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. `0.2 kg ms^-1`
B. `-0.2 kg ms^-1`
C. `0.1 kg m s^-1`
D. `-0.4 kg ms^-1`

Answer

A

Explanation

If a constant force `F` is applied on a body for a short interval of time `Delta t`, then impulse of this force is `F xx Delta t`. From Newton's second law.
`F = ma = m (Delta v)/(Delta t)`
`rArr F Delta t = m Delta v = Delta p`
`:.` Impulse, `I = Delta_p = m (Delta x)/(Delta t)`
Given, `m = 0.1 kg, (Delta x)/(Delta t) = (4)/(2) m//s`
`:. I = 0.1 xx (4)/(2) = 0.2 kg ms^-1`.