Answer to Question #128744 in Physics for Sulaiman Alsalamat

Question

. A car of mass M traveling with velocity v strikes a car of mass M that is at rest. The two cars’ bodies mesh in the collision. The loss of the kinetic energy the moving car undergo in the collision is

Answer 1

Using the momentum conservation law, we can write:

$$Mv + 0 = (M+M)V = 2MV$$

where $Mv$ and momentums of the first and second car before the collision respectively, and $V$ is a speed of the joined cars after the collision. Expressing $V$, obtain:

$$V = \dfrac{v}{2}$$

The initial kinetic energy of the first car is:

$$K_1 = \dfrac{Mv^2}{2}$$

The final kinetic energy of the first car is:

$$K_2 = \dfrac{MV^2}{2} = \dfrac{Mv^2}{2\cdot 4} = \dfrac{Mv^2}{8}$$

The loss of the kinetic energy is:

$$1-\dfrac{K_2}{K_1} =1- \dfrac14 = \dfrac34$$

Thus, 3/4 of the initial kinetic energy was lost in the collision.

Answer. 3/4.