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:

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. $9~\text{J}$