A vertical spring with a force constant \(k\) is fixed on a table. A ball of mass \(m\) at a height \(h\) above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance \(d\). The net work done in the process is:
1. \(mg(h+d)+\frac{1}{2}kd^2\)
2. \(mg(h+d)-\frac{1}{2}kd^2\)
3. \(mg(h-d)-\frac{1}{2}kd^2\)
4. \(mg(h-d)+\frac{1}{2}kd^2\)

When an object is shot from the bottom of a long, smooth inclined plane kept at an angle of \(60^\circ\) $$with horizontal, it can travel a distance \(x_1\) along the plane. But when the inclination is decreased to \(30^\circ\) $$and the same object is shot with the same velocity, it can travel \(x_2\) distance. Then \(x_1:x_2\) will be:

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 m dropped from a height h reaches the ground with a speed of 1.4$\sqrt{\mathrm{gh}}$. The work done by air drag is:

1. –0.2mgh 

2. –0.02mgh

3. –0.04mgh 

4. mgh