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

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(B) Let $m = 0.01\,kg$ be the mass of the bullet,$M = 2\,kg$ be the mass of the block,$u = 500\,m/s$ be the initial speed of the bullet,$v$ be the fin

Vedclass · Accepted Answer

$220$

Vedclass · Answer

(B) Let $m = 0.01\,kg$ be the mass of the bullet,$M = 2\,kg$ be the mass of the block,$u = 500\,m/s$ be the initial speed of the bullet,$v$ be the final speed of the bullet,and $V$ be the velocity of the block immediately after the collision.Applying the Conservation of Linear Momentum $(COLM)$:$m u = m v + M V$$0.01 	imes 500 = 0.01 v + 2 V$$5 = 0.01 v + 2 V \quad ...(1)$Applying the Conservation of Mechanical Energy $(COME)$ for the block rising to height $h = 0.1\,m$:$\frac{1}{2} M V^2 = M g h$$V = \sqrt{2 g h} = \sqrt{2 	imes 10 	imes 0.1} = \sqrt{2} \approx 1.414\,m/s$ (Using $g = 10\,m/s^2$)Substituting $V$ into equation $(1)$:$5 = 0.01 v + 2(1.414)$$5 = 0.01 v + 2.828$$0.01 v = 2.172$$v = 217.2\,m/s \approx 220\,m/s$ (Rounding to the nearest provided option).

$A$ bullet of mass $0.01\,kg$ and travelling at a speed of $500\,m/s$ strikes a block of mass $2\,kg$ which is suspended by a string of length $5\,m$. The centre of gravity of the block is found to rise a vertical distance of $0.1\,m$. What is the speed of the bullet after it emerges from the block?

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