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Two identical balls with completely smooth surfaces are moving uniformly in free space without rotation. At some moment, they undergo a perfectly elastic glancing collision. After the collision, the angle (in degrees) between the two vectors of velocities is

(a) 30

(b) 45

(c) 90

(d) Impossible to answer without knowledge of the angle between the two velocities before the collision.

The answer is apparently (c), but I thought that it would be 180 degrees, since the balls are glancing off each other, and clearly by drawing the plane of incidence of 1 ball one sees that the other should barely change its angle... Can someone please explain?

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The problem must assume that one of the balls is stationary before impact. This is a classic billiard ball problem, and the balls will always move in orthogonal directions after impact. To prove the reference frame matters, analyze the problem using the center of mass between the two balls as the reference frame, in which case it's easy to see that your idea that the resulting velocities will be 180 degrees apart is correct.

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  • $\begingroup$ The answer merely cites an already existing comment. $\endgroup$ – Benjamin Rogers-Newsome Mar 13 '18 at 0:37
  • $\begingroup$ Yup, hadn't seen your comment yet. $\endgroup$ – Tom B. Mar 13 '18 at 0:51
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So you have ball A moving towards B.enter image description here

The ball moving to the right barely touches the other ball, and only just glances the top. This pushed the bottom ball ever so slightly down (which in turn pushed the other ball slightly up but we will ignore this).

The point is that the closer the collision becomes to being glancing, the 'more vertical' the bottom ball is pushed downwards, and the less upwards the other ball is pushed.

This tends to make the balls move at 90 degrees to each other.

Is this clear with the diagram?

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  • $\begingroup$ This answer works only for a reference frame attached to one ball. $\endgroup$ – Tom B. Mar 13 '18 at 0:23
  • $\begingroup$ @TomB. It does, but then again, the answer is only 90 degrees for a reference frame attached to one ball, it is 180 if you think about both balls moving. I am guessing this is the logic that the questions wanted - depends on the context of the question I suppose? $\endgroup$ – Benjamin Rogers-Newsome Mar 13 '18 at 0:25
  • $\begingroup$ sorry, wrote my answer while you were writing +1! $\endgroup$ – Tom B. Mar 13 '18 at 0:39
  • $\begingroup$ And isn't it only 180 for the frame of the center of mass? $\endgroup$ – Tom B. Mar 13 '18 at 1:00
  • $\begingroup$ @TomB. Yea I think so $\endgroup$ – Benjamin Rogers-Newsome Mar 13 '18 at 1:01

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