NEET Kinetic Energy and Momentum MCQs & PYQs

Practice NEET Kinetic Energy and Momentum Questions

Question 21:

easy

The kinetic energy acquired by a mass \(m\) in travelling distance \(d\), starting from rest, under the action of a constant force is directly proportional to:

(1994)

Concept: Work-Energy Theorem. Formula: \(W = Fd = \Delta KE\). Starting from rest, \(KE_i = 0\). So, the final kinetic energy \(KE_f = Fd\). If the force \(F\) and distance \(d\) are constant, the work done \(Fd\) is constant. Therefore, the kinetic energy acquired is independent of mass \(m\), meaning it is proportional to \(m^0\).

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Question 22:

easy

Two masses of \(1\text{ g}\) and \(9\text{ g}\) are moving with equal kinetic energies. The ratio of the magnitudes of their respective linear momenta is:

(1993)

Concept: Relation between kinetic energy and momentum. Formula: \(KE = \frac{p^2}{2m}\). Given \(KE_1 = KE_2\), so \(\frac{p_1^2}{2m_1} = \frac{p_2^2}{2m_2}\). This implies \(\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}\). Substituting \(m_1 = 1\text{ g}\) and \(m_2 = 9\text{ g}\), we get \(\frac{p_1}{p_2} = \sqrt{\frac{1}{9}} = \frac{1}{3}\). Thus the ratio is \(1:3\).

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Question 23:

moderate

A particle of mass \(M\) is moving in a horizontal circle of radius \(R\) with uniform speed \(v\). When it moves from one point to a diametrically opposite point, its:

(1992)

Concept: Momentum and kinetic energy in uniform circular motion. Formula: Momentum \(p = Mv\), Kinetic Energy \(KE = \frac{1}{2}Mv^2\). Since speed \(v\) is uniform, KE remains constant (\(\Delta KE = 0\)). At diametrically opposite points, the direction of velocity reverses. If initial momentum is \(\vec{p_1} = M\vec{v}\), then final momentum is \(\vec{p_2} = -M\vec{v}\). The change in momentum is \(\Delta\vec{p} = \vec{p_2} - \vec{p_1} = -2M\vec{v}\). The magnitude of the change is \(2Mv\).

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