Hey people, question about energy transfer here! :)?

Question

If a ball flying forward in space at a constant speed due to momentary application of force collides with a stationary ball, is the kinetic energy of the moving ball's movement divided equally between the two balls? That's all I wanna know.

Zero friction concerned here, and the ball is travelling torwards the stationary ball where their centres of gravity lie on the same plane constantly. Both balls need not be of equal mass.

Answer 1

NO!! If people would stop (mis)quoting half-remembered precepts here, and do a spot of THINKING, instead, we might get somewhere.

1. If a moving spherical ball directly hits an identical free, stationary ball head on in an elastic collision, the moving ball STOPS, and the previously stationary one heads off with the original velocity of the other. So ALL of the moving ball's K.E. has been transferred to the other ball.

Here are two ways to see this:

1a. The device known somewhat indelicately to British schoolboys as "Newton's balls" (but in squeamish America as "The Swinging Wonder") provides an excellent table-top demonstration of this. Most undergraduate physics instructors must haul this out from time to time.

1b. You can easily solve it analytically, but a thought experiment will also demonstrate it. In the originally stationary ball's frame, let the other ball approach with speed v. Now shift (with an appropriately directed speed of v/2) to the frame in which the two equal balls are simply approaching each other, each with a speed in this frame of that same v/2 (oppositely directed, of course)! Clearly, from the symmetry of the situation, they'll bounce off each other with the same (oppositely directed) speeds in this frame, still of v/2. Now go back to the original frame: the previously moving ball is now completely stationary, while the originally stationary one is haring off with speed v. (You may need to make little sketches of what is happening as viewed successively from each of these frames, to see this; I admit that that's what I did.)

2a. A similar approach can be used to solve the general case for an arbitrary ratio of masses, m (initially moving)/M (initially stationary). Hint: have the balls approach one another in their center of mass frame with equal but opposite momenta of mv and MV. They'll clearly bounce off one another in such a frame, retaining the same values of their respective (but now oppositely directed) speeds. Then you can do the other frame-jugglings from there.

However, this analytical approach, done accurately, is a tedious mess. So, to appreciate the general problem more physically, consider just the following:

2b. The case of m << M.

Consider the "equal but opposite momentum" frame for these two balls, in which MV = mv. This means that if ball M is VERY MASSIVE, the speed it will have in such a frame, V = (m/M)v; that is, it's VERY SMALL compared to little puny m's high speed.

So, transforming to the frame in which M itself (the big guy) is initially stationary takes only a speed shift of O(m/M)v. (In fact identically so.) This only affects the speed of puny m minutely, so m's initial K.E. is still of order O(m v^2). (There's no need to do picky constant book-keeping for these purposes.)

Now after the collision: In the original frame, Big M is now moving (and in fact at 2V if you think about it.) But, being loose about the constants, M's K.E is only O(MV^2).

So, the ratio (m's K.E.)/(M's K.E.) ~ (mv^2)/(MV^2)

= (m/M)(M/m)^2 (since mv = MV)

= M/m.

BINGO! The little bugger still carries by far the larger amount of the kinetic energy. The massive ball (i) moves sluggishly, and (ii) because K.E. involves both mass but also the SQUARE of the speed, the TWO POWERS of (small) speed more than cancel out the effect of its (large) mass.

This is a well-known fact in particle collision theory, that little particles basically bounce off big ones.

You could also say that it's well-known in "space billiards": when spacecraft head close to Jupiter to receive a "gravitational sling-shot effect" that sends them off to the outermost planets.

It's more complicated because of 3-D effects there, but analogous frame-shifting methods can be used to show that the spacecraft can pick up considerable speed AND change the direction of its path to send it in a more desirable direction.

Thank goodness that this is so; otherwise the Enterprise would soon have exhausted Scotty's supply of lithium dichloride crystal propellant.

Live long and prosper.

Answer 2

No. It depends upon the angle of collision with respect to the direction of travel. You've seen novelties of a row of balls sups pended from wires or strings. One ball is drawn back and released. When it makes contact with the row of balls, only the ball on the end is knocked off. Consider a pool table. It depends upon the direction of movement with respect to the angle of impact.

Answer 3

it would not quite make it back to half the speed bof the original ball. the reason being you would have a loss of energy between the two due to friction but it would be very close

Answer 4

Is anything really "stationary" in space, since even the universe is expanding and every point is getting farther from every other point? Maybe not a question that can be answered without alot of "assuming that....".

Answer 5

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Answer 6

Yes if it is a completely elastic collision.