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(B) Since there is no external horizontal force acting on the trolley-water system,the horizontal momentum remains conserved.Let $M_0 = 4000\, kg$ be

Vedclass · Accepted Answer

$44.44$

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(B) Since there is no external horizontal force acting on the trolley-water system,the horizontal momentum remains conserved.Let $M_0 = 4000\, kg$ be the initial mass and $v_0 = 40\, m/s$ be the initial velocity.The rate of mass loss is $\frac{dm}{dt} = 8\, kg/s$.After time $t = 50\, s$,the mass of water lost is $\Delta m = (8\, kg/s) 	imes (50\, s) = 400\, kg$.The final mass of the trolley is $M = M_0 - \Delta m = 4000 - 400 = 3600\, kg$.Since the water falls vertically,it carries no horizontal momentum with it. Thus,the horizontal velocity of the trolley remains constant.$P_{initial} = P_{final} \implies M_0 v_0 = M v$$4000 	imes 40 = 3600 	imes v$$v = \frac{160000}{3600} = \frac{1600}{36} = \frac{400}{9} \approx 44.44\, m/s$.

$A$ trolley containing water has a total mass of $4000\, kg$ and is moving at a speed of $40\, m/s$. Water starts leaking from a hole at the bottom of the trolley at a rate of $8\, kg/s$. What is the speed of the trolley after $50\, s$?

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