A bullet of mass 0.01 ,kg travelling at a spe | vedclass

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(B) Given:Mass of bullet,$m = 0.01 \,kg$Initial speed of bullet,$u = 500 \,ms^{-1}$Mass of block,$M = 2 \,kg$Vertical rise of block,$h = 0.1 \,m$Let $

Vedclass · Accepted Answer

$220$

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(B) Given:Mass of bullet,$m = 0.01 \,kg$Initial speed of bullet,$u = 500 \,ms^{-1}$Mass of block,$M = 2 \,kg$Vertical rise of block,$h = 0.1 \,m$Let $v_b$ be the speed of the bullet after emerging from the block and $V$ be the velocity of the block immediately after the collision.Applying the law of conservation of energy to the block after the collision:$\frac{1}{2} M V^2 = Mgh$$V = \sqrt{2gh} = \sqrt{2 	imes 9.8 	imes 0.1} = \sqrt{1.96} = 1.4 \,ms^{-1}$Applying the law of conservation of linear momentum during the collision:$m u = m v_b + M V$$0.01 	imes 500 = (0.01 	imes v_b) + (2 	imes 1.4)$$5 = 0.01 v_b + 2.8$$0.01 v_b = 5 - 2.8 = 2.2$$v_b = \frac{2.2}{0.01} = 220 \,ms^{-1}$Thus,the speed of the bullet after it emerges from the block is $220 \,ms^{-1}$.

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