A ball of mass 3 kg,moving with a speed of 1 | vedclass

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(A) The angle given is with the wall,so the angle with the normal is $	heta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.Change in momentum $\Delta p$ occ

Vedclass · Accepted Answer

$1500\sqrt{3} \ N$

Vedclass · Answer

(A) The angle given is with the wall,so the angle with the normal is $	heta = 90^{\circ} - 60^{\circ} = 30^{\circ}$.Change in momentum $\Delta p$ occurs only in the direction perpendicular to the wall.The component of velocity perpendicular to the wall is $v_{\perp} = v \sin(60^{\circ}) = v \cos(30^{\circ})$.Initial momentum component perpendicular to the wall: $p_i = m v \cos(30^{\circ})$.Final momentum component perpendicular to the wall: $p_f = -m v \cos(30^{\circ})$.Change in momentum: $\Delta p = |p_f - p_i| = 2 m v \cos(30^{\circ})$.Substituting the values: $\Delta p = 2 	imes 3 	imes 100 	imes \frac{\sqrt{3}}{2} = 300\sqrt{3} \ kg \cdot m/s$.The force exerted is $F = \frac{\Delta p}{\Delta t} = \frac{300\sqrt{3}}{0.2} = 1500\sqrt{3} \ N$.

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