Questions

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

detailed solution

Correct option is q

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 $\mathrm{v}=38\times \mathrm{sin}{60}^{0}-\left(9.8\right)\times 10=-13.13\mathrm{m}/\mathrm{sec}$

So change in momentum in vertical direction is$=\left(0.5\times 98\times \sqrt{3}/2\right)-\left[-\left(0.5\times 13.13\right)\right]$

$=42.434+6.56=48.997\approx 49$

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detailed solution

Correct answer is 49

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 $\mathrm{v}=38\times \mathrm{sin}{60}^{0}-\left(9.8\right)\times 10=-13.13\mathrm{m}/\mathrm{sec}$

So change in momentum in vertical direction is$=\left(0.5\times 98\times \sqrt{3}/2\right)-\left[-\left(0.5\times 13.13\right)\right]$

$=42.434+6.56=48.997\approx 49$

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