How Is Impulse Calculated for a Tennis Ball Hitting a Wall?

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Homework Help Overview

The problem involves calculating the impulse experienced by a tennis ball of mass 0.087 kg and speed 43 m/s as it strikes a wall at a 45° angle and rebounds with the same speed at the same angle. The focus is on understanding the concept of impulse in relation to momentum change during the collision.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of impulse and its relation to momentum, with initial attempts to compute impulse based on mass and speed. Questions arise regarding the change in momentum, particularly whether the same speed implies no change in momentum. The distinction between speed and velocity is also explored, leading to discussions about the components of momentum in different directions.

Discussion Status

The discussion is active, with participants providing insights into the components of momentum and questioning the assumptions made about speed and direction. Some guidance has been offered regarding the calculation of momentum components, but no consensus has been reached on the final interpretation of impulse.

Contextual Notes

Participants are navigating the complexities of vector components and the implications of directionality in momentum change. The problem setup includes specific angles and speeds, which are critical to the discussion but may not be fully resolved.

amazondog
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Homework Statement


A tennis ball of mass m = 0.087 kg and speed v = 43 m/s strikes a wall at a 45° angle and rebounds with the same speed at 45° (Fig. 7-29). What is the impulse given the wall?



Homework Equations



Impulse = M x V
Sin (theta) = opp/hyp

The Attempt at a Solution



Impulse = .087(43) = 3.741 (at 45 degrees)

Sin(45) = x / 3.741
x = 3.1832 Is the impulse given the wall. Is this corect?
 
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Impulse is change in momentum. Momentum is mass times velocity. What you've found is momentum in the y direction.
 
Last edited:
Is there change in momentum though? Cause its going the same velocity.
 
Same speed or same velocity?
 
same speed, different velocity. But because the angles, speeds, and masses are the same don't they cancel?
 
Which velocities will cancel and which ones won't? You've got two components.
 
So the x and y components are both 30.4. That would be the x component cancels (30.4-30.4) but the y component would double because its going upwards..?
 
Actually it's the other way around. Let's call up and right positive, and say the ball hits the wall traveling to the the right. It hits the wall and continues up
\Delta p_y = mv_fy - mv_iy = m(30.4-30.4) = 0
while in the x direction it rebounds and goes the other way
\Delta p_x = mv_fx - mv_ix = m(30.4-(-30.4))= m*60.8
 

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