Find the magnitude of the momentum change of the ball?

AI Thread 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: 192
  • Screenshot 2023-02-11 at 9.27.05 AM.png
    Screenshot 2023-02-11 at 9.27.05 AM.png
    8.4 KB · Views: 131
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!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top