A particle having a mass 0.5kg is projected under gravity with a speed of 98m/s at an angle of  60° to the horizontal.  The magnitude of the change in momentum (in N-sec) of the particle after 10seconds is

# A particle having a mass 0.5kg is projected under gravity with a speed of 98m/s at an angle of  60° to the horizontal.  The magnitude of the change in momentum (in N-sec) of the particle after 10seconds is

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### Solution:

There is no change in horizontal velocity, hence no change in momentum in horizontal direction. The vertical velocity at t=10sec is
So change in momentum in vertical direction is  $\mathrm{v}=38×\mathrm{sin}{60}^{0}-\left(9.8\right)×10=-13.13\mathrm{m}/\mathrm{sec}$
So change in momentum in vertical direction is$=\left(0.5×98×\sqrt{3}/2\right)-\left[-\left(0.5×13.13\right)\right]$
$=42.434+6.56=48.997\approx 49$

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