Does the direction of the incoming ball affect the resulting exit velocity?

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The discussion centers on the effect of the incoming ball's direction on its exit velocity when struck by a tennis racket. Peter G. calculated the exit velocities for a 0.07 kg tennis ball under two scenarios: one where the ball is at rest and another where it approaches at -10 m/s. The calculations yielded an exit velocity of 40 m/s from rest and 30 m/s when struck while moving towards the player. The conversation highlights the importance of reference frames in velocity calculations and notes that the elastic properties of the tennis ball can further complicate the analysis.

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Peter G.
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Hi,

I was trying to figure whether a ball would have a larger exiting velocity if it started from rest or if it was coming towards me at a given velocity.

My first thought is that it would have a greater exiting velocity if it started at rest. When I hit an incoming ball it will change direction and, since momentum is a vector, I believed the exiting speed would be smaller.

I tried using some numbers from an old exercise question:

0.07 kg Ball
Time of Contact between Ball and Tennis Racket: 0.1 second
Maximum force exerted by Tennis Player 28 N
Incoming Speed: -10 m/s

The change in momentum the Tennis player can produce: F x Δt = 2.8 kgms

(I set the incoming velocity as negative)

2.8 = 0.07v - (0.07 * (-10))
2.8 = 0.07v + 0.7
v = 30 ms

Now, if the ball started at rest:

2.8 = 0.07v - 0
v = 40 ms

But if I set the incoming velocity as positive for example:

2.8 = (0.07 * -v) - 0.7
3.5 = (0.07 * -v)
v = - 50 ms

And this would change the whole thing!

I am probably making a stupid mistake somewhere, I am confused :confused:

Thanks,
Peter G.
 
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Peter G. said:
Hi,

I was trying to figure whether a ball would have a larger exiting velocity if it started from rest or if it was coming towards me at a given velocity.

My first thought is that it would have a greater exiting velocity if it started at rest. When I hit an incoming ball it will change direction and, since momentum is a vector, I believed the exiting speed would be smaller.

I tried using some numbers from an old exercise question:

0.07 kg Ball
Time of Contact between Ball and Tennis Racket: 0.1 second
Maximum force exerted by Tennis Player 28 N
Incoming Speed: -10 m/s

The change in momentum the Tennis player can produce: F x Δt = 2.8 kgms

(I set the incoming velocity as negative)

2.8 = 0.07v - (0.07 * (-10))
2.8 = 0.07v + 0.7
v = 30 ms

Now, if the ball started at rest:

2.8 = 0.07v - 0
v = 40 ms

But if I set the incoming velocity as positive for example:

2.8 = (0.07 * -v) - 0.7
3.5 = (0.07 * -v)
v = - 50 ms


And this would change the whole thing!

I am probably making a stupid mistake somewhere, I am confused :confused:

Thanks,
Peter G.

Using the reference frame for the velocity that you are using in the first equation, a positive velocity would imply the ball is moving away from you and negative means its coming towards you.

In the bold equation you have effective reversed this reference frame so a -ve velocity is moving away from you and +ve is coming towards you.

Also a tennis ball is an elastic collision so it makes it more difficult to calculate the true value as this can depend on the temperature, string tension, age of the ball etc. as these will change the elasticity of the tennis ball.

AL
 
Sorry, I am still a bit confused. I thought the +ve and -ve were just a convention, can it alter the result? So, even if we use the same ball we can't determine in which scenario it would come out faster?
 

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