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Laws of Motion MCQs for NEET — Physics Questions with Answers

Practice free Laws of Motion (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

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A ball of mass 0.5 kg moving with a velocity of 2 m/sec strikes a wall normally and bounces back with the same speed. If the time of contact between the ball and the wall is one millisecond, the average force exerted by the wall on the ball is

2000 N

1000 N

5000 N

125 N

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Explanation

$F_{a v} = \frac{Δ p}{Δ t} = \frac{m v - (- m v)}{Δ t} = \frac{2 m v}{Δ t} = \frac{2 \times 0 .5 \times 2}{10^{- 3}}$ = 2000 N

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A particle moves in the xy-plane under the action of a force F such that the components of its linear momentum p at any time t are $p_{x} = 2 \cos t$ , $p_{y} = 2 \sin t$ . The angle between F and p at time t is

90°

0°

180°

30°

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Explanation

Given that $\vec{p} = p_{x} \hat{i} + p_{y} \hat{j} = 2 \cos t \hat{i} + 2 \sin t \hat{j}$

∴ $\vec{F} = \frac{d \vec{p}}{d t} = - 2 \sin t \hat{i} + 2 \cos t \hat{j}$

Now, $\vec{F} . \vec{p} = 0$ i.e. angle between $\vec{F} and \vec{p}$ is 90°.

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A satellite in force-free space sweeps stationary interplanetary dust at a rate $d M / d t = α v$ where M is the mass, v is the velocity of the satellite and $α$ is a constant. What is the deacceleration of the satellite

$- 2 α v^{2} / M$

$- α v^{2} / M$

$+ α v^{2} / M$

$- α v^{2}$

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Explanation

$F = \frac{d p}{d t} = v (\frac{d M}{d t}) = α v^{2}$

∴ $a = \frac{F}{M} = \frac{α v^{2}}{M}$

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Swimming is possible on account of

First law of motion

Second law of motion

Third law of motion

Newton's law of gravitation

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Explanation

Swimming is a result of pushing water in the opposite direction of the motion.

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On a stationary sail-boat, air is blown at the sails from a fan attached to the boat. The boat will

Remain stationary

Spin around

Move in a direction opposite to that in which air is blown

Move in the direction in which the air is blown

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Explanation

The force exerted by the air of fan on the boat is internal and for motion external force is required.

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A man is standing at a spring platform. Reading of spring balance is 60 kg wt. If man jumps outside platform, then reading of spring balance

First increases then decreases to zero

Decreases

Increases

Remains same

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Explanation

For jumping he presses the spring platform, so the reading of spring balance increases first and finally it becomes zero.

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A player caught a cricket ball of mass 150 gm moving at a rate of 20 m/s. If the catching process be completed in 0.1 s, then the force of the blow exerted by the ball on the hands of the player is

0.3 N

30 N

300 N

3000 N

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Explanation

Force exerted by the ball on hands of the player

$= \frac{m d v}{d t} = \frac{0 .15 \times 20}{0 .1} = 30 N$

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A rocket has a mass of 100 kg. 90% of this is fuel. It ejects fuel vapours at the rate of 1 kg/sec with a velocity of 500 m/sec relative to the rocket. It is supposed that the rocket is outside the gravitational field. The initial upthrust on the rocket when it just starts moving upwards is

Zero

500 N

1000 N

2000 N

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Explanation

$F = u (\frac{d m}{d t}) = 500 \times 1 = 500 N$

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In which of the following cases forces may not be required to keep the

Particle going in a circle

Particle going along a straight line

The momentum of the particle constant

Acceleration of the particle constant

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Explanation

If momentum remains constant then force will be zero because $F = \frac{d P}{d t}$

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A ball of mass 150g starts moving with an acceleration of 20 m/s². When hit by a force, which acts on it for 0.1 sec. The impulsive force is

0.5 N-s

0.1 N-s

0.3 N-s

1.2 N-s

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Explanation

Impulse = Force × time = m a t

$= 0 .15 \times 20 \times 0 .1 = 0 .3 N - s$

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