Question

A retarding force $F=-2v$ is acting on a body of mass 10gram. Find out time taken for its velocity to reduce to 37% of its initial value if initial velocity is 1m/s.
A. 2.5ms
B. 5ms
C. 7.5ms
D. 8ms

Answer 1

As the first step, you could recall Newton’s equation of motion and then express acceleration in it as the time derivative of velocity and then equate this with the expression for force given. Now you could integrate on both sides by using the given conditions and thus you will be able to find the required time.
Formula used:
Newton’s Second of motion,
$F=ma$

Complete answer:
In the question, we are given that a retarding force of $F=-2v$ is acting on a mass of 10gram. We are also given its initial velocity as 1m/s and we are asked to find the time taken for the velocity to reduce to 37% of its initial value.
We have Newton’s second of motion given by,
$F=ma$
We know that acceleration is the time derivative of velocity, so,
$F=ma=\left( \dfrac{10}{1000} \right)\dfrac{dv}{dt}$ …………………………… (1)
But we are given,
$F=-2v$ …………………………. (2)
Equating RHS of (1) and (2),
$\left( \dfrac{1}{100} \right)\dfrac{dv}{dt}=-2v$
$\Rightarrow \dfrac{dv}{v}=-200dt$
Now, we could integrate on both sides to get,
$\int\limits_{100}^{37}{\dfrac{dv}{v}}=-\int\limits_{0}^{t}{200dt}$
$\Rightarrow \left[ \ln v \right]_{100}^{37}=-200t$
$\Rightarrow \ln \left( \dfrac{37}{100} \right)=-200t$
$\therefore t=-\dfrac{\ln \dfrac{37}{100}}{200}=0.5\times {{10}^{-2}}s=5ms$
Therefore, we found the time taken for the mass’s velocity to reduce to 37% of its initial value to be 5ms.

Option B is correct.

Note:
In most of the problems you will have to convert the quantities into its SI units when all the other quantities are given in SI units and thus we have converted the mass given in grams to kilograms. Then we are given the limits for integration by the given point that the mass’s velocity is reduced from 100% to 37%.