Question

A particle possessing a momentum $\overline{P_0}$ at t = O is subjected to a time varying force $\overline{F}=\overline{a}(t-\frac{t^2}{6})$ where $\overline{a}$ is a constant vector and t is in second. the momentum of the particle at t = 18 s. is then ( Assume that force is not acting after t = 6s)

• A. 6$\overline{a}+{\overline{P}}_0$
• B. ${\overline{P}}_0-162\overline{a}$
• C. ${\overline{P}}_0+162\overline{a}$
• D. ${\overline{P}}_0+18\overline{a}$

Answer 1

The correct option is A 6$\overline{a}+{\overline{P}}_0$

The particle with initial momentum $\overrightarrow{P_o}$ is acted by a force

$\overrightarrow{F}=\overrightarrow{a}(t-\frac{t^2}{6})$ time $t=0$ to bs

From Newton's second laws

$\overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}$ [Rate of change of momentum]

It given,

$\frac{d\overrightarrow{p}}{dt}=\overrightarrow{a}(t-\frac{t^2}{6})\Rightarrow d\overrightarrow{p}=\overrightarrow{a}(t-\frac{t^2}{9})dt$

now, if momentum changes to $\overrightarrow{P}$ in time $t=0\to 6$, integrate:

${\int }_{\overrightarrow{P_0}}^{\overrightarrow{{}_P}}d\overrightarrow{p}=\overrightarrow{a}{\int }_0^6(t-\frac{t^2}{6})dt=\overrightarrow{a}{[\frac{t^2}{2}-\frac{t^3}{18}]}_0^6$

$\Rightarrow \overrightarrow{P}-\overrightarrow{P_0}=\overrightarrow{a}\left(6\right)\Rightarrow \overrightarrow{P}=\overrightarrow{P_0}+6\overrightarrow{a}$

also, no force acts after $t>6s$ so final momentum $\left(\overrightarrow{P}\right)$ remains conserved thus at $t=18s,\overrightarrow{P}=\overrightarrow{P_0}+6\overrightarrow{a}$