# Video: Motion of Objects after an Inelastic Collision

Two cars of equal masses approach a frictionless perpendicular intersection. Car A travels north at 22 m/s and car B travels east. The cars collide and stick together, traveling at 17° north of east. What was the initial speed of car B?

03:37

### Video Transcript

Two cars of equal masses approach a frictionless perpendicular intersection. Car A travels north at 22 meters per second and car B travels east. The cars collide and stick together, traveling at 17 degrees north of east. What was the initial speed of car B?

We can label car B’s initial speed 𝑣 sub B and begin our solution by drawing a sketch of the situation. In this scenario, initially, we have two cars approaching one another at an intersection. Car A is traveling north at a speed given as 22 meters per second, and car B is traveling to the east at an unknown speed. At the center of the intersection, the cars collide and, finally, they move together as one mass in a direction given as 17 degrees north of east. Based on this information, we want to solve for the initial speed of car B.

We’ll use the principle that the total initial momentum in a system is equal to the total final momentum in that system. In our scenario, initially and finally, we have momenta pointing in two different directions. In the northern direction, we have 𝑚 sub A times 𝑣 sub A. And in the eastern direction, orthogonal to the north–south axis, we have 𝑚 sub B times 𝑣 sub B.

If we then consider the final momentum in our system, again both in the north and eastern directions separately, to the north we have our combined mass 𝑚 sub A plus 𝑚 sub B, since the cars have stuck together, times their final speed, whatever that is, we’ll call it 𝑣 sub 𝑓, multiplied by the sine of the angle 𝛳. And to the east, we have the same expression except now we use the cosine of that angle to point in the direction east of north.

Applying our conservation principle, we can write in the northern direction that 𝑚 sub A𝑣 sub A equals 𝑚 sub A plus 𝑚 sub B times 𝑣 sub 𝑓 times the sine of 𝛳. And separately, in the eastern direction, 𝑚 sub B𝑣 sub B is equal to the sum of the two masses times their final velocity times the cosine of 𝛳. We recall that we want to solve for 𝑣 sub B, the initial speed of car B. And very importantly, we’re told in the problem statement that the masses of the two cars are equal. 𝑚 sub A is equal to 𝑚 sub B. So, if we replace every 𝑚 sub A and 𝑚 sub B we see in our two equations simply with the mass 𝑚, we see then that that mass 𝑚 cancels out from both sides of both equations.

Looking at the equation in the northern direction, we can use the fact that 𝑣 sub A and 𝛳 are given information to solve for 𝑣 sub 𝑓, the final speed of the combined car mass. 𝑣 sub 𝑓 is equal to 𝑣 sub A over two times the sine of 𝛳, or 22 meters per second divided by two times the sine of 17 degrees. If we substitute this expression for 𝑣 sub 𝑓 into 𝑣 sub 𝑓 in our equation for 𝑣 sub B and enter our value of 17 degrees for 𝛳. Then we see our expression simplifies a bit and that factors of two cancel and the cosine of one angle divided by the sine of the same angle is equal to the cotangent of that angle.

So, entering these values on our calculator, we find a result for 𝑣 sub B which, to two significant figures, is 72 meters per second. That’s the initial speed of car B in this scenario.