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

• A. $mg(h_{1} - h_{2})$
• B. $m(\sqrt{2gh_{1}} + \sqrt{2gh_{2}})$
• C. $m\sqrt{2g(h_{1} + h_{2})}$
• D. $m\sqrt{2g}(h_{1} + h_{2})$

Answer 1

Answer: (B)

$m(\sqrt{2gh_{1}} + \sqrt{2gh_{2}})$

Answer 2

When ball falls vertically downward from height $h _ { 1 }$ its velocity and its velocity after collision

Change in momentum

$\Delta \vec { P } = m \left( \vec { v } _ { 2 } - \vec { v } _ { 1 } \right) = m \left( \sqrt { 2 g h _ { 1 } } + \sqrt { 2 g h _ { 2 } } \right)$

(because $\vec { v } _ { 1 }$ and $\vec { v } _ { 2 }$ are opposite in direction)