Question

A car traveling north at 10.0 m/s crashes into a car traveling east at 15 m/s at an unexpectedly icy intersection. The cars lock together as they skid on the ice. The two cars have the same mass. What is their combined speed after collision?

Answer 1

When dealing with collisions, always look to conservation of momentum.

Momentum is defined as mass times velocity: p = mv

The cars are separate before colliding and then stick together and continue at the same speed.

So we can write:
p(before) = m1*v1 + m2*v2
p(after) = (m1 + m2)*v12

We can set these equal and argue that momentum is conserved:
m1*v1 + m2*v2 = (m1+m2)*v12

Masses of the cars are the same (m1=m2), so we can simplify:
m*v1 + m*v2 = 2*m*v12
v1 + v2 = 2 * v12

Now we have a way to relate the velocities of the cars to one another. This problem is slightly more complicated, because we need to consider the components of each car's velocity. However, this complication is not too bad here since the directions of the cars are perpendicular. This means that car1 will contribute all of the momentum in the North direction and car2 will contribute all of the momentum in the East direction. This will result in both cars sliding some direction between North and East.

We can write an equation (modeled after the one we derived) for each direction:

North:
10 + 0 = 2 * v12
v12 = 5 m/s North

East:
0 + 15 = 2 * v12
v12 = 7.5 m/s East

So now we have the velocities in each direction. We can use the pythagorean theorem to find the net velocity of the cars stuck together:

(v12)^2 = (5)^2 + (7.5)^2
(v12)^2 = 81.25

v12 = 9.01 m/s <-- this is the final speed of the cars