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(D) Let the mass of water flowing in time $t$ be $m = \rho L t$.The initial momentum of the water in the vertical direction is $p_i = m V = \rho L t V

Vedclass · Accepted Answer

$\sqrt{2} \rho VL$

Vedclass · Answer

(D) Let the mass of water flowing in time $t$ be $m = \rho L t$.The initial momentum of the water in the vertical direction is $p_i = m V = \rho L t V$ (upward).The final momentum of the water in the vertical direction is $p_f = 0$.Change in momentum in the vertical direction: $\Delta p_y = p_f - p_i = -\rho L V t$.The initial momentum of the water in the horizontal direction is $p_i = 0$.The final momentum of the water in the horizontal direction is $p_f = m V = \rho L t V$ (horizontal).Change in momentum in the horizontal direction: $\Delta p_x = p_f - p_i = \rho L V t$.The net change in momentum in time $t$ is $\Delta p = \sqrt{(\Delta p_x)^2 + (\Delta p_y)^2} = \sqrt{(\rho L V t)^2 + (-\rho L V t)^2} = \sqrt{2} \rho L V t$.The force exerted by the water on the corner is equal to the rate of change of momentum:$F = \frac{\Delta p}{t} = \sqrt{2} \rho L V$.

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