Subject: Thanks.
Author:
Posted on: 2013-04-12 05:58:00 UTC
Derp. That was a big mistake.
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Off-topic physics question. by
on 2013-04-11 20:35:00 UTC
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So, I've been doing some tutoring, and I just covered momentum. I was thinking about it for a bit, and now I've got a question I can't figure out.
So, let's start with two objects. One is 1 kg, the other is 10 kg, and moving toward the first at 1 m/s. They collide in an elastic collision, and Conservation of Momentum says the 1 kg object is now moving at 10 m/s, and the 10 kg is still. However, the Kinetic Energy of the 10 kg when it was moving was 5 J, but the KE of the 1 kg is 50 J after the collision. Where does the 45 J of KE come from? -
Got it. by
on 2013-04-11 22:26:00 UTC
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Realistically speaking, the 10kg ball could not transfer all of its momentum to the 1 kg ball: the mass ratio is far too big. Instead, the 10 kg ball will keep on rolling forward at something just under its initial velocity and the 1 kg ball will be sent forward at about 2*v1.
Source: Hyperphysics website. It's essentially everything you need to know about high school/introductory university physics. Bookmark it!
Here is the information relevant to the question. -
Darn, ninja'd by hS. (nm) by
on 2013-04-11 22:27:00 UTC
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Error in the working. by
on 2013-04-11 22:11:00 UTC
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CoM says that (duh) momentum is conserved - it doesn't say that a 1kg ball can stop a 10kg ball dead in its tracks. Balls of equal mass exchange velocities in that way; unequal masses don't.
Running via wiki, I find a pair of equations (under 'Elastic collisions') allowing you to determine the final velocity, v(1,2), from the initial velocities u(1,2) and the masses m(1,2). For your 10kg mass, the equation reads:
v1 = ((m1-m2)/(m1+m2))*u1 + ((2*m2)/(m1+m2))*u2
Given that u2 (initial velocity of the smaller ball) = 0, and inputting the other figures, that gives:
v1 = (9/11) * 1 = ~0.82 m/s.
k.e. = 1/2 m1*v1^2 = 3.35 J.
The equation for m2 is the same, except with the subscripts flipped. Running the numbers gives us:
v2 = (20/11) * 1 = ~ 1.82 m/s
k.e. = 1.65 J.
And, of course,
3.35 + 1.65 = 5 J.
Does that answer your question?
(Short version: a bowling ball rolled at a ping-pong ball won't fire the ping-pong ball off into orbit)
hS