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A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall at an angle of $30^\circ$ with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 seconds, the average force acting on the wall is :-
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A
48 N
B
24 N
C
12 N
D
96 N
Detailed Solution
Resolve the velocity along the wall and perpendicular to the wall. The angle with the wall is $30^\circ$.
Component along the wall $= v\cos 30^\circ$; it is unchanged by the collision.
Component perpendicular to the wall $= v\sin 30^\circ$; it is reversed by the collision.
Change in momentum of the ball $= mv\sin 30^\circ - (-mv\sin 30^\circ) = 2mv\sin 30^\circ$
$\Delta p = 2 \times 0.5 \times 12 \times \dfrac{1}{2} = 6$ kg m/s
Average force $F = \dfrac{\Delta p}{\Delta t} = \dfrac{2mv\sin 30^\circ}{0.25} = \dfrac{6}{0.25}$
$F = 24$ N
Component along the wall $= v\cos 30^\circ$; it is unchanged by the collision.
Component perpendicular to the wall $= v\sin 30^\circ$; it is reversed by the collision.
Change in momentum of the ball $= mv\sin 30^\circ - (-mv\sin 30^\circ) = 2mv\sin 30^\circ$
$\Delta p = 2 \times 0.5 \times 12 \times \dfrac{1}{2} = 6$ kg m/s
Average force $F = \dfrac{\Delta p}{\Delta t} = \dfrac{2mv\sin 30^\circ}{0.25} = \dfrac{6}{0.25}$
$F = 24$ N