Question

Force acting on a body varies with time as shown below. if initial momentum of the body is p , then the time taken by the body to retain its momentum p again is

Answer 1

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

y=mx+c→F1(t)=t2

Let m be mass of body. From Newton`s second Law of motion.→a(t)=1m×t2

Acceleration →a≡d→vdt∴→v(t)=1m∫t2dt⇒→v(t)=1m(t24+C)

where C is constant of integration.

At t=0, →v(t)=→pm⇒→pm=C→v(t)=1m(t24+→p) .....(1)

Velocity at t=2 is found as →v(2)=1m(1+→p) ......(2)

Force equation for t=2 and thereafter becomes→F2(t)=−12t+2→a2(t)=−1m(t2−2)→v2(t)=−1m∫(t2−2)dt→v2(t)=−1m(t24−2t+C1)where C1 is constant of integration

Using (2) to find out C1→v2(2)=−1m(−3+C1)=1m(1+→p)⇒(−3+C1)=−1−→p⇒C1=2−→p

∴→v2(t)=−1m(t24−2t+2−→p) ......(3)

Imposing the given condition and solving for t−→p=t24−2t+2−→pt2−8t+8=0

Solution of the quadratic gives ust=8±√64−4×1×82t=4±2√2

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

Answer 2

Hopefully you can make something out of my messy writing :)