Q. No.1. \(2:1\)
2. \(4:1\)
3. \(1:4\)
4. \(1:2\)
1. \(M=m\)
2. \(M=2m\)
3. \(M\ll m\)
4. \(M\gg m\)
1. \(3:1\)
2. \(1:4\)
3. \(1:1\)
4. \(1:3\)
| Assertion (A): | When a firecracker (rocket) explodes in mid-air, its fragments fly in such a way that they continue moving in the same path, which the firecracker would have followed, had it not exploded. |
| Reason (R): | Explosion of cracker (rocket) occurs due to internal forces only and no external force acts for this explosion. |
| 1. | Both (A) and (R) are true and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are true but (R) is not the correct explanation of (A). |
| 3. | (A) is true but (R) is false. |
| 4. | (A) is false but (R) is true. |
Body \(\mathrm{A}\) of mass \(4m\) moving with speed \(u\) collides with another body \(\mathrm{B}\) of mass \(2m\) at rest. The collision is head-on and elastic in nature. After the collision, the fraction of energy lost by the colliding body \(\mathrm{A}\) is:
| 1. | \(\dfrac{5}{9}\) | 2. | \(\dfrac{1}{9}\) |
| 3. | \(\dfrac{8}{9}\) | 4. | \(\dfrac{4}{9}\) |
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 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 |
Two identical balls \(A\) and \(B\) having velocities of \(0.5~\text{m/s}\) and \(-0.3~\text{m/s}\), respectively, collide elastically in one dimension. The velocities of \(B\) and \(A\) after the collision, respectively, will be:
| 1. | \(-0.5~\text{m/s}~\text{and}~0.3~\text{m/s}\) |
| 2. | \(0.5~\text{m/s}~\text{and}~-0.3~\text{m/s}\) |
| 3. | \(-0.3~\text{m/s}~\text{and}~0.5~\text{m/s}\) |
| 4. | \(0.3~\text{m/s}~\text{and}~0.5~\text{m/s}\) |
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. | \(\frac{1}{2}m_1^2u_1^2+\frac{1}{2}m_2^2u_2^2+E = \frac{1}{2}m_1^2v_1^2+\frac{1}{2}m_2^2v_2^2\) |
On a frictionless surface, a block of mass \(M\) moving at speed \(v\) collides elastically with another block of the same mass \(M\) which is initially at rest. After the collision, the first block moves at an angle \(\theta\) to its initial direction and has a speed \(\frac{v}{3}\). The second block’s speed after the collision will be:
| 1. | \(\frac{2\sqrt{2}}{3}v\) | 2. | \(\frac{3}{4}v\) |
| 3. | \(\frac{3}{\sqrt{2}}v\) | 4. | \(\frac{\sqrt{3}}{2}v\) |
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