What angle does it scatter at if the yellow ball is scattered

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Homework Help Overview

The problem involves an elastic glancing collision between two balls of different masses, where the initial conditions include the speed of the white ball and the rest state of the yellow ball. Participants are tasked with determining the scattering angle of the yellow ball after the collision.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore momentum conservation equations and question the application of these principles in both x and y directions. There are discussions about the correct interpretation of elastic collisions and the conservation of kinetic energy.

Discussion Status

Some participants have provided hints and guidance regarding the use of conservation laws, while others express confusion about the calculations and the equations being used. There is an ongoing exploration of the correct values and methods to apply in the context of the problem.

Contextual Notes

Participants note the importance of direction in momentum conservation and the implications of the elastic nature of the collision. There is mention of specific values derived during the discussion that may not be universally agreed upon.

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Homework Statement



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?


Homework Equations



mva=mvacos(@)+mvbcos(@)

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.
 
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anyone please, a hint?
 


drkidd22 said:
mva=mvacos(@)+mvbcos(@)
That's momentum conservation in one direction. What about the other? And what about the fact that the collision is elastic?

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.
 


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.
 


drkidd22 said:
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.)
 


ok, so I think .58 was incorrect.

mava+mbvb = mava'+mbvb'

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

Right?
 


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

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


drkidd22 said:
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?
 


KE is also conserved
 
  • #10


drkidd22 said:
KE is also conserved
Right! Use that to determine the speed of the yellow ball after the collision.
 
  • #11


3.0276 = 1.8769 + .5(v^2)
(3.0276 - 1.8769)/(.5) = v^2
Vb' = 1.51 m/s
 
  • #12


drkidd22 said:
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.)
 
  • #13


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.
 

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