# Elastic Collision of Two Objects (2D)

1. Nov 16, 2008

### Hermit Solmu

1. The problem statement, all variables and given/known data

A 17g object moving to the right at 32 cm/s overtakes and collides elastically with a 35g object moving in the same direction at 12 cm/s. Find the velocity of the slower object after the collision. Answer in cm/s.

B) Find the velocity of the faster object after the collision. Answer in cm/s.

2. Relevant equations

(ma1*va1)+(mb1*vb1)=(ma2*va2)+(mb2*mb2)

.5((ma1*va1)^2+(mb1*vb1)^2)=.5((ma2*va2)^2+(mb2*mb2)^2)

3. The attempt at a solution

(17*32)+(35*12)=(17*x)+(35*y)

And since I have two variables, I'm not sure how to solve for either of them. I tried solving for one in terms of the other...

(928-17x)/35=y

But that wasn't the answer my instructor was looking for. Thanks so much for any help you can provide.

2. Nov 16, 2008

### Staff: Mentor

OK.

You have typos in this one. Each term should be ½mv², not ½(mv)².

So far, so good.

Now use your 2nd equation, the one for energy conservation. (2 unknowns requires 2 equations. Luckily you have 2 equations.)

3. Nov 16, 2008

### Hermit Solmu

So you're saying my second equation should be .5m^2+.5v^2=.5m2^2+.5v2^2?

And then just set the Y equal to each other, then do the same thing with the X?

4. Nov 16, 2008

### Staff: Mentor

No, it should be:
0.5ma*(va1)^2+0.5mb*(vb1)^2 = 0.5ma*(va2)^2+ 0.5mb*(vb2)^2

(Of course, you call va2 = X and vb2 = Y.)

If you substitute the expression you found from the first equation, which gave you Y in terms of X, into this equation, then you can solve for X. And then use it to get Y.

5. Nov 16, 2008

### phynoob

after we have this equation: lets call in 1 ->(17*32)+(35*12)=(17*x)+(35*y)
we also use (V1i-V2i)=-(v1f-v2f) => 32-12+v1f=v2f -(2)

sub equation 2 into 1 and we get the final answer
am i correct?

6. Nov 17, 2008

### Staff: Mentor

Yes, you can definitely use that relationship. The relative velocity is reversed in an elastic collision. Note that this relationship is derived from momentum and energy conservation. It's certainly quicker to use this formula, since much of the work has been done for you.

But some courses do not cover it, so I didn't want to bring it up. (You can always derive it for yourself, of course. )