Question

a ball of mass 200g moving at a speed of 22m/s hits a wall and rebounds at a speed of 17m/s what percentage of mechanical energy is lost in collision

Answer 1

Hope this helps for the future.

Answer 2

Mass of ball, $\sf{m=0.2\ kg.}$

Initial velocity of ball, $\sf{u=22\ m\,s^{-1}}$

Final velocity of ball, $\sf{v=-17\ m\,s^{-1}}$ since the ball rebounds after collision.

Here the mechanical energy is stored in the body as its kinetic energy since the ball is in motion,

Initial kinetic energy is,

$\longrightarrow\sf{E=\dfrac{1}{2}mu^2}$

$\longrightarrow\sf{E=\dfrac{1}{2}\times0.2\,(22)^2}$

$\longrightarrow\sf{E=48.4\ J}$

The loss in mechanical energy is,

$\longrightarrow\sf{\Delta E=\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2}$

$\longrightarrow\sf{\Delta E=\dfrac{1}{2}m\left(v^2-u^2\right)}$

$\longrightarrow\sf{\Delta E=\dfrac{1}{2}\times0.2\left((-17)^2-(22)^2\right)}$

$\longrightarrow\sf{\Delta E=\dfrac{1}{2}\times0.2\left(289-484\right)}$

$\longrightarrow\sf{\Delta E=\dfrac{1}{2}\times0.2\times-195}$

$\longrightarrow\sf{\Delta E=-19.5\ J}$

Hence the percentage loss of mechanical energy is,

$\longrightarrow\sf{\delta E=\dfrac{\Delta E}{E}\times100}$

$\longrightarrow\sf{\delta E=\dfrac{-19.5}{48.4}\times100}$

$\longrightarrow\sf{\underline{\underline{\delta E=-40.3\%}}}$