A railway wagon (open at the top) of mass M₁ is moving with speed v₁ a

A railway wagon (open at the top) of mass M₁ is moving with speed v₁ along a straight track. As a result of rain, after some time it gets partially filled with water so that the mass of the wagon becomes M₂ and speed becomes v₂. Taking the rain to be falling vertically and water stationary inside the wagon, the relation between the two speeds v₁ and v₂ is :

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[amp_mcq option1=”v₁ = v₂” option2=”1/2 M₁v₁² < 1/2 M₂v₂²" option3="M₁v₁ = M₂v₂" option4="M₁v₁ < M₂v₂" correct="option3"]

This question was previously asked in

UPSC NDA-1 – 2023

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As the rain falls vertically, it has no horizontal momentum. The horizontal momentum of the system (wagon + water) is conserved because there are no external horizontal forces acting on it (assuming negligible friction and air resistance).

The initial horizontal momentum of the wagon is $P_1 = M_1 v_1$. After the rain has collected, the total mass is $M_2$ and the speed is $v_2$. The final horizontal momentum is $P_2 = M_2 v_2$. By conservation of horizontal momentum, $P_1 = P_2$, so $M_1 v_1 = M_2 v_2$.

This is an example of an inelastic collision/process where mass is added to the system. Kinetic energy is not conserved in this case; the kinetic energy of the system decreases because the incoming water had zero horizontal kinetic energy relative to the ground.

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