A bullet of mass 0.01 , kg travelling at a sp | 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,$V_b$ be

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,$V_b$ be the final speed of the bullet,and $V_B$ be the velocity of the block immediately after the impact.Applying the Law of Conservation of Linear Momentum $(COLM)$ during the collision:$m \cdot u = m \cdot V_b + M \cdot V_B$$0.01 	imes 500 = 0.01 	imes V_b + 2 	imes V_B$$5 = 0.01 \cdot V_b + 2 \cdot V_B \quad \dots(1)$Applying the Law of Conservation of Mechanical Energy $(COME)$ for the block after the collision:$\frac{1}{2} M V_B^2 = Mgh$$V_B = \sqrt{2gh} = \sqrt{2 	imes 9.8 	imes 0.1} = \sqrt{1.96} = 1.4 \, m/s$Substitute $V_B = 1.4 \, m/s$ into equation $(1)$:$5 = 0.01 \cdot V_b + 2(1.4)$$5 = 0.01 \cdot V_b + 2.8$$0.01 \cdot V_b = 2.2$$V_b = \frac{2.2}{0.01} = 220 \, m/s$.

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