Bumper car A (281 kg) moving +2.82 m/s makes an elastic collision with bumper car B (209 kg) moving +1.72 m/s. What is the velocity of car A after the collision?

Answer 1

Hi Tamia L.,

You always approach these kinds of kinematics problems with a conservative mind -- not politically, mind you, but regarding momentum.

So you calculate the movement of the center-of-mass (CM) of the system. Add the momenta of each object, then divide by the total mass. That will yield a velocity -- the velocity of the CM. That momentum is conserved. In fact, viewed in the frame of reference of the CM, the two objects simply approach "it" and bounce off again, at the same velocity in-and-out.

So let's see how that works. We have A bringing 281 kg at +2.82 m/s, that's +792.42 kg m/s. (in a stationary frame of reference, or FOR).

And also, we have B bringing 209 kg at +1.72 m/s, that's +359.48 kg m/s. (in stationary FOR)

That's a total of 1151.9 kg m/s in all. (in FOR)

Since there are 281+209 = 490 kg in the system,

the CM is moving at 1151.9/490 = 2.35081632653 m/s . (in FOR)

Now, mass A was bashing along at 2.82 m/s, that is 0.46918367346 m/s faster than the CM. That means that it rebounds from the CM at the same relative speed. So it drops back to 2.35081632653-0.46918367346 = +1.88163265307 m/s. (That's your answer, in short.)

Similarly, mass B was puttering at 1.72 m/s, and the CM is moving up on it at 0.63081632653 m/s. So after the collision, mass B is bopped away from the CM at 0.63081632653 m/s, for a total velocity of 0.63081632653+2.35081632653 = 2.98163265306 m/s.

Now for the payoff for this question: What you should have learned, is that the CM acts like an infinite-mass but springy rock in elastic-rebound questions. Everything comes in, hits it, and bounces out again, at the same speed relative to it. Even if the collision were NOT elastic, the CM would act like that, but the objects would variously lose kinetic energy on the rebound, but still keep unchanged total momentum. You can't ever get rid of total momentum. But if you do questions like this with the CM as your reference, you never will miscalculate your velocities in an attempt to preserve kinetic energy -- that's automatically taken care of by the calculation method.

The next stage of complication beyond this problem has the masses colliding with various directions -- and in that case, you may calculate the momentum in the X- and Y-directions independently -- it's preserved in each orthogonal direction independently.

--Cheers, --Mr. d.