An open water-tight railway wagon of mass 5 t | vedclass

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(C) Given: Initial mass of the wagon,$m = 5 	imes 10^3 	ext{ kg}$.Initial velocity of the wagon,$v = 1.2 	ext{ m/s}$.After collecting rain drops,th

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(C) Given: Initial mass of the wagon,$m = 5 	imes 10^3 	ext{ kg}$.Initial velocity of the wagon,$v = 1.2 	ext{ m/s}$.After collecting rain drops,the total mass of the wagon becomes $m' = m + 10^3 	ext{ kg} = 6 	imes 10^3 	ext{ kg}$.Since the rain drops fall vertically,they have no horizontal component of momentum. Therefore,the horizontal momentum of the system is conserved.According to the law of conservation of linear momentum: $m 	imes v = m' 	imes v'$.Substituting the values: $(5 	imes 10^3) 	imes 1.2 = (6 	imes 10^3) 	imes v'$.$v' = \frac{5 	imes 10^3 	imes 1.2}{6 	imes 10^3}$.$v' = \frac{5 	imes 1.2}{6} = 1 	ext{ m/s}$.

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