Laws of Motion

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Laws of Motion — solved practice questions

8 NEET Physics questions with step-by-step solutions. Attempt each, then reveal the answer.

1. Q1easy

   Newton's first law of motion is essentially a statement about which physical quantity?

   • AMomentum
   • BEnergy
   • CInertia
   • DImpulse
   Show answer & solution

   Correct answer: (C) Inertia

   The first law defines inertia — the tendency of a body to resist any change in its state of rest or uniform motion. It also introduces the concept of an inertial frame.

2. Q2medium

   A body of mass 5 kg is acted upon by a horizontal force of 20 N on a surface where the coefficient of kinetic friction is 0.2. Taking g = 10 m/s^2, the acceleration of the body is:

   • A1 m/s^2
   • B2 m/s^2
   • C3 m/s^2
   • D4 m/s^2
   Show answer & solution

   Correct answer: (B) 2 m/s^2

   Kinetic friction f = mu m g = 0.2 x 5 x 10 = 10 N. Net force = 20 - 10 = 10 N, so a = F_net/m = 10/5 = 2 m/s^2.

3. Q3medium

   In an Atwood machine, two masses 3 kg and 2 kg are connected by a light inextensible string over a frictionless pulley. Taking g = 10 m/s^2, the acceleration of the system is:

   • A2 m/s^2
   • B4 m/s^2
   • C5 m/s^2
   • D10 m/s^2
   Show answer & solution

   Correct answer: (A) 2 m/s^2

   For an Atwood machine, a = (m1 - m2)g/(m1 + m2) = (3 - 2)(10)/(3 + 2) = 10/5 = 2 m/s^2.

4. Q4medium

   For the Atwood machine with masses 3 kg and 2 kg over a frictionless pulley (g = 10 m/s^2), the tension in the string is:

   • A12 N
   • B18 N
   • C20 N
   • D24 N
   Show answer & solution

   Correct answer: (D) 24 N

   Tension T = 2 m1 m2 g/(m1 + m2) = 2 x 3 x 2 x 10/(3 + 2) = 120/5 = 24 N.

5. Q5medium

   A gun of mass 2 kg fires a bullet of mass 20 g with a muzzle velocity of 200 m/s. The recoil velocity of the gun is:

   • A1 m/s
   • B2 m/s
   • C4 m/s
   • D20 m/s
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   Correct answer: (B) 2 m/s

   By conservation of momentum, m_bullet v_bullet = m_gun V. V = (0.02 x 200)/2 = 4/2 = 2 m/s.

6. Q6medium

   A car goes round a circular track of radius r banked at an angle theta with no friction. The speed for which the banking is designed satisfies:

   • Atan theta = r g / v^2
   • Bsin theta = v^2 / (r g)
   • Ctan theta = v^2 / (r g)
   • Dcos theta = v^2 / (r g)
   Show answer & solution

   Correct answer: (C) tan theta = v^2 / (r g)

   For a frictionless banked road, the horizontal component of the normal force provides the centripetal force while the vertical component balances gravity: N sin theta = m v^2/r and N cos theta = m g. Dividing gives tan theta = v^2/(r g).

7. Q7medium

   A person of mass 50 kg stands in a lift that accelerates upward at 2 m/s^2. Taking g = 10 m/s^2, the apparent weight (normal reaction) on the person is:

   • A400 N
   • B490 N
   • C500 N
   • D600 N
   Show answer & solution

   Correct answer: (D) 600 N

   For upward acceleration, N = m(g + a) = 50(10 + 2) = 50 x 12 = 600 N.

8. Q8medium

   A block is placed on a rough inclined plane whose angle of inclination is gradually increased. The block just begins to slide when the angle is theta. The coefficient of static friction between block and plane is:

   • Atan theta
   • Bsin theta
   • Ccos theta
   • Dcot theta
   Show answer & solution

   Correct answer: (A) tan theta

   At the angle of repose, mg sin theta = mu mg cos theta, so mu = tan theta. The angle at which sliding just begins is the angle of repose.

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