A rubber ball of mass 50$\mathrm { g }$ falls from a height of $\mathrm { m }$ and rebounds to a height of .5$\mathrm { m }$ . The impulse between the ball and the ground if the time during which they are in contact was $0,1$ sec is

Given,

mass of ball is 50 g

Height is 1 m

Rebound at height 0.5 m

When ball is falling from height 1 m then initial velocity will be zero. when it strike on ground then Final velocity is ,

$\begin{array}{l} v^{2}=u^{2}+2 g h \\ v^{2}=o^{2}+2 \times 9.8 \times 1 \\ v=\sqrt{19.6} \\ v=4.4 \mathrm{~m} / \mathrm{s} \end{array}$

Now when ball rebound then final velocity will be zero therefore initial velocity is

$\begin{array}{l} \mathrm{v}^{2}=\mathrm{u}^{2}-2 \mathrm{gh} \\ \mathrm{o}^{2}=\mathrm{u}^{2}-2 \times 9.8 \times 0.5 \\ \mathrm{u}=\sqrt{9.8} \\ \mathrm{u}=3.13 \mathrm{~m} / \mathrm{s} \end{array}$

Time of contact( t ) is o. 1 sec

Impulse is change in momentum

$\begin{array}{l} \mathrm{I}=\mathrm{m}(\mathrm{v}-\mathrm{u}) \\ \mathrm{I}=\frac{50}{1000}(4.4-(-3.13)) \\ \mathrm{I}=\frac{50}{1000}(4.4+3.13) \\ \mathrm{I}=0.376 \mathrm{~N}-\mathrm{s} \end{array}$

Force exerted is $\mathrm{F}=\frac{0.376}{0.1}=3.76 \mathrm{~N}$