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. \(\frac{\sqrt{3}}{2}v\)

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:

A force of 5 N making an angle $\theta$ with the horizontal acting on an object displaces it by 0.4 m along the horizontal direction. If the object gains kinetic energy of 1 J then the component of the force is:

A block of mass M is attached to the lower end of a vertical spring. The spring is hung from the ceiling and has a force constant value of k. The mass is released from rest with the spring initially unstretched. The maximum extension produced along the length of the spring will be:

A ball is thrown vertically downwards from a height of 20 m with an initial velocity v_o. It collides with the ground, loses 50% of its energy in a collision and rebounds to the same height. The initial velocity v_o is: (Take g = 10 ms^-2)

1. 14 ms^-1
2. 20 ms^-1
3. 28 ms^-1
4. 10 ms^-1

A pump ejects 12,000 kg of water at a speed of 4 m/s in 40 s. The average rate at which the pump is working is:

1.  0.24 kW

2.  2.4 W

3.  2.4 kW

4.  24 W

A block of mass 2 kg moving with a velocity of 10 m/s on a smooth surface hits a spring of force constant $80\times {10}^3$ N/m as shown. The maximum compression in the spring will be:
               

A body is displaced from (0, 0) to (1m, 1m) along the path x=y by a force $F=\left(x^2\hat{j}+y\hat{i}\right)N$. The work done by this force will be :

1. $\frac{4}{3}J$

2. $\frac{5}{6}J$

3. $\frac{3}{2}J$

4. $\frac{7}{5}J$

A block of mass m initially at rest, is dropped from a height h onto a spring of force constant k. If the maximum compression in the spring is x, then:     
         

1. $mgh=\frac{1}{2}k{x}^2$

2. $mg(h+x)=\frac{1}{2}k{x}^2$

3. $mgh=\frac{1}{2}k{(x+h)}^2$

4. $mg(h+x)=\frac{1}{2}k{(x+h)}^2$

The relationship between force and position is shown in the given figure (in a one-dimensional case). The work done by the force in displacing a body from \(x = 1\) cm to \(x = 5\) cm is:

1.   \(20\) ergs
2.   \(60\) ergs
3.   \(70\) ergs
4.   \(700\) ergs