Momentum and Exit Velocity

1. Jan 6, 2012

Peter G.

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

Thanks,
Peter G.

2. Jan 6, 2012

rollcast

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

3. Jan 6, 2012

Peter G.

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?