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. \(A/B\)