Question

The momentum of a moving particle given by $p = t\left( {\ln t} \right)$. The net force acting on this particle is defined by the equation $F = \dfrac{{dp}}{{dt}}$. The net force acting on the particle is zero at time:
(A) $t = 0$
(B) $t = \dfrac{1}{e}$
(C) $t = \dfrac{1}{{{e^2}}}$
(D) $t = 1$

Answer 1

Hint: In this question, we are given the value of the momentum in the terms of $t$ and then it is given that the force is defined by the equation $F = \dfrac{{dp}}{{dt}}$. So we differentiate the value of momentum by time to find the force. Then to find the time when the force is equal to zero we equate $F = 0$ and find the answer in terms of $t$.
Formula used: In the solution, we will be using the following formula,
$ F = \dfrac{{dp}}{{dt}}$
where $F$ is the force, $p$ is the momentum and $t$ is the time.

Complete step by step answer:
We are given the value of momentum $p = t\left( {\ln t} \right)$. And we are told in the question that the force is given by the formula $F = \dfrac{{dp}}{{dt}}$.
So to find the force we have to differentiate the momentum in terms of the time. Therefore we get,
$\Rightarrow F = \dfrac{d}{{dt}}\left[ {t\left( {\ln t} \right)} \right]$
This differentiation will be done using the chain rule,
$\Rightarrow \dfrac{d}{{dt}}uv = u\dfrac{{dv}}{{dt}} + v\dfrac{{du}}{{dt}}$
Here we take $u = t$ and $v = \ln t$
So by substituting in the formula we get,
$\Rightarrow \dfrac{d}{{dt}}\left[ {t\left( {\ln t} \right)} \right] = t\dfrac{{d\left( {\ln t} \right)}}{{dt}} + \left( {\ln t} \right)\dfrac{{dt}}{{dt}}$
Now, $\dfrac{{d\left( {\ln t} \right)}}{{dt}} = \dfrac{1}{t}$
So we get,
$\Rightarrow \dfrac{d}{{dt}}\left[ {t\left( {\ln t} \right)} \right] = t \times \dfrac{1}{t} + \ln t$
Hence we get,
$\Rightarrow \dfrac{d}{{dt}}\left[ {t\left( {\ln t} \right)} \right] = 1 + \ln t$
So the value of force is
$\Rightarrow F = 1 + \ln t$
Now we have to find the time when the force is equal to zero. So by equating the above value to zero we get,
$\Rightarrow F = 1 + \ln t = 0$
Therefore by taking $\ln t$ to the R.H.S we get,
$\Rightarrow \ln t = - 1$
To remove $\ln $ we take exponential to both the sides and get,
$\Rightarrow {e^{\ln t}} = {e^{ - 1}}$
the exponential of any logarithm function gives back the variable, so we get
$\Rightarrow t = {e^{ - 1}}$
Hence we can write,
$\Rightarrow t = \dfrac{1}{e}$
Therefore the time when the force will be zero is $t = \dfrac{1}{e}$
So the correct option is (B); $t = \dfrac{1}{e}$.

Note:
In the question, we are told that the force is given by $F = \dfrac{{dp}}{{dt}}$. But we have known $F = ma$. This can easily be derived as,
$\Rightarrow F = m\dfrac{{dv}}{{dt}}$ where we can write the acceleration as the derivative of the velocity.
$ \Rightarrow F = \dfrac{{d\left( {mv} \right)}}{{dt}}$
We can take the mass inside the derivative since mass remains constant with time. The momentum is given by $p = mv$. Hence we can write,
$\Rightarrow F = \dfrac{{dp}}{{dt}}$