A piece of wood of mass 0.03, kg is dropped f | vedclass

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(C) Let the mass of the wood be $m_1 = 0.03\, kg$ and the mass of the bullet be $m_2 = 0.02\, kg$. The total mass is $M = m_1 + m_2 = 0.05\, kg$.At th

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$40$

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(C) Let the mass of the wood be $m_1 = 0.03\, kg$ and the mass of the bullet be $m_2 = 0.02\, kg$. The total mass is $M = m_1 + m_2 = 0.05\, kg$.At the instant of collision,the wood has fallen from the top $(y = 100\, m)$ and the bullet has risen from the ground $(y = 0)$. Since they are dropped/fired at the same time,they collide when the bullet reaches the wood. However,the problem implies the center of mass $(COM)$ approach.The position of the center of mass from the ground is $Y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} = \frac{0.03 	imes 100 + 0.02 	imes 0}{0.05} = \frac{3}{0.05} = 60\, m$.The velocity of the center of mass is $V_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{0.03 	imes 0 + 0.02 	imes 100}{0.05} = \frac{2}{0.05} = 40\, m/s$.After the collision,the system moves as a single particle with initial velocity $V_{cm} = 40\, m/s$ at a height of $60\, m$.The maximum height $H_{max}$ reached by the center of mass from the ground is given by $H_{max} = Y_{cm} + \frac{V_{cm}^2}{2g} = 60 + \frac{40^2}{2 	imes 10} = 60 + \frac{1600}{20} = 60 + 80 = 140\, m$.The height above the top of the building is $140\, m - 100\, m = 40\, m$.

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