Answer to Question #128744 in Physics for Sulaiman Alsalamat

Question #128744

. 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


1
Expert's answer
2020-08-06T16:44:50-0400

Using the momentum conservation law, we can write:


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

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


V=v2V = \dfrac{v}{2}

The initial kinetic energy of the first car is:


K1=Mv22K_1 = \dfrac{Mv^2}{2}

The final kinetic energy of the first car is:

K2=MV22=Mv224=Mv28K_2 = \dfrac{MV^2}{2} = \dfrac{Mv^2}{2\cdot 4} = \dfrac{Mv^2}{8}

The loss of the kinetic energy is:


1K2K1=114=341-\dfrac{K_2}{K_1} =1- \dfrac14 = \dfrac34

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


Answer. 3/4.


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