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

Options

A. `0.2kgm//sec`
B. `-0.2kgm//sec`
C. `-0.1kgm//sec`
D. `-0.4kgm//sec`

Answer

B

Explanation

Velocity between `t=0` and `t=2sec`
implies `v_(1)=(dx)/(dt)=(4)/(2)=2m//s`
Velocity at `t=2sec` , `v_(f)=0`
Impulse=Change in momentum `=m(v_(f)-v_(t))`
`0.1(0-2)=-0.2kgmsec^(-1)` .