Q. 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 height of 20 m above the ground. The velocity attained by the ball is
A) $40\ {ms}^{- 1}$
B) $20\ {ms}^{- 1}$
C) $10{ms}^{- 1}$
D) $10\sqrt{30}\ {ms}^{- 1}$
Explanation
(a) According to conservation of energy,
[image:6_image78.emf]
$\ \ \ \ \ \ \ \ \ \ \ mgH = \frac{1}{2}mv^{2} + mgh_{2}$
Or $mg\left( H - h_{2} \right) = \frac{1}{2}mv^{2}$
Or $v = \sqrt{2g(100 - 20)}$
Or v=$\sqrt{2 \times 10 \times 80} = 40ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.