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 $u_1 = 250 \ ms^{-1}$.The mass of the block is $m_2 = 0.23 \ kg$ and i

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 $u_1 = 250 \ ms^{-1}$.The mass of the block is $m_2 = 0.23 \ kg$ and its initial velocity is $u_2 = 0 \ ms^{-1}$.After the impact,the bullet and block move together with a common velocity $v$. By the principle of conservation of linear momentum:$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 = 0.25 v$$v = \frac{5}{0.25} = 20 \ ms^{-1}$Now,the combined mass $M = m_1 + m_2 = 0.25 \ kg$ moves on a rough surface and comes to rest after covering a distance $d = 40 \ m$. The work done by the frictional force is equal to the change in kinetic energy:$W = \Delta K$$-f_k \cdot d = 0 - \frac{1}{2} M v^2$$\mu M g d = \frac{1}{2} M v^2$$\mu = \frac{v^2}{2 g d}$Substituting the values:$\mu = \frac{(20)^2}{2 	imes 9.8 	imes 40} = \frac{400}{784} \approx 0.51$

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