Kinetic Energy and Momentum - NEET Physics Chapterwise MCQs & PYQs
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NEET Kinetic Energy and Momentum MCQs & PYQs
Practice NEET Kinetic Energy and Momentum Questions
Question 11:
easy
Assertion (A): Kinetic energy of a system can be increased without applying any external force on the system.
Reason (R): If external forces are absent then work done by internal forces is equal to change in kinetic energy.
According to the work-energy theorem, `\(W_{net} = \Delta KE\)`. If external forces are absent, the net work done on the system is only due to internal forces, i.e., `\(W_{int} = \Delta KE\)`. Thus, internal forces can increase kinetic energy, for example, in an explosion. Both assertion and reason are true, and the reason correctly explains the assertion.
Arun has \(\left(\frac{1}{3}\right)^{\text{rd}}\) of the kinetic energy as that of Raunak when they both run. If Raunak has half the mass of Arun, then the relationship between the speed of Arun \(v’\) and speed of Raunak \(v\) is
Let \(m_A = 2m_R\). Given \(K_A = \frac{1}{3}K_R ⇒ \frac{1}{2}m_A v'^2 = \frac{1}{3}\left(\frac{1}{2}m_R v^2\right)\). Substituting \(m_A = 2m_R\), we get \(2v'^2 = \frac{1}{3}v^2 ⇒ v' = \frac{v}{\sqrt{6}}\).
A particle of mass \(0.4\text{ kg}\) is moving with a velocity of \((6\hat{i} – 8\hat{j})\text{ m/s}\), then the kinetic energy of the particle is
The speed squared is \(v^2 = 6^2 + (-8)^2 = 100\text{ m}^2\text{/s}^2\). The kinetic energy is \(KE = \frac{1}{2}mv^2 = \frac{1}{2}(0.4)(100) = 20\text{ J}\).
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 \(1:2\).
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 \(E_1 < E_2\).