A ball is dropped from a height of \(5\) m. If it rebounds up to a height of \(1.8\) m, then the ratio of velocities of the ball after and before the rebound will be:
1. $\frac{3}{5}$

2. $\frac{2}{5}$

3. $\frac{1}{5}$

4. $\frac{4}{5}$ 

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:
1. \(1:2\sqrt{3}\)
2. \(1:\sqrt{2}\)
3. \(\sqrt{2}:1\)
4. \(1:\sqrt{3}\)

A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB = D. The height h is equal to: 
            

1. $\frac{3}{2}D$

2. D

3. $\frac{7}{4}D$

4. $\frac{5}{4}D$

If a stone is projected vertically upward from the ground at a speed of 10 m/s, then it's: (g = 10 $\mathrm{m}/{\mathrm{s}}^2$)

1.  Potential energy will be maximum after 0.5 s

2.  Kinetic energy will be maximum again after 1 s

3.  Kinetic energy = potential energy at a height of 2.5 m from the ground

4.  Potential energy will be minimum after 1 s

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 spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It slides down a smooth surface to the ground, then climbs up another hill of height 30 m and finally slides down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is: 

The principle of conservation of energy implies that:

1.  the total mechanical energy is conserved.

2.  the total kinetic energy is conserved.

3.  the total potential energy is conserved.

4.  the sum of all types of energies is conserved.

The potential energy of a 1 kg particle free to move along the x-axis is given by:

$\mathrm{U}\left(\mathrm{x}\right) = \left(\frac{{\mathrm{x}}^4}{4}-\frac{{\mathrm{x}}^2}{2}\right)\mathrm{J}$

The total mechanical energy of the particle is 2J. Then, the maximum speed (in ms^-1) will be
1. \(3 \over \sqrt{2} \)
2. \(\sqrt{2}\)
3. \(1 \over \sqrt{2}\)
4. 2

A chain of length L and mass m is placed upon a smooth surface. The length of BA is (L–b). What will be the velocity of the chain when its end A reaches B?
        

1. \( \sqrt{\frac{2 g \sin \theta}{L}\left(L^2-b^2\right)} \)
2. \( \sqrt{\frac{g \sin \theta}{2 L}\left(L^2-b^2\right)} \)
3. \( \sqrt{\frac{g \sin \theta}{L}\left(L^2-b^2\right)}\)
4. None of these