Conservation of Momentum of a ball

[ximsooriginalx]

Homework Statement

A ball is thrown against a fixed wall where it bounces elastically. The mass of the ball is M and the velocity just before it hits the wall is U. Ignore the force of gravity in this question.
a) Does the Principle of Conservation of Momentum apply to this situation?
b) Obtain an expression for the change in momentum of the ball in terms of M and U

Homework Equations

momentum before = momentum after
change in momentum = m(v-u)
change in momentum= force x time

The Attempt at a Solution

I think that the situation does not follow the principle of conservation of momentum, because the situation is just before the ball hits the wall, so there has not been a transfer of momentum yet, but I am still not sure.

 

[ShayanJ]

The momentum of any system not subject to a net force, is conserved. Here it seems we have such a system!
Note that its regardless of the fact that the wall gets some of the ball's momentum or not.

 

[ximsooriginalx]

Thank you so much! Do you have any idea how to do the second question? I'm guessing that the initial and final velocity will be the same because the ball bounces elastically, but I don't really know.

 

[ShayanJ]

You don't need to write any equations. The wall is fixed and so gets no momentum and momentum is conserved. So its only the ball that can have all the momentum after the collision.

 

[ximsooriginalx]

Thank you!

 

[PeroK]

You don't need to write any equations. The wall is fixed and so gets no momentum and momentum is conserved. So its only the ball that can have all the momentum after the collision.

Are you sure about that?

The question asks about the change in momentum of the ball. So, you don't have to worry about the wall's momentum.

 

[ShayanJ]

Are you sure about that?

The question asks about the change in momentum of the ball. So, you don't have to worry about the wall's momentum.

I don't understand your objection but I'm sure about what I said.

 

[PeroK]

I don't understand your objection but I'm sure about what I said.

The question (part a) is ambiguous, as it's not clear what is meant by "this situation". If you take the system to be the ball and the wall, then momentum is not conserved.

 

[ShayanJ]

The question (part a) is ambiguous, as it's not clear what is meant by "this situation". If you take the system to be the ball and the wall, then momentum is not conserved.

The problem says we should neglect the Earth's gravity!

 

[PeroK]

The problem says we should neglect the Earth's gravity!

Gravity is out of the equation.

If the ball changes direction and the wall does not move (which I how I interpreted the question), then is momentum conserved?

It would be a better question if it asked for an explanation of why momentum is conserved, or not.

 

[Orodruin]

You don't need to write any equations. The wall is fixed and so gets no momentum and momentum is conserved. So its only the ball that can have all the momentum after the collision.

This would mean the ball would continue through the wall.

 

[ShayanJ]

Consider mass m with velocity v hitting a stationary object with mass M. After the collision, the speed of m is u and the speed of M is U. Conservation of momentum and conservation of kinetic energy(collision is elastic), give us:
<<Moderator note: Full expressions removed>>
Which in the limit [itex] M \rightarrow \infty [/itex], give [itex] U=0 [/itex] and [itex] u=-v [/itex].

 

[Orodruin]

The velocity of the heavy object goes to zero. Its momentum goes to 2mv.

 

[PeroK]

Consider mass m with velocity v hitting a stationary object with mass M. After the collision, the speed of m is u and the speed of M is U. Conservation of momentum and conservation of kinetic energy(collision is elastic), give us:
<<Moderator note: Full expressions removed>>
Which in the limit [itex] M \rightarrow \infty [/itex], give [itex] U=0 [/itex] and [itex] u=-v [/itex].

In which case, momentum is only conserved if ##\infty \cdot 0 = 2mv##

 

[ShayanJ]

Yeah...didn't see that coming! What does it mean to have an stationary object with momentum?

 

[Orodruin]

The point is it is not stationary. For every M you have a non-zero velocity, it is just the limit that is zero.

Another observation of interest for this situation is: The final momentum of the wall will increase with its mass.

Edit: You could just as well ask what it means to have an infinitely massive object.

 

[ShayanJ]

The point is it is not stationary. For every M you have a non-zero velocity, it is just the limit that is zero.

Another observation of interest for this situation is: The final momentum of the wall will increase with its mass.

Edit: You could just as well ask what it means to have an infinitely massive object.

Well, this is how I see it: We throw the ball against the wall. But the wall is rigidly attached to Earth and so can only move with earth. This means we're effectively hitting the ball with the Earth itself. Comparing the mass of the ball with the mass of the earth, we can easily see what it means to have an infinitely massive object.
So this all means that when we do such an experiment, the Earth gains a momentum equal to 2mv. But its velocity is 2 (m/M) v which is effectively zero because of the very very very small mass ratio in the formula.

 

[Orodruin]

Yet it is not zero and the Earth mass is not infinite, so the Earth velocity does change although by a very minuscule and undetectable amount. This is the thing with limits. All limits are going to depend on the conditions. If an object has a given momentum, its velocity will naturally be lower if its mass is larger.

However, I suspect the intention of the problem is to consider the ball only, which is subject to an external force and thus momentum non-conservation.

 

[ShayanJ]

Yet it is not zero and the Earth mass is not infinite, so the Earth velocity does change although by a very minuscule and undetectable amount. This is the thing with limits. All limits are going to depend on the conditions. If an object has a given momentum, its velocity will naturally be lower if its mass is larger.

Of course but there is always an amount of largeness or smallness that resembles the limit very well.(Assuming continuity!). That was what I meant.

However, I suspect the intention of the problem is to consider the ball only, which is subject to an external force and thus momentum non-conservation.

The problem tells us to ignore gravity so there is no external force and the momentum is conserved!

 

[Orodruin]

The problem tells us to ignore gravity so there is no external force and the momentum is conserved!

There is a force from the wall on the ball when it bounces, which is when the momentum of the ball is not conserved. The origin of this force is electromagnetic and has nothing to do with gravity.

Edit: I removed the full expressions from post #12 as this is part of the problem that should be solved by the OP.

 

[ShayanJ]

There is a force from the wall on the ball when it bounces, which is when the momentum of the ball is not conserved. The origin of this force is electromagnetic and has nothing to do with gravity.

I was considering ball-wall(better to say ball-earth) system as the system with conserved momentum. Of course if we take the ball alone, its momentum is not conserved. But why should we do that?

 

[Orodruin]

I would say that it is the typical thing this kind of problems refer to. In particular when the wall is considered "fixed".