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newTonn
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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.newTonn said:Can anybody explain how the momentum is conserved in the case of a rubber ball (bounced back),when it was throwned to a wall?
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.Doc Al said: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.
no it is not.ank_gl said: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?
yes can you please explain and please consider the iron ball is (not falling) but moving with uniform velocityank_gl said:is this what you are talking about?
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.)newTonn said: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,
You assume the final momentum of the wall is 0. Wrong! The final momentum of the wall equals 2mv (to an very good approximation).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?
ok ,thanks .agreed.Can you please tell me what is the force exerted by cotton ball on iron ball?ank_gl said: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(please note that Vi_horizontal ll be very small, ie. Vi_horizontal=2McUc/Mi ) and vertical momentum remains unchanged(of the iron ball ofcourse)
rate of change in one second or rate of change during contact period?ank_gl said: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
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.newTonn said: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 ?
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?russ_watters said: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.
newTonn said:is the given datas not enough to find the force?
newTonn said:is the given datas not enough to find the force?
what are the other datas required and how it is contributing to the force?ank_gl said:no it is not:grumpy:
It strains credulity to believe that you could have such trouble with such a simple concept and you have made assertions that deny Newton's 1st law in addition to arguing against definitions, in addition to making repeated simple math errors and insisting the errors are in others' understanding. Going around in circles and refusing to accept Newton's first law and good explanations is why your first thread was closed. You are doing the same thing again.newTonn said: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?
As I stated in my post above, you need to know the initial and final momentum and duration of the collision. From this you can calculate the average force*. Did you not read that part?what are the other datas required and how it is contributing to the force?
russ_watters said:you have made assertions that deny Newton's 1st law.
When a rubber ball is thrown against a wall, it compresses upon impact due to the force of the throw. This compression causes the ball to store potential energy, which is then released as kinetic energy when the ball bounces off the wall. The ball's elasticity allows it to quickly regain its original shape, resulting in a bounce.
The ball's bounce off the wall can be affected by a few factors, including the force of the throw, the angle at which the ball hits the wall, and the surface of the wall. A harder throw, a more perpendicular angle, and a smoother wall surface all tend to result in a higher bounce.
When a rubber ball bounces off a surface, it loses a small amount of energy due to factors such as air resistance and friction. This means that with each bounce, the ball has slightly less energy to rebound with, resulting in a decrease in the height of the bounces over time.
Yes, temperature can affect the ball's bounce off the wall. As the temperature increases, the air inside the ball expands, making the ball less dense. This can result in a lower bounce as there is less air resistance to help the ball regain its shape after compression.
The material of the ball can greatly impact its bounce off the wall. A rubber ball is able to store and release more energy upon impact due to its elasticity, resulting in a higher bounce compared to a ball made of a less elastic material, such as plastic. Additionally, different types of rubber can also affect the bounce, with some being more bouncy than others.