In the laboratory frame of reference, when a moving object collides elastically and obliquely with a stationary object of the same mass, the objects always move off at a right angle. The proof is very straightforward. However, is the opposite also true? (i.e. when two objects collide and move off at a right angle, they must be of the same mass and the collision must be elastic.) Answer to my textbook says yes, but I think that it is also possible for two objects of different mass to make an inelastic collision and move off at a right angle. Why is it not the case? Can anyone provide a proof for this?

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    $\begingroup$ Throw a ball at the ground at a $45$ degree angle. $\endgroup$ – mmesser314 Jun 16 '20 at 11:08
  • $\begingroup$ Can you explain more? $\endgroup$ – Learning Mathematics Jun 16 '20 at 11:19
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    $\begingroup$ You should try to answer this for yourself; you will learn more. Imagine a light ball such as a ping-pong ball bouncing off a heavy one such as a football. Can you imagine throwing the ping-pong ball so that after the collision the two velocities afterwards are at more than 90 degrees to each other? How about less? Why not 90 degrees then? $\endgroup$ – Andrew Steane Jun 16 '20 at 18:59

I have a spreadsheet which calculates the result of a 2 body collisions in 2D. I found that with equal masses, one starting at rest, and the two separating at 90 degrees, energy was conserved. With masses that are different it is almost always possible to choose an exit velocity (magnitude and direction) for one that will put it at 90 degrees to the other (and not conserve energy). (By the way, in an elastic collision, if you go into a center of mass system, the speed of each mass relative to the center of mass will be the same leaving as it was approaching the CM. Also, the two masses approach the CM along a line and leave along a (probably) different line.)

  • $\begingroup$ Can you provide an example for that? $\endgroup$ – Learning Mathematics Jun 17 '20 at 14:04
  • $\begingroup$ Example (90 degrees not conserving energy): A 10 gm mass comes in at 30 cm/sec and strikes a stationary 20 gm mass. The 10 gm mass bounces off at 10 cm/sec and 70.53 degrees from the x axis (original direction). The 20 gm mass recoils at 14.142 cm/sec and -19.47 degrees. The x component of momentum is conserved at 300 gm-cm/sec. Total y component is zero. Kinetic energy drops from 4500 to 2500 ergs. $\endgroup$ – R.W. Bird Jun 18 '20 at 16:48

As @mmesser314 noted, even in a real scenario and by throwing a ball at 45° with respect to the ground you could achieve a collision from which a 90° angle results, although it involves both two drastically different masses---the one of a ball against the one of the Earth---and energy dissipation.

So no, 90° is not a sufficient condition to conclude the collision was elastic and/or between two identical masses.

  • $\begingroup$ I don't understand this, what I mean by 90° is the angle between the subsequent path of the ground and the ball, how throwing a ball at 45 degree can make that angle 90°? If the ground is frictionless, then wouldn't the angle between them to be 135°? (because the ground moves vertically downward) $\endgroup$ – Learning Mathematics Jun 16 '20 at 23:18
  • $\begingroup$ The "ricochet" angle is measured with respect to the inciding ball's original trajectory. Imagine a horizontal line (the ground), and on it put a square flipped $45^\circ$. The two sides closer to the ground represent the trajectory of the ball. You will certainly agree that the angle between these two sides is a right angle (it is a square), and since $90 = 180 / 2$, it obviously follows that the ball bounces forming a $90^\circ$ angle with respect to its original trajectory (this is, with the line formed by extending out of the square the side the ball followed before hitting the ground). $\endgroup$ – torito verdejo Jun 17 '20 at 9:29
  • $\begingroup$ No this is not answering my question. The angle between the orginal and new trajectory being 90° and both objects having the same mass doesn't mean that the collision is elastic $\endgroup$ – Learning Mathematics Jun 17 '20 at 10:52
  • $\begingroup$ Your question says: "does a 90 degree angle imply the colliding masses are equal and the collision is elastic?" My answer says "No". $\endgroup$ – torito verdejo Jun 17 '20 at 11:07
  • $\begingroup$ Yes but I disagree with your explanation $\endgroup$ – Learning Mathematics Jun 17 '20 at 11:08

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