question_answer

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 [AIEEE 2005]

A)                 10 m/s

B)                    \[10\sqrt{30}\]m/s

C)                 40 m/s 

D)                 20 m/s

Correct Answer: C

Solution :

      Ball starts from the top of a hill which is 100 m high and finally rolls down to a horizontal base which is      20 m above the ground so from the conservation of energy \[mg\,({{h}_{1}}-{{h}_{2}})=\frac{1}{2}m{{v}^{2}}\]                 Þ \[v=\sqrt{2g({{h}_{1}}-{{h}_{2}})}=\sqrt{2\times 10\times (100-20)}\]                 \[=\sqrt{1600}=40\,m/s\].