A ball bounces to 80% of its original height. Calculate the change in momentum?

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Solution

the velocity of ball before hitting the ball

$v_{1} = \sqrt{2 g h}$

after bouncing to $80$ % of original height

$v_{2} = \sqrt{2 g \times 0.8 h} = 0.9 \sqrt{2 g h}$

the change in the momentum will be,

$\frac{m v_{2}}{m v_{1}} = \frac{0.9}{1} = 90$ %

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A ball bounces to 80% of its original height. Calculate the change in momentum?

the velocity of ball before hitting the ballv1=√2ghafter bouncing to 80% of original heightv2=√2g×0.8h=0.9√2ghthe change in the momentum will be,mv2mv1=0.91=90%

the velocity of ball before hitting the ball

$v_{1} = \sqrt{2 g h}$

after bouncing to $80$ % of original height

$v_{2} = \sqrt{2 g \times 0.8 h} = 0.9 \sqrt{2 g h}$

the change in the momentum will be,

$\frac{m v_{2}}{m v_{1}} = \frac{0.9}{1} = 90$ %