NEET Kinetic Energy and Momentum MCQs & PYQs

Q: A ball of mass \(2\text{ kg}\) and another ball of mass \(4\text{ kg}\) are dropped together from a \(60\text{ feet}\) tall building. After a fall of \(30\text{ feet}\) each towards earth, their respective kinetic energies will be in the ratio of: (2004)

When objects are dropped, potential energy is converted to kinetic energy. For a fall of height \(h\), the kinetic energy gained is \(KE = mgh\). For the two balls, \(KE_1 = m_1gh\) and \(KE_2 = m_2gh\). The ratio is \(\frac{KE_1}{KE_2} = \frac{m_1gh}{m_2gh} = \frac{m_1}{m_2} = \frac{2\text{ kg}}{4\text{ kg}} = \frac{1}{2}\). So, the ratio is \(1:2\).

Q: A stationary particle explodes into two particles of masses \(m_1\) and \(m_2\) which move in opposite directions with velocities \(v_1\) and \(v_2\). The ratio of their kinetic energies \(E_1/E_2\) is: (2003)

By conservation of momentum, since the initial particle is stationary, \(m_1v_1 = m_2v_2\) (in magnitude). This means their momenta are equal in magnitude: \(p_1 = p_2 = p\). Kinetic energy \(E = \frac{p^2}{2m}\). So \(E_1 = \frac{p^2}{2m_1}\) and \(E_2 = \frac{p^2}{2m_2}\). The ratio \(\frac{E_1}{E_2} = \frac{p^2/(2m_1)}{p^2/(2m_2)} = \frac{m_2}{m_1}\).

Q: If kinetic energy of a body is increased by \(300\%\) then percentage change in momentum will be: (2002)

Kinetic energy \(KE = \frac{p^2}{2m}\), so momentum \(p = \sqrt{2mKE}\). If \(KE_i\) is initial KE, then \(KE_f = KE_i + 300\%\ KE_i = 4KE_i\). So, \(p_f = \sqrt{2m(4KE_i)} = 2\sqrt{2mKE_i} = 2p_i\). Percentage change in momentum is \(\frac{p_f - p_i}{p_i} \times 100\% = \frac{2p_i - p_i}{p_i} \times 100\% = 100\%\).

Q: Two bodies of masses \(m\) and \(4m\) are moving with equal kinetic energies. The ratio of their linear momenta is: (1998, 97, 89)

Kinetic energy \(KE = \frac{p^2}{2m}\), so momentum \(p = \sqrt{2mKE}\). Given \(KE_1 = KE_2 = KE\). For the two bodies, \(p_1 = \sqrt{2mKE}\) and \(p_2 = \sqrt{2(4m)KE}\). The ratio of their linear momenta is \(\frac{p_1}{p_2} = \frac{\sqrt{2mKE}}{\sqrt{8mKE}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\). So, the ratio is \(1:2\).

Q: 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 \(m^0\).

Q: 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 \(1:3\).

Practice NEET Kinetic Energy and Momentum Questions

Question 11:

easy

A particle of mass \(m_1\) is moving with a velocity \(v_1\) and another particle of mass \(m_2\) is moving with a velocity \(v_b\). Both of them have the same momentum but their different kinetic energies are \(E_1\) and \(E_2\) respectively. If \(m_1 > m_2\), then:

(2004)

Given \(p_1 = p_2 = p\). Kinetic energy \(E = \frac{p^2}{2m}\). So, \(E_1 = \frac{p^2}{2m_1}\) and \(E_2 = \frac{p^2}{2m_2}\). The ratio is \(\frac{E_1}{E_2} = \frac{p^2/(2m_1)}{p^2/(2m_2)} = \frac{m_2}{m_1}\). Since \(m_1 > m_2\), \(\frac{m_2}{m_1} < 1\), which implies \(E_1 < E_2\).

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Question 12:

easy

A ball of mass \(2\text{ kg}\) and another ball of mass \(4\text{ kg}\) are dropped together from a \(60\text{ feet}\) tall building. After a fall of \(30\text{ feet}\) each towards earth, their respective kinetic energies will be in the ratio of:

(2004)

When objects are dropped, potential energy is converted to kinetic energy. For a fall of height \(h\), the kinetic energy gained is \(KE = mgh\). For the two balls, \(KE_1 = m_1gh\) and \(KE_2 = m_2gh\). The ratio is \(\frac{KE_1}{KE_2} = \frac{m_1gh}{m_2gh} = \frac{m_1}{m_2} = \frac{2\text{ kg}}{4\text{ kg}} = \frac{1}{2}\). So, the ratio is \(1:2\).

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Question 13:

easy

A stationary particle explodes into two particles of masses \(m_1\) and \(m_2\) which move in opposite directions with velocities \(v_1\) and \(v_2\). The ratio of their kinetic energies \(E_1/E_2\) is:

(2003)

By conservation of momentum, since the initial particle is stationary, \(m_1v_1 = m_2v_2\) (in magnitude). This means their momenta are equal in magnitude: \(p_1 = p_2 = p\). Kinetic energy \(E = \frac{p^2}{2m}\). So \(E_1 = \frac{p^2}{2m_1}\) and \(E_2 = \frac{p^2}{2m_2}\). The ratio \(\frac{E_1}{E_2} = \frac{p^2/(2m_1)}{p^2/(2m_2)} = \frac{m_2}{m_1}\).

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Question 14:

easy

If kinetic energy of a body is increased by \(300\%\) then percentage change in momentum will be: (2002)

Kinetic energy \(KE = \frac{p^2}{2m}\), so momentum \(p = \sqrt{2mKE}\). If \(KE_i\) is initial KE, then \(KE_f = KE_i + 300\%\ KE_i = 4KE_i\). So, \(p_f = \sqrt{2m(4KE_i)} = 2\sqrt{2mKE_i} = 2p_i\). Percentage change in momentum is \(\frac{p_f - p_i}{p_i} \times 100\% = \frac{2p_i - p_i}{p_i} \times 100\% = 100\%\).

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Question 15:

easy

Two bodies of masses \(m\) and \(4m\) are moving with equal kinetic energies. The ratio of their linear momenta is:

(1998, 97, 89)

Kinetic energy \(KE = \frac{p^2}{2m}\), so momentum \(p = \sqrt{2mKE}\). Given \(KE_1 = KE_2 = KE\). For the two bodies, \(p_1 = \sqrt{2mKE}\) and \(p_2 = \sqrt{2(4m)KE}\). The ratio of their linear momenta is \(\frac{p_1}{p_2} = \frac{\sqrt{2mKE}}{\sqrt{8mKE}} = \sqrt{\frac{1}{4}} = \frac{1}{2}\). So, the ratio is \(1:2\).

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Question 16:

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 17:

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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