A bullet of mass 0.02 kg travelling horizont | vedclass

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(C) Let the mass of the bullet be $m_1 = 0.02 \ kg$ and its initial velocity be $u_1 = 250 \ ms^{-1}$.Let the mass of the block be $m_2 = 0.23 \ kg$ a

Vedclass · Accepted Answer

$0.51$

Vedclass · Answer

(C) Let the mass of the bullet be $m_1 = 0.02 \ kg$ and its initial velocity be $u_1 = 250 \ ms^{-1}$.Let the mass of the block be $m_2 = 0.23 \ kg$ and its initial velocity be $u_2 = 0 \ ms^{-1}$.By the law of conservation of linear momentum,the momentum before impact equals the momentum after impact:$m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$$0.02 	imes 250 + 0.23 	imes 0 = (0.02 + 0.23)v$$5 = 0.25v$$v = \frac{5}{0.25} = 20 \ ms^{-1}$Now,the combined mass $M = m_1 + m_2 = 0.25 \ kg$ moves with initial velocity $v = 20 \ ms^{-1}$ and comes to rest after covering a distance $d = 40 \ m$ due to friction.The work done by friction equals the change in kinetic energy:$W = \Delta K$$f_k \cdot d = \frac{1}{2} M v^2$$(\mu M g) \cdot d = \frac{1}{2} M v^2$$\mu = \frac{v^2}{2gd} = \frac{20^2}{2 	imes 9.8 	imes 40}$$\mu = \frac{400}{784} \approx 0.51$

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