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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 0.1kg is shown. The impulse at t=2 sec is

A

`"0.2 kg m sec"^(-1)`

B

`-0.2 kg m sec"^(-1)`

C

`"0.1 kg m sec"^(-1)`

D

`"-0.4 kg m sec"^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
Impulse = change in momentum
`=m Delta V=(m Delta x)/(Delta t)=0.1xx(4-0)/(2-0)=0.2" kg m sec"^(-1)`
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