Question

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

(a) 40 m/s

(b) 20 m/s

(c) 10 m/s

(d) 10√30 m/s

Answer 1

Correct option (a) 40 m/s
Explanation:

According to conservation of energy, potential energy at height H is sum of kinetic energy and potential energy at h₂.

Answer 2

As the ball moves on smooth plane no dissipation of initial KE occurs here.

Here net fall of height h = 100- 20=80m

So to this fall  decrease in PE will be = 20×10×80J

If This energy  is converted to KE and the ball gains velocity Vm/s then

1/2×20×V^2= 20×10×80

=> V= 40m/s which is option (a)