Force and Simple Dynamics: Dropped Tennis Ball

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SUMMARY

A 0.3 kg tennis ball dropped from a height of 3.6 m reaches a speed of approximately 8.4 m/s upon contact with the floor, calculated using the formula |v| = sqrt(2*9.81*3.6). During contact with the floor, the ball compresses by 0.6 cm, and the force exerted by the floor can be determined using the work-energy principle, where the work done equals the change in kinetic energy. The average force can be calculated as F = W / 0.006 m, where W represents the work done to bring the ball to rest.

PREREQUISITES
  • Understanding of one-dimensional kinematics with constant acceleration
  • Knowledge of gravitational acceleration (g = 9.81 m/s²)
  • Familiarity with the work-energy principle
  • Basic concepts of force and mass (Newton's Second Law)
NEXT STEPS
  • Calculate the average force exerted by the floor on the tennis ball using F = W / d
  • Explore the relationship between kinetic energy and work done during collisions
  • Learn about impulse and momentum in the context of contact forces
  • Investigate the effects of different materials on the compression and rebound of balls
USEFUL FOR

Physics students, educators, and anyone interested in understanding dynamics, particularly in relation to collisions and energy transfer in sports physics.

Naeem
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A 0.3 kg tennis ball is dropped from rest at a height of 3.6 m onto a hard floor.
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a) What is the speed of the ball at the instant of contact with the floor?
|v| = m/s *
sqrt( 2*9.81*3.6) OK

HELP: This is just a problem in one-dimensional kinematics with constant acceleration. What is the acceleration of the ball while it is in the air?


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A flash photograph shows that the ball is compressed a maximum of 0.6 cm when it strikes the floor.
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b) Assuming that the acceleration of the ball is constant during its contact with the floor, what force does the floor exert on the ball?
|f| = N

HELP: Again, this is a problem in one-dimensional kinematics with constant acceleration. What is the acceleration of the ball when it is in contact with the floor?


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c) Over what time does the force act in bringing the ball to rest?
D t = s

Request some guidance as how to proceed with part b and c.

Thanks,
 
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What is the only force acting on the ball,while in the air.What is the force acting on the ball,while in contact with the ground?

What it the definition of average force?

Daniel.
 
Force due to gravity while in air , and on contact the balls acceleration.

Ft = mv

know m , v ... but don't know time
 
I know part c. once I find part b.
 
You know the kinetic energy of the ball when it is about to hit the ground.

You know that this energy was countered by the ground with a constant force that lasted x = 0.6cm.

You can find the force :)
 
Naeem said:
HELP: This is just a problem in one-dimensional kinematics with constant acceleration. What is the acceleration of the ball while it is in the air?

Well, the force responsible for the moton is gravity, so the acceleration is g

A flash photograph shows that the ball is compressed a maximum of 0.6 cm when it strikes the floor.

W = E_k^{END} - E_k^{BEGINNING}

W is work and in this case equal to 0.6*F and the change in kinetic energy is just the transferred kinetic energy when the ball hits the gorund (this is easy to calculate)...Solve this equation for F. the question really is : do you understand what this formula means physically ? Do you know what is happening ?

c) Over what time does the force act in bringing the ball to rest?

You can find this yourself, like you said

marlon
 
W is work and in this case equal to 0.6*F and the change in kinetic energy is just the transferred kinetic energy when the ball hits the gorund

Careful, 0.6*F isn't the correct equation. The ball was compressed 0.6cm, so 0.006m*F.
 
whozum said:
Careful, 0.6*F isn't the correct equation. The ball was compressed 0.6cm, so 0.006m*F.

Thanks for the correction whozum

marlon
 

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