If kinetic energy of a body is increased by \(300\%\) then percentage change in momentum will be: (2002)
1. \(100\%\)
2. \(150\%\)
3. \(265\%\)
4. \(73.2\%\)
View Answer
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 \(\frac{p_f - p_i}{p_i} \times 100\% = \frac{2p_i - p_i}{p_i} \times 100\% = 100\%\).
A child is sitting on a swing. Its minimum and maximum heights from the ground is \(0.75\text{ m}\) and \(2\text{ m}\) respectively, its maximum speed will be:
(2001)
1. \(10\text{ m/s}\)
2. \(5\text{ m/s}\)
3. \(1\text{ m/s}\)
4. \(15\text{ m/s}\)
View Answer
Maximum speed occurs at minimum height, where potential energy is lowest and kinetic energy is highest. Minimum speed (zero) occurs at maximum height. By conservation of mechanical energy: \(mgh_{max} + \frac{1}{2}mv_{min}^2 = mgh_{min} + \frac{1}{2}mv_{max}^2\). With \(v_{min}=0\), \(mg(2) = mg(0.75) + \frac{1}{2}mv_{max}^2\). \(2g - 0.75g = \frac{1}{2}v_{max}^2 \Rightarrow 1.25g = \frac{1}{2}v_{max}^2\). Using \(g = 10\text{ m/s}^2\), \(v_{max}^2 = 2.5 \times 10 = 25\), so \(v_{max} = 5\text{ m/s}\).
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)
1. \((\sqrt{2}+1)\text{ m/s}\)
2. \((2+\sqrt{2})\text{ m/s}\)
3. \(2(\sqrt{2}+2)\text{ m/s}\)
4. None
View Answer
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)
1. \(1 : 2\)
2. \(1 : 4\)
3. \(4 : 1\)
4. \(1 : 1\)
View Answer
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 potential energy between two atoms, in a molecule, is given by \(U(x) = \frac{a}{x^{12}} – \frac{b}{x^6}\) where \(a\) and \(b\) are positive constants and \(x\) is the distance between the atoms. The atom is in stable equilibrium, when:
(1995)
1. \(x = (2a/b)^{1/6}\)
2. \(x = (11a/5b)^{1/6}\)
3. \(x = 0\)
4. \(x = (a/2b)^{1/6}\)
View Answer
For equilibrium, the force is zero: \(F = -\frac{dU}{dx} = 0\). Given \(U(x) = ax^{-12} - bx^{-6}\), then \(\frac{dU}{dx} = -12ax^{-13} + 6bx^{-7}\). Setting \(\frac{dU}{dx} = 0\) gives \(\frac{12a}{x^{13}} = \frac{6b}{x^7} \Rightarrow 12a = 6bx^6 \Rightarrow x^6 = \frac{12a}{6b} = \frac{2a}{b}\). So, \(x = (\frac{2a}{b})^{1/6}\). For stable equilibrium, \(\frac{d^2U}{dx^2} > 0\), which holds true for this value of \(x\).
A particle is released from height \(S\) from the surface of the Earth. At a certain height its kinetic energy is three times its potential energy. The height from the surface of Earth and the speed of the particle at that instant are respectively:
(2021)
1. \(h = \frac{S}{4}, v = \sqrt{\frac{3gS}{2}}\)
2. \(h = \frac{S}{2}, v = \sqrt{\frac{3gS}{2}}\)
3. \(h = \frac{S}{4}, v = \frac{3gS}{4\sqrt{2}}\)
4. \(h = \frac{S}{4}, v = \frac{3gS}{4\sqrt{2}}\)
View Answer
Let the initial height be \(S\). At height \(h\), \(KE = 3PE = 3mgh\). By conservation of energy, \(mgS = mgh + 3mgh = 4mgh\), so \(h = S/4\). Also, \(KE = \frac{1}{2}mv^2 = 3mgh\), substituting \(h\) gives \(v^2 = 6g(S/4) = \frac{3gS}{2}\), so \(v = \sqrt{\frac{3gS}{2}}\).
The potential energy of particle in a force field is \(U = \frac{A}{r} – \frac{B}{r^2}\) where \(A\) and \(B\) are positive constants and \(r\) is the distance of particle from the centre of the field. For stable equilibrium, the distance of the particle is:
(2012 Pre)
1. \(B/2A\)
2. \(2A/B\)
3. \(A/B\)
4. \(B/A\)
View Answer
For equilibrium, the force is zero: \(F = -\frac{dU}{dr} = 0\). Given \(U = \frac{A}{r} - \frac{B}{r^2}\), if the question implicitly means \(U = \frac{A}{r^2} - \frac{B}{r}\), then \(\frac{dU}{dr} = -\frac{2A}{r^3} + \frac{B}{r^2}\). Setting \(\frac{dU}{dr} = 0\) gives \(\frac{2A}{r^3} = \frac{B}{r^2} \Rightarrow r = \frac{2A}{B}\). Checking stability, \(\frac{d^2U}{dr^2} = \frac{6A}{r^4} - \frac{2B}{r^3}\), which is positive at \(r = \frac{2A}{B}\).
The potential energy of a system increases if work is done:
(2011 Pre)
1. Upon the system by a nonconservative force
2. By the system against a conservative force
3. Upon the system by a conservative force
4. Upon the system by a nonconservative force
View Answer
Potential energy is associated with conservative forces. Work done by a conservative force is \(W_c = -\Delta U\). Therefore, if potential energy increases (\(\Delta U > 0\)\), then \(W_c\) must be negative. This happens when work is done by the system against a conservative force (e.g., lifting an object against gravity).
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)
1. \(243\text{ J}\)
2. \(486\text{ J}\)
3. \(564\text{ J}\)
4. \(388\text{ J}\)
View Answer
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)
1. \(E_1/E_2 = m_1/m_2\)
2. \(E_1 > E_2\)
3. \(E_1 = E_2\)
4. \(E_1 < E_2\)
View Answer
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 \(E_1 < E_2\).