Two cars of equal mass collide inelastically and stick together after the collision. Before the collision, their speeds are V1 and V2 . What is the speed of the two-car system after the collision?
a) V1+V2
b) V1-V2
c) V2-V1
d)(V1V2)^(1/2)
e) (1/2)(V1+V2)
f)(V1+V2)^(1/2)
g)The answer depends on the directions in which the cars were moving before the collision.

In a perfectly inelastic collision two masses or objects will hit each other (or collide) and be stuck together. The idea here is that energy is not conserved, there is energy loss but the momentum of objects is conserved.

The general formula is as follow

[tex]m_{1} v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v_{f}[/tex]

Where

[tex]m_{1}[/tex] is mass of first object

[tex]m_{2}[/tex] is mass of second object

[tex]v_{1}[/tex] is velocity of first object (before collision)

[tex]v_{2}[/tex] is velocity of second object (before collision)

[tex]v_{f}[/tex] is velocity after collision

since we know that the masses are equal in this case we try to simplify this formula

[tex]m_{a} v_{1} + m_{a}v_{2} = (m_{a} + m_{a})v_{f}\\ m_{a}(v_{1} + v_{2} ) = 2m_{a} v_{f}\\ v_{1} + v_{2} = 2v_{f}\\ \frac{v_{1}+ v_{2}}{2} = v_{f}[/tex]

hence the answer is e