• Support PF! Buy your school textbooks, materials and every day products Here!

Momentum collision of balls

  • #1
1,728
13

Homework Statement


The problem is given in the picture: (please refer to the 2 pictures below, thank you!)

I'm having trouble with the last part...

Homework Equations


Conservation of momentum in both x and y - directions.


The Attempt at a Solution



From symmetry it can be seen that ball D would be directly below ball C, with ball C moving at speed of (2/5)sqrt(3)*u at an angle of 30 degrees below horizontal.

Here's my attempt using the x-y direction conservation of momentum

Taking downwards and leftwards as positive,

Horizontally:
There would be no transfer in horizontal momentum to ball D, hence ball C still would have a horizontal velocity of: (3/5)u

Vertically (downwards as positive):
sqrt(3)/5 * u = vc cosθ + vD

Energy equation:
(12/25)u2 = (vc)2 + (vD)2


That leaves me with 2 equations with 3 unknowns, vc, vD and θ.


I tried using a new approach, but would like to ask if this obeys the laws of physics:

Horizontal speed of c: unchanged = (3/5)u

Considering collision "vertically":
can i simply do the momentum equation by solely considering only the vertical velocities?

(1/5)sqrt(3)* u = Vc,y + VD

Again only considering energy in the vertical direction: ( i know this is wrong as energy is not a vector!):

(9/25)u2 = (vc,y)2 + (vD )2


This allows me to solve the problem, but is this the right approach?

Here's a quick, not-so-rigorous proof: vx2 + vy2 = v'2, multiplying (1/2)m throughout.... gives us the split energy equation!)
 

Attachments

Last edited:

Answers and Replies

  • #2
gneill
Mentor
20,792
2,770
...can i simply do the momentum equation by solely considering only the vertical velocities?
Yes. Momentum is conserved separately and collectively for the components of the motion.
 
  • #3
BruceW
Homework Helper
3,611
119
It is wrong to consider energy in the vertical direction. So far, you've got 2 equations, and 3 unknowns (as you said):

Energy equation:
(12/25)u2 = (vc)2 + (vD)2

Vertically (downwards as positive):
sqrt(3)/5 * u = vc cosθ + vD

And there is one more equation: the horizontal momentum. This will involve u, vc and theta. So once you write that out, you have 3 equations and 3 unknowns, which you can then solve.
 
  • #4
BruceW
Homework Helper
3,611
119
The reason that you get the correct answer by considering only the components of KE from the y direction is because the speeds of the objects in the x direction happen to be the same before and after collision. So in this case, you could consider the vertical components of KE, but this only works by coincidence. In most problems, doing this will get you a nonsense answer.
 
  • #5
1,728
13
It is wrong to consider energy in the vertical direction. So far, you've got 2 equations, and 3 unknowns (as you said):

Energy equation:
(12/25)u2 = (vc)2 + (vD)2

Vertically (downwards as positive):
sqrt(3)/5 * u = vc cosθ + vD

And there is one more equation: the horizontal momentum. This will involve u, vc and theta. So once you write that out, you have 3 equations and 3 unknowns, which you can then solve.
I tried to consider the horizontal momentum, but that simply gives me:

(3/5)u = (3/5)u

since ball D only moves downwards and has 0 horizontal momentum.
 
  • #6
BruceW
Homework Helper
3,611
119
Well, you've written that the final vertical speed of ball c is vc cos(theta) So what will the horizontal speed of ball c be?
 

Related Threads on Momentum collision of balls

Replies
5
Views
2K
Replies
2
Views
4K
Replies
5
Views
1K
Replies
38
Views
7K
Replies
2
Views
3K
  • Last Post
Replies
1
Views
657
  • Last Post
Replies
14
Views
1K
  • Last Post
Replies
12
Views
3K
Replies
17
Views
1K
Replies
3
Views
3K
Top