Elastic Collisions

This is the ideal bounce, so there is no loss of kinetic energy. (Some people say perfectly elastic for the ideal case, and a lesser bounce is still considered elastic. Most people use elastic for the ideal case only. Be sure to get the terms right for your particular course.) It can be shown that for a head-on collision, conservation of energy and momentum lead to the conclusion that the relative speed of approach before collision is equal to the relative speed of recession after for the ideal bounce. If you have velocities v₁ and v₂ before and v₁' and v₂' after, then
v₁ - v₂ = v₂' - v₁'. These are vector differences. For example, two very bouncy balls have a head-on. Let's say they are moving toward each other with speeds 2 m/s and 3 m/s respectively. Then the relative speed of approach is 5 m/s. Formally, along the motion axis their velocities are 2 and -3 m/s, so the left-hand side of our velocity equation is 2 - (-3) m/s or 5 m/s.

After collision, their velocities might be -1 and 4, for example, and 4 - (-1) = 5. (If they were the same mass, it turns out that they just trade velocities.) You may have two unknowns, and you have the velocity equation (above) and the momentum equation to play with. (There is also the energy equation, but that is more trouble. Two of the three are independent equations, and you should use the two easy ones.)

Non-head-on Elastic Collisions

In this case you need to choose a convenient coordinate system, then write your momentum equations, one equation for each coordinate (sum of the x components of momentum before collision = sum of x components after...). Energy is not a vector so you have one energy equation. Now if it is a numerical problem in a book or on a test, the number of unknowns will match the number of equations, 3. If it is a real-life problem, maybe not.

A special case: equal masses. It is easy to show that if the masses are equal and one is initially at rest, after the collision they go off at right angles to each other. If you run into this on a test, prove it: Write the energy equation, cancel (1/2)m, and notice the Pythagorean relationship. The laboratory work for this should take place in your local pool hall. Here is an example.

In the case of unequal masses, if both angles after the collision are known, the algebra required to do the problem is not too bad. In the sketch below, m hits m' which was at rest. Define the direction of v_o to be the x direction, and write the three equations for this case. Don't peek below.

. . . . . BEFORE. . . . . . . . . . . . . . . . . . . . . . . . AFTER

(I said don’t peek.) The x components of momentum are the same before and after, so
mv_o = mvcosf + m'v'cosq . The y components likewise, so

0 = mvsinf - m'v'sinq , if you take q to be a positive angle. Now the energy equation is

(1/2)mv_o² = (1/2)mv² + (1/2)m'v'². (Cancel the 1/2.)

With 3 independent equations you can find 3 unknowns, but it could be a lot of trouble, especially if the angles are unknown. In the unknown angle(s) case, you can save yourself some trouble by drawing the momentum vector sum diagram (Let momentum = p for simplicity, so p_o = p' + p), then write the law of cosines equation. Try it, assuming f is known. Let's say m = 2.0 kg, m' = 3.0 kg, v_o =4.0 m/s and f = 30^o. Calculate the other stuff. Don't peek below.

The law of cosines is
p'² = p² + p_o² - 2pp_ocosf . This combines nicely with the energy equation, which is
p_o²/(2m) = p'²/(2m') + p²/(2m). Isolate p² from the energy equation and substitute it into the law of cosines equation. Then plug and chug using the numbers above, get
p' = 8.3 kg m/s, so v' = 2.8 m/s. Then go to the energy equation and get
p = 4.2 m/s. Finally, use the law of sines or cosines and get
q = 15^o.

If both angles are unknown and you choose to use the component method, write the two momentum equations, put q and f terms on opposite sides, square both equations, add them, simplify using
sin² q + cos² q = 1, and same for f . At this point you have derived the law of cosines equation. Congratulations.

One more bit of advice: If you expect a problem of the above type, you might be tempted to have it more or less completed ahead of time so you can just plug in some numbers. Don't do that. Your prof wants to see your knowledge of basic principles. Also (s)he could foil that so easily by having both move before collision or have some energy loss during collision. Or here is a problem with 3 equal mass balls, and after the collision the two that were at rest have the same speed v and the other one has speed v_f. Find v and v_f in terms of v_o. Don't peek at the end of this page.

. . . . . BEFORE . . . . . . . . . . . . . .. . . . . .. . .. . .. . AFTER

The Slingshot Effect

NASA uses this effect to get energy from planets for their various space probes. The idea is this: if a planet and a probe are heading approximately toward each other for a near miss (not to enter the planet's atmosphere, but near in astronomical terms), you can rig it so that after the encounter with the planet, the probe is traveling approximately in the opposite direction from its original velocity. Then the equations are the same as in the elastic collision case. Try bouncing a ping-pong ball off a bowling ball moving toward you. The bowling ball, like the planet, will have about the same speed after the collision. The ping-pong ball and the probe get quite a boost in speed. Let's say the planet has velocity v₁ and the probe v₂. Then after the interaction, the planet has velocity v₁' = v₁, and the probe has v₂'. Solve the equation in paragraph 1 for v₂' and get 2 v₂ - v₁ and note that v₂ was in the opposite direction from v₁, so we have a big boost in speed. For example if they are headed toward each other with the same speed v, afterward the probe will have speed 3v, and hence 9 times the kinetic energy.

That problem just above the slingshot effect:
v = (2/3)v_o , v_f = (1/3)v_o.
To get these answers, write momentum and energy equations and do a lot of algebra. Then go to the local pool hall to check it out experimentally. Notice how in the picture they look like they are in the same x location after the collision? They would be if they had negligible radii, because they have the same v_x! But the middle one should be behind the others by 2rcos30^o. The picture is a little off.

Don't click on back or you will be propelled into deep cyberspace with no way out. (Well actually it's back to the momentum page.)

Or meander to one of my main mantras:

Mechanics
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 Questions, comments: fredrick.gram at tri-c.edu (remove spaces and replace at with @. This is my defense against spammer software that gets email addresses that are listed on the web).