Two identical balls \(\mathrm{A}\) and \(\mathrm{B}\) having velocities of \(0.5~\text{m/s}\) and \(-0.3~\text{m/s}\) respectively collide elastically in one dimension. The velocities of \(\mathrm{B}\) and \(\mathrm{A}\) after the collision respectively will be:
1. \(-0.5 ~\text{m/s}~\text{and}~0.3~\text{m/s}\) 
2. \(0.5 ~\text{m/s}~\text{and}~-0.3~\text{m/s}\)
3. \(-0.3 ~\text{m/s}~\text{and}~0.5~\text{m/s}\)
4. \(0.3 ~\text{m/s}~\text{and}~0.5~\text{m/s}\)

A body of mass 1 kg begins to move under the action of a time dependent force \(F = 2 t\) \(\hat{i} + 3 t^{2}\ \hat{j}\) N, where \(\hat{i}\) and \(\hat{j}\) are unit vectors along X and Y axis, What power will be developed by the force at the time (t) ?

1. \(\left(2 t^{2} + 4 t^{4}\right) W\)         

2. \(\left(2 t^{3} + 3 t^{4}\right) W\)

3. \(\left(2 t^{3} + 3 t^{5}\right) W\)         

4. \(\left(2 t + 3 t^{3}\right) W\)

What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?

1. $\sqrt{2\mathrm{gR}}$

2. $\sqrt{3gR}$

3. $\sqrt{5gR}$

4. $\sqrt{gR}$
 

A block of mass \(10\) kg, moving in the \(x\text-\)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:

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) $\sqrt{\left(\frac{mk}{2}\right)}{t}^{-1/2}$

(2) $\sqrt{\left(mk\right)}{t}^{-1/2}$

(3) $\sqrt{\left(2mk\right)}{t}^{-1/2}$

(4) $\frac{1}{2}\sqrt{\left(mk\right)}{t}^{-1/2}$

Two particles of masses m₁,m₂ move with initial velocities u₁ and u₂. On collision, one of the particles get excited to higher level, after absorbing energy $\epsilon$. If final velocities of particles be v₁ and v₂, then we must have

(a)m₁²u₁+m₂²u₂-$\epsilon$=m₁²v₁+m₂²v₂

(b)$\frac{1}{2}$m₁u₁²+$\frac{1}{2}$m₂u₂=$\frac{1}{2}$m₁v₁²+$\frac{1}{2}$m₂v₂²-$\epsilon$

(c)$\frac{1}{2}$m₁u₁²+$\frac{1}{2}$m₂u₂²-$\epsilon$=$\frac{1}{2}$m₁v₁²+$\frac{1}{2}$m₂v₂²

(d)$\frac{1}{2}$m₁²u₁²+$\frac{1}{2}$m₂²u₂²+$\epsilon$=$\frac{1}{2}$m₁²v₁²+$\frac{1}{2}$m₂²v₂²

A ball is thrown vertically downwards 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 \(g = 10~\text{m/s}^2\))
1. \(14~\text{m/s}\)
2. \(20~\text{m/s}\)
3. \(28~\text{m/s}\)
4. \(10~\text{m/s}\)

Two particles A and B. move with constant velocities v₁ and v₂. At the initial moment, their position vectors are $r_1$ and $r_2$ respectively. The condition for particles A and B for their collision is-


(1) $\frac{r_1-r_2}{\left|r_1-r_2\right|}=\frac{v_2-v_1}{\left|v_2-v_1\right|}$

(2) $r_1·v_1=r_2·v_2$

(3) _{$r_1\times v_1=r_2\times v_2$}

(4) r₁-r₂=v₁-v₂

On a frictionless surface, a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle θ to its initial direction and has a speed v/3. The second block's speed after the collision is:

(1)2√2v/3
 

(2)3v/4
 

(3)3v/√2
 

(4)√3v/2

The force F acting on a particle of mass m is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to 8 s is:
     
1. 24 Ns

2. 20 Ns

3. 12 Ns

4. 6 Ns