Question

Force acting on a body varies with time as shown in the graph given below. If initial momentum of body is $\overrightarrow{P}$, then the time taken by the body to retain its momentum $\overrightarrow{P}$ again is:

• A. $8\text{s}$
• B. $(4+2\sqrt{2})\text{s}$
• C. $6\text{s}$
• D. can never obtain

Answer 1

The correct option is B $(4+2\sqrt{2})\text{s}$


From the $F\text{vs}t$ graph,

$\tan \theta =\frac{1-0}{4-2}=\frac{1}{2}$

From the lower triangle,

$\tan \theta =\frac{F_0}{t_0-4}$

$\Rightarrow \frac{1}{2}=\frac{F_0}{t_0-4}$

$\Rightarrow F_0=\frac{t_0-4}{2}$

Here $t_0$ is the time at which $\overrightarrow{P}$ is again achieved.
In order to retain the momentum again,

$\Delta \overrightarrow{P}=\overrightarrow{P_f}-\overrightarrow{P_i}=0$

As we know that,
Change in momentum = area under $F\text{vs}t$ curve

Thus, area above $t-$axis must cancel with the area below $t-$axis.

$\therefore \left|\text{Postive area under curve}\right|=\left|\text{Negative area under curve}\right|$

$\Rightarrow \frac{1}{2}\times 4\times 1=\frac{1}{2}\times (t_0-4)\times F_0$

putting the value of $F_0$

$\Rightarrow 2=\frac{(t_0-4)}{2}\times \frac{(t_0-4)}{2}$

$\Rightarrow {(t_0-4)}^2=8$
$\Rightarrow t_0=4+2\sqrt{2}\text{or}{t}_0=4-2\sqrt{2}$

From the $F-t$ curve we can infer that the momentum will be retained only after $t=4\text{s}$ because force is acting in reverse direction on particle in that region$(t>4\text{s})$, hence it will lead to achieving same momentum after producing deceleration in motion.

Therefore neglecting the value of $t<4\text{s}$

$\therefore t_0=(4+2\sqrt{2})\text{s}$

Hence, option $\left(b\right)$ is correct.

$\begin{array}{c}\text{Why this question}?\\ \text{Tip: From Newton's second law we know that}\\ F=\frac{dp}{dt}\text{, thus dp = Fdt. Integrating on}\\ \text{both sides we get}\Delta P=\int Fdt,\text{it represents}\\ \text{change in momentum is equal to area under}F-t\text{curve.}\end{array}$