Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति
JEE Main & Advanced
Physics
Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति
Question Bank
Conservation of Energy and Momentum
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 \[v=\sqrt{2g({{h}_{1}}-{{h}_{2}})}=\sqrt{2\times 10\times (100-20)}\] \[=\sqrt{1600}=40\,m/s\].