Is energy conserved is this problem?

[Rikudo]

Homework Statement

Two identical frictionless slides A and B each of height h and mass M are placed on a horizontal frictionless floor. A small disc C of mass m much less than M is released on the top of the slide A. Find expressions for the speeds eventually acquired by both the slides when interaction of the slides with the bead ceases. (see fig. 1)

Relevant Equations

Work- energy

Firstly, what I'm about to do is to find the velocity of the mass soon after it doesn't touch the slide A anymore.
using momentum conservation, I got $0 = mv' + MV'$ with $v'$ and $V'$ are the velocity for mass C and slide A respectively immediately after they lost interaction.

Now, I need one more equation. However, I doubt that energy is conserved here because the slide is free to move, and this will cause the contact force between them does Work. But, when I watch the solution video, I'm surprised to know that energy is conserved if we consider slide A.B and mass C is one system. I wonder why...

 

[malawi_glenn]

You have both "disc" and "bead", or is it just one object that is moving on the slides?

I doubt that energy is conserved here because the slide is free to move, and this will cause the contact force between them does Work

force between "them"? Means?

Is the bead/disk thingy released at rest?

Consider this. When the bead/disc whatever thing is leaving the left slide. What is the velocity of the slide? Hint: momentum conservation. See pic below.

 

[kuruman]

Why the two slides? Would the answer be any different if you had just one?

 

[malawi_glenn]

Why the two slides? Would the answer be any different if you had just one?

The bead/disc whatever is going to move back and forward until it can no longer catch up one of the slides.

 

[Rikudo]

You have both "disc" and "bead", or is it just one object that is moving on the slides?

Sorry,I have no idea either. That is what the book says. So, I think it is fine to assume that both refer to the same thing ( mass C).

Is the bead/disk thingy released at rest?

I think so. It is not indicated in the book,though.

force between "them"? Means?

Contact force between slide A and mass C.

Consider this. When the bead/disc whatever thing is leaving the left slide. What is the velocity of the slide? Hint: momentum conservation. See pic below

$v_2= -m v_1/M$

 

[Lnewqban]

And what is the relation between h and Vc at its lowest point?

 

[malawi_glenn]

Contact force between slide A and mass C.

It said frictionless.

Great,

what is v1 in terms of h then?

Then work out the velocity of the right slide when the bead/disc is on its highest point.

What will the velocity of the right slide be here?

 

[jbriggs444]

However, I doubt that energy is conserved here because the slide is free to move, and this will cause the contact force between them does Work. But, when I watch the solution video, I'm surprised to know that energy is conserved if we consider slide A.B and mass C is one system. I wonder why...

By adding all of the moving objects into the "system", the external forces between the objects become internal forces.

Crucially, the internal forces are contact forces with no motion between the mating surfaces. It follows that any work done by the force of A on B is equal and opposite to the work done by the force of B on A.

Kinetic Energy (implicitly including rotational kinetic energy) is conserved for internal contact forces between rigid objects without slipping.

One does have to avoid impulsive collisions where energy can be lost to an internal contact force between rigid objects. But we do not have any impulsive collisions here.

Edit: I was thinking in terms of rolling without slipping here. @malawi_glenn correctly notes that we are actually talking about slipping without friction. Which also works out to conserve kinetic energy.

 

[Rikudo]

The bead/disc whatever is going to move back and forward until it can no longer catch up one of the slides.

Now that I think about it... It is pointed out that the slide is much massive than the mass. Then, I think this will not happen if energy is conserved?

 

[erobz]

The bead/disc whatever is going to move back and forward until it can no longer catch up one of the slides.

Surely this just rocks back and forth...forever. The CoM must stay put? What do you mean until it can't catch up to one of the slides?

 

[malawi_glenn]

Now that I think about it... It is pointed out that the slide is much massive than the mass. Then, I think this will not happen if energy is conserved?

When you realease the bead from h, you have some mechanical energy no?
At some point, you will only have kinetic energy for the two slides and the bead

 

[malawi_glenn]

Surely this just rocks back and forth...forever. The CoM must stay put? What do you mean until it can't catch up to one of the slides?

At some point, perhaps the left slide will move to the left with speed V when the beadslides down from the right slide with speed v<V. Then the bead will not be able to catch up with the left slide.

 

[Rikudo]

Crucially, the internal forces are contact forces with no motion between the mating surfaces. It follows that any work done by the force of A on B is equal and opposite to the work done by the force of B on A.

Is this always true if the it is a contact force?

Consider this simple scenario. I have an ordinary inclined plane that is free to move and I place a block on it. Soon after I released it, I think the work from the contact force on the mass and on the plane will not cancel out. (This will make the energy not conserved)

EDIT : Ah. Nevermind. You are correct.

 

[malawi_glenn]

work from the contact force

Can the normal force perform work here?
It is the friction force that performs (negative) work. If the surface is frictionless, there is no force of friciton. Hence no work done by a non-conservative force, and mechanical energy is preserved.

Two identical frictionless slides A and B each of height h and mass M are placed on a horizontal frictionless floor.

 

[Lnewqban]

Surely this just rocks back and forth...forever. The CoM must stay put? What do you mean until it can't catch up to one of the slides?

The bead can only push each slide at a time, but it can’t fully recover its initial potential energy because can’t pull on them.

The slides will spread apart forever.

 

[kuruman]

This can be solved as a one-dimensional protracted explosion of a mass into three fragments, $M$, $M$ and $m<M$. Fragment $m$ is between the other two. After the explosion is fully completed, fragment $m$ moves with the same velocity as one of the other fragments. The energy released in the explosion should be obvious.

 

[Orodruin]

Can the normal force perform work here?

Yes. If the ramps are moving, the normal force will act over a non-zero distance. This is transferring energy from the disc to the ramp.

This can be solved as a one-dimensional protracted explosion of a mass into three fragments, $M$, $M$ and $m<M$. Fragment $m$ is between the other two. After the explosion is fully completed, fragment $m$ moves with the same velocity as one of the other fragments. The energy released in the explosion should be obvious.

This is the way. You can of course delve deep down into the details but the end result will be this and the argument is sound (apart from that $m$ can have any velocity in between that of the two ramps, what matters is that it is bounded by the ramp velocities and the corresponding kinetic energy and momenta are negligible compared to those of the ramps).

 

[Steve4Physics]

Surely this just rocks back and forth...forever. The CoM must stay put? What do you mean until it can't catch up to one of the slides?

A and B get pushed apart when C slides up/down them. A and B get accelerated outwards so there is an increasing gap between them.

C’s maximum speed and height get reduced because it gradually transfers its energy to A and B.

Eventually C leaves one of the slides but C’s speed is insufficient for it to catch up with the other slide.

(There is a very simple - though approximate - solution as long as we use m<<M.)

 

[malawi_glenn]

Yes. If the ramps are moving, the normal force will act over a non-zero distance. This is transferring energy from the disc to the ramp.This is the way. You can of course delve deep down into the details but the end result will be this and the argument is sound (apart from that $m$ can have any velocity in between that of the two ramps, what matters is that it is bounded by the ramp velocities and the corresponding kinetic energy and momenta are negligible compared to those of the ramps).

Yeah i forgot about that the ramp can move.

 

[malawi_glenn]

It is pointed out that the slide is much massive than the mass. Then, I think this will not happen if energy is conserved?

That m << M will just make your calculations and the final expression a whole lot simpler.

 

[kuruman]

. . . apart from that $m$ can have any velocity in between that of the two ramps . . .

I assumed that the velocity of the bead must be equal to the velocity of one of the ramps because it looks like the problem cannot be solved if there are three unknowns unless one worries about the details and sums some kind of series.

 

[haruspex]

I assumed that the velocity of the bead must be equal to the velocity of one of the ramps because it looks like the problem cannot be solved if there are three unknowns unless one worries about the details and sums some kind of series.

I agree, and I believe this is why it mentions m<<M. At each interaction, the speed difference only goes down a little.
Otherwise, the final difference will depend on the relationship between m and M in an interesting way… likely involving ceil/floor functions.

 

[Orodruin]

I assumed that the velocity of the bead must be equal to the velocity of one of the ramps because it looks like the problem cannot be solved if there are three unknowns unless one worries about the details and sums some kind of series.

Well, it is a three-body problem so it is not kinematically constrained to a particular set of resulting velocities. The point is that with $m \ll M$ and $|v| \leq |V|$ the energy and momenta of the bead are negligible - effectively resulting in a two-body problem.

 

[haruspex]

Well, it is a three-body problem so it is not kinematically constrained to a particular set of resulting velocities. The point is that with $m \ll M$ and $|v| \leq |V|$ the energy and momenta of the bead are negligible - effectively resulting in a two-body problem.

True, but that leaves a certain ambiguity.
There's the somewhat complicated exact solution, which we can rule out; an approximation which assumes the final speed of the bead matches that of one of the ramps (but which one? maybe they're near enough the same); and the less accurate approximation which neglects the final speed of the ball.
It is unclear which degree of approximation is intended.

 

[Orodruin]

and the less accurate approximation which neglects the final speed of the ball.
It is unclear which degree of approximation is intended.

I disagree. It is perfectly clear from the problem statement ($m \ll M$) that the approximation where the bead momentum and energy are neglected is a valid one.

I also disagree that assuming the ball has the same speed as one of the ramps is in any way more accurate. As you mentioned there is an ambiguity in which ramp has the same velocity as the bead. The two cases represent the extreme cases of the range in which the bead velocity can be. However, the bead energy and momentum are still negligible and taking the limit of $m \to 0$ will give the same result. The results all differ by at most a contribution $\mathcal O(m/M)$ so they all work. Neither is necessarily more accurate than the other.

All three approximations have the same degree of approximation. They are valid to corrections of $\mathcal O(m/M)$.

 

[malawi_glenn]

But, when I watch the solution video, I'm surprised to know that energy is conserved if we consider slide A.B and mass C is one system. I wonder why...

Questions for OP to think about:
Is there any heat generated in the process when the bead is sliding down the incline and when the incline is sliding on the floor?
The normal forces will exert work on each object respectively, but will heat be generated?
Will there be any deformation of the objects causing internal energies to be altered?

 

[haruspex]

Is there any heat generated in the process when the bead is sliding down the incline and when the incline is sliding on the floor?
The normal forces will exert work on each object respectively, but will heat be generated?
Will there be any deformation of the objects causing internal energies to be altered?

No heat, and no rotation - it's all frictionless.

 

[malawi_glenn]

No heat, and no rotation - it's all frictionless.

It was meant as a question for the OP to think about...
perhaps I should have been more clear.
I have solved this problem long time ago

 

[haruspex]

I disagree. It is perfectly clear from the problem statement ($m \ll M$) that the approximation where the bead momentum and energy are neglected is a valid one.

I also disagree that assuming the ball has the same speed as one of the ramps is in any way more accurate. As you mentioned there is an ambiguity in which ramp has the same velocity as the bead. The two cases represent the extreme cases of the range in which the bead velocity can be. However, the bead energy and momentum are still negligible and taking the limit of $m \to 0$ will give the same result. The results all differ by at most a contribution $\mathcal O(m/M)$ so they all work. Neither is necessarily more accurate than the other.

All three approximations have the same degree of approximation. They are valid to corrections of $\mathcal O(m/M)$.

Unfortunately it is yet another case of guessing intent from subtle detail in the wording.

Assuming the bead matches the velocity of one slide allows two slide speeds to be deduced, the one the bead last encountered and the one it is vainly chasing.
What we cannot do is say which is A and which is B. They will take it in turns to be the faster.

The question says "Find expressions for the speeds". I note that a) it says speeds, not speed, and b) it is not entirely clear that it requires us to identify which slide has which speed.
I'm fairly confident that such a solution would be more accurate than $\mathcal O(m/M)$. I tried to start with the exact solution then approximate, but I get a 3x3 transition matrix, so it's a bit messy. There's probably an easier way.

As to which is intended, I toss a coin.

 

[malawi_glenn]

I note that a) it says speeds

Problem also says "disc" and "bead".
I would not worry too much about grammar if they can't stick to a single noun for naming the object with mass m.

That is what the book says

Just out of curiosity, which book?

 

[jbriggs444]

Apologies for the length. I wrote some simple code. Rather than getting fancy with a 3 x 3 transition matrix and some kind of Markov chain that I think @haruspex was going for, I simply ran the simulation.

It seems the the idea of a single resulting speed is a correct heuristic. [As conservation of momentum demands].

The number of bounces seems to scale roughly with the square root of the mass ratio.

For 1000 kg ramps with a 1 kg bead one gets:

After 35 bounces the bead is moving at -0.722076525808794, the new target ramp is moving at 1.00023061595632 and the bounced-from ramp is moving at 0.999508539430512
The bead/disc can no longer catch up. Total bounces: 35
Initial total ke = 1000
Final total ke = 1000
Final total momentum = -3.33066907387547e-013

Note that the final bounce left the bead moving in the same direction as the bounced-from ramp. This is not unusual for the final bounce. [Roughly 50/50 chance].

As expected, quadrupling the ramp height does not change the number of bounces or the velocity ratios.

After 35 bounces the bead is moving at -1.44415305161759, the new target ramp is moving at 2.00046123191264 and the bounced-from ramp is moving at 1.99901707886102
The bead/disc can no longer catch up. Total bounces: 35
Initial total ke = 4000
Final total ke = 4000
Final total momentum = -6.66133814775094e-013

Increasing the mass ratio to 1,000,000 : 1 increases the number of bounces substantially and seems to equalize the ramp velocities rather nicely.

After 1110 bounces the bead is moving at 0.0279794771352128, the new target ramp is moving at 0.0632455361191364 and the bounced-from ramp is moving at 0.0632455640986138
The bead/disc can no longer catch up. Total bounces: 1110
Initial total ke = 4000
Final total ke = 4000
Final total momentum = -2.53513265935368e-010

The lead up to that final bounce was:

After 1105 bounces the bead is moving at 0.6604285866624, the new target ramp is moving at 0.0632434988675383 and the bounced-from ramp is moving at 0.0632441592961247
After 1106 bounces the bead is moving at 0.533940394558343, the new target ramp is moving at 0.0632441592961247 and the bounced-from ramp is moving at 0.0632446932365195
After 1107 bounces the bead is moving at 0.407451134574564, the new target ramp is moving at 0.0632446932365195 and the bounced-from ramp is moving at 0.0632451006876539
After 1108 bounces the bead is moving at 0.280961059689331, the new target ramp is moving at 0.0632451006876539 and the bounced-from ramp is moving at 0.0632453816487138
After 1109 bounces the bead is moving at 0.15447042288254, the new target ramp is moving at 0.0632453816487138 and the bounced-from ramp is moving at 0.0632455361191364
After 1110 bounces the bead is moving at 0.0279794771352128, the new target ramp is moving at 0.0632455361191364 and the bounced-from ramp is moving at 0.0632455640986138

Roughly speaking, each bounce saps twice the receding ramp's velocity from the bead's speed.

Just for grins, at a mass ratio of a billion (1,000,000,000) to one we get ten sig fig agreement on ramp speeds:

After 35124 bounces the bead is moving at -0.00170532064285094, the new target ramp is moving at 0.00200000000048912 and the bounced-from ramp is moving at 0.0019999999987838
The bead/disc can no longer catch up. Total bounces: 35124
Initial total ke = 4000
Final total ke = 3999.99999999991
Final total momentum = -3.35048848647962e-009

#!/usr/bin/perl

my $ke = 4000;		# Initial KE in Joules
my $m = 1;		# Mass of small bead/disc
my $M = 1000000000;	# Mass of each of the two ramps

my $v_right = 0;

my $trips = 0;

# Initial launch will distribute KE to launch ramp (left ramp) and bead in inverse ratio of their masses.
my $bead_ke = $M * $ke / ( $m+$M );
my $ramp_ke = $m * $ke / ( $m+$M );
$v_left =  sqrt(2*$ramp_ke/$M);
$v_bead =  sqrt(2*$bead_ke/$m);
print "Initial total momentum = ", $M*$v_right - $M*$v_left + $m*$v_bead, "\n";	# Bead is currently on rightward convention.# Adopt the simplification that the bead is always moving to the "right". We'll flip the directions every time through the loop.

print "Begin with $m kg bead moving rightward at $v_bead, left $M kg ramp at $v_left and right $M kg ramp at rest\n";

while ( $v_bead > $v_right ) {

	# The center-of-momentum frame is moving to the right at (m*v_bead + M*v_right)/(m+M)
	# The collision is elastic. So the velocity toward the CM before the collision becomes
	# the velocity away from the CM afterward. A little algebra then yields:

	my $v_cm = ( $m * $v_bead + $M * $v_right ) / ( $m+$M );

	$v_bead = 2 * $v_cm - $v_bead;		# Result will almost certainly be negative (leftward)
	$v_right = 2 * $v_cm - $v_right;	# Result will certainly be positive (rightward)

	$trips++;

	$v_bead = - $v_bead;	# Shift bead/disc to leftward convention;

	print "After $trips bounces the bead is moving at $v_bead, the new target ramp is moving at $v_left and the bounced-from ramp is moving at $v_right\n";

	# Shift ramps to leftward convention
	my $temp = $v_right;
	$v_right = $v_left;
	$v_left = $temp;

};
print "The bead/disc can no longer catch up. Total bounces: $trips\n";
print "Initial total ke = $ke\n";
print "Final total ke = ", $M*$v_right**2/2 + $M*$v_left**2/2 + $m*$v_bead**2/2, "\n";
print "Final total momentum = ", $M*$v_right - $M*$v_left + $m*$v_bead, "\n";

 

[haruspex]

Apologies for the length. I wrote some simple code. Rather than getting fancy with a 3 x 3 transition matrix and some kind of Markov chain that I think @haruspex was going for, I simply ran the simulation.

It seems the the idea of a single resulting speed is a correct heuristic. [As conservation of momentum demands].

The number of bounces seems to scale roughly with the square root of the mass ratio.

For 1000 kg ramps with a 1 kg bead one gets:

Note that the final bounce left the bead moving in the same direction as the bounced-from ramp. This is not unusual for the final bounce. [Roughly 50/50 chance].

As expected, quadrupling the ramp height does not change the number of bounces or the velocity ratios.

Increasing the mass ratio to 1,000,000 : 1 increases the number of bounces substantially and seems to equalize the ramp velocities rather nicely.

The lead up to that final bounce was:

Roughly speaking, each bounce saps twice the receding ramp's velocity from the bead's speed.

Just for grins, at a mass ratio of a billion (1,000,000,000) to one we get ten sig fig agreement on ramp speeds:

#!/usr/bin/perl

my $ke = 4000;        # Initial KE in Joules
my $m = 1;        # Mass of small bead/disc
my $M = 1000000000;    # Mass of each of the two ramps

my $v_right = 0;

my $trips = 0;

# Initial launch will distribute KE to launch ramp (left ramp) and bead in inverse ratio of their masses.
my $bead_ke = $M * $ke / ( $m+$M );
my $ramp_ke = $m * $ke / ( $m+$M );
$v_left =  sqrt(2*$ramp_ke/$M);
$v_bead =  sqrt(2*$bead_ke/$m);
print "Initial total momentum = ", $M*$v_right - $M*$v_left + $m*$v_bead, "\n";    # Bead is currently on rightward convention.# Adopt the simplification that the bead is always moving to the "right". We'll flip the directions every time through the loop.

print "Begin with $m kg bead moving rightward at $v_bead, left $M kg ramp at $v_left and right $M kg ramp at rest\n";

while ( $v_bead > $v_right ) {

    # The center-of-momentum frame is moving to the right at (m*v_bead + M*v_right)/(m+M)
    # The collision is elastic. So the velocity toward the CM before the collision becomes
    # the velocity away from the CM afterward. A little algebra then yields:

    my $v_cm = ( $m * $v_bead + $M * $v_right ) / ( $m+$M );

    $v_bead = 2 * $v_cm - $v_bead;        # Result will almost certainly be negative (leftward)
    $v_right = 2 * $v_cm - $v_right;    # Result will certainly be positive (rightward)

    $trips++;

    $v_bead = - $v_bead;    # Shift bead/disc to leftward convention;

    print "After $trips bounces the bead is moving at $v_bead, the new target ramp is moving at $v_left and the bounced-from ramp is moving at $v_right\n";

    # Shift ramps to leftward convention
    my $temp = $v_right;
    $v_right = $v_left;
    $v_left = $temp;

};
print "The bead/disc can no longer catch up. Total bounces: $trips\n";
print "Initial total ke = $ke\n";
print "Final total ke = ", $M*$v_right**2/2 + $M*$v_left**2/2 + $m*$v_bead**2/2, "\n";
print "Final total momentum = ", $M*$v_right - $M*$v_left + $m*$v_bead, "\n";

So my belief that the bead's speed would diminish so slowly that it would end up close to that of the ramps was mistaken. Interesting.

 

[jbriggs444]

So my belief that the bead's speed would diminish so slowly that it would end up close to that of the ramps was mistaken. Interesting.

It's the sapping of twice the ramp speed at each bounce that ends up dominating, once the ramps are up to near their final speeds.

 

[haruspex]

It's the sapping of twice the ramp speed at each bounce that ends up dominating, once the ramps are up to near their final speeds.

Indeed, and it should happen nearly half the time that the bead finishes chasing the ramp it last bounced off.

 

[Lnewqban]

Your work shown in post #31 is a jewel, @jbriggs444 !
Thank you very much.