Question

Two cars of the same mass collide at an intersection. Just before the collision, one car is traveling east at 80.0 km/h and the other car is traveling south at 60.0 km/h. If the collision is completely inelastic, so the two cars move as one object after the collision, what is the speed of the cars immediately after the collision?

Answer 1

`v_(f)=70(km)/(h)`

Step-by-step explanation:

In a completely inelastic collision the momentum P is conserved, so can be expressed as:

`P_(f)-P_(i)=0`

As P is defined as:
`P=m.v`

Replacing:

`m_(1).v_(1f)+m_(2).v_(2f)=m_(1).v_(1i)+m_(2).v_(2i)` (Eq. 1)

As the problem says that the two cars have the same mass:

`m_(1)=m_(2)=m`

And in a completely inelastic collision the velocities after the collision are equal, so:

`v_(1f)=v_(2f)=v_(f)`

So replacing in Eq. 1:

`m.v_(f)+m.v_(f)=m.v_(1i)+m.v_(2i)`

`2m.v_(f)=m.(v_(1i)+v_(2i))`

Solving for
`v_(f)`:

`v_(f)=((v_(1i)+v_(2i)))/(2)`

And replacing the values for the velocity:

`v_(f)=((80.0(km)/(h)+60.0(km)/(h)))/(2)`

`v_(f)=70(km)/(h)`