Question

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

Answer 1

u=square root of 2gh
V= square root of 2h×0.8h
change in momentum= mv-mu=m×(v-u)

Answer 2

The change in momentum of the ball is 90 %

Explanation:

Percentage of ball bounces from its original position = 80 %

Calculate the change in momentum

Momentum is defined as the product of mass and velocity.Unit of momentum is  $\frac{kg -s^2}{m}$

Momentum (p) = mass (m) × velocity (v)  $\frac{kg -s^2}{m}$

The initial velocity of the ball before hitting is

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

The velocity of the ball after bounces to 80 % of its  original height is

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

$v_2 =0.9 \sqrt{2 g h}$

$\text{The change in momentum of the ball} = \frac{m v_{2}}{m v_{1}}$

$\frac{m v_{2}}{m v_{1}}=\frac{0.9\sqrt{2gh} }{1\sqrt{2gh} }$

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

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

Therefore the change in momentum of the ball after bounces to 80 % from its original position is 90 %

To Learn More ...

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2) A ball of mass 0.2 kg is thrown against the wall . the ball strikes the wall normally with velocity of 30m/sand rebounds with velocity of 20m/s. calculate the impulse of the force exerted by the ball on the wall

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