Q. No.A particle of mass 'm' is projected at an angle ' with the horizontal, with an initial velocity 'u'. The work done by gravity during the time it reaches its highest point is:
1.
2.
3.
4.
1. \(10.9~\text{m}^{3}\)
2. \(0.45~\text{m}^{3}\)
3. \(1.8~\text{m}^{3}\)
4. \(0.25~\text{m}^{3}\)
The position-time graph of a particle of mass \(2\) kg is shown in the figure. Total work done on the particle from \(t=0\) to \(t=4\) s is:
1. \(8\) J
2. \(4\) J
3. \(0\) J
4. can't be determined
Five balls are placed one after another along a straight line as shown in the figure. Initially, all the balls are at rest. Then the second ball is projected with speed towards the third ball. Mark the correct statement(s). (Assume all collisions to be head-on and elastic):
1. The total number of collisions in the process is 5.
2. The velocity of separation between the first and fifth ball after the last possible collision is
3. Finally, three balls remain stationary.
4. All of the above are correct.
| 1. | \(200~\text{J/s}\) | 2. | \(40~\text{J/s}\) |
| 3. | \(140~\text{J/s}\) | 4. | \(170~\text{J/s}\) |
A body of mass \(2\) kg moving with a velocity of \(3\) m/s collides with a body of mass of \(1\) kg moving with a velocity of \(4\) m/s in the opposite direction. If the collision is head-on and completely inelastic, then the wrong statement is:
| 1. | Both bodies move together with a velocity \((2/3)\) m/s. |
| 2. | The momentum of the system is \(2\) kg-m/s throughout. |
| 3. | The momentum of the system is \(10\) kg-m/s. |
| 4. | The loss of KE for the system is \((49/3)\) J. |
The power supplied to a particle of mass 2 kg varies with time as Watt, where t is in seconds. If the velocity of a particle at t = 0 is v = 0, then the velocity of the particle at t = 2 s will be:
| 1. | \(1 \mathrm{~m} / \mathrm{s} \) | 2. | \(4 \mathrm{~m} / \mathrm{s} \) |
| 3. | \(2 \mathrm{~m} / \mathrm{s} \) | 4. | \(2 \sqrt{2} \mathrm{~m} / \mathrm{s}\) |
The potential energy of a 1 kg particle free to move along the x-axis is given by:
The total mechanical energy of the particle is 2J. Then, the maximum speed (in ms-1) will be
1. \(3 \over \sqrt{2} \)
2. \(\sqrt{2}\)
3. \(1 \over \sqrt{2}\)
4. 2
Block A moves on a smooth surface and collides with block B at rest. The maximum energy stored in the spring will be:
| 1. | \(\frac{1}{8} m v^2 \) | 2. | \(\frac{1}{4} \mathrm{~m} v^2 \) |
| 3. | \(\frac{1}{3} m v^2 \) | 4. | \(\frac{1}{2} m v^2\) |
1. \(\dfrac{2 um}{m}\)
2. \(\dfrac{2 um}{M}\)
3. \(\dfrac{2 u}{1 + {m \over M}}\)
4. \(\dfrac{2 u}{1 + {M \over m}}\)
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