Question

A car of mass 10 metric ton rolls at 2 m/s along a level track and collides with a loaded car of the mass of 20 metric ton, standing at rest. If the cars couple together, the common speed after the collision is
(A) 1 m/s
(B) 0.67 m/s
(C) 0.5 m/s
(D) 0.3 m/s

Answer 1

Hint:We are given a two-body problem in which masses are given in which one body is initially moving while the second body is at rest. They collide with each other and after the collision, the two bodies stick to each other and move with a common velocity. We can use the law of conservation of linear momentum here.

Complete step by step answer:
The law of conservation of momentum states that the momentum of an isolated system remains constant.If no external force is acting on the system, so the linear momentum of the collision must be conserved.
Let the bodies be A and B.
\[{{M}_{a}}=\]10 mT
\[{{M}_{b}}=\]20 mT
Initial velocities,
\[{{u}_{a}}=\]2 m/s
\[{{u}_{b}}=\]0 m/s
Since both after collision stick together, let their common velocity be V.
Now applying the law of conservation of momentum, \[{{M}_{a}}{{u}_{a}}+{{M}_{b}}{{u}_{b}}=({{M}_{a}}+{{M}_{b}})V\]
Putting the values back into it, \[(10\times 2)+0=(10+20)\times V\]
\[20+0=30\times V\]
So V=0.67 m/s
Hence the two bodies after collision stick to one another and move with the common speed of 0.67 m/s.
So,option (B) is the correct answer.

Note: Here the masses were given in metric ton, we did not change into kg because in the equation the mass was there on both the sides, so ultimately they cancel out each other. This is an example of an elastic collision. Also, the system was isolated.