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NEET Laws of Motion MCQs with answers

290 questions on Laws of Motion for NEET 2027. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easy

1. A ball of mass 450 g strikes a wall normally at 6 m/s and rebounds with the same speed. The magnitude of the impulse on the ball is:

  1. 2.7 N s
  2. 10.8 N s
  3. 0 N s
  4. 5.4 N s
Show answer and explanation

Answer: (D) 5.4 N s

Impulse = change in momentum = m(v − (−v)) = 2mv = 2 × 0.45 × 6 = 5.4 N s.

medium

2. A 3 kg block on a horizontal floor (coefficient of kinetic friction 0.4) is pulled horizontally with a force of 37 N. Taking g = 10 m/s², its acceleration is:

  1. 8.33 m/s²
  2. 12.3 m/s²
  3. 16.3 m/s²
  4. 4 m/s²
Show answer and explanation

Answer: (A) 8.33 m/s²

Kinetic friction = μmg = 0.4 × 3 × 10 = 12 N. a = (F − f)/m = (37 − 12)/3 = 8.33 m/s².

hard

3. A car is driving along a level circular track of radius 80 m. If the speed of the car increases at a constant tangential rate of 1.5 m/s² and the coefficient of static friction is 0.25, the speed at which the car just begins to slide is (take g = 10 m/s²):

  1. 12.6 m/s
  2. 10 m/s
  3. 14.1 m/s
  4. 16.0 m/s
Show answer and explanation

Answer: (A) 12.6 m/s

The total acceleration is a = √[a_t² + a_c²] = √[a_t² + (v²/r)²]. Slipping starts when the total acceleration exceeds μ·g = 0.25 × 10 = 2.5 m/s². Thus, (2.5)² = (1.5)² + (v²/80)², which gives (v²/80)² = 6.25 - 2.25 = 4.0. Hence v²/80 = 2.0, so v² = 160, yielding v = √160 ≈ 12.65 m/s.

medium

4. A 8 kg block on a horizontal floor (coefficient of kinetic friction 0.5) is pulled horizontally with a force of 50 N. Taking g = 10 m/s², its acceleration is:

  1. 5 m/s²
  2. 11.3 m/s²
  3. 6.25 m/s²
  4. 1.25 m/s²
Show answer and explanation

Answer: (D) 1.25 m/s²

Kinetic friction = μmg = 0.5 × 8 × 10 = 40 N. a = (F − f)/m = (50 − 40)/8 = 1.25 m/s².

easy

5. A motor vehicle moves along a circular path on a level road with a speed of 10 m/s. If the coefficient of static friction is 0.5, what is the minimum radius of the curve so that the vehicle does not skid? (Take g = 10 m/s²)

  1. 10 m
  2. 20 m
  3. 25 m
  4. 15 m
Show answer and explanation

Answer: (B) 20 m

The maximum speed is v = √(μ_s · R · g), which gives R_min = v² / (μ_s · g). Substituting the given values: R_min = (10)² / (0.5 × 10) = 100 / 5 = 20 m.

hard

6. A car negotiates an unbanked circular curve of radius R at a constant speed v. If the distance between the two wheels on the same axle is 2d and the centre of gravity of the car is at a height h above the ground, the condition to prevent overturning outward is:

  1. v² ≤ 2(g · R · d) / h
  2. v² ≤ (g · R · h) / d
  3. v² ≥ (g · R · d) / h
  4. v² ≤ (g · R · d) / h
Show answer and explanation

Answer: (D) v² ≤ (g · R · d) / h

Overturning occurs when the inner wheels lose contact with the road. Taking torque about the outer wheel contact point: mg · d ≥ (mv²/R) · h for no toppling. This simplifies directly to v² ≤ (g · R · d) / h.

medium

7. Given below are two statements: Statement I: On a level circular road, the centripetal force is provided solely by the component of normal reaction parallel to the road. Statement II: On a rough banked road, static friction can act either up or down the inclined surface depending on whether the vehicle's speed is lower or higher than the optimum speed. In the light of the above statements, choose the most appropriate answer from the options given below:

  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: (C) Statement I is incorrect but Statement II is correct

On a level road, normal reaction is perpendicular to the road and balances weight; centripetal force is provided entirely by static friction, making Statement I false. Statement II is correct because friction points up the incline if v < v₀ and down the incline if v > v₀.

medium

8. A car takes a turn on a frictionless road of radius R banked at an angle θ at its optimum speed v₀. The normal force exerted by the road on the car of mass m is:

  1. mg / cos θ
  2. mg cos θ
  3. mg sin θ
  4. mg / sin θ
Show answer and explanation

Answer: (A) mg / cos θ

Resolving forces along the vertical axis for an unaccelerated vertical state: N cos θ = mg. Therefore, the normal force exerted by the road is N = mg / cos θ.

medium

9. A 10 kg block on a horizontal floor (coefficient of kinetic friction 0.2) is pulled horizontally with a force of 60 N. Taking g = 10 m/s², its acceleration is:

  1. 6 m/s²
  2. 8 m/s²
  3. 4 m/s²
  4. 2 m/s²
Show answer and explanation

Answer: (C) 4 m/s²

Kinetic friction = μmg = 0.2 × 10 × 10 = 20 N. a = (F − f)/m = (60 − 20)/10 = 4 m/s².

medium

10. A 500 g block on a frictionless surface experiences a net force of 2 N. Its acceleration is:

  1. 2 m/s²
  2. 0.25 m/s²
  3. 4 m/s²
  4. 1 m/s²
Show answer and explanation

Answer: (C) 4 m/s²

Unit check: Convert to SI, then a = F/m = 2/0.5 = 4 m/s².

medium

11. Consider the following two statements regarding a system of interacting particles: Statement I: The linear momentum of an isolated system remains conserved even if non-conservative forces act between its constituent particles. Statement II: Internal forces between interacting particles always occur in action-reaction pairs and cancel out in the expression for net force on the system. Choose the correct option:

  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: (A) Both Statement I and Statement II are correct

The conservation of linear momentum requires only that the net external force on the system is zero (isolated system). Even if internal forces are non-conservative (like friction or inelastic deformations), internal forces sum to zero by Newton's third law. Therefore, both statements are correct.

medium

12. A cricket ball of mass 150 g moving horizontally with a speed of 20 m/s is caught by a player in 0.1 s. The average force exerted by the player's hands on the ball is:

  1. 15 N
  2. 300 N
  3. 30 N
  4. 150 N
Show answer and explanation

Answer: (C) 30 N

Initial momentum p₁ = 0.15 kg × 20 m/s = 3 kg·m/s. Final momentum p₂ = 0. The magnitude of the average force F = Δp / Δt = (3 - 0) / 0.1 = 30 N.

easy

13. Three coplanar forces of magnitudes 5 N, 12 N, and 13 N maintain a particle in static equilibrium. What is the angle between the 5 N force and the 12 N force?

  1. 45°
  2. 90°
  3. 60°
  4. 120°
Show answer and explanation

Answer: (B) 90°

For three forces in equilibrium, any one force must be equal and opposite to the resultant of the other two. Here, 5² + 12² = 25 + 144 = 169 = 13². Hence, the 5 N and 12 N forces are perpendicular to each other, making the angle between them 90°.

medium

14. A small body of weight W is placed on a smooth inclined plane of inclination angle α. To keep the body in equilibrium, a force F is applied horizontally into the incline. The magnitude of the normal reaction N exerted by the inclined plane on the body is:

  1. W / cos α
  2. W cos α
  3. W sin α
  4. W / sin α
Show answer and explanation

Answer: (A) W / cos α

Resolving forces along and perpendicular to the inclined plane: Along the incline, F cos α = W sin α, which gives F = W tan α. Perpendicular to the incline, N = W cos α + F sin α. Substituting F = W tan α gives N = W cos α + (W sin α / cos α) sin α = W (cos² α + sin² α) / cos α = W / cos α.

medium

15. Masses of 3 kg and 1 kg hang from the two ends of a light string over a frictionless pulley. Taking g = 10 m/s², the tension in the string is:

  1. 15 N
  2. 10 N
  3. 30 N
  4. 20 N
Show answer and explanation

Answer: (A) 15 N

a = (m₁ − m₂)g/(m₁ + m₂) = 5 m/s². T = 2m₁m₂g/(m₁ + m₂) = 2 × 3 × 1 × 10/4 = 15 N.

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