Finding Final Velocities in head-on Collision

[Cait602]

Homework Statement

Two balls move toward each other. Ball one is moving in the positive direction, has a mass of m1, and a velocity of v initially. While ball two has a mass of m1/2, is moving in the negative direction, and has initial velocity of v2. Once the collision takes place ball one will have a final velocity equal to v1,i/2, at an angle of 90 degrees downward. Ball two will have a final velocity of v2,f at an angle of 30 degrees with respect to the positive x axis, seen as going upward and to the right. Find the v2 final in relation to v1 initial.

Seen as something like this

Homework Equations

Pox=Pfx, Poy=Pfy

The Attempt at a Solution

In terms of the y direction, Initially we would have a momentum of 0, so
m1v1,i + m1/2 v2,i = m1v1,f+ m1/2 v2,f
0 + 0 = m1v1,f+ m1/2 v2,f
0= -m1 (v1, i /2) + m1/2 (v2 sin(30))
m1 (v1, i /2) = m1/2 (v2 sin(30))[/B]
m1 (v1, i /2)/ m1/2 = v2 sin (30)
I'm not sure how this would cancel out..
m1/2 (v1, i/2) / sin(30) = v2

But I'm really not sure if this is right at all

 

[haruspex]

m1 (v1, i /2)/ m1/2 = v2 sin (30)
I'm not sure how this would cancel out..
m1/2 (v1, i/2) / sin(30) = v2

You did not cancel correctly. The final equation is dimensionally inconsistent, having a mass factor on one side but not the other.

 

[Cait602]

wouldn't we have to have only a velocity on one side to find the final velocity of ball 2? how would I solve for the ball 2's final velocity with the mass on that side with it? Was I correct up until that point?

 

[haruspex]

wouldn't we have to have only a velocity on one side to find the final velocity of ball 2?

Of course, but there should be no residual mass dimension on the other side. This is not the same as saying there should be no mass terms on the other side, only that when all the dimensional cancellations are done there should be no remaining mass dimension. E.g. it would be ok to have a ratio of two masses in there.

 

[Cait602]

m1 (v1, i /2) = m1 /2 (v2 sin(30))
Math is not my strongest area, and neither is doing math with only variables..
Should I just leave it so.. if i divided the m1 /2 out of the right side.. on the left side m1 would cancel and the /2 would cancel with the /2 from the v1, i /2? I might still totally be wrong.. they have to cancel out because I still have to divide sin(30) over..

 

[haruspex]

m1 (v1, i /2) = m1 /2 (v2 sin(30))
Math is not my strongest area, and neither is doing math with only variables..
Should I just leave it so.. if i divided the m1 /2 out of the right side.. on the left side m1 would cancel and the /2 would cancel with the /2 from the v1, i /2? I might still totally be wrong.. they have to cancel out because I still have to divide sin(30) over..

Sounds good. What do you end up with?

 

[Cait602]

v1, i / sin(30) = v2

If m1 was say like m1 /5 or something that didn't divide out as easily, can you show me what that type of situation would look like for future reference? Would I just end up with a mess of fractions? m1 would still cancel, but we would be left with everything else divided by say 5, then I would still have to divide the sin(30) over. Might be a silly question.

 

[Cait602]

I mean m2 because m2= m1 /5 in the above problem.

 

[haruspex]

v1, i / sin(30) = v2

Yes.

If m2 was say like m1 /5 or something that didn't divide out as easily, can you show me what that type of situation would look like for future reference?

Umm.. you don't know how to divide by 1/5?

 

[Cait602]

I do... But we don't know what mass 1 is, so I was assuming that that isn't a certain 1/5, rather that mass 1 may be 15 kg or some mass divided by 5. But, I guess it's 1/5 then.

 

[haruspex]

I do... But we don't know what mass 1 is, so I was assuming that that isn't a certain 1/5, rather that mass 1 may be 15 kg or some mass divided by 5. But, I guess it's 1/5 then.

Just treat is as two separate operations. m/5 is the same as m times 1/5. So divide both sides by m, then divide both sides by 1/5. (Or, equivalently, multiply both sides by 5.)