Impulse of a Ball - Greatest Change in Momentum

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The discussion centers on the impulse of a tennis ball when thrown against a wall, with a focus on which scenario—throwing, bouncing, or catching—results in the greatest change in momentum. It is concluded that the bounce generates the greatest change in momentum, calculated as -2m*v, since the ball reverses direction upon impact. In contrast, when the ball is caught, the change in momentum is -m*v, and when thrown, it is m*v, both of which are smaller than the bounce scenario. The reasoning remains valid even if kinetic energy is lost during the interactions. Overall, the bounce is confirmed as having the greatest change in momentum.
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Homework Statement



If a tennis player were to throw his ball against a wall, which impulse would be the greatest? The throw, the bounce, or the catch?

Homework Equations



I = F * t
P = m * v

The Attempt at a Solution



I believe the bounce is the greatest, as it has the greatest change in momentum, but I can't explain my thought...
 
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yes you are right. bounce has the greatest change in momentum. let's say that velocity before the impact is \vec{v} , then velocity after the bounce would be
-\vec{v}. so initial momentum is m\vec{v} and final momentum is
-m\vec{v}. so the change in momentum would be -m\vec{v}-m\vec{v} which is -2m\vec{v}. but when the ball is caught, final momentum is zero, so the change in momentum is 0-m\vec{v} which is -m\vec{v}
same argument can be made for the case when the the ball is thrown.

Edit: I am assuming above that the kinetic energy is not lost. Even if the kinetic energy is lost, the reasoning is not affected.
 
IssacNewton said:
yes you are right. bounce has the greatest change in momentum. let's say that velocity before the impact is \vec{v} , then velocity after the bounce would be
-\vec{v}. so initial momentum is m\vec{v} and final momentum is
-m\vec{v}. so the change in momentum would be -m\vec{v}-m\vec{v} which is -2m\vec{v}. but when the ball is caught, final momentum is zero, so the change in momentum is 0-m\vec{v} which is -m\vec{v}
same argument can be made for the case when the the ball is thrown.

Edit: I am assuming above that the kinetic energy is not lost. Even if the kinetic energy is lost, the reasoning is not affected.

Hey! Thanks for the reply! It makes sense, and I assume when the ball is thrown it is the reverse? The initial momentum is zero?

Thank you!
 
yes, when the ball is thrown, the initial momentum is zero. and final momentum is
m\vec{v} so the change in momentum is m\vec{v} which is a smaller change in absolute terms than in the case of bouncing ball.
 
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