NEET Laws of Motion PYQs & Practice Questions

10 free NEET Physics questions on Laws of Motion, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Friction on Inclined Plane · Impulse-Momentum Theorem · Newton's First Law and Inertia · Newton's Second Law and F = ma · Newton's Third Law · Friction – Static and Kinetic · Equilibrium and Free Body Diagrams · Circular Motion and Centripetal Force · Translatory Motion · Newton's Laws · Conservation of Momentum · Impulse and Force

1. Q1NEET 2026Banking of Roads

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

   • A. tan⁻¹(2/5)
   • B. tan⁻¹(√3/2)
   • C. tan⁻¹(2√3)
   • D. tan⁻¹(1/5)
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   Answer: C — tan⁻¹(2√3)

   Wear and tear on tyres is minimum when no frictional force is needed, i.e., at the optimum (design) speed for the banked angle: tanθ = v²/(rg). Substituting v=10 m/s, r=50 m, g=10 m/s²: tanθ = (10)²/(50×10) = 100/500 = 1/5. So θ = tan⁻¹(1/5).

2. Q2NEET 2026Resultant of perpendicular forces

   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:

   • A. 2 m/s²; tan⁻¹(3/4) with 8 N force
   • B. 2 m/s²; tan⁻¹(4/3) with 8 N force
   • C. 2 m/s²; tan⁻¹(3/4) with 6 N force
   • D. 20 m/s²; tan⁻¹(4/3) with 8 N force
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   Answer: —

   Resultant force=√(8²+6²)=10N, a=F/m=10/5=2 m/s², angle=tan⁻¹(6/8)=tan⁻¹(3/4) with the 8N force.

3. Q3NEET 2025Friction on Inclined Plane

   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

   • A. 0.25
   • B. 0.40
   • C. 0.5
   • D. 0.75
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   Answer: C — 0.5

   a1 = g sin45° − μg cos45° = (g/√2)(1−μ) on rough surface; a2 = g sin45° = g/√2 on smooth surface. Using L = ½a1t1² = ½a2t2² with t1 = 2t2: (1−μ)(2t)² = t² ⟹ (1−μ)×4 = 1 ⟹ μ = 3/4 = 0.75.

4. Q4NEET 2025Impulse-Momentum Theorem

   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)

   • A. 21 NS
   • B. 7 NS
   • C. 0
   • D. 84 NS
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   Answer: —

   Vf = √(2×9.8×10) = 14 m/s upward; Vi = √(2×9.8×40) = 28 m/s downward. ΔV = 14−(−28) = 42 m/s. Impulse I = mΔV = 0.5×42 = 21 N·s.

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5. Q5NEET 2024Circular Motion - Tension vs Angular Speed

   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:

   • A. T/4
   • B. √2 T
   • C. T
   • D. 4T
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   Answer: C — T

   Since r and m are constant, T ∝ ω². So T1/T2 = ω²/(2ω)² = 1/4, giving T2 = 4T.

6. Q6NEET 2024Contact Force Between Blocks

   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: [Figure: F=10N applied to block A (2kg) pushing block B (3kg), both on frictionless surface]

   • A. 6 N
   • B. 10 N
   • C. zero
   • D. 4 N
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   Answer: —

   Common acceleration a = F/m = 10/5 = 2 m/s². Force on B from A: N = m_B × a = 3 × 2 = 6 N.

7. Q7NEET 2023Deceleration - Bullet Through Block

   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 u/3. Then its 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:

   • A. 24 cm
   • B. 28 cm
   • C. 30 cm
   • D. 27 cm
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   Answer: C — 30 cm

   Using v²=u²-2a(2×24): (u/3)²=u²-48a, giving a=u²/54. The block ends when v=0: 0=u²-2a·d, so d=u²/(2a)=u²/(2×u²/54)=27cm.

8. Q8NEET 2023Maximum Acceleration with Friction

   Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is 0.15 (g=10 ms^-2).

   • A. 150 ms^-2
   • B. 1.5 ms^-2
   • C. 50 ms^-2
   • D. 1.2 ms^-2
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   Answer: A — 150 ms^-2

   For the body to remain stationary, the pseudo force must not exceed the maximum static friction: ma_max = μmg, so a_max = μg = 0.15×10 = 1.5 m/s².

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9. Q9NEET 2022Explosion and Conservation of Momentum

   A shell of mass m is at rest initially. It explodes into three fragments having equal mass in the ratio 2:2:1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is:

   • A. √2 v
   • B. 2√2 v
   • C. 3√2 v
   • D. v
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   Answer: A — √2 v

   Net linear momentum after explosion must be zero. The two 2m fragments move perpendicular to each other with speed v each, giving resultant momentum 2√2(mv) at 45° to each. This must be balanced by the m fragment: mv' = 2√2 mv ⟹ v' = 2√2 v.

10. Q10NEET 2022Conservation of Momentum

    A shell of mass m is at rest initially. It explodes into three fragments having mass in the ratio 2 : 2 : 1. If the fragments having equal mass fly off along mutually perpendicular directions with speed v, the speed of the third (lighter) fragment is

    • A. v
    • B. √2v
    • C. 2√2v
    • D. 3√2v
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    Answer: B — √2v

    Two fragments of mass 2m/5 each move perpendicular with speed v; resultant momentum = √2×(2m/5)v. This equals (m/5)×v3, so v3 = 2√2 v.