10/20/2016
https://www.lhup.edu/~dsimanek/ideas/bounce.htm
Ball bouncing from massive wall. Most physics textbooks consider the case of a ball bouncing from a massive object, say the floor, or a wall. They consider the case where the collision is nearly or totally elastic. In the totally elastic collision, the ball loses no kinetic energy in the collision, so its speed after collision is the same as before the collision. The student thinks, "Of course that must be the case, because of conservation of energy." Seldom does the textbook, or the student, consider how conservation of momentum is satisfied in this problem. They should, for the analysis is instructive. instructive. Analyzing this may also give some insight into why energy, momentum and the conservation laws took so long to be formulated, since the concepts are subtle. Consider a ball of mass m colliding elastically with a stationary object of larger mass M. Draw the picture before and after the collision. The conservation equations are: [1] mv1 = mv2 + MV2 (1/2)mv12 = (1/2)mv22 + (1/2)MV2
[2]
where v1 is the initial velocity of the smaller ball, v 2 is its final velocity after collision, and V2 is the velocity of the larger mass after the collision. Multiply the energy equation by 2 to eliminate the (1/2) factors. mv1 2 = mv22 + MV2
[3]
Divide this by m on both sides. v12 = v2 2 + (M/m)V22
[4]
Rearrange. v12 - v22 = (M/m)V22
[5]
10/20/2016
https://www.lhup.edu/~dsimanek/ideas/bounce.htm
Divide the momentum equation by m on both sides. v1 = v2 + (M/m)V2
[6]
Rearrange and square both sides. (v1 - v2)2 = (M/m)2V2 2
[7]
Multiply Eq. [5] by (M/m) and combine with [7] to eliminate V 22. (M/m)(v12 - v22) = (v1 - v2)2
[8]
Multiply both sides by (m/M) (v12 - v22 ) = (m/M)(v1 - v2)2
[9]
Take the limit as (m/M) goes to zero. (v12 - v22 ) = 0
[10]
So one solution of this is v 1 = -v2. Another solution is v 1 = v2, corresponding to the case where the two objects do not collide at all. One can graph the values of v 2 and V2 against (m/M) and show that as (m/M) goes to zero, the values of the final velocities do indeed smoothly go to the limiting case values. In words, the reason this can happen is that kinetic energy is a scalar, and momentum is a vector, and kinetic energy varies as the square of the speed, while momentum varies as only the first power of speed. Therefore the quantity momentum/energy varies with speed as (1/v). So when v goes to zero, the momentum/energy can be infinite. Those who have had calculus will recognize that this is a case where an indeterminate form arises when you take the limit of the value of momentum of a body whose mass increases to infinity.
10/20/2016
https://www.lhup.edu/~dsimanek/ideas/bounce.htm
Bouncing at an angle. A related problem is that of a ball bouncing elastically from a massive wall incident at an angle. If the ball is incident to the wall at some angle, the rebound is at the same angle, provided the ball had no spin before or after the collision. That's actually a rather unlikely situation, for even if the ball had no spin initially, it usually does after the collision and the incident and rebound angles will not be equal. The problem can therefore get quite messy. But in the textbook discussions of classical kinetic theory one often begins with an idealized situation of point particles bouncing elastically from the infinitely massive walls of a box enclosure. The assertion is then made that the particle's momentum component parallel to the wall is unchanged (and therefore this component of velocity is unchanged), but the momentum and velocity components perpendicular to the wall are reversed. This is equivalent to saying that (a) angular momentum is zero for a point particle and (b) the wall exerts a force on the particle that is perpendicular to the wall, so the impulse the ball experiences is perpendicular to the wall. Therefore the change in the particle's momentum is perpendicular to the wall, affecting only the particle's velocity component perpendicular to the wall. —Donald E. Simanek, March 5, 2005
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