Calculating Average Force on a Tennis Ball During a Serve

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Homework Help Overview

The discussion revolves around calculating the average force exerted on a tennis ball during a serve, using the ball's mass, speed, and contact time with the racket. The subject area includes concepts from mechanics and momentum.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the application of momentum and impulse concepts, questioning the calculations involved in determining the average force. Some participants express uncertainty about their results and seek clarification on the correct approach.

Discussion Status

There is ongoing exploration of the calculations related to the average force on the tennis ball. Some participants have provided hints and suggestions for applying the momentum-impulse theorem, while others are questioning the validity of their calculations and assumptions.

Contextual Notes

Participants discuss unit conversions and the implications of the calculated force in relation to lifting a specified weight, indicating a need for clarity on the definitions and relationships involved in the problem.

Rapta3
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How do you slove for Mb from this equation (MaVa+MbVb)/Ma+Mb=Vab
 
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HINT: Multiply both sides by [itex]M_a + M_b[/itex]
 
K got it, ty, got another one for you tho.

This one is HARD.

A 13 g bullet traveling 218 m/s penetrates a 2.0 kg block of wood and emerges going 161 m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?
 
Try Conservation of Momentum.
 
Change the units into kg first...
13g = 0.013kg

so...

[tex] 0.013(218) = 0.013(161) + 2v[/tex]

..

[tex] v = \frac{0.013(218-161)}{2} = 0.3705 m/s[/tex]
 
Yea converting the g to kg was the missing step, thanks.
 
No problems - I used to make careless mistakes like that all the time.
 
A tennis ball may leave the racket of a top player on the serve with a speed of 62.0 m/s. If the ball's mass is 0.0600 kg and it is in contact with the racket for 0.0200 s, what is the average force on the ball?

Would this force be large enough to lift a 60 kg person?
large enough or
not large enough

I got .0600x62=3.72

.0200/3.72=.00537

I don't think that is right because that seems way to small but I don't know what I am missing...if it is incorrec that is.
 
Try momentum-impluse theorem:
[tex]F\Delta t = m\Delta v[/tex]
 
  • #10
Rapta3 said:
A tennis ball may leave the racket of a top player on the serve with a speed of 62.0 m/s. If the ball's mass is 0.0600 kg and it is in contact with the racket for 0.0200 s, what is the average force on the ball?

Would this force be large enough to lift a 60 kg person?
large enough or
not large enough

I got .0600x62=3.72

.0200/3.72=.00537

I don't think that is right because that seems way to small but I don't know what I am missing...if it is incorrec that is.

You're last calculation is wrong.As smb else said,apply the definition of force wrt to changes in momentum (the integrated form of Newton's second law) correctly.
You should be getting round about 19 kgf which would not be enough to lift 60kg off the ground,since the force needed to do that is obviously 60kgf.
Divagation ( culture useful):
The greatest tennis player ever (to me ond not only,though he didn't win Rolland Garros),Pete Sampras (i don't believe the bull**** they say about Federer),used to have the strings of his racket streched to a tension of 34kgf.It was the most "tensioned" racket in the history of the sport.And even so,it broke frequently.He was famous for breaking an impressive number of rackets (strings) during serve,which meant that (at a maximum speed of 62m/s -he never served more than 220 kmph/137mph) for light balls (the ones used on grass) the time of impact between the racket and the ball was at the order of microseconds.Amazing,really!That's the reason for those huge and perfect forehands crosscourt and down the line.Alongside genious,of course.
I almost cried when he retired. :cry:
 
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