Elastic Glancing Collision-HELP

  • Thread starter Thread starter psychfan29
  • Start date Start date
  • Tags Tags
    Elastic
Click For Summary
In an elastic glancing collision involving a 1 kg white ball and a 2 kg yellow ball, the white ball initially moves at 1.68 m/s and after the collision, its speed is 1.24 m/s. The yellow ball, initially at rest, is scattered at an angle of 280 degrees. The user attempted to apply conservation of kinetic energy and momentum equations but encountered a domain error when calculating the vertical component of momentum. The discussion raises questions about the validity of the calculations and whether the mass relationship affects the outcome. The problem highlights the complexities of solving elastic collision scenarios in physics.
psychfan29
Messages
9
Reaction score
0
Elastic Glancing Collision-HELP!

A white ball, mass of 1 kg has a speed of 1.68 m/s and a yellow ball, mass of 2kg, is at rest prior to an elastic glancing collision. After the collision the white ball has a speed of 1.24 m/s. To the nearest tenth of a degree, measured counterclockwise from east, what angle does it scatter at if the yellow ball is scattered at 280degrees?

I don't know what equations to use for this problem!
 
Physics news on Phys.org


I used this equation:

1/2mv_1i^2 = 1/2mv_1f^2 + 1/2mv_2f^2 (the 1i, 1f, and 2f are subscripts indicating which velocities)

When I plugged in the numbers from the problem, I got that the final velocity for the yellow ball is 0.801m/s.

I then used the following equation:

Py=1kg(1.24m/s)sintheta+2kg(vyellow)sin280

But sin becomes greater than 1, so it's a domain error.

Am I doing something completely wrong?
Or is it possible that for this problem to work, the mass of the white ball must be greater than that of the yellow ball?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
10
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K