A ball is thrown vertically downward from a height of \(20\) 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, \(g=10~\mathrm{ms^{-2}}\)]
1. \(14\) ms^-1
2. \(20\) ms^-1
3. \(28\) ms^-1
4. \(10\) 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. \(W_P<W_Q;~W_P<W_Q\)

A block of mass \(10\) kg, moving in the x-direction with a constant speed of \(10\) ms^-1 is subjected to a retarding force \(F=0.1x\) J/m during its travel from \(x = 20\) m to \(30\) m. Its final kinetic energy will be:
1. \(475\) J
2. \(450\) J
3. \(275\) J
4. \(250\) J

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 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 m₁ and m₂ move with initial velocities u₁ and u₂ respectively. On collision, one of the particles gets excited to a higher level, after absorbing energy E. If the final velocities of particles are v₁ and v₂, then we must have:

1. ${\mathrm{m}}_1^2{\mathrm{u}}_1+{\mathrm{m}}_2^2{\mathrm{u}}_2-\mathrm{E}={\mathrm{m}}_1^2{\mathrm{v}}_1+{\mathrm{m}}_2^2{\mathrm{v}}_2$

2. $\frac{1}{2}{\mathrm{m}}_1{\mathrm{u}}_1^2+\frac{1}{2}{\mathrm{m}}_2{\mathrm{u}}_2^2=\frac{1}{2}{\mathrm{m}}_1{\mathrm{v}}_1^2+\frac{1}{2}{\mathrm{m}}_2{\mathrm{v}}_2^2$

3. $\frac{1}{2}{\mathrm{m}}_1{\mathrm{u}}_1^2+\frac{1}{2}{\mathrm{m}}_2{\mathrm{u}}_2^2-\mathrm{E}=\frac{1}{2}{\mathrm{m}}_1{\mathrm{v}}_1^2+\frac{1}{2}{\mathrm{m}}_2{\mathrm{v}}_2^2$

4. $\frac{1}{2}{\mathrm{m}}_1^2{\mathrm{u}}_1^2+\frac{1}{2}{\mathrm{m}}_2^2{\mathrm{u}}_2^2+\mathrm{E}=\frac{1}{2}{\mathrm{m}}_1^2{\mathrm{v}}_1^2+\frac{1}{2}{\mathrm{m}}_2^2{\mathrm{v}}_2^2$

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. $\frac{\sqrt{3}}{2}v$

A uniform force of \((3 \hat{i} + \hat{j})\) newton acts on a particle of mass \(2\) kg. Hence the particle is displaced from position \((2 \hat{i} + \hat{k})\) meter to position \((4 \hat{i} + 3 \hat{j} - \hat{k})\) meter. The work done by the force on the particle is:
1. \(6\) J
2. \(13\) J
3. \(15\) J
4. \(9\) J

The potential energy of a particle in a force field is \(U=\)$\frac{A}{r^2}-\frac{B}{r}$ where \(A\) and \(B\) are positive constants and \(r\) is the distance of the particle from the center of the field. For stable equilibrium, the distance of the particle is:
1. \(B/A\)
2. \(B/2A\)
3. \(2A/B\)
4. \(A/B\)