In the figure given below, the position ...

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

`- 0.2kgms^(-1)`

`0.2kgms^(-2)`

`0.1 kgms^(-1)`

`-0.4 kgms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:

`vecI=Deltavecp=vecp_(f)-vecp_(i)`
`=0.-0.1=0.2ks ms^(-1)`

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