Centre of Mass 2 Question 3
4. A particle of mass $m$ is moving in a straight line with momentum $p$. Starting at time $t=0$, a force $F=k t$ acts in the same direction on the moving particle during time interval $T$, so that its momentum changes from $p$ to $3 p$. Here, $k$ is a constant. The value of $T$ is (JEE Main 2019, 11 Jan Shift II)
(a) $\sqrt{\frac{2 p}{k}}$
(b) $2 \sqrt{\frac{p}{k}}$
(c) $\sqrt{\frac{2 k}{p}}$
(d) $2 \sqrt{\frac{k}{p}}$
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Answer:
Correct Answer: 4. (b)
Solution:
- Here,
$$ F=k t $$
When $t=0$, linear momentum $=p$
When $t=T$, linear momentum $=3 p$
According to Newton’s second law of motion,
applied force, $F=\frac{d p}{d t}$
or $d p=F \cdot d t$
or $\quad d p=k t \cdot d t$
Now, integrate both side with proper limit
$$ \begin{aligned} & \qquad \int _p^{3 p} d p=k \int _0^{T} t d t \text { or }[p] _p^{3 p}=k \frac{t^{2}}{2} \\ & \text { or }(3 p-p)=\frac{1}{2} k\left[T^{2}-0\right] \\ & \text { or } \quad T^{2}=\frac{4 p}{k} \text { or } T=2 \sqrt{\frac{p}{k}} \end{aligned} $$
