CBSE Class 10 Physics Force and Laws of Motion MCQs Set 03

Answer: (d) reaction of the stone

Explanation:
1. Your foot pushes the stone, and the stone pushes back on the foot with an equal force (third law).

Teacher's Note:
1. The force of the reaction is what hurts.

Answer: (a) move away from the shore

Explanation:
1. The man pushes the boat backward (away from the shore) as he walks.
2. The boat pushes him forward with an equal and opposite force, so the boat moves away from the shore.

Teacher's Note:
1. Total momentum of the man-boat system stays zero.
2. Both move in opposite directions.

Answer: (a) 1.5 N

Explanation:
1. The hooks exert equal and opposite forces on each other (third law).
2. The tension through both balances is the same, so B also reads 1.5 N.

Teacher's Note:
1. The two readings are not added.

Answer: (c) momentum

Explanation:
1. Burnt gases are thrown backward at high speed.
2. To keep the total momentum constant, the rocket gains forward momentum.

Teacher's Note:
1. Third law and conservation of momentum together explain propulsion.

Answer: (a) \( \dfrac{AB}{A+C} \)

Explanation:
1. Momentum before = A x B.
2. After the bullet embeds, total mass = A + C.
3. \( V = AB/(A+C) \).

Teacher's Note:
1. Use conservation of momentum for a perfectly inelastic collision.

Answer: (c) a heavy bat, because it will recoil less

Explanation:
1. A heavier bat gets a smaller velocity change from the ball's impact (momentum is conserved), so it recoils less.

Teacher's Note:
1. Momentum gained by the bat equals that lost by the ball.
2. A larger mass means a smaller recoil speed.

Answer: (d) 2/3 m/s

Explanation:
1. Taking A's direction as positive: total momentum = 2 x 3 - 1 x 4 = 2 kg m/s.
2. Common velocity = 2 / (2 + 1) = 2/3 m/s.

Teacher's Note:
1. Opposite directions have opposite signs.
2. The result is in the direction of A.

Answer: (b) law of conservation of momentum

Explanation:
1. The third law gives equal and opposite impulses in a collision, and the second law relates force to rate of change of momentum.
2. Together they lead to conservation of momentum.

Teacher's Note:
1. Total momentum of an isolated system does not change.

Answer: (d) both (b) and (c)

Explanation:
1. The floor's force exceeds mg when the acceleration is upward.
2. Going up and speeding up has upward acceleration.
3. Going down and slowing down also has upward acceleration.

Teacher's Note:
1. Slowing down while moving down means acceleration is directed upward.

Answer: (a) 640 N

Explanation:
1. For a lift moving down with acceleration a, apparent weight = m(g - a).
2. = 80 x (10 - 2) = 640 N.

Teacher's Note:
1. The person feels lighter when the lift accelerates downward.

Answer: (c) g

Explanation:
1. The lift is not accelerating, so it is an inertial frame.
2. The coin falls with the acceleration due to gravity, g.

Teacher's Note:
1. Constant velocity of the frame does not change the observed acceleration.

Answer: (d) 1.96 m s\( ^{-2} \) upwards

Explanation:
1. Apparent weight 4.8 kgf is greater than 4 kgf, so the acceleration is upward.
2. m(g + a) = 4.8 g, so a = 0.2 g = 0.2 x 9.8 = 1.96 m/s\( ^2 \) upward.

Teacher's Note:
1. The weighing scale reads more when the lift accelerates upward.

Answer: (c) When a person walks on a rough surface, the frictional force exerted by the surface on the person is in the direction of his motion

Explanation:
1. The foot pushes the ground backward.
2. By the third law the ground exerts a forward static friction on the person.
3. It is static friction because the foot does not slip.

Teacher's Note:
1. Without friction walking would be impossible, as on slippery ice.

Answer: (a) 2.8 N

Explanation:
1. Maximum static friction = 0.4 x 2 x 9.8 = 7.84 N.
2. The applied 2.8 N is smaller, so the block does not move.
3. Static friction adjusts to equal the applied force: 2.8 N.

Teacher's Note:
1. Static friction is only as large as needed, up to its limit.

Answer: (c) not move

Explanation:
1. Limiting friction = 0.4 x 10 = 4 N.
2. The applied 3 N is less than 4 N, so the block stays at rest.

Teacher's Note:
1. The 3.5 N figure applies only after motion has started.
2. Compare the applied force with the limiting friction.

Answer: (b) 0.98 N

Explanation:
1. The block is at rest, so friction balances its weight.
2. f = mg = 0.1 x 9.8 = 0.98 N.
3. This is less than the limiting value 0.5 x 5 = 2.5 N.

Teacher's Note:
1. Do not use 0.5 x 5 here: that is only the maximum possible friction.

Answer: (d) 0.375

Explanation:
1. Limiting friction = 75 N = \( \mu N \), with N = mg = 200 N.
2. \( \mu = 75/200 = 0.375 \).

Teacher's Note:
1. The force that just starts motion equals the limiting friction.

Answer: (c) bumps and cracks on them

Explanation:
1. Microscopic irregularities on the two surfaces interlock and resist sliding.

Teacher's Note:
1. Even smooth-looking surfaces have bumps at the microscopic level.

Answer: (d) limiting friction

Explanation:
1. The largest static friction a pair of surfaces can provide before sliding starts is the limiting friction.

Teacher's Note:
1. \( f_s \le \mu_s N \). The equality gives the limiting value.