A particle of mass \(m\) is driven by a machine that delivers a constant power of \(k\) watts. If the particle starts from rest, the force on the particle at the time \(t\) is:

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:

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:

A body of mass (\(4m\)) is lying in the x-y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (\(m\)) move perpendicular to each other with equal speeds (\(u\)). The total kinetic energy generated due to explosion is:

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. \(9~\text{J}\)

The potential energy of a system increases if work is done:

Force \(F\) on a particle moving in a straight line varies with distance \(d\) as shown in the figure. The work done on the particle during its displacement of \(12\) m is: