A rubber ball thrown to a wall

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Can anybody explain how the momentum is conserved in the case of a rubber ball (bounced back),when it was throwned to a wall?
 

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  • #2
HallsofIvy
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If you throw the ball at a wall with velocity v, so that its momentum is mv, then, in a perfectly elastic collision with the wall, it bounces back with momentum -mv (in reality, slightly less) and so have momentum change 2mv. That is "balanced" by the change in momentum of molecules in the wall.
 
  • #3
Doc Al
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Can anybody explain how the momentum is conserved in the case of a rubber ball (bounced back),when it was throwned to a wall?
If you chose to look at the ball itself, then momentum is obviously not conserved. Why should it be? After all, there's an external force acting on it.

But if you look at the ball + "wall & attached Earth" as a composite system, then there are no external forces and momentum is conserved. As HallsofIvy explained, the change in the ball's momentum is -2mv (assuming an elastic collision), while the change in momentum of the "wall & attached Earth" must be +2mv.
 
  • #4
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you need to realise that the force is impulsive as well.
Therefore on the ball alone there is basically a force acting. This is resulting in a increase in the momentum of the ball.
 
  • #5
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* decrease
 
  • #6
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obviously you dont observe the earth moving in the other direction with a momentum change of 2mv because of massive mass of earth
 
  • #7
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If you chose to look at the ball itself, then momentum is obviously not conserved. Why should it be? After all, there's an external force acting on it.

But if you look at the ball + "wall & attached Earth" as a composite system, then there are no external forces and momentum is conserved. As HallsofIvy explained, the change in the ball's momentum is -2mv (assuming an elastic collision), while the change in momentum of the "wall & attached Earth" must be +2mv.
let Mass of "wall " is M(consider only gravity is holding the wall in position) ;Initial and final velocity of wall is zero:if mass of ball is m,and let the ball was thrown with a velocity u and let v be the velocity with which the ball bounce back.
substittuting the values in the equation,

m1u1 + m2u2 = m1v1 + m2v2(conservation of momentum)

we will get M*0 + m*u = M*0 + m*v

considering elastic collision, v should be equal to u;but in this case
v = -u
so how the momentum is conserved?
 
  • #8
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who says walls final velocity is 0??
m1u1 + m2u2 = m1v1 + m2v2
v2=m1(u1 - v1)/m2 (here u1 and v1 are in opposite direction, so they ll add up, remember vector addition??)
now put m2 = mass of earth and get v2, is it even noticable???
 
  • #9
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elastic collision means the conservation of energy and momentum. it does not say that initial and final velocities are same
 
  • #10
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who says walls final velocity is 0??
m1u1 + m2u2 = m1v1 + m2v2
v2=m1(u1 - v1)/m2 (here u1 and v1 are in opposite direction, so they ll add up, remember vector addition??)
now put m2 = mass of earth and get v2, is it even noticable???
no it is not.
Ok let us try to get rid of this earth.
A huge iron ball is falling vertically down.a cotton ball of same volume hit it horizontally with a velocity u and bounce back with a velocity -u.
can you please explain the momentum conservation in this case
 
  • #11
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is this what you are talking about????:confused::confused:
 

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  • #12
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is this what you are talking about????:confused::confused:
yes can you please explain and please consider the iron ball is (not falling) but moving with uniform velocity
 
  • #13
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ok if you consider that the cotton ball strikes head on with the heavy ball, n bounces back along its initial line of motion withits change in momentum = 2McUc.
the heavy iron ball ll alter its line of motion. it ll now move at an angle to the vertical direction so that its horizontal component is 2McUc(plz note that Vi_horizontal ll be very small, ie. Vi_horizontal=2McUc/Mi ) and vertical momentum remains unchanged(of the iron ball ofcourse)
 

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  • #14
Doc Al
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let Mass of "wall " is M(consider only gravity is holding the wall in position) ;Initial and final velocity of wall is zero:if mass of ball is m,and let the ball was thrown with a velocity u and let v be the velocity with which the ball bounce back.
substittuting the values in the equation,
OK. Assume the heavy wall is resting on frictionless ice, if you like. (Gravity doesn't hold the wall in position--the ice does; but there's no horizontal force from the ground on it, unlike the actual case where the wall is attached to something. No big deal.)

The final velocity of the wall is not exactly zero, just close to it. But realize that a huge mass M multiplied by a small speed is still a finite momentum.

m1u1 + m2u2 = m1v1 + m2v2(conservation of momentum)

we will get M*0 + m*u = M*0 + m*v

considering elastic collision, v should be equal to u;but in this case
v = -u
so how the momentum is conserved?
You assume the final momentum of the wall is 0. Wrong! The final momentum of the wall equals 2mv (to an very good approximation).

Try this: Apply both conservation of energy and momentum and solve for the final speed of the wall.
 
  • #15
rcgldr
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The wall is effectively connected to the earth. The final velocity of the wall and earth will be changed very little by a bouncing ball.
 
  • #16
russ_watters
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If at all - the energy has to transfer through the wall to the floor and due to the large mass and inelasticity of the wall, most of the energy would likely just be absorbed and converted to heat at the impact site....

....none of which in any way contradicts the laws of motion.
 
  • #17
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ok if you consider that the cotton ball strikes head on with the heavy ball, n bounces back along its initial line of motion withits change in momentum = 2McUc.
the heavy iron ball ll alter its line of motion. it ll now move at an angle to the vertical direction so that its horizontal component is 2McUc(plz note that Vi_horizontal ll be very small, ie. Vi_horizontal=2McUc/Mi ) and vertical momentum remains unchanged(of the iron ball ofcourse)
ok ,thanks .agreed.Can you please tell me what is the force exerted by cotton ball on iron ball?
 
  • #18
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duh.... force is the rate of change of momentum.
do you see momentum of cotton ball changing???
if yes, that is the force on the iron ball
 
  • #19
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duh.... force is the rate of change of momentum.
do you see momentum of cotton ball changing???
if yes, that is the force on the iron ball
rate of change in one second or rate of change during contact period?
If contact period,how can i get the contact period(just assumption or?)?
what is the time interval ,in which the cotton ball ,attained its final velocity ?
 
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  • #20
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well obviously it is the rate of change during the contact period. and it depends upon the physical properties of the ball like the extent of deformation they undergo when they collide(elasticity)[but that is a complex case, because if you consider that part, you ll also have to take into account the energy lost in the deformation], for simplicity and considering it as an ideal case, the reversal of velocity is considered instantaneous
 
  • #21
russ_watters
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rate of change in one second or rate of change during contact period?
If contact period,how can i get the contact period(just assumption or?)?
what is the time interval ,in which the cotton ball ,attained its final velocity ?
In this example, it is very difficult to answer those questions because it is a relatively difficult hypothetical (a piece of cotton deforms easily and does not necessarily return to its original shape in a collision). When you can't even understand a simple scenario, intentionally making more complex ones is not going to help you gain an understanding of the issue.

In your previous thread on exactly the same subject, which was locked because you appear not to be making an effort to learn, an example was given in which it was very easy to actually measure the contact period with a high speed camera: a golf club striking a golf ball.

newTonn, we're covering ground already covered here. You are making the staff wonder if you are really making a serious effort to learn or just trolling us with the same simple questions over and over again, asked in different ways to see how long we'll entertain them. Please improve your effort here. Our patience is not infinite.

Since your problem appears to be a flat refuslal to accept a very simple physical reality (learned in minutes by millions of teenagers every year), I don't believe any amount of explaining the same thing over and over is going to convince you. My advice to you, if you really want to learn, is to get yourself a golf ball and perform some tests and calculations on it to verify for yourself what we have been telling you. All of the data required is already out there (in the previous thread or on the net), but you can do things like piling a lot of weight on the ball to see just how much force is required to deform it by the amount specified in the thread. Then you can apply that to the motion equations (with the contact time estimated and mass of the ball known) and calculate for yourself how fast the ball is moving when it leaves the club face.

On the fact that an object in motion requires no outside force to stay in motion, this should be intuitive. If it isn't, consider some examples of the phenomena, such as objects traveling through space or large pendulums (which transfer energy from kinetic to potential and back thousands of times with little energy loss). These examples should cause immediate acceptance of this physical reality.
 
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  • #22
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In this example, it is very difficult to answer those questions because it is a relatively difficult hypothetical (a piece of cotton deforms easily and does not necessarily return to its original shape in a collision). When you can't even understand a simple scenario, intentionally making more complex ones is not going to help you gain an understanding of the issue.

In your previous thread on exactly the same subject, which was locked because you appear not to be making an effort to learn, an example was given in which it was very easy to actually measure the contact period with a high speed camera: a golf club striking a golf ball.

newTonn, we're covering ground already covered here. You are making the staff wonder if you are really making a serious effort to learn or just trolling us with the same simple questions over and over again, asked in different ways to see how long we'll entertain them. Please improve your effort here. Our patience is not infinite.

Since your problem appears to be a flat refuslal to accept a very simple physical reality (learned in minutes by millions of teenagers every year), I don't believe any amount of explaining the same thing over and over is going to convince you. My advice to you, if you really want to learn, is to get yourself a golf ball and perform some tests and calculations on it to verify for yourself what we have been telling you. All of the data required is already out there (in the previous thread or on the net), but you can do things like piling a lot of weight on the ball to see just how much force is required to deform it by the amount specified in the thread. Then you can apply that to the motion equations (with the contact time estimated and mass of the ball known) and calculate for yourself how fast the ball is moving when it leaves the club face.

On the fact that an object in motion requires no outside force to stay in motion, this should be intuitive. If it isn't, consider some examples of the phenomena, such as objects traveling through space or large pendulums (which transfer energy from kinetic to potential and back thousands of times with little energy loss). These examples should cause immediate acceptance of this physical reality.
this reply is a prejudiced one.In this thread ,i never denied anything.I just want to know what are the factors controlling the contact period.Or in other words,is the given datas not enough to find the force?
 
  • #23
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ok from next time, be clear what you are asking and be ready to accept concepts or rather understand them.
hmmm.. contact period depends on the physical structure. try colliding two steel balls and then two rubber balls, you ll see the difference.
actually what you are trying to ask(as i m getting it) is the practical sequence of the events in TWO BALLS COLLIDING event. i guess this is your question. well that is a very complex chain of events if you want to see it microscopically but with no real advantages. and in such situation you ll have to undertake many othe other concepts to solve this problem along with the principle(or principal, duhhh... i always get confused in this) of conservation of momentum and energy

and i saw this in someone's signature
The whole problem with the world is that fools and fanatics are always so certain of themselves, and wiser people so full of doubts.--Bertrand Russell
so dont be sad:)
 
  • #24
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is the given datas not enough to find the force?

no it is not
 
  • #25
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is the given datas not enough to find the force?

no it is not:grumpy:
 

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