 1 N s
  2. 4 N s
  3. 0 N s
  4. 2 N s
Show answer and explanation

Answer: (D) 2 N s

Impulse = change in momentum = m(v − (−v)) = 2mv = 2 × 0.1 × 10 = 2 N s.

easy

11. A ball of mass 400 g strikes a wall normally at 18 m/s and rebounds with the same speed. The magnitude of the impulse on the ball is:

  1. 14.4 N s
  2. 0 N s
  3. 28.8 N s
  4. 7.2 N s
Show answer and explanation

Answer: (A) 14.4 N s

Impulse = change in momentum = m(v − (−v)) = 2mv = 2 × 0.4 × 18 = 14.4 N s.

easy

12. A time-dependent force F = (20t + 5) N acts on a particle of mass 2 kg for a duration from t = 0 to t = 2 s. The magnitude of the impulse imparted to the particle during this time interval is:

  1. 40 N·s
  2. 50 N·s
  3. 25 N·s
  4. 30 N·s
Show answer and explanation

Answer: (B) 50 N·s

Impulse is given by the integral of F dt from t = 0 to t = 2 s. J = ∫(20t + 5) dt = [10t² + 5t] from 0 to 2 = 10(4) + 5(2) = 40 + 10 = 50 N·s.

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