Question

A man standing on a trolley pushes another identical trolley (both trolleys are at rest on a rough surface), so that they are set in motion and stop after some time. If the ratio of mass of first trolley with man to mass of second trolley is 3, then find the ratio of the stopping distances of the second trolley to that of the first trolley. (Assume coefficient of friction to be the same for both the trolleys).

Answer 1

Hint
To find the ratio of the stopping distances of both the trolleys, we need to know about the velocity, momentum, and acceleration of both the trolleys. Also, we need to understand about the deceleration due to frictional force.
Formula used: Momentum $p = mv$ where m is the mass and v is the velocity of the body.

Complete step by step answer
It is given in the question that, the ratio of mass of the first trolley with man to the mass of the second trolley is 3. Then, let us consider the mass of the second trolley be m.
Then, $\dfrac{{{\text{mass of 1st trolley}}}}{m} = 3$
So, the mass of the first trolley becomes 3m.
Since, the man pushes the second trolley, the momentums of both the trolleys are the same and it is denoted as p. The velocities of the first and second trolleys are $v_1$ and $v_2$ respectively.
$\Rightarrow p = m{v_2} = 3m{v_1}$
is the momentum of the trolleys.
From the above equation, the velocities can be obtained as,
$\Rightarrow {v_1} = \dfrac{p}{{3m}}$
And ${v_2} = \dfrac{p}{m}$
Now, we know the equations of motion very well. One of the formulas involving initial, final velocity and distance of a body is,
$\Rightarrow {v^2} = {u^2} + 2as$
$\Rightarrow {v^2} - {u^2} = 2as$
We know that the trolleys were at rest before moving. Thus, the initial velocity becomes 0.
$\Rightarrow {v^2} = 2as$
For the first trolley, the distance $s_1$ is,
$\Rightarrow {s_1} = \dfrac{{{v_1}^2}}{{2a}}$
And for the second trolley, the distance $s_2$ is,
$\Rightarrow {s_2} = \dfrac{{{v_2}^2}}{{2a}}$
Now, it is given in the question that the coefficient of friction is the same for both the trolleys. Thus, the frictional force is the same. The acceleration of the body comes as,
$\Rightarrow F - \mu mg = ma$
$\Rightarrow a = \dfrac{{(F - \mu mg)}}{m}$
Here, the force F is zero, as there is no external force applied. Thus, the equation becomes,
$\Rightarrow \begin{array}{l}a = \dfrac{{(0 - \mu mg)}}{m}\\ \Rightarrow a = - \dfrac{{\mu mg}}{m}\end{array}$
Thus, $a = - \mu g$
The body decelerates after a certain time and stops. This deceleration is the same for both the trolleys as their coefficients of friction are the same.
Thus, the ratio of the stopping distances of first and second trolley be,
$\Rightarrow \dfrac{{{s_1}}}{{{s_2}}} = \dfrac{{\dfrac{{{v_1}^2}}{{2a}}}}{{\dfrac{{{v_2}^2}}{{2a}}}}$
$ \Rightarrow \dfrac{{{s_1}}}{{{s_2}}} = \dfrac{{{v_1}^2}}{{{v_2}^2}}$
Then, substitute the values of velocities to get the ratio.
$ \Rightarrow \dfrac{{{s_1}}}{{{s_2}}} = \dfrac{{{{\left( {\dfrac{p}{{3m}}} \right)}^2}}}{{{{\left( {\dfrac{p}{m}} \right)}^2}}}$
$ \Rightarrow \dfrac{{{s_1}}}{{{s_2}}} = \dfrac{{\dfrac{{{p^2}}}{{9{m^2}}}}}{{\dfrac{{{p^2}}}{{{m^2}}}}}$
$ \Rightarrow \dfrac{{{s_1}}}{{{s_2}}} = \dfrac{1}{9}$
Hence, the ratio of the stopping distance of the second trolley to the stopping distances of the first trolley becomes,
$ \Rightarrow \dfrac{{{s_2}}}{{{s_1}}} = 9$.

Note
While working out with these types of problems, examine the motion of the body properly. The state of the body before motion should be understood correctly to understand the initial velocity and the forces applied on it.