A railway wagon of mass 5 times 10^3 kg is m | vedclass

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(C) Using the Conservation of Linear Momentum $(COCM)$ in the horizontal direction:$m_1 v_1 = (m_1 + m_2) v'$$(5 	imes 10^3) 	imes 1.2 = (5 	imes 1

Vedclass · Accepted Answer

$600$

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(C) Using the Conservation of Linear Momentum $(COCM)$ in the horizontal direction:$m_1 v_1 = (m_1 + m_2) v'$$(5 	imes 10^3) 	imes 1.2 = (5 	imes 10^3 + 10^3) 	imes v'$$6 	imes 10^3 = 6 	imes 10^3 	imes v'$$v' = 1 \ m/s$Initial kinetic energy $(KE_i)$:$KE_i = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} 	imes (5 	imes 10^3) 	imes (1.2)^2 = 0.5 	imes 5000 	imes 1.44 = 3600 \ J$Final kinetic energy $(KE_f)$:$KE_f = \frac{1}{2} (m_1 + m_2) v'^2 = \frac{1}{2} 	imes (6 	imes 10^3) 	imes (1)^2 = 3000 \ J$Change in kinetic energy $(\Delta KE)$:$\Delta KE = KE_i - KE_f = 3600 - 3000 = 600 \ J$

$A$ railway wagon of mass $5 \times 10^3 \ kg$ is moving with an initial velocity of $1.2 \ m/s$ on frictionless rails. Rain falls vertically into the wagon. When $10^3 \ kg$ of water has collected in the wagon,what is the change in the kinetic energy of the wagon in $J$?

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