Motion of two objects connected by a string

[songoku]

Homework Statement

As shown in Figure 1, two objects A and B of the same mass m are moving on a horizontal plane without friction along x axis. The two objects are connected by a string of length l. The mass of the string is negligible, and the length is constant. At time t = 0, A is to the right of B at a distance of l / 2, (the string remains flabby), and A and B are moving in the same direction at velocities v_A and v_B, respectively (v_A > v_B). When the distance between A and B becomes l, the string is stretched tightly to its full length. Immediately afterwards, A and B approach each other, and then collide. Suppose that no energy is lost when the string is stretched to its full length, and the collision between A and B is elastic. Answer the following questions:
(a) Estimate the elapsed time T₁ from t = 0 to the moment when the string is stretched to its full length.
(b) Find the velocities of A and B immediately after the string is stretched to its full length for the first time
(c) Estimate the elapsed time T₂ from t = 0 to the first collision between A and B
(d) Find the velocities of A and B immediately after the first collision
(e) Calculate the time elapsed T₃ from t = 0 to the moment when the string is stretched to its full length for the second time
(f) Draw the graph of v_B(t) from t = 0 and t = T₃

Homework Equations

kinematics
momentum

The Attempt at a Solution

(a) T₁ = L / (2(v_A - v_B))
(b) I am not sure about this one. Will the speed be the same but opposite direction for A? So the velocity of A is - v_A and velocity of B still v_B?

Thanks

 

[John Park]

Part (a) looks right.
What is the speed of the centre of mass the the two blocks? Are there any forces that would change this?

 

[gneill]

Hint: Is there any difference (in principle) between two blocks colliding and two blocks "hitting" the string limit?

 

[songoku]

Part (a) looks right.
What is the speed of the centre of mass the the two blocks? Are there any forces that would change this?

The force that can change the speed is the tension when it is taut. The speed of center of mass of block A and B is v_A and v_B respectively. Should I find the location of center of mass of the two blocks?

Hint: Is there any difference (in principle) between two blocks colliding and two blocks "hitting" the string limit?

I am not sure. Let me try:
The total momentum when the string is flabby = m (v_A + v_B)
The total momentum when the string is taut = m (v_A' + v_B')

m (v_A + v_B) = m (v_A' + v_B')
v_A + v_B = v_A' + v_B'

Then, can I use relative speed of approach = relative speed of separation thinking the system "hit" string limit being the same as elastic collision?

Thanks

 

[John Park]

The force that can change the speed is the tension when it is taut. The speed of center of mass of block A and B is vA and vBrespectively. Should I find the location of center of mass of the two blocks?

I meant the (single) speed of the (single) centre of mass of the two blocks taken together. I was trying to lead you towards the view you're asking about at the end,

Then, can I use relative speed of approach = relative speed of separation thinking the system "hit" string limit being the same as elastic collision?

.

I think that is what you're expected to do.

Note that in later parts of the problem there are actual collisions; I think you're expected to assume that energy is conserved in them too, though the problem doesn't say so.

 

[gneill]

The total momentum when the string is flabby = m (v_A + v_B)
The total momentum when the string is taut = m (v_A' + v_B')

m (v_A + v_B) = m (v_A' + v_B')
v_A + v_B = v_A' + v_B'

Then, can I use relative speed of approach = relative speed of separation thinking the system "hit" string limit being the same as elastic collision?

Right. So assuming a light, inextensible string the event will behave just like a perfectly elastic collision between the blocks. You might think of the string as providing a "conduit" for Newton's 3rd law forces between the blocks.

 

[songoku]

I meant the (single) speed of the (single) centre of mass of the two blocks taken together. I was trying to lead you towards the view you're asking about at the end, .

I think that is what you're expected to do.

Note that in later parts of the problem there are actual collisions; I think you're expected to assume that energy is conserved in them too, though the problem doesn't say so.

Right. So assuming a light, inextensible string the event will behave just like a perfectly elastic collision between the blocks. You might think of the string as providing a "conduit" for Newton's 3rd law forces between the blocks.

(b) Because their masses are the same and the type of "collision" is elastic, their speed will interchange. So velocity of A = - v_B and velocity of B = v_A

(c) The time for collision (t) = L/(v_A + v_B). Total time = T₁ + t = L / (2(v_A - v_B)) + L/(v_A + v_B)

(d) This question will be the same as (b), their speed will interchange once more. So velocity of A = v_A and velocity of B = - v_B

(e) The time taken for collision will be the same as the time taken for the string to be stretched again, so the total time will be L / (2(v_A - v_B)) + 2L/(v_A + v_B)

Am I correct?

Thanks

 

[PeroK]

(b) Because their masses are the same and the type of "collision" is elastic, their speed will interchange. So velocity of A = - vB and velocity of B = vA

That looks like a guess, rather than something you calculated. In any case, it's not right.

You should note the comment about centre of mass in post #2. As there are no external forces acting on the system, the velocity of the centre of mass cannot change. The tension in the string is not an external force, as it acts (by Newton's third law) equal and opposite on the two blocks. That's an internal force that is part of the system of two blocks.

In fact, analysing the problem in the centre-of-mass reference frame would not be a bad idea.

If you're not sure how to do that, then you'll have to continue from where you got in post #6:

m (vA + vB) = m (vA' + vB')
vA + vB = vA' + vB'

Now, you need to use conservation of energy.

 

[songoku]

That looks like a guess, rather than something you calculated. In any case, it's not right.

You should note the comment about centre of mass in post #2. As there are no external forces acting on the system, the velocity of the centre of mass cannot change. The tension in the string is not an external force, as it acts (by Newton's third law) equal and opposite on the two blocks. That's an internal force that is part of the system of two blocks.

In fact, analysing the problem in the centre-of-mass reference frame would not be a bad idea.

If you're not sure how to do that, then you'll have to continue from where you got in post #6:
Now, you need to use conservation of energy.

By conservation of energy, do you mean that the total kinetic energy of the system when the string slacks will be equal to the total kinetic energy when the string is taut?

If yes, then I think the result will be the same compared to using relative speed of approach = relative speed of separation
v_A - v_B = v_B' - v_A'

Using elimination, I get v_A' = v_B and v_B' = v_A.

The question states that A and B approach each other when the string taut so I think A will move backward then I add minus sign to the velocity, v_A' = v_B

If you don't mean the conservation of kibetic energy, then I do not have idea how to continue.

The velocity of center of mass = (v_A + v_B) / 2. There is no external force means that the acceleration of center of mass is zero but I do not know how to continue from there

 

[PeroK]

By conservation of energy, do you mean that the total kinetic energy of the system when the string slacks will be equal to the total kinetic energy when the string is taut?

If yes, then I think the result will be the same compared to using relative speed of approach = relative speed of separation
vA - vB = vB' - vA'

Using elimination, I get vA' = vB and vB' = vA.

That looks right.

The question states that A and B approach each other when the string taut so I think A will move backward then I add minus sign to the velocity, vA' = vB

Why do you think A moves backward? If $v_A > v_B$, then after the first "collision" where the string goes tight, B will be catching up with A.

 

[songoku]

That looks right.
Why do you think A moves backward? If $v_A > v_B$, then after the first "collision" where the string goes tight, B will be catching up with A.

The question states that A and B approach "each other" so I think they move toward each other. Seems I interpret the question wrongly.

(c) the total time will be L/(2(v_A - v_B)) + L/(v_A - v_B) = 3/2 × L/(v_A - v_B)

Is that correct?

Thanks

 

[PeroK]

The question states that A and B approach "each other" so I think they move toward each other. Seems I interpret the question wrongly.

(c) the total time will be L/(2(v_A - v_B)) + L/(v_A - v_B) = 3/2 × L(v_A - v_B)

Is that correct?

Thanks

Yes.

If you go back to the original situation where they are $l/2$ apart and assume that $V_B > V_A$, then they will collide despite both moving in the same direction.

 

[PeroK]

@songoku would you like to see how to use the COM frame?

 

[songoku]

@songoku would you like to see how to use the COM frame?

Of course I want to see it

 

[PeroK]

Of course I want to see it

As the blocks are of equal mass, in the COM frame they are moving away from each other with equal and opposite velocity (initially). Let's call this speed $v$, we can work it out later.

Now the picture is simple: they start $l/2$ apart and both are moving away from the COM at a speed of $v$. And, when the rope goes tight, by conservation of momentum and energy, they must both reverse their direction of motion. So, they then move towards each other with the same speed $v$. And, after colliding, they bounce apart with the same speed $v$, which is back to the original situation. And, so on.

This method would be even better if the masses were of different sizes, as the motion is again simple in the COM frame.

To calculate the speed of the COM, you can use:

$v_{COM} = \frac{mv_A + mv_B}{2m} = \frac{v_A + v_B}{2}$

And, in that frame, we have:

$v'_A = v_A - v_{COM} = \frac{v_A - v_B}{2}$

$v'_B = v_B - v_{COM} = \frac{v_B - v_A}{2} = -v'_A$

 

[songoku]

I am very sorry for my late reply

As the blocks are of equal mass, in the COM frame they are moving away from each other with equal and opposite velocity (initially)

I already don't understand this part

The location of the COM will be in the middle of the two blocks so the left block will move toward the COM and the right block will move away from COM?

Thanks

 

[PeroK]

I am very sorry for my late replyI already don't understand this part

The location of the COM will be in the middle of the two blocks so the left block will move toward the COM and the right block will move away from COM?

Thanks

The COM reference frame is a reference frame moving with respect to the original with a velocity $v_{COM}$ that I calculated for you in post #15. In the COM frame, the COM is stationary.

 

[songoku]

The COM reference frame is a reference frame moving with respect to the original with a velocity $v_{COM}$ that I calculated for you in post #15. In the COM frame, the COM is stationary.

I think I get the idea a little bit. I'll go searching and reading about COM frame first and then go back here to understand your explanation in post #15.

Thanks

 

[songoku]

As the blocks are of equal mass, in the COM frame they are moving away from each other with equal and opposite velocity (initially). Let's call this speed $v$, we can work it out later.

Now the picture is simple: they start $l/2$ apart and both are moving away from the COM at a speed of $v$. And, when the rope goes tight, by conservation of momentum and energy, they must both reverse their direction of motion. So, they then move towards each other with the same speed $v$. And, after colliding, they bounce apart with the same speed $v$, which is back to the original situation. And, so on.

This method would be even better if the masses were of different sizes, as the motion is again simple in the COM frame.

To calculate the speed of the COM, you can use:

$v_{COM} = \frac{mv_A + mv_B}{2m} = \frac{v_A + v_B}{2}$

And, in that frame, we have:

$v'_A = v_A - v_{COM} = \frac{v_A - v_B}{2}$

$v'_B = v_B - v_{COM} = \frac{v_B - v_A}{2} = -v'_A$

I think I get it

Thank you very much for the help