Find the magnitude of the momentum change of the ball?

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SUMMARY

The discussion centers on calculating the change in momentum of a tennis ball during an elastic collision. The participant correctly identifies that the momentum in the y-direction remains unchanged while the momentum in the x-direction changes due to the opposite velocities. The final formula derived for the change in momentum is Δp = -2mvcos(θ), confirming that the correct answer to the homework question is (D) 2mv cos θ.

PREREQUISITES
  • Understanding of 2D momentum concepts
  • Knowledge of elastic collisions
  • Familiarity with vector diagrams in physics
  • Proficiency in using conservation of momentum equations
NEXT STEPS
  • Study the principles of elastic and inelastic collisions in detail
  • Learn how to analyze momentum changes using vector diagrams
  • Explore advanced topics in momentum conservation in two dimensions
  • Practice problems involving momentum change in various collision scenarios
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This discussion is beneficial for physics students, educators, and anyone interested in mastering concepts related to momentum, particularly in the context of elastic collisions.

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!
 

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