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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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If a molecule in a gas container hits a horizontal wall and rebounds with the same speed but an angle of 30° with the normal, is momentum conserved in the collision at the wall, considering a single molecule and the wall?

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Explanation

The question asks 'Is momentum conserved in the collision?' If we consider just the molecule, its momentum changes in direction, so its momentum isn't conserved. However, for the system of the molecule and the wall, momentum is conserved. The NCERT implicitly refers to the 'system' being conserved. When a molecule hits a wall and rebounds, the wall experiences an impulse equal and opposite to the impulse on the molecule, ensuring conservation of momentum for the molecule-wall system. The text for the example problem 5.14 doesn't explicitly state the answer for the question posed within the problem statement, but general principle implies system momentum conservation.

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The Law of Conservation of Momentum can be derived directly from which of Newton's Laws?

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Explanation

The NCERT text clearly states: 'The second and third laws of motion lead to an important consequence: the law of conservation of momentum.'

NEET 2023
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A football player is moving southward and suddenly turns eastward with the same speed to avoid an opponent. The force that acts on the player while turning is:

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Explanation

$\Delta\vec{v} = \vec{v}_f - \vec{v}_i$ points north-east (east minus south $=$ east + north). Force is along $\Delta\vec{v}$, i.e., north-east.

NEET 2023
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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 $0.15$ ($g = 10\ \text{m s}^{-2}$).

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Explanation

$a_{\max} = \mu_s g = 0.15 \times 10 = 1.5\ \text{m s}^{-2}$.

NEET 2024
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A bob is whirled in a horizontal plane by means of a string with an initial speed of $\omega$ rpm. The tension in the string is $T$. If speed becomes $2\omega$ while keeping the same radius, the tension in the string becomes:

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Explanation

$T \propto \omega^2$, so tension becomes $4T$.

NEET 2024
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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:

F = 10 N A 2 kg B 3 kg
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Explanation

$a = 10/5 = 2$ m/s²; force on B $= 3 \times 2 = 6$ N.

NEET 2025
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There are two inclined surfaces of equal length ($L$) and same angle of inclination $45^\circ$ 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 the rough surface than on the smooth surface. The coefficient of kinetic friction $(\mu_k)$ between the object and the rough surface is close to:

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Explanation

For equal $L$: $a_{smooth}t_1^2 = a_{rough}t_2^2$ with $t_2 = 2t_1\Rightarrow a_{smooth} = 4a_{rough}$. $g\sin45^\circ = 4g(\sin45^\circ-\mu\cos45^\circ)\Rightarrow \mu = \tfrac{3}{4}\tan45^\circ = 0.75$.

NEET 2025
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A bob of heavy mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal velocity $v_0$ as shown in figure. If the string gets slack at some point $P$ making an angle $\theta$ from the horizontal, the ratio of the speed $v$ of the bob at point $P$ to its initial speed $v_0$ is:

O l m v₀ P θ
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Explanation

String slack at $P$: $mg\sin\theta = \dfrac{mv^2}{l}\Rightarrow v^2 = gl\sin\theta$. Energy from bottom to $P$ (height $l(1+\sin\theta)$): $v^2 = v_0^2 - 2gl(1+\sin\theta)$. Hence $v_0^2 = gl(2+3\sin\theta)$ and $\dfrac{v}{v_0} = \left(\dfrac{\sin\theta}{2+3\sin\theta}\right)^{1/2}$.

NEET 2025
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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 $g = 9.8\ \text{m/s}^2$):

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Explanation

$v_{down} = \sqrt{2g(40)} = 28\ \text{m/s}$, $v_{up} = \sqrt{2g(10)} = 14\ \text{m/s}$. Impulse $= m(v_{up}+v_{down}) = 0.5(14+28) = 21\ \text{N·s}$.

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