Work Energy And Power Question 105
Question: A ball of mass m falls vertically to the ground from a height h1 and rebound to a height $ h_2 $ . The change in momentum of the ball on striking the ground is [AMU (Engg.) 1999]
Options:
A) $ mg(h_1-h_2) $
B) $ m(\sqrt{2gh_1}+\sqrt{2gh_2}) $
C) $ m\sqrt{2g(h_1+h_2)} $
D) $ m\sqrt{2g}(h_1+h_2) $
Show Answer
Answer:
Correct Answer: B
Solution:
When ball falls vertically downward from height
$ h_1 $ its velocity $ {{\overrightarrow{v}}_1}=\sqrt{2gh_1} $
and its velocity after collision $ {{\overrightarrow{v}}_2}=\sqrt{2gh_2} $
Change in momentum $ \Delta \vec{P}=m({{\overrightarrow{v}}_2}-{{\overrightarrow{v}}_1})=m(\sqrt{2gh_1}+\sqrt{2gh_2}) $ (because $ {{\overrightarrow{v}}_1} $
and $ {{\overrightarrow{v}}_2} $ are opposite in direction)
