Answer to Question #161898 in Mechanics | Relativity for Mercy

Question

A body of mass 50g falls from a height of 30cm on to a horizontal surface and rebounds to a height of 30cm. If the time of collision is 0.05s. Calculate the average force experienced by the body during the impact, the kinetic energy lost by the body on impact and the coefficient of restitution between the two bodies

Answer 1

$\text{find the speed of the body when it reaches the surface}$

$h_1 - \text{body height above the surface}$

$V_1=\sqrt{2gh_1}=\sqrt{2*9.8*0.3}\approx2.42$

$\text{the rate of rebound of the body from the surface}$

$h_2=\text{body height after rebound}$

$V_2=-\sqrt{2gh_2}=-\sqrt{2*9.8*0.3}\approx-2.42$

$\text{average force experienced by the body during the impact:}$

$F=\frac{\Delta{p}}{t}=\frac{mV_1-mV_2}{t}=\frac{0.05*2.42-0.05*(-2.42)}{0.05}=4.48$

$\text{kinetic energy lost}$

$\Delta{E}=E_1-E_2=\frac{mV_1^2}{2}-\frac{mV_2^2}{2}=0$

$\text{coefficient of restitution }$

$e=-\frac{V_2}{V_1}=1$

Answer:$F=4.48N;\Delta{E}=0;e=1$