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
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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?
 

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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 .
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