A bag of sand of mass 9.8 , kg is suspended b | vedclass

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(B) Initial momentum of the system: $P_i = m_b v_b + m_s v_s = 0.2 \, kg 	imes 10 \, ms^{-1} + 9.8 \, kg 	imes 0 = 2 \, kg \cdot ms^{-1}$.Final mome

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$9.8$

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(B) Initial momentum of the system: $P_i = m_b v_b + m_s v_s = 0.2 \, kg 	imes 10 \, ms^{-1} + 9.8 \, kg 	imes 0 = 2 \, kg \cdot ms^{-1}$.Final momentum of the system: $P_f = (m_b + m_s) v = (0.2 + 9.8) \, kg 	imes v = 10 \, kg 	imes v$.By the law of conservation of linear momentum: $P_i = P_f \implies 2 = 10v \implies v = 0.2 \, ms^{-1}$.Initial kinetic energy: $K_i = \frac{1}{2} m_b v_b^2 = \frac{1}{2} 	imes 0.2 	imes (10)^2 = 0.1 	imes 100 = 10 \, J$.Final kinetic energy: $K_f = \frac{1}{2} (m_b + m_s) v^2 = \frac{1}{2} 	imes 10 	imes (0.2)^2 = 5 	imes 0.04 = 0.2 \, J$.Loss in kinetic energy: $\Delta K = K_i - K_f = 10 - 0.2 = 9.8 \, J$.

$A$ bag of sand of mass $9.8 \, kg$ is suspended by a rope. $A$ bullet of mass $200 \, g$ travelling with a speed of $10 \, ms^{-1}$ gets embedded in it. The loss of kinetic energy will be $... J$.

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