A ball of mass 0.2 kg rests on a vertical po | vedclass

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(D) Let the mass of the bullet be $m = 0.01 \ kg$ and the mass of the ball be $M = 0.2 \ kg$. The ball is initially at rest at a height $h = 5 \ m$. S

Vedclass · Accepted Answer

$500 \ m/s$

Vedclass · Answer

(D) Let the mass of the bullet be $m = 0.01 \ kg$ and the mass of the ball be $M = 0.2 \ kg$. The ball is initially at rest at a height $h = 5 \ m$. Since the motion is horizontal projectile motion,the time $t$ taken to reach the ground is given by $h = \frac{1}{2} g t^2$. Taking $g = 10 \ m/s^2$,we have $5 = \frac{1}{2} 	imes 10 	imes t^2$,which gives $t = 1 \ s$. After the collision,the horizontal velocity of the ball is $v_b = \frac{	ext{distance}}{	ext{time}} = \frac{20 \ m}{1 \ s} = 20 \ m/s$. The horizontal velocity of the bullet is $v_u = \frac{100 \ m}{1 \ s} = 100 \ m/s$. By the principle of conservation of linear momentum in the horizontal direction: $m V = m v_u + M v_b$ $0.01 	imes V = (0.01 	imes 100) + (0.2 	imes 20)$ $0.01 	imes V = 1 + 4 = 5$ $V = \frac{5}{0.01} = 500 \ m/s$.

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