Ball Free Fall Homework: Speed & Energy Calculations

AI Thread Summary
The discussion revolves around a physics homework problem involving a ball dropped from a height of 2.5 m that bounces back to 1.4 m, with air resistance considered. The key calculations involve determining the speed of the ball upon impact and the energy lost during the collision. The initial gravitational potential energy (GPE) converts to kinetic energy (KE) as the ball falls, but not all energy is recovered in the bounce due to losses, which are attributed to factors like air resistance and inelastic collisions. The final height of 1.4 m is crucial for calculating the energy absorbed during the impact, leading to a conclusion that a percentage of energy (44%) is lost. The discussion emphasizes using gravitational potential energy for these calculations rather than kinetic energy formulas.
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Homework Statement


When the ball dropped from a height of 2.5 m on the floor, it bounces back to 1.4 m in height.
Air resistance is taken into account. g= 9.81 m/s2

Homework Equations


a) the speed at which the ball hits the floor?
b) How much of the energy consumed in collision with a change of direction?

The Attempt at a Solution


a)
V2-V02=2gh
V=\sqrt{}(2gh)
V=\sqrt{}2*9.81*2.5
V=7m/s
i'm wondering why they gave us the information that the ball will bounce 1.4m high...

b)
E=1/2*m*V2
we don't have the mass of the ball
 
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in responce to your question about why they gave you the final height of the ball..

At the begiing you start with a certain amount of energy, it could be sound, could be heat could be anything. In this situation it is gravitational potential energy which is equated as m*g*h. because the ball wants to go down, but you are holding it up.

The ball is then let go and falls to the ground. Energy is always conserved so all this energy must turn into Kinetic Energy jusssttt as it hits the ground.

let me explain:
KE = 0.5*m*v^2 and GPE = m*g*h

as you let go of the ball, h gets smaller and smaller. so GPE gets smaller
ALSO the ball gets faster and gaster, so KE gets faster.

And it just so happens, that the initial GPE = final KE when the ball touches the ground

Ok so the ball touches the ground, you have no GPE, only KE. Obviously the ball can not move faster, so it hits the ground and slows down dramatically. All this KE turns into ELASTIC POTENTIAL ENERGY because obviously the ball has some elasticity to it. so the ball squishes until it can't squish anymore than BAM it un squishes itself and shooots back into the air.

now if there where no losses or work done by the system, the ball should return to its original height. However it didnt!
therefore if you compare the initial GPE at 2.5m and the final GPE 1.4m, obviously they won't match up as they have different heights. they should be the same, but they arent cause all the enregy that would be used to push the ball back to its original height has been lost somewhere along the system :)
 
You can use the height at which the ball bounced back up, 1.4m, to determine the answer to part b, you don't need the mass, the mass will cancel out, simply compare the energy of the ball at the start of its free fall and then the energy immediately as it hits the ground and begins to bounce back upward and you'll see the change in energy must be due to energy absorbed in the collision, if there's no air resistance.

Hint: don't use the kinetic energy formula, (1/2)mv2 for this, it's much easier to compare gravitational potential energies at the very start and very end: mgh. A lot will cancel out.
 
faymalaka said:
in responce to your question about why they gave you the final height of the ball..

At the begiing you start with a certain amount of energy, it could be sound, could be heat could be anything. In this situation it is gravitational potential energy which is equated as m*g*h. because the ball wants to go down, but you are holding it up.

The ball is then let go and falls to the ground. Energy is always conserved so all this energy must turn into Kinetic Energy jusssttt as it hits the ground.

let me explain:
KE = 0.5*m*v^2 and GPE = m*g*h

as you let go of the ball, h gets smaller and smaller. so GPE gets smaller
ALSO the ball gets faster and gaster, so KE gets faster.

And it just so happens, that the initial GPE = final KE when the ball touches the ground

Ok so the ball touches the ground, you have no GPE, only KE. Obviously the ball can not move faster, so it hits the ground and slows down dramatically. All this KE turns into ELASTIC POTENTIAL ENERGY because obviously the ball has some elasticity to it. so the ball squishes until it can't squish anymore than BAM it un squishes itself and shooots back into the air.

now if there where no losses or work done by the system, the ball should return to its original height. However it didnt!
therefore if you compare the initial GPE at 2.5m and the final GPE 1.4m, obviously they won't match up as they have different heights. they should be the same, but they arent cause all the enregy that would be used to push the ball back to its original height has been lost somewhere along the system :)

Great explanation, thank you :)
 
soothsayer said:
You can use the height at which the ball bounced back up, 1.4m, to determine the answer to part b, you don't need the mass, the mass will cancel out, simply compare the energy of the ball at the start of its free fall and then the energy immediately as it hits the ground and begins to bounce back upward and you'll see the change in energy must be due to energy absorbed in the collision, if there's no air resistance.

Hint: don't use the kinetic energy formula, (1/2)mv2 for this, it's much easier to compare gravitational potential energies at the very start and very end: mgh. A lot will cancel out.

Actually they said that the air resistance is to take account of it...
I don't get what they ask in part B, the result should 44%
and if we compare (substract) Energies at different heights, we won't get that 44%...
However i could find that 44% but still don't get what they ment..
 
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