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4-1.Newton's Laws of Motion
hard
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 ........... $s$
A
$8$
B
$(4+2 \sqrt{2})$
C
$6$
D
Can never obtain
Solution
(b)
$\tan \theta=\frac{1}{2}=\frac{F_0}{t_0-4}$
$\Rightarrow F_0=\frac{t_0-4}{2}$
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 $t_0$.
$\frac{1}{2}(4)(1)=\frac{1}{2}\left(t_0-4\right) F_0$
$8=\frac{\left(t_0-4\right)}{2} \cdot \frac{\left(t_0-4\right)}{2}$
$\left(t_0-4\right)^2=32$
$t_0=4+2 \sqrt{2}$
Standard 11
Physics
