Answer to Question #141518 in Mechanics | Relativity for Varylan

Question

A ball of mass 0.2kg falls from a height of 45m. On striking the ground, it rebounds in 0.1s with ⅔ of the velocity with which it struck the ground. Calculate (i) the momentum change on hitting the ground (ii) the force on the ball due to the impact

Answer 1

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.

$mgh=\frac{1}2m{v_1}²\\ v_1=\sqrt{2gh}\\$

$\begin{aligned} \textsf{where }&m=\textsf{mass of the ball}\\ &g=\textsf{acceleration due to gravity}\\&h=\textsf{height from which the ball fell}\\&v_1=\textsf{velocity with which the ball struck the ground}\\\end{aligned}$ $v_1=\sqrt{2gh}\\ v_1=\sqrt{2×9.81×45}\\ v_1=9.4ms^{-1}\\ \textsf{Considering the direction,} \overrightarrow{v_1} = -9.4ms^{-1}.\\ \textsf{Let } v_2 \textsf{represent the velocity with which the ball rebounds}\\ \overrightarrow{v_2}=-\frac{2}3\overrightarrow{v_1}\\ \overrightarrow{v_2}=-\frac{2}3(-9.4ms^{-1})\\ \overrightarrow{v_2}=6.26ms^{-1}\\$

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

$\begin{aligned} \Delta P&=m\overrightarrow{v_2}-m\overrightarrow{v_1}\\&=m(\overrightarrow{v_2}-\overrightarrow{v_1})\\&=0.2kg(6.26ms^{-1}-(-9.4ms^{-1}))\\\Delta P&=3.132Ns\end{aligned}$

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

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