Force acting on a body varies with time as shown below. If the initial momentum of body is p then time taken to retain its momentum p again is\t8s\t4+2root2 s\t6s

For t=0 to t=2, Force equation can be written in the standard formy=mx+c→F1(t)=t2Let m be mass of body. From Newton`s second Law of motion.→a(t)=1m×t2Acceleration →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 integrationUsing (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=0Solution of the quadratic gives ust=8±√64−4×1×82t=4±2√2Taking only −ve sign for original question. For magnitude of momentum we have two solutions as above.