question_answer

A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It roils 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)  40m/s  

B)                        20 m/s

C)  10 m/s                

D)        10\[qv{{R}^{2}}\] m/s

Correct Answer: A

Solution :

According to conservation of energy: \[v_{f}^{2}=v_{i}^{2}+\frac{2G{{m}_{e}}}{{{R}_{e}}}\left( 1+\frac{1}{10} \right)\] or   \[v_{f}^{2}=v_{i}^{2}+\frac{2G{{m}_{e}}}{{{R}_{e}}}\left( 1-\frac{1}{10} \right)\] or  \[v_{f}^{2}=v_{i}^{2}+\frac{2Gm}{{{R}_{e}}}\left( 1-\frac{1}{10} \right)\] or \[\mathbf{\vec{v}}\]