The Juggler Problem

[essentialsaltes]

Really?
A one line line response lacking any sort of detail is clear enough you have nothing to offer in refuting the maths.

It doesn't take much effort to see that the situation described is not that of a perfectly inelastic collision where the two bodies stick together. Nor does my analysis make any such assumption, so you bringing it up is a non sequitur.

 

[sjastro]

The elastic ball, explained.

We hold a perfectly elastic ball about 1.25 meters above the surface of the earth. It is at rest. As is the earth.

We gently release the ball. At the moment of release it is still motionless, as is the earth.

The initial momentum of the ball is zero.

It begins to fall. Its momentum is no longer zero.

The change in its momentum is nonzero.

Its momentum is not conserved.

Nothing could be so obvious.

But, as we know, momentum is conserved in an isolated system. How are we to explain this? Obviously, the ball is not isolated. It is being acted on by an external force. There is gravity. The earth.

As sjastro has correctly described but incorrectly calculated, if we treat the earth as part of the system, then there are only internal forces, and momentum is conserved. But we must actually treat the earth as part of the system. If there is an internal force of the earth on the ball, then by Newton's Third Law, there is an internal force of the ball on the earth. And they are equal and opposite.

Consider the ball (of mass m) after it has fallen for a time t

The force acting on it is the force of gravity = -mg (I shall take up to be positive)

Its acceleration obviously is -g

Its velocity is -gt

Its momentum is -mgt

Now the earth (of mass M) falls up to meet it.

The force on it is equal and opposite to the force on the ball, so that is +mg

To calculate the acceleration of the earth a(E)

F = Ma(E) = mg

Therefore a(E) = (m/M)g

and v(E) = (m/M)gt

and the momentum of the earth is Mv(E) = M * (m/M)gt = mgt

Miraculously, the momentum of the earth is equal and opposite to the momentum of the falling ball at every moment during the fall. Thus the total momentum of the system is zero. Exactly the same as when we released the ball, when both the ball and the earth were at rest.

After about 0.5 s (let's call that moment t=T) the ball hits the earth and collides elastically. For both momentum and KE to be conserved (as in an elastic collision), it is obvious that both the ball and the earth flip the sign of their respective velocities/momenta.

Before the collision:
v(ball) = -gT
v(E) = (m/M)gT

Afterwards:
v(ball) = +gT
v(E) = -(m/M)gT

Now, this situation is (as sjastro again correctly points out, but misunderstands) very like the juggler situation. The difference between the initial and final situations is that the velocity of the ball/pin has changed direction (and therefore so has the momentum).

sjastro focuses only on the pin, or ball, and somehow suggests that the momentum has not changed, when in fact it has changed by twice its amount.

initial momentum of the ball is -mgT
final momentum of the ball is +mgT

Change in momentum is 2mgT

This is no cause for concern, because we need to account for the earth. Its momentum flips from mgT to -mgT, changing by -2mgT. So the total change of the system is zero. Momentum is conserved and there is joy throughout the universe.

Now imagine if we had chosen to release the ball from a height of 0.6 meters or 1 meter. It would take longer to fall to the earth, and would reach a greater velocity. And the same applies to the earth. To conserve momentum, it too would reach a higher speed. When they collide, there would be a greater change in the momentum of each (though the total as we see is always zero).

That change in momentum is the impulse described up-thread, which is related to the force between the two objects. In the one case, the force of contact between the ball and the ground. In the other, the force between the juggler's hand and the pin (which is then transmitted to the bridge via the juggler's feet). Choosing to throw the pin higher than the previous throw can be achieved by increasing the amount of force used, which will then be transmitted to the bridge. Since the juggler has control over the amount of force applied, the amount of force transmitted to the bridge is not fixed, as implied by choice (E).

Again, it's simple to think about this in terms of recoil. Throwing a pin up is like firing a bullet up. If you fire a bullet up with a greater velocity, you're giving it more momentum, and there will be a greater recoil. If you throw the pin up with greater velocity, there will be a greater recoil force directed into the bridge.

The only part of your post worth commenting on is your explanation in trying to justify vector subtraction instead of addition.
All your other comments have been refuted in previous posts and are not going to be repeated.

Here is the correct method after which I will show two fatal flaws in your explanation.
Since the process involves an elastic collision both KE and momentum are conserved.
Before collision:

Initial KE = 0.5mv² + 0.5MVi²

m is the mass of the pin and v its velocity, M is the mass of the Earth and Vi its velocity.
The Earth is stationary the instant before collision since the KE of the ball (=0.5mv²) reaches a maximum value, therefore Vi=0.
Hence:

Initial KE = 0.5mv².

Immediately after collision:

Final KE = 0.5mv² + 0.5MVf² where Vf is the velocity of the Earth after collision.

Since KE is conserved, initial KE = final KE.

0.5mv² = 0.5mv² + 0.5MVf²

This can only be true if Vf = 0.

The conservation of momentum can be proved using the impulse-momentum equation and Newton’s third law.
The impulse J1 applied on the ball due to the Earth = (mv – mu) where at this stage v and u are unknown and represent velocities before and after collision.
The impulse J2 applied on the Earth due to the ball = (MVf – MVi)

Since Vi = 0 requires Vf =0 for KE to be conserved the equation reduces to:

J2 = (MVf – MVi)=0

By Newton’s third law J2=-J1

Therefore (mv –mu) =0

This can only be true if u=v, e.g. (mv –mv) = 0 or in vector addition form mv + m(-v) = 0.
Note the difference (mv – mv) =0, not 2mv (or -2mv).

This method also addresses your ongoing lying that I did not consider the Earth’s momentum.
Since Vi = Vf = 0, the Earth’s momentum is zero during an elastic collision and therefore does not turn up in the conservation equations.

As for the fatal flaws in your explanation in trying to justify vector subtraction………….

After about 0.5 s (let's call that moment t=T) the ball hits the earth and collides elastically. For both momentum and KE to be conserved (as in an elastic collision), it is obvious that both the ball and the earth flip the sign of their respective velocities/momenta.

The first fatal flaw is this statement is nonsensical.
For the Earth’s momentum to flip during the collision process the velocity goes from an initial value of V≠0 to –V.
This contradicts the fact the Earth is stationary immediately before collision where the initial velocity Vi=0.
Your explanation is physically impossible.
Even for the scenario of a stationary Earth set in motion by a colliding ball contradicts an elastic collision since the Earth’s initial KE is zero in which case the ball must lose KE.

The second fatal flaw as pointed out, since the Earth’s momentum is zero, the difference between the initial and final momentum of the ball cannot be 2mv or (-2mv) otherwise momentum is not conserved.

If you had used the conservation equations you would have avoided the pitfalls you fell into.

 

[sjastro]

It doesn't take much effort to see that the situation described is not that of a perfectly inelastic collision where the two bodies stick together.
[

Another example of your dishonesty.
If it didn’t “take much effort” then why not explain it in detail?
Given what you served up in your last post where you were blissfully unaware that your explanation was physically impossible, the collision could not be elastic, and momentum could not be conserved, is a sure sign you are not capable of doing so.

Nor does my analysis make any such assumption, so you bringing it up is a non sequitur.

Of course you didn’t make any such assumption as you were clueless in not realizing your “analysis” was describing a 100% inelastic process in the first place.

It is pointless continuing this thread as your botched explanation of the elastic ball was the last nail in the coffin plus your ongoing intellectual dishonesty and personal attacks is getting rather tiring.

 

[essentialsaltes]

The impulse J1 applied on the ball due to the Earth = (mv – mu) where at this stage v and u are unknown and represent velocities before and after collision.

But we agree u = -v. As you agree up is not the same as down.

This can only be true if u=v

But even you know better. the velocity after the collision is not the same as before. One is up and one is down. It seems you've made a mistake.

This method also addresses your ongoing lying that I did not consider the Earth’s momentum.
Since Vi = Vf = 0, the Earth’s momentum is zero during an elastic collision and therefore does not turn up in the conservation equations.

Hilarious. You consider it by declaring it 0 by fiat both before and after. This is why you fail.

As for the fatal flaws in your explanation in trying to justify vector subtraction………….

I'm using the same formula as you -- delta p = (mv – mu). And you have agreed that u = -v

The first fatal flaw is this statement is nonsensical.
For the Earth’s momentum to flip during the collision process the velocity goes from an initial value of V≠0 to –V.
This contradicts the fact the Earth is stationary immediately before collision where the initial velocity Vi=0.

This is not a fact. This is a result of your (poor) choice of coordinate frame. These problems are much easier in the Center-of-mass frame. Which is why I framed the problem as I did with the ball and earth at rest at the start. So that as the ball falls, the earth also moves (slightly) to meet it. With the center of mass at rest, the problem is very simple.

In your set-up with the earth at rest at the time of the impact, then the earth will get knocked backward, and Vf will most certainly not be 0. Your bizarrely geocentric assumption that the earth is immovable explains your silly result that u = v. This is the trivial case where the ball misses the earth and does not hit it. It and the earth maintain their velocities.

Even for the scenario of a stationary Earth set in motion by a colliding ball contradicts an elastic collision since the Earth’s initial KE is zero in which case the ball must lose KE.

In the COM reference frame, the earth's initial KE is not zero.

 

[essentialsaltes]

The impulse J1 applied on the ball due to the Earth = (mv – mu) where at this stage v and u are unknown and represent velocities before and after collision.
The impulse J2 applied on the Earth due to the ball = (MVf – MVi)

It is refreshing that you now agree that the impulse (aka the change in momentum) is obtained by subtraction. Given your agreement that up is different from down, that should settle the issue of whether the impulse on the ball is zero or not.

 

[essentialsaltes]

Since sjastro prefers the frame of reference where the earth is at rest at the moment of collision, here is the analysis in that frame.

initial velocity of the ball = v
initial velocity of the earth = 0

In sjastro's treatment of the problem, the final velocities are:
"This can only be true if u=v"
"Since Vi = Vf = 0"

The ball continues moving down with constant velocity. The earth continues not moving. As I described before, this is the trivial solution to the equations where there is actually no collision at all. Somehow the ball has missed the earth and continued its straight-line constant speed motion.

sjastro's objection to my analysis "Even for the scenario of a stationary Earth set in motion by a colliding ball contradicts an elastic collision since the Earth’s initial KE is zero in which case the ball must lose KE." does not apply to my calculation in the center of mass frame. However, it does to the problem in this new frame where the earth is initially at rest. And, of course, we can calculate the loss in speed of the ball.

The constraints on the problem from conservation of KE (which involves v^2) is a quadratic. Thus it produces two solutions. The trivial one where nothing happens. And the actual one.

the final velocity of the ball is neither v nor -v, but rather = (m-M)/(m+M)v
the final velocity of the earth is 2mv/(m+M)

For convenient shorthand, let Q = (m+M)

Momentum of ball = m(m-M)v/Q = (m^2 - mM)v/Q
Momentum of earth = M(2mv)/Q = 2mMv/Q

Total momentum at end = [(m^2 - mM) + 2mM]v/Q = [m^2 + mM]v/Q = m(m+M)v/Q = mQv/Q = mv

Momentum is conserved.

KE of ball = 1/2(v/Q)^2 * m(m-M)^2 = 1/2(v/Q)^2 * m(m^2 - 2mM + M^2) = 1/2(v/Q)^2
KE of Earth = 1/2(v/Q)^2 * M(2m)^2 = 1/2(v/Q)^2 * m(4mM)

Total KE = 1/2(v/Q)^2 * [m(m^2 - 2mM + M^2) + m(4mM)]

= 1/2(v/Q)^2 * m * (m^2 - 2mM + M^2 + 4mM)
= 1/2(v/Q)^2 * m * (m^2 + 4mM + M^2)
= 1/2(v/Q)^2 * m * (m + M)^2
= 1/2(v/Q)^2 * m * Q^2
= 1/2*v^2 * m
= 1/2mv^2

KE is conserved.

Obviously, this is more complicated than doing it in the COM system, but the two are related, because the velocity of the center of mass is mv/(m+M)

If we 'shift over' by that amount, then the initial velocity of the earth is -mv/Q, and the final velocity is +mv/Q. It swaps signs as I indicate above in that analysis. The same for the ball itself. [the 'velocity' is no longer v because the two situations are described slightly differently. In one case v is the total relative velocity between the two objects, but in the center of mass v is the velocity of the ball, while the earth also has a velocity as it rushes up to hit the ball.]

 

[sjastro]

Since sjastro prefers the frame of reference where the earth is at rest at the moment of collision, here is the analysis in that frame.

initial velocity of the ball = v
initial velocity of the earth = 0

In sjastro's treatment of the problem, the final velocities are:
"This can only be true if u=v"
"Since Vi = Vf = 0"

The ball continues moving down with constant velocity. The earth continues not moving. As I described before, this is the trivial solution to the equations where there is actually no collision at all. Somehow the ball has missed the earth and continued its straight-line constant speed motion.

sjastro's objection to my analysis "Even for the scenario of a stationary Earth set in motion by a colliding ball contradicts an elastic collision since the Earth’s initial KE is zero in which case the ball must lose KE." does not apply to my calculation in the center of mass frame. However, it does to the problem in this new frame where the earth is initially at rest. And, of course, we can calculate the loss in speed of the ball.

The constraints on the problem from conservation of KE (which involves v^2) is a quadratic. Thus it produces two solutions. The trivial one where nothing happens. And the actual one.

the final velocity of the ball is neither v nor -v, but rather = (m-M)/(m+M)v
the final velocity of the earth is 2mv/(m+M)

For convenient shorthand, let Q = (m+M)

Momentum of ball = m(m-M)v/Q = (m^2 - mM)v/Q
Momentum of earth = M(2mv)/Q = 2mMv/Q

Total momentum at end = [(m^2 - mM) + 2mM]v/Q = [m^2 + mM]v/Q = m(m+M)v/Q = mQv/Q = mv

Momentum is conserved.

KE of ball = 1/2(v/Q)^2 * m(m-M)^2 = 1/2(v/Q)^2 * m(m^2 - 2mM + M^2) = 1/2(v/Q)^2
KE of Earth = 1/2(v/Q)^2 * M(2m)^2 = 1/2(v/Q)^2 * m(4mM)

Total KE = 1/2(v/Q)^2 * [m(m^2 - 2mM + M^2) + m(4mM)]

= 1/2(v/Q)^2 * m * (m^2 - 2mM + M^2 + 4mM)
= 1/2(v/Q)^2 * m * (m^2 + 4mM + M^2)
= 1/2(v/Q)^2 * m * (m + M)^2
= 1/2(v/Q)^2 * m * Q^2
= 1/2*v^2 * m
= 1/2mv^2

KE is conserved.

Obviously, this is more complicated than doing it in the COM system, but the two are related, because the velocity of the center of mass is mv/(m+M)

If we 'shift over' by that amount, then the initial velocity of the earth is -mv/Q, and the final velocity is +mv/Q. It swaps signs as I indicate above in that analysis. The same for the ball itself. [the 'velocity' is no longer v because the two situations are described slightly differently. In one case v is the total relative velocity between the two objects, but in the center of mass v is the velocity of the ball, while the earth also has a velocity as it rushes up to hit the ball.]

As much as I intended to quit this thread your post has to be addressed.
The mistake I made in my last post was failing to impose the condition the velocities v and u are in opposite directions.
The correct formula for J1 is therefore.

J1= m(v)-m(-u) (not J1 = m(v) – m(u))
= m(v)+m(u).

Since J1 is zero as previously described this can only occur if u = -v.

Whether you like it or not, in the observer’s frame of reference, the Earth’s momentum is zero before collision has occurred.
As a result your COM reference frame makes no sense whatsoever.
For elastic collisions the velocity of the COM is simply v which is subtracted from the ball and Earth velocities in the observer's frame before collision and added to the ball and Earth velocities in the observer’s frame after collision.
This leads to the simple conclusion in the COM’s frame of reference the ball remains stationary before and after collision while the Earth hits and rebounds at velocities -v and v respectively.

The problem displayed in this thread is that you seem to think momentum is always subtracted.
While this is true for impulse it is definitely not the case when calculating the conservation of momentum or a resultant force.
For example two forces of equal magnitude acting on a mass m in opposite directions produces a resultant force R = 0.
Not to you however:

R = ΣF = d/dt(mv-m(-v)) = d/dt(2mv) = 2ma = 2R.
Instead of the correct R = ΣF = d/dt(mv+m(-v)) = d/dt(0) = 0.

As long as you hold on to this cherished belief you can throw any red herring you want such as the COM’s frame of reference, the end result is you will always be comprehensively wrong.

 

[essentialsaltes]

The mistake I made in my last post was failing to impose the condition the velocities v and u are in opposite directions.

It's not something to impose. It's something to solve for.

The correct formula for J1 is therefore.

J1= m(v)-m(-u) (not J1 = m(v) – m(u))
= m(v)+m(u).

But anyone can look at the page for Impulse and see that the impulse is the momenta of the same object at different times subtracted.

If v2 and v1 have opposite signs (regardless of whether you call them u or -u or Barney the Dinosaur) then J cannot be zero.

As a result your COM reference frame makes no sense whatsoever.

We are always free to choose any inertial reference frame we like. I chose a convenient one. But I can use yours as well, as above.

For elastic collisions the velocity of the COM is simply v

No, it isn't. Don't bring up the earth, if you're only going to ignore it.

Velocity of center of mass

For our case with the earth at rest, velocity of the center of mass is (mv + M(0))/(m + M) = mv/(m+M) as I correctly wrote above.

The problem displayed in this thread is that you seem to think momentum is always subtracted.
While this is true for impulse

Well, then why did you claim your formula was a mistake?

[You seem to be saying that the final velocity is not u, but -u. But if that's the case, when you conclude that u = -v, that again means that the final velocity (which you are now calling -u) is v. Your ball is missing the earth again, moving with constant velocity.]

 

[essentialsaltes]

Off for the cruise. So just to boil down the most recent conversations into a nutshell for sjastro to ponder.

"you seem to think momentum is always subtracted.
While this is true for impulse"
"the velocities v and u are in opposite directions."

If you subtract things with different signs, the answer cannot be zero.

"Since J1 is zero"

It can't be.

 

[Nithavela]

Without reading the thread: e.

Normal gravitation is aproximately 10 m/s², so when the pins are falling with 10 m/s, when caught they are essentially exerting double their weight onto the juggler and thus onto the rope. At the same time, when the juggler throws the pins into the air, he again exerts additional force upwards and thus is pushed the same force in the opposite direction. The average weight is equal to the weight he had when holding the pins in hand, although it would vary during the juggling process.

 

[sjastro]

It's not something to impose. It's something to solve for.



But anyone can look at the page for Impulse and see that the impulse is the momenta of the same object at different times subtracted.

If v2 and v1 have opposite signs (regardless of whether you call them u or -u or Barney the Dinosaur) then J cannot be zero.






Well, then why did you claim your formula was a mistake?

[You seem to be saying that the final velocity is not u, but -u. But if that's the case, when you conclude that u = -v, that again means that the final velocity (which you are now calling -u) is v. Your ball is missing the earth again, moving with constant velocity.]

It amazes me you cannot grasp such a simple concept.
Hopefully this example will make things clearer.
(Try http://members.iinet.net.au/~sjastro/astrophysics/Impulse.jpg if image doesn't download)

In the first diagram the larger ball of mass M has a velocity U, the smaller ball has a mass m with velocity u.
Since U > u the larger mass catches up and collides with the smaller ball.
After collision the velocities are V and v respectively.
The impulse on the smaller ball due to the larger ball is J1 = mv – mu which you are familiar with.

In the second diagram the balls approach each other and collide; the impulse on the smaller ball is J2 = mv – m(-u) since the small ball changes direction after collision.
mv – m(-u) = mv + mu is still defined as an impulse even though vector addition occurs.
If u = -v the impulse is zero.
To be so incredulous indicates a lack of understanding of basic vector algebra.

We are always free to choose any inertial reference frame we like. I chose a convenient one. But I can use yours as well, as above.
No, it isn't. Don't bring up the earth, if you're only going to ignore it.

Velocity of center of mass

For our case with the earth at rest, velocity of the center of mass is (mv + M(0))/(m + M) = mv/(m+M) as I correctly wrote above.

Your “free” selection of an inertial free frame is so ridiculously wrong I don’t know where to start.
All you have done is to calculate the total momentum and KE after the collision in the COM frame and to then boldly declare both are conserved when you haven’t even bothered to consider the momentum and KE before collision in the same frame is totally illogical.
The total momentum in the COM frame of reference before and after collision is zero not mv.

You don’t seem to comprehend either when you transform the COM frame back into the observer’s frame it needs to reflect the physical process occurring in that frame.
To transform the momentum back into observer’s frame after collision add the velocity of the COM to both the ball’s and Earth’s velocities and then multiply by their masses.
This gives –mv and 0 respectively which is expected for an elastic collision with a stationary Earth.
Your values of m(m-M)v/(M+m) and M(2mv)/(M+m) on the other hand in the COM frame after collision do not transform back to –mv and 0 in the observer’s frame.

 

[JackRT]

This problem turned up a few years ago in the selection process for high school students to compete in the Physics Olympiad.

An expert juggler, carrying five juggling pins, has to cross a swing bridge which has a maximum safe load rating of 50 kg. The juggler weighs 47 kg and each of his five pins weighs 2 kg. He believes he can make it across safely in one trip by juggling the pins, so that he is never holding more than one pin. His skill enables him to juggle smoothly without any jerking.
He is
A. incorrect – more information is needed.
B. correct – the total weight will never exceed 49 kg
C. correct – no jerking means no extra weight.
D. correct – the total weight can be made to exceed 49 kg by an arbitrarily small amount.
E. incorrect – the total weight is 47 kg + 5x2 kg = 57 kg.

The challenge here is not only to select to correct answer but to use physics to explain the selection.
Good luck!

If you are about to be mugged by a bunch of clowns, go for the juggler.

 

[sjastro]

Off for the cruise. So just to boil down the most recent conversations into a nutshell for sjastro to ponder.

"you seem to think momentum is always subtracted.
While this is true for impulse"
"the velocities v and u are in opposite directions."

If you subtract things with different signs, the answer cannot be zero.

"Since J1 is zero"

It can't be.

If true then the total of two forces of equal magnitude and opposite direction on an object cannot be zero either.
Funny isn't it how you consistently avoid this inconvenient truth while perpetrating your nonsense.

 

[essentialsaltes]

It amazes me you cannot grasp such a simple concept.

Well, you're still struggling with the idea that up and down are different, so there's amazement on both sides.

You will notice that the larger object does not have the same velocity before and after the collision.

In the second diagram the balls approach each other and collide; the impulse on the smaller ball is J2 = mv – m(-u) since the small ball changes direction after collision.

Here you are putting the minus sign in by hand. You are just calling the final velocity -u instead of u. So when you conclude that "If u = -v the impulse is zero." you are saying that the final velocity, which is -u, is v. That is, no collision has taken place. The ball goes on its way.

Look, momentum is always conserved. Not just at some two points in the journey. The velocity of the falling ball is gt. Its momentum is mgt. It is a function of time. If you only look at the ball's momentum, it's clearly not constant. So its change is (in general) not zero.

This gives –mv and 0 respectively which is expected for an elastic collision with a stationary Earth.

Again with your geocentrism. The Earth is not an immovable object. If you treat it as one, then gravity is an external force. If you don't treat it as one, then the earth is in motion after the collision.

The one-dimensional elastic collision of two objects is a solved problem. We can write the formula in closed form.

The collision is fully specied given the two initial velocities and masses of the colliding objects. Combining the above equations gives a solution to the final velocities for an elastic collision of two objects:
vf1 = [(m1 - m2)·vi1 + 2 m2·vi2]/(m1 + m2)

vf2 = [2 m1·vi1 - (m1 - m2)·vi2]/(m1 + m2)

In your preferred frame, the initial velocity of the ball (m1 = m) is v, and the initial velocity of the Earth (m2 = M) is zero.

Final velocity of the ball is vf1 = [(m1 - m2)·vi1 + 2 m2·vi2]/(m1 + m2)
= [(m - M)·v + 2 M·(0)]/(m + M)
= (m-M)/(m+M)v

Final velocity of the Earth is vf2 = [2 m1·vi1 - (m1 - m2)·vi2]/(m1 + m2)
= [2 m·v - (m - M)·(0)]/(m + M)
= 2mv/(m + M)

Either you treat the earth as fixed and gravity as an external force (in which case momentum is not conserved and your argument is wrong).
Or you treat the earth as movable and gravity as an internal force (in which case momentum is conserved and your final velocities are wrong).

If one momentum is up and the other is down, and you get deltap by subtraction, then deltap cannot be zero.

 

[essentialsaltes]

If true then the total of two forces of equal magnitude and opposite direction on an object cannot be zero either.
Funny isn't it how you consistently avoid this inconvenient truth while perpetrating your nonsense.

No, because total means add, while it "is true for impulse" that we subtract.

 

[JackRT]

Well, you're still struggling with the idea that up and down are different, so there's amazement on both sides.



You will notice that the larger object does not have the same velocity before and after the collision.





Here you are putting the minus sign in by hand. You are just calling the final velocity -u instead of u. So when you conclude that "If u = -v the impulse is zero." you are saying that the final velocity, which is -u, is v. That is, no collision has taken place. The ball goes on its way.

Look, momentum is always conserved. Not just at some two points in the journey. The velocity of the falling ball is gt. Its momentum is mgt. It is a function of time. If you only look at the ball's momentum, it's clearly not constant. So its change is (in general) not zero.



Again with your geocentrism. The Earth is not an immovable object. If you treat it as one, then gravity is an external force. If you don't treat it as one, then the earth is in motion after the collision.

The one-dimensional elastic collision of two objects is a solved problem. We can write the formula in closed form.

The collision is fully specied given the two initial velocities and masses of the colliding objects. Combining the above equations gives a solution to the final velocities for an elastic collision of two objects:
vf1 = [(m1 - m2)·vi1 + 2 m2·vi2]/(m1 + m2)

vf2 = [2 m1·vi1 - (m1 - m2)·vi2]/(m1 + m2)

In your preferred frame, the initial velocity of the ball (m1 = m) is v, and the initial velocity of the Earth (m2 = M) is zero.

Final velocity of the ball is vf1 = [(m1 - m2)·vi1 + 2 m2·vi2]/(m1 + m2)
= [(m - M)·v + 2 M·(0)]/(m + M)
= (m-M)/(m+M)v

Final velocity of the Earth is vf2 = [2 m1·vi1 - (m1 - m2)·vi2]/(m1 + m2)
= [2 m·v - (m - M)·(0)]/(m + M)
= 2mv/(m + M)

Either you treat the earth as fixed and gravity as an external force (in which case momentum is not conserved and your argument is wrong).
Or you treat the earth as movable and gravity as an internal force (in which case momentum is conserved and your final velocities are wrong).

If one momentum is up and the other is down, and you get deltap by subtraction, then deltap cannot be zero.

Well said!

 

[sjastro]

Well, you're still struggling with the idea that up and down are different, so there's amazement on both sides.



You will notice that the larger object does not have the same velocity before and after the collision.





Here you are putting the minus sign in by hand. You are just calling the final velocity -u instead of u. So when you conclude that "If u = -v the impulse is zero." you are saying that the final velocity, which is -u, is v. That is, no collision has taken place. The ball goes on its way.

Look, momentum is always conserved. Not just at some two points in the journey. The velocity of the falling ball is gt. Its momentum is mgt. It is a function of time. If you only look at the ball's momentum, it's clearly not constant. So its change is (in general) not zero.

This is ridiculous.
In the example given if u and v have the same sign before and after collision it means the direction of the ball hasn’t changed; otherwise how do you differentiate the case when the direction has changed.
Where u = -v is an elastic collision against a stationary (massive) object.
In this case the impulse on the large ball is zero since it remains stationary throughout the collision where U = V = 0.
By Newton’s third law if J1 = mv – m(-u) and J2 = MV – MU then J1 = -J2 = 0 since J2 = 0.
Hence J1 = mv – m(-u) = 0 can only be true if u = -v where u and v are of the same magnitude but in opposite directions.
Needless to state the impulse has to be zero in this case otherwise momentum is not conserved as there is an external force.

Again with your geocentrism. The Earth is not an immovable object. If you treat it as one, then gravity is an external force. If you don't treat it as one, then the earth is in motion after the collision.

Either you treat the earth as fixed and gravity as an external force (in which case momentum is not conserved and your argument is wrong).
Or you treat the earth as movable and gravity as an internal force (in which case momentum is conserved and your final velocities are wrong).

This is so profoundly wrong.
The reason why momentum is conserved in both elastic and inelastic collisions is due to Newton’s third law.
On collision both the ball and Earth exert forces that are equal on each other but opposite in direction.
As a result the external force is zero and momentum is always conserved in the observer’s frame of reference.
Here is a simple explanation for the connection between conservation of momentum, impulse and Newton’s third law.

The one-dimensional elastic collision of two objects is a solved problem. We can write the formula in closed form.

The collision is fully specied given the two initial velocities and masses of the colliding objects. Combining the above equations gives a solution to the final velocities for an elastic collision of two objects:
vf1 = [(m1 - m2)·vi1 + 2 m2·vi2]/(m1 + m2)

vf2 = [2 m1·vi1 - (m1 - m2)·vi2]/(m1 + m2)

In your preferred frame, the initial velocity of the ball (m1 = m) is v, and the initial velocity of the Earth (m2 = M) is zero.

Final velocity of the ball is vf1 = [(m1 - m2)·vi1 + 2 m2·vi2]/(m1 + m2)
= [(m - M)·v + 2 M·(0)]/(m + M)
= (m-M)/(m+M)v

Final velocity of the Earth is vf2 = [2 m1·vi1 - (m1 - m2)·vi2]/(m1 + m2)
= [2 m·v - (m - M)·(0)]/(m + M)
= 2mv/(m + M)


If one momentum is up and the other is down, and you get deltap by subtraction, then deltap cannot be zero.

Your finest effort so far in this thread in terms of obfuscation was to throw in a COM frame of reference as a red herring to confuse the issue when you were asked to address the elastic collision in the observer’s frame of reference.
Now the goalposts have been changed again as you have been caught out making claims the total momentum in a COM frame is non zero.
You seem to be blissfully unaware that your cut and paste job based on the observer’s frame of reference is acknowledging momentum is conserved; a point you have vehemently opposed apparently until now!!
When m2 is very large vf1 → -vi1 and vf2 → 0, which is exactly the condition one would expect of an elastic ball hitting a stationary Earth which remains stationary while the ball rebounds with the same magnitude initial velocity but in the opposite direction.
Did you actually try using the calculator?
If you plug in very large values for m2 with vfi = 0 there are no surprises that m2 remains stationary and vf1 = -vi1

The irony is if you got the maths right for the COM frame in the first place, it would have confirmed the Earth’s momentum in the observer’s frame remains zero throughout the elastic collision.

To set the record straight let me correct your mistake about the total momentum in the COM frame being mv instead of zero.

The velocity of the COM = mv/(M+m).
In the COM frame using the transformation v → v – mv/(M+m) :
Initial momentum of ball = m(v-mv/(M+m)) = mv - m²v/(M+m)
Initial momentum of Earth = -Mmv/(M+m)
Total momentum before collision = mv - m²v/(M+m) - Mmv/(M+m)
= v(m - (m² + Mm)/(M+m))
= v(m - m(M+m)/(M+m)) =0
= v(m – m) =0
Hence total momentum in the COM frame is zero before collision.

In the COM frame using the transformation -v → -v + mv/(M+m)
Final momentum of ball = m(-v+mv/(M+m)) = -mv + m²v/(M+m)
Final momentum of Earth = Mmv/(M+m)
Total momentum after collision = -mv + m²v/(M+m) + Mmv/(M+m)
= v(-m + (m² + Mm)/(M+m))
= v(-m + m(M+m)/(M+m)) =0
= v(-m + m) =0
Hence total momentum in the COM frame is zero after collision.

Your incorrect COM frame would result in the Earth acquiring momentum before the collision has even occurred in the observer’s frame of reference.

 

[sjastro]

No, because total means add, while it "is true for impulse" that we subtract.

Still running away from the problem for two forces of equal magnitude but in opposite directions.
R = ΣF = d/dt(mv+m(-v)) = d/dt(0) = 0.

The problem you have consistently demonstrated in this thread in not knowing when to add or subtract momentum as highlighted by your criticism of the above formula without understanding its physical significance.
By your “logic” the formula should read R = ΣF = d/dt(mv-m(-v)) = d/dt(2mv) ≠ 0.

Your mishandling of the elastic ball problem like everything else you have thrown up in this thread is my sign off.
I am not going to be baited into any further participation in this thread and that includes any effort on your part in deliberately misrepresenting my posts which has been done on a regular basis.

 

[essentialsaltes]

This is ridiculous.
In the example given if u and v have the same sign before and after collision it means the direction of the ball hasn’t changed

Correct. But u and v are what we call variables. They can take on both positive and negative values.

; otherwise how do you differentiate the case when the direction has changed.

Because we get the result that u = -v. They have opposite signs! If v is positive, u is negative. And vice versa. And when you subtract them you can't get zero.

Now the goalposts have been changed again as you have been caught out making claims the total momentum in a COM frame is non zero.

Actually, as I wrote in post #60: "Miraculously, the momentum of the earth is equal and opposite to the momentum of the falling ball at every moment during the fall. Thus the total momentum of the system is zero. Exactly the same as when we released the ball, when both the ball and the earth were at rest."

When m2 is very large vf1 → -vi1

Yes, it certainly does. As I keep saying, u = -v. u and v have opposite signs. And therefore when you subtract them for the deltap, we can't get zero.

To set the record straight let me correct your mistake about the total momentum in the COM frame being mv instead of zero.

I don't believe I said any such thing. As above, I said it was zero.

 

[essentialsaltes]

Hence total momentum in the COM frame is zero after collision.

Your incorrect COM frame would result in the Earth acquiring momentum before the collision has even occurred in the observer’s frame of reference.

It's hard to tell what you're saying since you talk about both the COM frame and the observer's frame.

If we're talking about the COM frame:

Riddle me this, Batman.
If the total momentum in the COM frame is zero.
And the earth has no momentum before the collision (since you object to it acquiring momentum).
How is the total momentum zero, since the ball clearly has momentum?

If we're talking about the observer's frame:

No, because as you can see the speed of the COM and the initial speed of the earth are the same (practically by definition). When you transform back to the observer's frame, the earth is at rest (and therefore has no momentum) before the collision.