Laws of Motion NEET PYQs (2005–2026) with Solutions

10 solved Laws of Motion PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026EasyFriction

   A box of mass 15 kg is kept on the floor of a stationary trolley. The coefficient of static friction between the box and the trolley is 0.12. Keeping the box in stationary state over the trolley, the maximum acceleration with which the trolley can be moved horizontally in m s⁻² is: (g = 10 m/s²)

   1. AOption A: 2.1
   2. BOption B: 1.8
   3. COption C: 1.5
   4. DOption D: 1.2

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   Correct answer: (D) 1.2

   Explanation

   For the box to remain at rest relative to the trolley, static friction must provide the required acceleration. Maximum static friction: fₘₐₓ = μₛN Since: N = mg Maximum acceleration occurs when: ma = μₛmg Therefore: a = μₛg Substitute: μₛ = 0.12, g = 10 m/s² ⇒ a = 0.12 × 10 ⇒ a = 1.2 m/s² Hence option (D) is correct.

2. Question 2 (NEET 2026)

   NEET 2026EasyEquilibrium of Particles

   The magnitude and direction of the acceleration produced in a body of mass 5 kg when two mutually perpendicular forces 8 N and 6 N act on it, are respectively:

   1. AOption A: 20 m s⁻²; tan⁻¹(4/3) with 8 N force
   2. BOption B: 2 m s⁻²; tan⁻¹(3/4) with 6 N force
   3. COption C: 2 m s⁻²; tan⁻¹(4/3) with 8 N force
   4. DOption D: 2 m s⁻²; tan⁻¹(3/4) with 8 N force

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   Correct answer: (D) 2 m s⁻²; tan⁻¹(3/4) with 8 N force

   Explanation

   Resultant force: F = √(8²+6²)=10 N. Acceleration: a=F/m=10/5=2 m s⁻². Direction with respect to 8 N force: tanθ=6/8=3/4, therefore θ=tan⁻¹(3/4). Hence option (D) is correct.

3. Question 3 (NEET 2026)

   NEET 2026EasyMotion of a car on a banked road

   A car travels on a circular racetrack of radius $50\ \mathrm{m}$, which is banked at an angle $\theta$. If the car travels at a speed $10\ \mathrm{ms^{-1}}$, then the wear and tear on its tyres is minimum. Taking the acceleration due to gravity to be $10\ \mathrm{ms^{-2}}$, the value of $\theta$ is:

   1. AOption A: $\tan^{-1}\left(\frac{1}{5}\right)$
   2. BOption B: $\tan^{-1}\left(\frac{2}{5}\right)$
   3. COption C: $\tan^{-1}(\sqrt{3}/2)$
   4. DOption D: $\tan^{-1}(2\sqrt{3})$

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   Correct answer: (A) $\tan^{-1}\left(\frac{1}{5}\right)$

   Explanation

   Wear and tear is minimum when no friction is needed, i.e. at the optimum speed of the banked road: $v^2 = rg\tan\theta$. $\tan\theta = \frac{v^2}{rg} = \frac{10^2}{50 \times 10} = \frac{1}{5}$. $\theta = \tan^{-1}\left(\frac{1}{5}\right)$, option (A).

4. Question 4 (NEET 2025)

   NEET 2025MediumFriction

   There are two inclined surfaces of equal length (L) and same angle of inclination 45° with the horizontal. One of them is rough and the other is perfectly smooth. A given body takes 2 times as much time to slide down on rough surface than on the smooth surface. The coefficient of kinetic friction (μk) between the object and the rough surface is close to

   1. AOption A: 0.25
   2. BOption B: 0.40
   3. COption C: 0.50
   4. DOption D: 0.75

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   Correct answer: (D) 0.75

   Explanation

   For smooth incline, acceleration a₁ = g sin45°. For rough incline, acceleration a₂ = g(sin45° − μk cos45°). Since time t ∝ 1/√a and rough surface takes twice the time, a₂ = a₁/4. Using sin45° = cos45°, we get 1 − μk = 1/4, hence μk = 3/4 = 0.75.

5. Question 5 (NEET 2025)

   NEET 2025EasyMomentum & Impulse

   A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take g = 9.8 m/s²)

   1. AOption A: 21 Ns
   2. BOption B: 7 Ns
   3. COption C: 0
   4. DOption D: 84 Ns

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   Correct answer: (A) 21 Ns

   Explanation

   Speed before collision: v₁ = √(2gh) = √(2×9.8×40) = 28 m/s downward. Speed after rebound: v₂ = √(2×9.8×10) = 14 m/s upward. Impulse equals change in momentum: J = m(v₂ − (−v₁)) = 0.5 × (14 + 28) = 21 Ns.

6. Question 6 (NEET 2025)

   NEET 2025MediumEquilibrium of Particles

   A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60° with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (g = 10 m/s²)

   1. AOption A: 100 N
   2. BOption B: 100√3 N
   3. COption C: 200 N
   4. DOption D: 200√3 N

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   Correct answer: (B) 100√3 N

   Explanation

   The rod makes 30° with the floor. Taking moments about the lower end: N × (5 sin30°) = mg × (2.5 cos30°). With mg = 200 N, the wall reaction N = 100√3 N. Horizontal equilibrium gives friction force equal to the wall reaction. Hence friction = 100√3 N.

7. Question 7 (NEET 2024)

   NEET 2024EasyCircular Motion Applications

   A bob is whirled in a horizontal plane by means of a string with an initial speed of ω rpm. The tension in the string is T. If speed becomes 2ω while keeping the same radius, the tension in the string becomes:

   1. AOption A: T/4
   2. BOption B: √2 T
   3. COption C: T
   4. DOption D: 4T

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   Correct answer: (D) 4T

   Explanation

   Centripetal force is provided by tension: $$T = \frac{mv^2}{r}$$ If speed becomes double: $$T' = \frac{m(2v)^2}{r} = \frac{4mv^2}{r} = 4T$$ Hence, option (D) is correct.

8. Question 8 (NEET 2024)

   NEET 2024MediumMomentum & Impulse

   Two bodies A and B of same mass undergo completely inelastic one dimensional collision. The body A moves with velocity v₁ while body B is at rest before collision. The velocity of the system after collision is v₂. The ratio v₁ : v₂ is:

   1. AOption A: 4 : 1
   2. BOption B: 1 : 4
   3. COption C: 1 : 2
   4. DOption D: 2 : 1

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   Correct answer: (D) 2 : 1

   Explanation

   In a completely inelastic collision, the bodies move together after collision. Using conservation of momentum: $$mv_1 + m(0) = (m+m)v_2$$ $$mv_1 = 2mv_2$$ $$v_1 = 2v_2$$ Therefore: $$v_1 : v_2 = 2 : 1$$ Hence, option (D) is correct.

9. Question 9 (NEET 2024)

   NEET 2024MediumForce, Inertia & Newton's Laws

   A horizontal force 10 N is applied to a block A as shown in figure. The mass of blocks A and B are 2 kg and 3 kg, respectively. The blocks slide over a frictionless surface. The force exerted by block A on block B is:

   1. AOption A: 6 N
   2. BOption B: 10 N
   3. COption C: zero
   4. DOption D: 4 N

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   Correct answer: (A) 6 N

   Explanation

   Total mass: $$M = 2 + 3 = 5\,kg$$ Acceleration of the system: $$a = \frac{F}{M} = \frac{10}{5} = 2\,m\,s^{-2}$$ Force exerted by A on B: $$F_{AB} = m_B a = 3 \times 2 = 6\,N$$ Hence, option (A) is correct.

10. Question 10 (NEET 2023)

    NEET 2023MediumRetardation and Motion in a Medium

    A bullet from a gun is fired on a rectangular wooden block with velocity $u$. When bullet travels 24 cm through the block along its length horizontally, velocity of bullet becomes $\frac{u}{3}$. Then it further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is:

    1. AOption A: 27 cm
    2. BOption B: 24 cm
    3. COption C: 28 cm
    4. DOption D: 30 cm

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    Correct answer: (A) 27 cm

    Explanation

    Assuming constant retardation inside the wooden block: Using equation of motion: $$v^2 = u^2 + 2as$$ After travelling 24 cm: $$\left(\frac{u}{3}\right)^2 = u^2 + 2a(24)$$ $$\frac{u^2}{9} - u^2 = 48a$$ $$a = -\frac{u^2}{54}$$ For the total penetration distance $S$before stopping:$$0 = u^2 + 2aS$$$$S = \frac{u^2}{-2a} = 27\text{ cm}$$ Hence the total length of the block is 27 cm.