Motion of masses connected by a string over a pulley

[songoku]

Homework Statement

Two particles A and B of masses m and 2m are attached to the ends of light inextensible string which passes over smooth fixed pulley. The particles are released from rest and when particle B has moved h meters, it hits the ground and does not rebound.
(i) Find the speed of the particles just before B hits the ground
(ii) Find the direction and magnitude of the impulse that the particle B exerting on the ground
(iii) In the subsequent motion, B will rest on the ground until it is jerked into motion again. Find the period of time when B is resting on the ground before it is first jerked into motion
(iv) Find the speed when B first leaves the ground
(v) After jerked into motion, B rises up and then drops to hit the ground again. Find the distance B travels between the first hit and the second hit on the ground
(vi) The motion continues with B hitting the ground and jerked into motion indefinitely. Find the total distance traveled by B from the instance that the particles are released from rest.

Homework Equations

v² = u² + 2ad
v = u + at
d = u.t + 1/2 at²
ΣF = m.a

The Attempt at a Solution

(i) acceleration of system = g/3

v² = u² + 2ad
= \sqrt {\frac{2gh}{3}}

(ii) Impulse on B = m . Δv = m . (0 - \sqrt {\frac{2gh}{3}})
= m . \sqrt {\frac{2gh}{3}}

So the impulse on ground is downwards

(iii) Period of time when B is resting = twice the period of time for A to travel (after B hits the ground) to max. height, so:
v = u - gt
0 = \sqrt {\frac{2gh}{3}} - gt
t = \sqrt {\frac{2h}{3g}}

(iv) Speed when B leaves the ground = speed when B almost hits the ground = \sqrt {\frac{2gh}{3}}

(v) distance of B = twice the distance when it accelerates upwards + decelerate to stop (when A hits the ground)

Distance when B decelerates to stop:
v² = u² - 2gd
0 = 2gh/3 - 2gd
d = h/3

So total distance = 4h + 2(h/3) = 14h/3Are answer from (i) to (v) correct?For (vi), the answer is infinite?

Thanks

 

[BvU]

Hi,

Don't forget the factor 2 in iii

I don't understand your argument for iv : B does not rebound and the momentum from B is lost. What happens when A pulls the string taut again ?

 

[songoku]

Hi BvU

Hi,

Don't forget the factor 2 in iii

Yes, my bad.

I don't understand your argument for iv : B does not rebound and the momentum from B is lost. What happens when A pulls the string taut again ?

Yeah, the system should be decelerating instead of accelerating . The deceleration = g/3

But I do not know how to find the speed of B when it leaves the ground. Won't it be the same as speed of A when the string becomes taut, which is \sqrt {\frac{2gh}{3}}?

Thanks

 

[haruspex]

But I do not know how to find the speed of B when it leaves the ground. Won't it be the same as speed of A when the string becomes taut, which is \sqrt {\frac{2gh}{3}}?

That's the speed of A just before the string becomes taut again. Yes, they will have the same speed as each other afterwards (what is the name for such an interaction?), but it won't be the same as the speed A had.

 

[songoku]

That's the speed of A just before the string becomes taut again. Yes, they will have the same speed as each other afterwards (what is the name for such an interaction?), but it won't be the same as the speed A had.

I do not know. Is there a name for such interaction? Action - reaction?

If speed of B when leaving ground ≠ speed of A when the string becomes taut again, how can we find speed of B? I don't think knowing the deceleration helps

Thanks

 

[Delta2]

You have to use conservation of momentum i think (though there are external forces which do not cancel out, still i think conservation of momentum can be applied cause the "collision" forces when string becomes suddenly taut are much more bigger than the external forces) to find the common speed of A and B when string becomes taut due to the movement of A.

 

[haruspex]

I do not know. Is there a name for such interaction?

If an interaction between moving bodies results in their moving together at the same velocity it is called a coalescence. What happens with these two masses when the string tautens is slightly different because they will move at the same speed but in opposite directions.

If speed of B when leaving ground ≠ speed of A when the string becomes taut again,

As I wrote, the speeds will be the same after it is taut again, but not before.

@Delta2 is correct, but as D2 notes it is not quite kosher just to say 'conservation of momentum'. I'll try to show you the authorised version:
As far as A is concerned, what makes it suddenly change speed? (It doesn't "know" about B.) Likewise, what makes B start moving again?

 

[songoku]

You have to use conservation of momentum i think (though there are external forces which do not cancel out, still i think conservation of momentum can be applied cause the "collision" forces when string becomes suddenly taut are much more bigger than the external forces) to find the common speed of A and B when string becomes taut due to the movement of A.

The external forces you mean are normal force and weight?

@Delta2 is correct, but as D2 notes it is not quite kosher just to say 'conservation of momentum'. I'll try to show you the authorised version:
As far as A is concerned, what makes it suddenly change speed? (It doesn't "know" about B.) Likewise, what makes B start moving again?

A suddenly changes speed because it is being pulled upwards by the taut string? And same reasoning why B starts moving again, because it is pulled upwards by the taut string?

I still do not know how to use conservation of momentum in this case. What is the system? The momentum between B and the string conserved?

Thanks

 

[haruspex]

A suddenly changes speed because it is being pulled upwards by the taut string? And same reasoning why B starts moving again, because it is pulled upwards by the taut string?

Yes. There is a sudden large tension in the string for a short time, leading to an impulse (change in momentum) of each mass.
What do you usually assume about tension in a string going over a pulley? What do you deduce about the two impulses?

 

[Delta2]

it should be clear that the two masses exchange momentum each time the string becomes taut due to the movement of A downwards.
The only technicality is that the two bodies have opposite velocities exactly after the string has become taut. You can bypass this and apply conservation of momentum like it was an inelastic collision and both bodies move with common velocity V just after the string has become taut.

 

[haruspex]

it should be clear that the two masses exchange momentum each time the string becomes taut

Ah, but they do not exchange momentum. That would imply momentum is conserved, which it is not.
I think it is instructive to go through what actually happens.

 

[Delta2]

Ah, but they do not exchange momentum. That would imply momentum is conserved, which it is not.
I think it is instructive to go through what actually happens.

Strictly speaking you are correct , momentum is not conserved because the two bodies have same speed but opposite velocities, however as i said we can bypass this technicality and treat the situation as an inelastic collision where both bodies move with the same velocity V after the collision. The direction of V for each body is not important and it doesn't play a role to what this exercise asks.

 

[BvU]

What happens when A pulls the string taut again ?

We are given that the string is inextensible. As far as I can see we (your helpers and you) don't have a common picture of what happens. But we agree on momentum conservation, so perhaps treating this events as a kind of collision (this time elastic) between an m with a certain speed and a 2m at rest would be a good way to proceed.

I wonder if the exercise author meant you to go through all this, I find it getting complicated...

 

[haruspex]

we agree on momentum conservation, so perhaps treating this events as a kind of collision (this time elastic)

Elastic? That would be work conservation, but I assure you that is not conserved.
Also, momentum is not conserved in the standard sense. There's a sign flip.

 

[ehild]

The situation can be considered as curvilinear motion along the string, clockwise or ant-clockwise. When the string is taut, the velocities are along the tangent of the string and both bodies move clockwise or anti-clockwise, with the same speed. You can consider anti-clockwise motion positive, clockwise motion negative, and the momenta of both add in this sense, and are conserved during the "collision" when the bodies just start to interact through the string.

 

[jbriggs444]

Elastic? That would be work conservation, but I assure you that is not conserved.

I doubt that the questioner intended for the string to come taut elastically. But neither is it ruled out by the problem statement. The adjective "inextensible" is not adequate to determine whether mechanical energy is conserved.

 

[haruspex]

I doubt that the questioner intended for the string to come taut elastically. But neither is it ruled out by the problem statement. The adjective "inextensible" is not adequate to determine whether mechanical energy is conserved.

Interesting...
In my experience, the standard intent in problems where an inextensible string becomes suddenly taut is that it is treated as a coalescence, i.e. the string is inelastic and the bodies attached immediately match speeds. But I had never noticed this is illogical and inconsistent with the way we treat collisions of incompressible bodies.
Indeed, inextensible ought to imply perfectly elastic. If so, the string only becomes taut momentarily (in the British English sense).

In the present problem that is going to complicate matters. It is not immediately apparent what will happen next - will the string become taut again before B hits the ground again? So I strongly suspect the intent is inelastic, but I will see what happens in the other view.

 

[jbriggs444]

Indeed, inextensible ought to imply perfectly elastic.

In my view, "inextensible" and "incompressible" are both orthogonal to "elastic". They provide no information either way. You end up with an indeterminate form, an infinite force integrated over a zero displacement either way. The whole point of elasticity or a coefficient of restitution is so that you can parameterize the loss of mechanical energy independent of the rigidity of the material.

At the risk of over-generalizing, I suspect that instructors have an intuition for steel balls which are both very rigid and also very elastic and sometimes use this to leap from the idea that since an incompressible substance cannot deform and dissipate non-zero energy under a finite force that it therefore cannot dissipate energy at all, even if it encounters the proverbial irresistable force.

Edit: By contrast, the intuition for inextensible cords tends to take hemp ropes as the model. Highly inextensible and highly inelastic.

 

[haruspex]

In my view, "inextensible" and "incompressible" are both orthogonal to "elastic". They provide no information either way. You end up with an indeterminate form, an infinite force integrated over a zero displacement either way. The whole point of elasticity or a coefficient of restitution is so that you can parameterize the loss of mechanical energy independent of the rigidity of the material.

At the risk of over-generalizing, I suspect that instructors have an intuition for steel balls which are both very rigid and also very elastic and sometimes use this to leap from the idea that since an incompressible substance cannot deform and dissipate non-zero energy under a finite force that it therefore cannot dissipate energy at all, even if it encounters the proverbial irresistable force.

We certainly have to assume one extreme or the other or there is insufficient data to solve the problem.
Because assuming perfect elasticity creates a complicated motion, I feel certain the intent here is inelastic.
I found an article on JSTOR, https://www.jstor.org/stable/3616194?read-now=1&seq=3#metadata_info_tab_contents, which argues the opposite interpretation, but I think I see a flaw in that. If I am right, it leads to your view that extensibility is orthogonal to elasticity.

The JSTOR article also corroborates my experience that such problems intend coalescence.

 

[BvU]

Elastic? That would be work conservation

At least | momentum |
conservation. But I wouldn't know where the kinetic energy should go... The pulley does no work.

 

[haruspex]

where the kinetic energy should go... The pulley does no work.

If the string is inelastic then that's where the energy goes.
Edit: @jbriggs444 started a private conversation with me so as not to overload this thread. I've tried to add you by using @BvU, but I don't know if that works. Maybe jb has to add you.

 

[Delta2]

if I judge by question (vi) where it asks for the total distance traveled by mass B, then the total energy of the system should decrease over time (loss of energy should happen when 1) mass B hits the ground but does not rebound) 2) when string gets taut) so that it eventually comes to a stop (even if that happens after infinite time).

 

[BvU]

f the string is inelastic

But it's not: It's inextensible. dp/dt for A and B are identical at all times during the 'collision'
I conclude the jerk when the string is taut pulls up both A and B according to Newton 3 -- as if it were a collision where A bounces back when it hits B . With momentum conservation

string is inelastic then that's where the energy goes

How ?

 

[haruspex]

dp/dt for A and B are identical at all times during the 'collision'

That only ensures equal changes in momentum (as opposed to conservation thereof). It says nothing about energy. I think jb has added you to the inbox conversation.

 

[haruspex]

if I judge by question (vi) where it asks for the total distance traveled by mass B, then the total energy of the system should decrease over time (loss of energy should happen when 1) mass B hits the ground but does not rebound) 2) when string gets taut) so that it eventually comes to a stop (even if that happens after infinite time).

That the total distance be finite only requires that one of the two processes loses energy, and certainly B colliding with the ground does that. The question makes sense whether we assume string tautening is work conserving or coalescent, but they will produce different answers, and the coalescent case is much easier to analyse.

 

[songoku]

I am really sorry for late reply.

Let me try again:
(iv)
The two particles will experience same impulses.
Impulse A = m . Δv_a = m . (-v + \sqrt{\frac{2gh}{3}} )

Impulse B = 2m . Δv_b = 2m . (v - 0)

Equating them:
m . (-v + \sqrt{\frac{2gh}{3}} ) = 2mv

v = \frac{1}{3} \sqrt{\frac{2gh}{3}}

Is this correct?

Thanks

 

[haruspex]

I am really sorry for late reply.

Let me try again:
(iv)
The two particles will experience same impulses.
Impulse A = m . Δv_a = m . (-v + \sqrt{\frac{2gh}{3}} )

Impulse B = 2m . Δv_b = 2m . (v - 0)

Equating them:
m . (-v + \sqrt{\frac{2gh}{3}} ) = 2mv

v = \frac{1}{3} \sqrt{\frac{2gh}{3}}

Is this correct?

Thanks

Looks good to me.

 

[ZeusPerseus]

Yes, conservation of momentum is to be applied to situation in which ball B is suddenly jerked upward. In this particular case, the absolute value of the momentum of ball A prior to the jerk will equal the sum of the absolute values of the momentum of ball A and B after the jerk. Absolute value is taken because of the effect of the pulley. Note that each time ball B hits the ground, both momentum and energy is lost to the system. The velocities of the two balls will be equal and opposite after the jerk. The string is inelastic - at the time of the jerk a very sharp force is applied by the string such that the impulse delta-F times delta-t is applied to both balls and give the change in momentum of each ball. The lost kinetic energy will appear as deformations and vibrations in the two balls.

Note also that the acceleration you calculate in i is incorrect - check your math.

 

[haruspex]

The string is inelastic

It is given as inextensible.
Several on this thread had a separate conversation and concluded that the two concepts are independent. But if there is any elasticity then the problem becomes much more difficult, so the question setter probably meant it to be taken to be inelastic as well.

the acceleration you calculate in i is incorrect

Looks right to me. A net force of mg, a total mass of 3m, so an acceleration of g/3.

 

[ZeusPerseus]

Duh, you'r right, I was equating forces on the two masses should be equal and opposite, versus accelerations on the two masses being equal and opposite. I got a little too sophisticated in my analysis.

 

[neilparker62]

ii) Impulse on ground when mass B impacts: 2mΔv ?

 

[songoku]

ii) Impulse on ground when mass B impacts: 2mΔv ?

Yes, sorry. The mass should be 2m.For (v) :
Distance of B = twice the distance when it accelerates upwards + decelerate to stop (when A hits the ground)

Speed of B when A hits the ground
v^2 = (\frac{1}{3} \sqrt{\frac{2gh}{3}})^2 + 2 . \frac{g}{3} . 2h
v^2 = \frac{38gh}{27}

Distance traveled by B when decelerating:
v^2 = u^2 - 2gd
0 = \frac{38gh}{27} - 2gd
d = \frac{19h}{27}

So total distance
= 4h + 2 . \frac{19h}{27}
= \frac{146h}{27}

For (vi), I think I need to use formula of sum to infinity of geometric progression but I can not find the ratio. The distance traveled by B:
a. when released from rest until hits the ground = h
b. between first hit and second hit = \frac{146h}{27}
c. between second hit and third hit (repeating the same process as (v)) = \frac{1298h}{243}

How to find the total distance?

Thanks