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

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(C) Let $m_b = 0.01\, kg$ be the mass of the bullet,$v_i = 500\, m/s$ be its initial velocity,and $v_f$ be its final velocity.Let $M = 2\, kg$ be the

Vedclass · Accepted Answer

$220$

Vedclass · Answer

(C) Let $m_b = 0.01\, kg$ be the mass of the bullet,$v_i = 500\, m/s$ be its initial velocity,and $v_f$ be its final velocity.Let $M = 2\, kg$ be the mass of the block and $V$ be the velocity of the block immediately after the bullet emerges.Using the principle of conservation of mechanical energy for the block,the kinetic energy gained by the block is equal to the potential energy gained: $\frac{1}{2} M V^2 = Mgh$.$V = \sqrt{2gh} = \sqrt{2 	imes 9.8 	imes 0.1} = \sqrt{1.96} = 1.4\, m/s$.Now,applying the principle of conservation of linear momentum in the horizontal direction:$m_b v_i = m_b v_f + M V$.$0.01 	imes 500 = 0.01 	imes v_f + 2 	imes 1.4$.$5 = 0.01 v_f + 2.8$.$0.01 v_f = 5 - 2.8 = 2.2$.$v_f = \frac{2.2}{0.01} = 220\, m/s$.

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