Find the magnitude of the momentum change of the ball?

Click For Summary
The discussion focuses on calculating the momentum change of a tennis ball during an elastic collision. The user identifies that in the y-direction, momentum remains unchanged, while in the x-direction, it changes due to opposite velocity vectors. They derive the momentum change as Δp = -2mvcosθ, indicating a significant change in momentum. The responses confirm that the calculations are correct. The final answer aligns with option (D) 2mv cos θ, confirming the user's understanding of the problem.
paulimerci
Messages
287
Reaction score
47
Homework Statement
A tennis ball of mass m rebounds from a racquet with the same speed v as it had
initially as shown. The magnitude of the momentum change of the ball is
(A) 0 (B) 2mv (C) 2mv sin theta (D) 2mv cos theta
Relevant Equations
Conservation of momentum
I understand that it is a 2D momentum problem with an elastic collision;
Looking at the vector diagrams below, I notice that the velocity vectors initial and final in the y direction are in the same direction, indicating that momentum does not change, whereas the velocity vectors initial and final in the x direction are opposite each other, indicating that momentum does change.
Therfore,
$$ \Delta p = p_f - p_i$$
$$ = -mvcos\theta -mvcos\theta$$
$$ \Delta p = -2mvcos\theta$$

Have I done it right?
 

Attachments

  • Screenshot 2023-02-11 at 9.27.16 AM.png
    Screenshot 2023-02-11 at 9.27.16 AM.png
    4.6 KB · Views: 204
  • Screenshot 2023-02-11 at 9.27.05 AM.png
    Screenshot 2023-02-11 at 9.27.05 AM.png
    8.4 KB · Views: 140
Physics news on Phys.org
paulimerci said:
Homework Statement:: A tennis ball of mass m rebounds from a racquet with the same speed v as it had
initially as shown. The magnitude of the momentum change of the ball is
(A) 0 (B) 2mv (C) 2mv sin theta (D) 2mv cos theta
Relevant Equations:: Conservation of momentum

I understand that it is a 2D momentum problem with an elastic collision;
Looking at the vector diagrams below, I notice that the velocity vectors initial and final in the y direction are in the same direction, indicating that momentum does not change, whereas the velocity vectors initial and final in the x direction are opposite each other, indicating that momentum does change.
Therfore,
$$ \Delta p = p_f - p_i$$
$$ = -mvcos\theta -mvcos\theta$$
$$ \Delta p = -2mvcos\theta$$

Have I done it right?
Looks good.
 
TSny said:
Looks good.
Thank you!
 
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 13 ·
Replies
13
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
55
Views
5K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
27
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 16 ·
Replies
16
Views
4K
  • · Replies 40 ·
2
Replies
40
Views
5K