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

Question

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

Infinity Learn · Accepted Answer

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

v = 38 \times \sin 60^{0} - (9 .8) \times 10 = - 13 .13 m / \sec

So change in momentum in vertical direction is

= (0 .5 \times 98 \times \sqrt{3} / 2) - [- (0 .5 \times 13 .13)]

= 42 .434 + 6 .56 = 48 .997 \approx 49

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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