Kinetic Energy and Momentum - NEET Physics Chapterwise MCQs & PYQs
Rankers Physics
One Stop for Your NEET Physics Preparation
NEET Kinetic Energy and Momentum MCQs & PYQs
Practice NEET Kinetic Energy and Momentum Questions
Question 1:
easy
A bomb of mass \(30\text{ kg}\) at rest explodes into two pieces of masses \(18\text{ kg}\) and \(12\text{ kg}\). The velocity of \(18\text{ kg}\) mass is \(6\text{ ms}^{-1}\). The kinetic energy of the other mass is:
(2005)
By conservation of momentum, \(m_1v_1 + m_2v_2 = 0\) (since initial momentum is zero). Given \(m_1 = 18\text{ kg}\), \(v_1 = 6\text{ m/s}\), and \(m_2 = 12\text{ kg}\). So, \(18 \times 6 + 12v_2 = 0 \Rightarrow 108 + 12v_2 = 0 \Rightarrow v_2 = -9\text{ m/s}\). The kinetic energy of the other mass is \(KE_2 = \frac{1}{2}m_2v_2^2 = \frac{1}{2}(12)(-9)^2 = 6 \times 81 = 486\text{ J}\).
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 \(E_1 < E_2\).
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 \(1:2\).
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 \(\frac{E_1}{E_2} = \frac{p^2/(2m_1)}{p^2/(2m_2)} = \frac{m_2}{m_1}\).
The K.E. of a person is just half of K.E. of a boy whose mass is just half of that person. If person increases its speed by \(1\text{ m/s}\), then its K.E. equals to that of boy then initial speed of person was:
(1999)
Let person's mass be \(M_p\) and speed \(v_p\). Boy's mass \(M_b = M_p/2\) and speed \(v_b\). Given \(KE_p = \frac{1}{2}KE_b\) and \(KE_p' = KE_b\) when \(v_p' = v_p+1\). From \(KE_p = \frac{1}{2}KE_b\), \(\frac{1}{2}M_p v_p^2 = \frac{1}{2} (\frac{1}{2} \frac{M_p}{2} v_b^2)\Rightarrow 4v_p^2 = v_b^2 \Rightarrow v_b = 2v_p\). From \(KE_p' = KE_b\), \(\frac{1}{2}M_p (v_p+1)^2 = \frac{1}{2}M_b v_b^2 = \frac{1}{2}(\frac{M_p}{2})(2v_p)^2 = \frac{1}{2}M_p (2v_p^2)\). So \((v_p+1)^2 = 2v_p^2 \Rightarrow v_p+1 = \sqrt{2}v_p\). \(1 = v_p(\sqrt{2}-1) \Rightarrow v_p = \frac{1}{\sqrt{2}-1} = \sqrt{2}+1\text{ m/s}\).
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 \(1:2\).
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 \(m^0\).
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 \(1:3\).
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 \(2Mv\).