Question

Force acting on a body varies with time as shown below. If the initial momentum of the body is `vecP` then the time taken by the body to retain its momentum `vecP` again is?

Answer 1

This is what I get.

Explanation:

For `t=0` to `t=2`, Force equation can be written in the standard form

`y=mx+c`
`vecF_1(t)=t/2`

Let `m` be mass of body. From Newton's second Law of motion.
`vec a(t)=1/mxxt/2`

Acceleration `veca-=(dvecv)/dt`
`:. vecv(t)=1/(m) int t/2dt`
`=>vecv(t)=1/m ( t^2/4+C)`
where `C` is constant of integration.

At `t=0`, `vecv(t)=vecp/m`
`=>vecp/m=C`
`vecv(t)=1/m ( t^2/4+vecp)` .....(1)

Velocity at `t=2` is found as
`vecv(2)=1/m(1+vecp)` ......(2)

Force equation for `t=2` and thereafter becomes

`vecF_2(t)=-1/2t+2`
`veca_2(t)=-1/m(t/2-2)`
`vecv_2(t)=-1/m int(t/2-2)dt`
`vecv_2(t)=-1/m ( t^2/4-2t+C_1)`
where `C_1` is constant of integration

Using (2) to find out `C_1`

`vecv_2(2)=-1/m (-3+C_1)=1/m(1+vecp)`
`=> (-3+C_1)=-1-vecp`
`=>C_1=2-vecp`

`:. vecv_2(t)=-1/m(t^2/4-2t+2-vecp)` ......(3)

Imposing the given condition and solving for `t`

`-vecp=t^2/4-2t+2-vecp`
`t^2-8t+8=0`

Solution of the quadratic gives us

`t=(8+-sqrt(64-4xx1xx8))/2`
`t=4+-2sqrt2`

Taking only `-ve` sign for original question.
For magnitude of momentum we have two solutions as above.