momemtum

[drkidd22]

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

The white ball (2kg) in the figure has a speed of 1.74 m/s and the yellow ball (1kg) is at rest prior to an elastic glancing collision. After the collision the white ball has a speed of 1.37 m/s. what angle does it scatter at if the yellow ball is scattered at 280 degrees?


2. Relevant equations

mva=mvacos(@)+mvbcos(@)

3. The attempt at a solution
2(1.37)Cos(@)a+0.58Cos280
2.74Cos@+.10
=92 degrees

I think I'm close, but not quite.

[drkidd22]

anyone please, a hint?

[Doc Al]

mva=mvacos(@)+mvbcos(@)
That's momentum conservation in one direction. What about the other? And what about the fact that the collision is elastic?

3. The attempt at a solution
2(1.37)Cos(@)a+0.58Cos280
2.74Cos@+.10
=92 degrees
I don't understand what you're doing here. I don't see the full equation being used. Where did you get '1.37' and '0.58'? Show all your steps.

[drkidd22]

1.37 is given as the speed of ball a after the collision.

0.58 is what I had found the speed of ball b to be after collision, but I think it's not correct as I'm not sure how to really do this problem. I can't really understand what the author of the book is trying to say on a similar problem.

[Doc Al]

1.37 is given as the speed of ball a after the collision.
OK.
0.58 is what I had found the speed of ball b to be after collision,
How did you get this? (Hint: That's where the fact that the collision is elastic will come in handy.)

[drkidd22]

ok, so I think .58 was incorrect.

mava+mbvb = mava'+mbvb'

2(1.74)+0=2(1.37)+vb'
.74m/s = vb'

Right?

[drkidd22]

When I put this in I still don't get the right answer.

2(1.37)Cos(@)+.74Cos280
2.74Cos@+.13
=92.72 degrees

[Doc Al]

ok, so I think .58 was incorrect.

mava+mbvb = mava'+mbvb'

2(1.74)+0=2(1.37)+vb'
.74m/s = vb'

Right?
No. That equation isn't valid. (Momentum is a vector--direction matters.)

Instead, make use of the fact that the collision is elastic. What does that mean?

[drkidd22]

KE is also conserved

[Doc Al]

KE is also conserved
Right! Use that to determine the speed of the yellow ball after the collision.

[drkidd22]

3.0276 = 1.8769 + .5(v^2)
(3.0276 - 1.8769)/(.5) = v^2
Vb' = 1.51 m/s

[Doc Al]

3.0276 = 1.8769 + .5(v^2)
(3.0276 - 1.8769)/(.5) = v^2
Vb' = 1.51 m/s
Looks good. (I get 1.52, when I round off.)

[drkidd22]

0 = 2(1.37)Sin@+1.52Sin280
0 = 2.74Sin@ - 1.50
= 33 degree

I think that's right.

Thanks a million. Took me while to understand it, but I got it. Thanks.