Question

1. A ball of mass 0.6 kg is dropped from a height of 1.8m onto a solid floor. Each time the ball
bounces on the floor it loses 10% of its speed.
a) Work out how much momentum was absorbed by the floor in the first bounce?
b) Show that the ball first fails to reach a height of 1 m after the third bounce.

Answer 1

$g\approx 10\frac{m}{s^2}$

$m= 0.6kg$

$h_1= 1.8m$

$v_n-\text{ball speed when reaching ground level}$

$v_n'\text{-ball velocity after hitting the ground}$

$a)$

$\text{According to the law of conservation of energy:}$

$mgh_1= \frac{mv_1^2}{2}$

$v_1=\sqrt{2gh_1}= \sqrt{2*10*1.8}= 6\frac{m}{s}$

$v_1' = v_1-v_1*10 \%=0.9v_1=0.9*6=5.4\frac{m}{s}$

$p_1 = mv_1$

$p'_1=mv'_1$

$\Delta p = p'_1-p=m(v'_1-v)=0.6*(6-5.4)= -0.36kg\frac{m}{s}$

$\text{Answer: }-0.36kg*\frac{m}{s}$

$b)$

$v_1'=0.9v_1$

$v_2 = v'_1$

$v_2'=0.9v_2=0.9^2v_1$

$v_3 = v'_2$

$v_3'=0.9v_3=0.9^3v_1=0.9^3*6=4.374\frac{m}{s}$

$mgh_3 = \frac{mv'^2_3}{2}$

$h_3 = \frac{v'^2_3}{2g}=\frac{4.374^2}{2*10}\approx0.96m$

$\text{Answer: }h_3 = 0.96m<1m$