Physics Friday 108

Last week, I illustrated an example of momentum and energy conservation under special relativity, with the example of Compton scattering. Now, let’s go back to classical mechanics, and examine perfectly elastic collisions (mechanical energy conserved) between a mass M, moving with velocity V in our laboratory frame, and a mass m stationary in our laboratory frame.

First, consider when the collision is direct enough so that the mass m has post-collision velocity in the direction of V, so that the problem is one-dimensional. Thus, we choose our direction so that V is positive. Then if the masses M and m have post-collison velocities Vf and vf, respectively, then momentum conservation tells us that
,
and mechanical energy conservation tells us that
.
Solving the first equation for the post-collision momentum of m and the second for the post-collision kinetic energy of m, we get
,
.
Squaring the first and multiplying the second by 2m, we get

and
.
Equating these,
.
The above is true (both sides equal zero) if V=Vf, but this is where there is no collision. Thus, for a collision, VVf, and we can cancel a factor of VVf from both sides to get
.

Note that this is negative when M<m; a lighter mass will bounce back when colliding elasticly with a larger one. Note also when the masses are equal, then Vf=0 (and vf=V), so that the motion transfers entirely from one mass to the other; consider Newton’s cradle.
What is most significant, though, is when the moving mass is the heavier one, M>m; then Vf is positive, and both masses will be moving forward post collision.

Now, consider the two-dimensional problem, where the collision can deflect the mass of at an angle. While for M<m, the mass can be deflected through any angle (except exactly zero), our above result indicates that when M>m, then the mass cannot be reflected directly back, and in fact there must be some angle θ<180° which is the maximum angle by which the collision can deflect M.


To determine this angle, let us first consider the center of mass. For two masses m1 and m2 at locations x1 and x2, respectively, then the center of mass is located at . If they are moving, then the velocity of the center of mass is
,
so for our colliding masses, the center of mass will have velocity in the direction of our initial velocity V, here chosen to be the positive x direction. Note that this will be the velocity of the center of mass both before and after the collision.
So, now let us consider coordinates moving relative to our laboratory frame with velocity equal to vcm. This is our center of mass frame, in which the center of mass is stationary. We will denote quatities in this frame with primes. Since this is a classical problem, we will be using the Galilean transformation between frames.

Now, in the center of mass frame, our mass M has initial velocity in the x‘ direction (same as the x direction) of

and an initial momentum of ,
while the mass M has initial velocity

and an initial momentum of ,
so that the total momentum is zero in this frame, as expected for a center of mass frame.
If the mass M is deflected by an angle θ‘ in this frame, then we see that the mass m must also be deflected by this angle, as the momenta must be equal in magnitude and opposite in direction after the collision. Combining this with mechanical energy conservation, we see that in the center of mass frame, the post-collision velocities must have the same magnitudes after the collision as before; only the directions change.



Thus, in the velocity space, the tip of the vector traces a circle of radius about the origin. The velocities in the lab frame are , and so trace an off-center circle in the lab frame velocity space. If M>m, then is greater than , and this circle does not contain the origin within its interior:


The largest possible deflection, which we see is not only less than 180°, but is less than 90°, is obtained when our lab velocity vector Vf is tangent to the circle.


Simple trigonometry then tells us that
.
So, for example, if M is twice as massive than m, then , and so θ=sin-1(1/2)=30°

Lastly, one should note that when M=m, then the circle of possible velocities in the lab frame passes through the origin. Secondly, since the momenta in the center of mass frame are opposite vectors, and the masses are the same, the center of mass frame velocity vectors and are equal in magnitude and opposite in direction; we see that they form a diameter of the circle. Thus, by the inscribed angle theorem (or more specifically, the special subcase known as Thales’ Theorem), we see that the lab frame post collision velocities and must form a right angle:


This result is well known in billiards.

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2 Responses to “Physics Friday 108”

  1. Physics Friday 109 « Twisted One 151's Weblog Says:

    […] Friday 109 By twistedone151 Continuing from last week, we consider the speed of the mass M after it undergoes a deflection by the maximum angle in a […]

  2. Physics Friday 114 « Twisted One 151's Weblog Says:

    […] . Note that each of the masses of our semicircle, the deflecting masses, must have mass . We found here the maximum angle by which a larger mass can be deflected by a smaller one in a perfectly elastic […]

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