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

Answer:

$v_{f}=70\frac{km}{h}$

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 $v_{f}$:

$v_{f}=\frac{(v_{1i}+v_{2i})}{2}$

And replacing the values for the velocity:

$v_{f}=\frac{(80.0\frac{km}{h}+60.0\frac{km}{h})}{2}$

$v_{f}=70\frac{km}{h}$