Question

A ball of mass $20$ gram hits a wall at an angle of ${45}^o$ with a velocity of $15m/s$. If the ball rebounds at ${90}^o$ to the direction of incidence, calculate the impulse received by the ball.

Answer 1

Given $m=20g$ and $v=15m/s$,
As shown in figure,
Before the particle strike,
Linear momentum perpendicular to the wall $mv\sin \left({45}^o\right)=\frac{mv}{\sqrt{2}}$
Linear momentum along to the wall $=mv\cos \left({45}^o\right)$
After the particle strike,
Linear momentum perpendicular to the wall $=-mv\sin \left({45}^o\right)=-\frac{mv}{\sqrt{2}}$
Linear momentum along to the wall $=mv\cos \left({45}^o\right)$
Impulse =change in linear momentum,
Linear Momentum changed perpendicular to the wall.
$\Rightarrow Ompulse=-\frac{mv}{\sqrt{2}}-\frac{mv}{\sqrt{2}}=-\sqrt{2}mv=-\sqrt{2}\times \frac{20}{1000}\times 15\times =-\frac{3\sqrt{2}}{10}Kgm/s$.