A force acting on a body varies with time as | vedclass

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(B) The change in momentum $\Delta \vec{p}$ is equal to the impulse,which is the area under the $F-t$ graph.For the body to regain its initial momentu

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$(4+2 \sqrt{2})$

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(B) The change in momentum $\Delta \vec{p}$ is equal to the impulse,which is the area under the $F-t$ graph.For the body to regain its initial momentum $\vec{p}$,the total change in momentum must be zero,meaning the net area under the $F-t$ graph must be zero.Let the graph cross the $t$-axis at $t = 4 \ s$. The area of the positive triangle (from $t=0$ to $t=4$) is $A_1 = \frac{1}{2} 	imes 4 	imes 1 = 2 \ N \cdot s$.For the total area to be zero,the area of the negative triangle (from $t=4$ to $t_0$) must be equal in magnitude to the positive area.Let the force at time $t_0$ be $-F_0$. From the slope of the line,$	an 	heta = \frac{1}{2} = \frac{F_0}{t_0-4}$,so $F_0 = \frac{t_0-4}{2}$.The area of the negative triangle is $A_2 = \frac{1}{2} 	imes (t_0-4) 	imes F_0 = \frac{1}{2} 	imes (t_0-4) 	imes \frac{t_0-4}{2} = \frac{(t_0-4)^2}{4}$.Setting $A_1 = A_2$,we get $2 = \frac{(t_0-4)^2}{4}$,which implies $(t_0-4)^2 = 8$.Thus,$t_0-4 = \sqrt{8} = 2\sqrt{2}$,so $t_0 = 4 + 2\sqrt{2} \ s$.

$A$ force acting on a body varies with time as shown below. If the initial momentum of the body is $\vec{p}$,then the time taken by the body to regain its momentum $\vec{p}$ again is ........... $s$.

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