# Homework Help: 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?