Perfectly elastic collision problem with no given values

Click For Summary
In a perfectly elastic collision involving two identical curling stones, the initial moving stone collides head-on with a stationary stone. The correct final speed of the initially moving stone is zero, while the stationary stone takes on the initial speed of the moving stone. The solution requires applying conservation of momentum and energy principles to derive the final velocities. A mass-independent relationship can simplify calculations, avoiding complex algebra. It is advised to type out equations for clarity and better understanding in discussions.
Balsam
Messages
226
Reaction score
8

Homework Statement


A curling stone with initial speed vi1 collides head-on with a second, stationary stone of identical mass, m. Calculate the final speeds of the two curling stones.

Homework Equations


See attached picture

The Attempt at a Solution


I solved for vf1 as shown, but my answer was wrong. I got vi1-vf2=vf1, but the correct answer is vf1=0. What did I do wrong?
IMG_6163.jpg
 
Physics news on Phys.org
Please type your equations here, as the rules require, rather than uploading a picture. I have a very difficult time seeing anything you wrote in step 3, which is where you went wrong. It looks like you butchered your algebra, but I can hardly read it.

However, in general, to approach this problem, use the conservation of energy and the conservation of momentum. You'll then have a system of two equations and can solve for the velocities.
 
I second @RedDelicious ' remarks about postings, and note that BvU has said the same in another of your threads. Please bear in mind that although typing costs you a bit of time it will save time for the many people who read it.

For perfectly elastic collisions, it can save a lot of working to use a simple mass-independent result relating initial and final velocities. This relationship can be derived from the two conservation laws, but using it avoids the quadratics that come from conservation of energy.
Look up coefficient of restitution, and consider the case where it is 1.
 
haruspex said:
I second @RedDelicious ' remarks about postings, and note that BvU has said the same in another of your threads. Please bear in mind that although typing costs you a bit of time it will save time for the many people who read it.

For perfectly elastic collisions, it can save a lot of working to use a simple mass-independent result relating initial and final velocities. This relationship can be derived from the two conservation laws, but using it avoids the quadratics that come from conservation of energy.
Look up coefficient of restitution, and consider the case where it is 1.
I just thought it would make more sense if I showed a picture of exactly what I did
 
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 10 ·
Replies
10
Views
4K
  • · Replies 14 ·
Replies
14
Views
11K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 3 ·
Replies
3
Views
882
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 13 ·
Replies
13
Views
1K
Replies
20
Views
2K