Question

A lorry and a car of mass ratio $4:1$ are moving with KE in the ratio $3:2$ on a horizontal road. Now brakes are applied and the braking force produced is in the ratio $1:2$. Then the ratio of stopping timing of lorry and car is
(A) $1:1$
(B) $1:3$
(C) $2\sqrt 6 :1$
(D) $6\sqrt 2 :1$

Answer 1

To solve this question, we have to use the formula of the kinetic energy and Newton's second law using which we can obtain the ratio of the initial speeds and the retardations of the lorry and the car. Then using the first equation of motion, we can use these ratios for calculating the required ratio of the stopping time.

Formula used: The formulae used to solve this question are given by
$K = \dfrac{1}{2}m{v^2}$, $K$ is the kinetic energy of a particle of mass $m$ moving with speed $v$.
$F = ma$, here $F$ is the force acting on a body of mass $m$ which moves with an acceleration of $a$.
$v = u + at$, here $v$ and $u$ are the final and initial velocities of an object moving with an acceleration $a$ in time $t$.

Complete step-by-step solution:
Let the mass of the lorry be ${m_1}$ and that of the car be ${m_2}$. Also, let ${K_1}$, ${F_1}$, ${t_1}$ and ${K_2}$, ${F_2}$, ${t_2}$ be the respective kinetic energies, braking force produced, and the stopping times of the lorry and the car.
According to the question, we have
$\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{4}{1}$..........................(1)
Also,
$\dfrac{{{K_1}}}{{{K_2}}} = \dfrac{3}{2}$ …………………….(2)
And
$\dfrac{{{F_1}}}{{{F_2}}} = \dfrac{1}{2}$.................(3)
Now, let the initial velocities of the lorry and the car be ${u_1}$ and ${u_2}$ respectively.
We know that the kinetic energy is given by the expression
$K = \dfrac{1}{2}m{v^2}$
So the kinetic energy of the lorry becomes
${K_1} = \dfrac{1}{2}{m_1}{u_1}^2$.....................(4)
And the kinetic energy of the car becomes
${K_2} = \dfrac{1}{2}{m_2}{u_2}^2$..........................(5)
Putting (4) and (5) in (2) we get
$\dfrac{{1/2{m_1}{u_1}^2}}{{1/2{m_2}{u_2}^2}} = \dfrac{3}{2}$
$\dfrac{{{m_1}}}{{{m_2}}}{\left( {\dfrac{{{u_1}}}{{{u_2}}}} \right)^2} = \dfrac{3}{2}$
Putting (1) in the above equation, we get
$4{\left( {\dfrac{{{u_1}}}{{{u_2}}}} \right)^2} = \dfrac{3}{2}$
$\Rightarrow {\left( {\dfrac{{{u_1}}}{{{u_2}}}} \right)^2} = \dfrac{3}{8}$
Taking square root both sides, we get
$\dfrac{{{u_1}}}{{{u_2}}} = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }}$ .............................(6)
Also, from the Newton’s second law of motion we know that
$F = ma$
Therefore, if the retardations of the lorry and the car are ${a_1}$ and ${a_2}$ respectively, then their breaking forces are given by
${F_1} = {m_1}{a_1}$..................(7)
${F_2} = {m_2}{a_2}$.....(8)
Putting (7) and (8) in (3) we get
$\dfrac{{{m_1}{a_1}}}{{{m_2}{a_2}}} = \dfrac{1}{2}$
Putting (1) in the above equation, we get
$4\dfrac{{{a_1}}}{{{a_2}}} = \dfrac{1}{2}$
$\Rightarrow \dfrac{{{a_1}}}{{{a_2}}} = \dfrac{1}{8}$ .....................(9)
Now, from the second kinematic equation of motion, we have
$v = u + at$
For the case of retardation, we replace $a$ by $- a$ in the above equation to get
$v = u - at$
Since both of the lorry and the car have been stopped, we substitute $v = 0$ in the above to get
$0 = u - at$
$\Rightarrow t = \dfrac{u}{a}$
Therefore, the stopping time for the lorry and the car are respectively given by
${t_1} = \dfrac{{{u_1}}}{{{a_1}}}$.........................(10)
${t_2} = \dfrac{{{u_2}}}{{{a_2}}}$..............(11)
Dividing (10) by (11) we get
$\dfrac{{{t_1}}}{{{t_2}}} = \dfrac{{{u_1}}}{{{u_2}}}\dfrac{{{a_2}}}{{{a_1}}}$
$\Rightarrow \dfrac{{{t_1}}}{{{t_2}}} = \dfrac{{{u_1}/{u_2}}}{{{a_1}/{a_2}}}$
Substituting (9) and (6) in the above equation, we finally get
$\dfrac{{{t_1}}}{{{t_2}}} = 8 \times \dfrac{{\sqrt 3 }}{{2\sqrt 2 }}$
$\Rightarrow \dfrac{{{t_1}}}{{{t_2}}} = 2\sqrt 6$
Thus, the ratio of the stopping time of the lorry and the car is equal to $2\sqrt 6 :1$.

Hence, the correct answer is option (3).

Note: The breaking force and hence the retardation of the vehicles are taken to be constant in this solution. This is because we are given the ratio of the braking force on the lorry and the car, which clearly indicates that it must be constant.