Is energy conserved is this problem?

[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.

 

[malawi_glenn]

1 kg bead and 1000 J initial energy, quite tall ramps :)

 

[jbriggs444]

1 kg bead and 1000 J initial energy, quite tall ramps :)

Yes. When I re-ran to look at elapsed time, separation distance and total bead travel, I scaled that back to a 1 meter launch height. Expected values (over a distribution of mass ratios) for elapsed time and separation distance probably diverge (integral of an approximately harmonic function). Typical elapsed times are modest and are dominated by the final trip. For instance, about 2.5 minutes total for the first 34 one way trips and about 15 minutes additional for the last.

[In order to get meaningful figures for elapsed time, one needs for the bounces to take non-zero time. Otherwise, the distance between the ramps is zero, all the trips take place instantly and the total elapsed time is zero. So I imputed a bounce time based on twice the free fall time based on the bead's share of the impact energy at each bounce. And counted only 1 times free fall time for the launch. So I needed a reasonable launch height]

 

[Rikudo]

Will there be any deformation of the objects causing internal energies to be altered?

Umm... By "internal energies" here, do you refer to the whole system's (slide A,B and mass C)?

Just out of curiosity, which book?

Pathfinder.

 

[malawi_glenn]

Umm... By "internal energies" here, do you refer to the whole system's (slide A,B and mass C)?
Pathfinder.

The Individual objects

Never heard of pathfinder, which author and publisher?

 

[jbriggs444]

Never heard of pathfinder, which author and publisher?

Google suggests: https://www.amazon.com/dp/B072QR5CWH/?tag=pfamazon01-20

Arvind Tiwari & Sachin Singh
Pearson India, 2016 - 376 pages

 

[Orodruin]

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.

The thing is, you cannot assume that the bead will have the same velocity as one of the ramps. These are extreme cases of the possible solutions. The range of solutions include the bead having any speed between these extreme solutions. You cannot know which without performing a series argument with a cutoff condition.

The assumption of assuming the bead to have the same speed as either block is therefore no more accurate than neglecting the bead momentum and energy. Even if the bead does have the same velocity as one of the blocks, you won’t know which - giving you an inherent uncertainty that goes as m/M.

 

[kuruman]

I modified my approach and assumed that the bead has a final speed that is a fraction of the final speed of the slide on the left, $v_{\text{bead}}=\beta ~V_{\text{L}}$ ($0 \leq\beta\leq 1.)$ I solved the momentum and energy conservation equations from the moment the bead starts moving at the top of the left slide until all masses are moving away from each other with final speeds $V_{\text{L}}$, $V_{\text{R}}$ and $\beta ~V_{\text{L}}.$

Then I assumed a mass ratio $\mu=m/M=10^{-4}$ and plotted the two speeds normalized to $\sqrt{2gh}$ as a function of $\beta$. The plot is shown below. The orange line is the speed of the right slide and the blue line is the speed of the right left slide. I will not post the equations because I think this homework problem is still live. However, it is fair to say that a Taylor series expansion of the difference gives a leading term $$\frac{V_{\text{R}}-V_{\text{L}}}{\sqrt{2gh}}=\frac{\mu^{3/2}}{\sqrt{2}}\beta.$$ In the limit $m<<M$, the speeds of the slides are, to a very good approximation, the same and this result is independent of $\beta$, i.e. the final speed of the bead.

I think that this holistic approach complements while I compliment the detailed work by @jbriggs444.

 

[Orodruin]

The orange line is the speed of the right slide and the blue line is the speed of the right slide.

I think one of those should be a left.

 

[haruspex]

you cannot assume that the bead will have the same velocity as one of the ramps

I see that now. In post #22 I gave a reason why I thought it would always be close to a ramp's speed, but, as I acknowledged in post #32, @jbriggs444 proved me wrong.

 

[kuruman]

I think one of those should be a left.

You're right. It's fixed, thanks.