A car and a lorry are moving with same momentum, if same braking force is applied, then:
A. Car comes to rest in shorter distance
B. Lorry comes to rest in shorter distance
C. Both travels same distance before coming to rest
D. None

Hint:Here the first thing that we are given is both car and lorry have the same momentum which is related with mass and velocity. Therefore we have to consider the differences in masses and velocities of both the vehicles. Secondly, braking force is also the same for both and so we will use the relation between force and mass as well as per the second law of motion of Newton.

Formulas used:
$P = mv$, where $P$ is momentum, $m$ is mass and $v$ is velocity
$F = ma$, where $F$ is force applied, $m$ is mass and $a$ is acceleration
${v^2} - {u^2} = 2as$ , where $v$ is the final velocity, $u$ is the initial velocity, $a$ is acceleration and $s$ is the distance covered

Complete step by step solution:
We know that momentum,
$P = mv \\
\Rightarrow v = \dfrac{P}{m}$
Here, the final velocity will be zero as both the vehicles come to stop and so putting the initial velocity $u$ in this equation, we get $u = \dfrac{P}{m}$. Now, as per the Newton’s second law of motion, $F = ma$. But, here we are applying brakes, therefore there is retardation. It means that the vehicle’s velocity will keep decreasing until it comes to rest.Therefore we will take braking force,
$
s = \dfrac{{{{\left( {\dfrac{P}{m}} \right)}^2}}}{{2\left( { - \dfrac{F}{m}} \right)}} \\
\Rightarrow s = \dfrac{{{P^2}}}{{2mF}} \\
\Rightarrow F = - ma \\
\Rightarrow a = - \dfrac{F}{m}$
Here, we need to find the relation between distance and mass of the bodies.
Let us consider the equation of motion,
${v^2} - {u^2} = 2as$
$ \Rightarrow s = \dfrac{{{v^2} - {u^2}}}{{2a}}$
We will take final velocity $v = 0m/s$ as the vehicles are going to stop after covering some distances.
$s = - \dfrac{{{u^2}}}{{2a}}$
Now, putting the values $u = \dfrac{P}{m}$ and $a = - \dfrac{F}{m}$
$
\Rightarrow s = \dfrac{{{{\left( {\dfrac{P}{m}} \right)}^2}}}{{2\left( { - \dfrac{F}{m}} \right)}} \\
\therefore s = \dfrac{{{P^2}}}{{2mF}}$
This means that the covered distance before the vehicle stops is inversely proportional to its mass.We know that lorry is heavier than a car, therefore it comes to rest in a shorter distance than the car.

Hence, option B is the right answer.

Note:Here, we have seen that the distance covered by both car and lorry is different according to their mass. However the amount of time required to come to the rest position will be equal for both the vehicles in this case.