Q. No.A bullet of mass \(10\) g moving horizontal with a velocity of \(400\) m/s strikes a wood block of mass \(2\) kg which is suspended by light inextensible string of length \(5\) m. As a result, the centre of gravity of the block is found to rise a vertical distance of \(10\) cm. The speed of the bullet after it emerges horizontally from the block will be:
1.
\(100\) m/s
2.
\(80\) m/s
3.
\(120\) m/s
4.
\(160\) m/s
A moving block having mass \(m\) collides with another stationary block having a mass of \(4m.\) The lighter block comes to rest after the collision. When the initial velocity of the lighter block is \(v,\) then the value of the coefficient of restitution \((e)\) will be:
1. \(0.5\)
2. \(0.25\)
3. \(0.8\)
4. \(0.4\)
A body initially at rest and sliding along a frictionless track from a height \(h\) (as shown in the figure) just completes a vertical circle of diameter \(\mathrm{AB}= D.\) The height \({h}\) is equal to:
1. \({3\over2}D\)
2. \(D\)
3. \({7\over4}D\)
4. \({5\over4}D\)
2. \(\left(2 t^3+3 t^3\right) ~\text W\)
3. \(\left(2 t^3+3 t^5\right) ~\text W\)
4. \(\left(2 t^3+3 t^4\right) ~\text W\)
A ball is thrown vertically downward from a height of \(20~\text m\) with an initial velocity \(v_0.\) It collides with the ground, loses \(50\%\) of its energy in a collision, and rebounds to the same height. The initial velocity \(v_0\) is:
(Take, \(g=10~\text{ms}^{-2}\))
1. \(14~\text{ms}^{-1}\)
2. \(20~\text{ms}^{-1}\)
3. \(28~\text{ms}^{-1}\)
4. \(10~\text{ms}^{-1}\)
Two similar springs \(P\) and \(Q\) have spring constants \(k_P\) and \(k_Q\), such that \(k_P>k_Q\). They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs \(W_P\) and \(W_Q\) are related as, in case (a) and case (b), respectively:
| 1. | \(W_P=W_Q;~W_P>W_Q\) |
| 2. | \(W_P=W_Q;~W_P=W_Q\) |
| 3. | \(W_P>W_Q;~W_P<W_Q\) |
| 4. | \(W_P<W_Q;~W_P<W_Q\) |
1. \(475\) J
2. \(450\) J
3. \(275\) J
4. \(250\) J
| 1. | \( \sqrt{\frac{m k}{2}} t^{-1 / 2} \) | 2. | \( \sqrt{m k} t^{-1 / 2} \) |
| 3. | \( \sqrt{2 m k} t^{-1 / 2} \) | 4. | \( \frac{1}{2} \sqrt{m k} t^{-1 / 2}\) |
Two particles of masses \(m_1\) and \(m_2\) move with initial velocities \(u_1\) and \(u_2\) respectively. On collision, one of the particles gets excited to a higher level, after absorbing energy \(E\). If the final velocities of particles are \(v_1\) and \(v_2\), then we must have:
| 1. | \(m_1^2u_1+m_2^2u_2-E = m_1^2v_1+m_2^2v_2\) |
| 2. | \(\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2= \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2\) |
| 3. | \(\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2-E= \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2\) |
| 4. | \(\frac{1}{2}m_1^2u_1^2+\frac{1}{2}m_2^2u_2^2+E = \frac{1}{2}m_1^2v_1^2+\frac{1}{2}m_2^2v_2^2\) |
A uniform force of \((3 \hat{i} + \hat{j})\) newton acts on a particle of mass \(2~\text{kg}.\) Hence the particle is displaced from the position \((2 \hat{i} + \hat{k})\) metre to the position \((4 \hat{i} + 3 \hat{j} - \hat{k})\) metre. The work done by the force on the particle is:
1. \(6~\text{J}\)
2. \(13~\text{J}\)
3. \(15~\text{J}\)
4. \(9~\text{J}\)
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