A bullet of mass 0.02, kg travelling horizont | vedclass

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(C) Step $1$: Conservation of linear momentum during the perfectly inelastic collision.Let $m_1 = 0.02\, kg$ be the mass of the bullet and $u_1 = 250\

Vedclass · Accepted Answer

$0.51$

Vedclass · Answer

(C) Step $1$: Conservation of linear momentum during the perfectly inelastic collision.Let $m_1 = 0.02\, kg$ be the mass of the bullet and $u_1 = 250\, ms^{-1}$ be its initial velocity.Let $m_2 = 0.23\, kg$ be the mass of the block and $u_2 = 0\, ms^{-1}$ be its initial velocity.Let $v$ be the common velocity of the bullet and block after the collision.According to the law of conservation of 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$$v = \frac{5}{0.25} = 20\, ms^{-1}$Step $2$: Work-Energy theorem to find the coefficient of friction.The kinetic energy of the combined system is dissipated by the work done by friction.Initial kinetic energy $K = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} 	imes 0.25 	imes (20)^2 = 0.125 	imes 400 = 50\, J$.The work done by friction $W = f_k \cdot d = \mu N \cdot d = \mu (m_1 + m_2) g \cdot d$.Equating work done to the initial kinetic energy:$50 = \mu 	imes 0.25 	imes 9.8 	imes 40$$50 = \mu 	imes 98$$\mu = \frac{50}{98} \approx 0.51$.

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