A bullet of mass 25 ,g moving horizontally at | vedclass

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(D) Let $m = 25 \,g = 0.025 \,kg$ be the mass of the bullet and $u = 250 \,m/s$ be its initial velocity.Let $M = 1 \,kg$ be the mass of the wooden blo

Vedclass · Accepted Answer

$170$

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(D) Let $m = 25 \,g = 0.025 \,kg$ be the mass of the bullet and $u = 250 \,m/s$ be its initial velocity.Let $M = 1 \,kg$ be the mass of the wooden block.Let $v_1$ be the final velocity of the bullet and $v_2$ be the velocity of the block immediately after the bullet emerges.First, we use the principle of conservation of mechanical energy for the block as it rises to a height $h = 20 \,cm = 0.2 \,m$:$\frac{1}{2} M v_2^2 = M g h$$v_2 = \sqrt{2 g h} = \sqrt{2 	imes 10 	imes 0.2} = \sqrt{4} = 2 \,m/s$Next, we apply the principle of conservation of linear momentum for the system (bullet + block) during the collision:$m u = m v_1 + M v_2$$0.025 	imes 250 = 0.025 	imes v_1 + 1 	imes 2$$6.25 = 0.025 v_1 + 2$$0.025 v_1 = 4.25$$v_1 = \frac{4.25}{0.025} = 170 \,m/s$Thus, the speed of the bullet as it emerges from the block is $170 \,m/s$.

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