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A body of mass 3 kg hits a wall at an angle of $60^\circ$ and returns at the same angle. The impact time was 0.2 s. Calculate the force exerted on the wall :
A
$150\sqrt{3}$ N
B
$50\sqrt{3}$ N
C
100 N
D
$75\sqrt{3}$ N
Detailed Solution
The velocity makes $60^\circ$ with the wall, i.e. $30^\circ$ with the normal to the wall.
[add image: velocity 10 m/s at $30^\circ$ to the normal before and after the impact]
The component along the wall ($v\sin 30^\circ$) is unchanged. The component along the normal ($v\cos 30^\circ$) is reversed.
Change in momentum $\Delta p = mv\cos 30^\circ - (-mv\cos 30^\circ) = 2mv\cos 30^\circ$
$F = \dfrac{\Delta p}{\Delta t} = \dfrac{2 \times 3 \times 10 \times \frac{\sqrt{3}}{2}}{0.2} = \dfrac{30\sqrt{3}}{0.2}$
$F = 150\sqrt{3}$ N
[add image: velocity 10 m/s at $30^\circ$ to the normal before and after the impact]
The component along the wall ($v\sin 30^\circ$) is unchanged. The component along the normal ($v\cos 30^\circ$) is reversed.
Change in momentum $\Delta p = mv\cos 30^\circ - (-mv\cos 30^\circ) = 2mv\cos 30^\circ$
$F = \dfrac{\Delta p}{\Delta t} = \dfrac{2 \times 3 \times 10 \times \frac{\sqrt{3}}{2}}{0.2} = \dfrac{30\sqrt{3}}{0.2}$
$F = 150\sqrt{3}$ N