A car and a lorry are moving with the same momentum, if the same braking force is applied, then:
A. car comes to rest in a shorter distance.
B. lorry comes to rest in shorted distance
C. both travels same distance before coming to rest
D. none of these

Hint: We will consider the differences between other quantities such as mass and velocity of the lorry and the car not that the momentum for both is given to be the same. Using them and Newton’s equation of motion, we will find which one will stop first.

Formula used:
Newton’s third equation of motion
${{v}^{2}}-{{u}^{2}}=2as$

Complete answer:
Let us assume P to be the momentum of both the car and the lorry. And M to be the mass of the lorry and m to be the mass of the car. And let F be the breaking force applied on the vehicles. Then we will get the initial velocities of both the car (u) and the lorry (U) as

$u=\dfrac{P}{m}$ and \[U=\dfrac{P}{M}\]. The breaking force will produce different accelerations on both the bodies which will be given as $a=\dfrac{F}{m}$ and \[A=\dfrac{F}{M}\]. Using these values in the third equation of motion and considering till the time they come to rest we will get.

${{v}^{2}}-{{u}^{2}}=2as\Rightarrow -{{u}^{2}}=2(-a)d\Rightarrow d=\dfrac{{{u}^{2}}}{2a}=\dfrac{{{\left( \dfrac{P}{m} \right)}^{2}}}{2\left( \dfrac{F}{m} \right)}=\dfrac{{{P}^{2}}}{2mF}$${{v}^{2}}-{{u}^{2}}=2as\Rightarrow -{{U}^{2}}=2(-A)D\Rightarrow D=\dfrac{{{U}^{2}}}{2A}=\dfrac{{{\left( \dfrac{P}{M} \right)}^{2}}}{2\left( \dfrac{F}{M} \right)}=\dfrac{{{P}^{2}}}{2MF}$
Here, d and D are the distances moved by the car and lorry respectively before coming to rest. As we can see they are inversely proportional to mass of the vehicle. As the car will have less mass, it will come to rest in a longer distance.

So, the correct answer is “Option B”.

Note:
Using the second law of motion it can be shown that although they come to rest in different distances, they will come to rest in the same amount of time after applying the braking force, the car covers a longer distance due to having higher speed in the beginning. If they had the same speed, the car would have come to rest first.