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Question

Physics Question on laws of motion

A rigid ball of mass mm strikes a rigid wall at 6060^� and gets reflected without loss of speed as shown in the figure. The value of impulse imparted by the wall on the ball will be

A

mm VV

B

2m2 \,m VV

C

mV2\frac{m V} {2}

D

mV3\frac{m V}{3}

Answer

mm VV

Explanation

Solution

Given, pip_{i} =pf=p_{f} =mV=m V
Change in momentum of the ball = pf\vec{p}_{f} pi-\vec{p}_{i}
=(pfxi^pfyj^)=\left(-p_{fx} \hat{i}-p_{fy} \hat{j}\right) (pixi^piyj^)-\left(p_{ix} \hat{i}-p_{iy} \hat{j}\right)
=i^(pfx+pix)j^=-\hat{i} \left(p_{fx}+p_{ix}\right)-\hat{j} (pfxpiy) \left(p_{fx}-p_{iy}\right)
=2pixi^=mVi^=-2 p_{ix} \hat{i}=-m V \hat{i} [pfypiy=0]\left[\because\quad p_{fy}-p_{iy}=0\right]
Here, pix=pfxp_{ix}=p_{fx} =picos60=p_{i} \, cos \, 60^{\circ} =mV2=\frac{m \,V}{2}
\because Impulse imparted by the wall = change in the momentum of the ball = mm VV