Question

A particle of mass is moving in a straight line with momentum p. Starting at the time $t=0$, a force $F=kt$ acts in the same direction on the moving particle during time interval T so that its momentum changes from $p$ to $3p$. Here k is constant. The value $T$ is:

Answer 1

the momentum of the object is given as the product of the mass and the velocity of the body. And the force can be defined as the rate of change of momentum of the body. By equating the momentum and value of force the time period of motion can be calculated.

As per the given data,
The value of force acting on the object is, $F=kt$
The momentum in the time period $0-T$ is in the range $(p-3p)$

Complete answer:
The momentum is mathematically expressed as the product of the mass and the velocity with which a body is moving. It is denoted by a symbol $P$.
Mathematically,
$P=mv$
Every object in the universe that is moving peruses some value of momentum. If the body is at rest the momentum of the object will be zero.
The rate of change of momentum in which the progressive time can be termed as a force.
$F=\dfrac{dP}{dt}$
As per the information given in the question, the force on the particle is equivalent to $kt$ .
$F=\dfrac{dP}{dt}=kt$
This could be rearranged as,
$dP=ktdt$
So integrating both the sides over their respective limits (range),
$\begin{aligned} & \int_{p}^{3p}{dP}=\int_{0}^{T}{ktdt} \\\ & \Rightarrow \left[ P \right]_{p}^{3p}=k\left[ \dfrac{{{t}^{2}}}{2} \right]_{0}^{T} \\\ & \Rightarrow 2p=k\dfrac{{{T}^{2}}}{2} \\\ \end{aligned}$
So by rearranging the equation (1), the value of time period $(T)$ can be given as,
$\begin{aligned} & {{T}^{2}}=\dfrac{4p}{k} \\\ & \Rightarrow T=\sqrt{\dfrac{4p}{k}} \\\ & \therefore T=2\sqrt{\dfrac{p}{k}} \\\ \end{aligned}$
So, the value of the time period $T$ for the given condition in the question is $2\sqrt{\dfrac{p}{k}}$. And this is the required answer to satisfy the question.

Note:
To change the momentum of the object an external force is needed without this force the momentum of the object will be constant on a frictionless path. This is due to the inertia of the object. According to Newton’s ${{1}^{st}}$ law of motion, an external force is needed to change the inertia of an object.