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In the figure given below, the position-...


In the figure given below, the position-time graph of a particle of mass 0.1 kg is shown. The impulse at `t=2` sec is

Text Solution

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`v=(dx)/(dt)=4/2=2m//s`
velocity between `t=0 sec "to" t=2sec=2 m//s`
velocity at ` 2sec = 0 m//s`
Impulse = change in linear momentum
`=m(V_(f)-V_(i))=0.1(0-2) =-0.2 kg ms^(-1)`
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