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A carriage of mass M moves with constant initial velocity u along a straight horizontal frictionless track. At t = 0 rain begins to fall vertically (vertical velocity u'), adding mass at the rate of m per second to the carriage. Find the speed of the carriage after time T.

1. M*u / (M + m*t)
2. (M*u + m*u') / (M + m*t)
3. (u + u') * (M + m) / M
4. M * (u + u') / (m*t)

Correct answer: M*u / (M + m*t)

Solution

Rain has no horizontal velocity, so horizontal momentum is conserved: M*u = (M + m*t)*v, giving v = M*u/(M + m*t).

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