# Answer in Mechanics | Relativity for Varylan #141518

Applying the principle of conservation of energy,

The potential energy of the ball before falling is equal to the kinetic energy of the ball right before it strucks the ground.

“mgh=frac{1}2m{v_1}u00b2\nv_1=sqrt{2gh}\”

“begin{aligned} textsf{where }&m=textsf{mass of the ball}\ &g=textsf{acceleration due to gravity}\&h=textsf{height from which the ball fell}\&v_1=textsf{velocity with which the ball struck the ground}\end{aligned}” “v_1=sqrt{2gh}\nv_1=sqrt{2u00d79.81u00d745}\nv_1=9.4ms^{-1}\ntextsf{Considering the direction,} overrightarrow{v_1} = -9.4ms^{-1}.\ntextsf{Let } v_2 textsf{represent the velocity with which the ball rebounds}\noverrightarrow{v_2}=-frac{2}3overrightarrow{v_1}\noverrightarrow{v_2}=-frac{2}3(-9.4ms^{-1})\noverrightarrow{v_2}=6.26ms^{-1}\”

(i) the momentum change on hitting the ground is,

“begin{aligned} Delta P&=moverrightarrow{v_2}-moverrightarrow{v_1}\&=m(overrightarrow{v_2}-overrightarrow{v_1})\&=0.2kg(6.26ms^{-1}-(-9.4ms^{-1}))\Delta P&=3.132Nsend{aligned}”

(ii) The change of momentum is equal to the impulse. The impulse is given by,

“I=Ft\nbegin{aligned}textsf{Where }&F=textsf{force due to impact}\&t=textsf{time of impact} end{aligned}\hspace{1cm}\Ft=Delta P\nF=frac{Delta P}t\nF=frac{3.132Ns}{0.1s}\nF=31.32N”

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