Force acting on a body varies with time as sh | vedclass

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Question

(b)

$	an 	heta=\frac{1}{2}=\frac{F_0}{t_0-4}$

$\Rightarrow F_0=\frac{t_0-4}{2}$

Total change in momentum should be zero, then only it will

Vedclass · Accepted Answer

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

Vedclass · Answer

(b)

$	an 	heta=\frac{1}{2}=\frac{F_0}{t_0-4}$

$\Rightarrow F_0=\frac{t_0-4}{2}$

Total change in momentum should be zero, then only it will retain its initial momentum.

So, positive area of $F-t$ curve should be equal to negative area of $F-t$ curve till time $t_0$.

$\frac{1}{2}(4)(1)=\frac{1}{2}\left(t_0-4ight) F_0$

$8=\frac{\left(t_0-4ight)}{2} \cdot \frac{\left(t_0-4ight)}{2}$

$\left(t_0-4ight)^2=32$

$t_0=4+2 \sqrt{2}$

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

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