Question

A ball falls vertically onto a floor, with momentum $p$, then bounces repeatedly. The coefficient of restitution is $\frac{1}{2}$, the total momentum imparted on the ball by the floor (due to multiple collisions) is

• A. $2p$
• B. $\frac{3p}{2}$
• C. $3p$
• D. $4p$

Answer 1

The correct option is C $3p$


Let $v_1$ be velocity after first collision
$v_2$ be velocity after second collision
During first collision, applying equation for $e$
$e=\frac{0-v_1}{-u-0}$
$v_1=eu$
During second collision, applying equation for $e$
$e=\frac{0-v_2}{-v_1-0}$
$v_2=e{v}_1$
$v_2=e^{2}u$

Initial momentum before first collision $p=-mu,$
momentum after first collision$p_1=m{v}_1$
Initial momentum before second collision $p_1=m{v}_1$,
momentum after second collision $p_2=m{v}_2$
Change in momentum during first collision $\Delta p_1=p_1-p=m{v}_1-(-mu)=m(v_1+u)$
Change in momentum during second collision }$\Delta p_2=p_2-p_1=m{v}_2-(-m{v}_1)=m(v_2+v_1)$


Total momentum imparted is
$\Delta p_{tot}=\Delta p_1+\Delta p_2+..........$
$=m[u+2v_1+2v_2+..........]$
$=m[u+2eu+2e^{2}u+.........]$
$=mu[1+2e(1+e+e^2+........\left)\right]$
$=p[1+2e\left(\frac{1}{1-e}\right)]$
$\Delta p_{tot}=p\left[\frac{1+e}{1-e}\right]$

Given $e=\frac{1}{2}$,

$\Delta p_{tot}=3p$