Collision involving blocks and pulley

[Tanya Sharma]

Homework Statement

Homework Equations

The Attempt at a Solution

When the particle of mass M/2 collides with the pan ,an impulsive tension acts on the (pan+particle) . Let us call this J . Tension acting in the string between the two blocks is T .

Speed of particle before collision = u = √(gl)
Speed of (pan+particle) after collision = v

Applying impulse momentum theorem on (pan+particle), Mu/2 -J = Mv/2

Applying impulse momentum theorem on block A, J = Mv

From the above two equations we get v = √(gl)/3 .

Is my reasoning correct ? How should I proceed ? I would be grateful if somebody could help me with the problem .

 

[andrevdh]

Since A and B is connected by a taught string shouldn't the magnitude of their acceleration be the same?
Problem is the tension in the string between A and B also changes during the collision.
So I am getting the impression that you need to consider the system as a whole.
Setup looks very much like a swing/pendulum on the left.

 

[Tanya Sharma]

Since A and B is connected by a taught string shouldn't the magnitude of their acceleration be the same?

Yes .

Problem is the tension in the string between A and B also changes during the collision.
So I am getting the impression that you need to consider the system as a whole.
Setup looks very much like a swing/pendulum on the left.

So , how should I proceed ?

 

[AlephNumbers]

So , how should I proceed ?

What would you expect to happen if you dropped a mass on the edge of a pan that is connected by a string to the ceiling?

 

[TSny]

Is my reasoning correct ? How should I proceed ? I would be grateful if somebody could help me with the problem .

I think you are correct so far. Draw a free body diagram for each mass M. Apply 2nd law to the vertical direction for each mass.

The vertical acceleration of the mass M on the left will consist of a superposition of two parts: one part from the change in length of the left portion of the string and one part due to the fact that the sideways motion makes the mass move in circular motion.

 

[Tanya Sharma]

Hello TSny ,

Thanks for replying .

Draw a free body diagram for each mass M. Apply 2nd law to the vertical direction for each mass.

The vertical acceleration of the mass M on the left will consist of a superposition of two parts: one part from the change in length of the left portion of the string and one part due to the fact that the sideways motion makes the mass move in circular motion.

Assuming tension T exists in the string connecting A and B after the collision and A moves upwards as well as towards left and B downwards .

For A , $T - Mg = M(a+\frac{v^2}{l})$ ,

For B , $Mg-T = Ma$

Does it make sense ?

 

[TSny]

Yes. You are letting "$a$" be the upward acceleration of A due to the changing length (and therefore the downward acceleration of B). Looks good.

 

[TSny]

Do you think that B will move downward?

 

[Tanya Sharma]

But this gives $M(2a+\frac{v^2}{l}) = 0$ ? Does this mean , my assumption that A going upwards was wrong . Instead B moves upwards with acceleration g/18 .

Is that so ?

 

[TSny]

But this gives $M(2a+\frac{v^2}{l}) = 0$ ? Does this mean , my assumption that A going upwards was wrong . Instead B moves upwards with acceleration g/18 .

Is that so ?

I think so. I also get that B accelerates upward at g/18.

 

[TSny]

Here’s another way to look at it. Treat masses A and B as one system. The gravity force on each side cancels. So, the net force accelerating the system is just the “centrifugal” force due to the swinging of A.

So, for the system: $F = ma$ gives $Mv^2/l = (2M)a$. So $a = v^2/(2l)$ where $v^2 = gl/9$.

 

[Tanya Sharma]

I think so. I also get that B accelerates upward at g/18.

The answer given is option b) i.e g/9

 

[TSny]

The answer given is option b) i.e g/9

Good. We must still have something to learn! I don't see a mistake, so we can hope that someone will set us straight.

 

[Tanya Sharma]

Here’s another way to look at it. Treat masses A and B as one system. The gravity force on each side cancels. So, the net force accelerating the system is just the “centrifugal” force due to the swinging of A.

So, for the system: $F = ma$ gives $Mv^2/l = (2M)a$. So $a = v^2/(2l)$ where $v^2 = gl/9$.

How does gravity cancels if we treat A and B as one system ? It is acting downwards on both the blocks .

 

[TSny]

Weight of A tries to make the system move counterclockwise while weight of B acts clockwise.

 

[Tanya Sharma]

We get g/9 as answer if we consider only the centripetal acceleration of block A .

For block A , $T - Mg = M\frac{v^2}{l}$ . So, T = (10/9)Mg

Now for block B , acceleration = net force /mass . So acc = (T - Mg) / M = g/9 .

But the problem I see is that if this is the case , both the blocks then accelerate upwards with g/9 . Doesn't look convincing .

I saw the solution given , and this is how the option b) is obtained .

 

[TSny]

Doesn't look convincing .

I agree. This solution appears to neglect the downward acceleration of A due to the changing length of the string on the left.

 

[Tanya Sharma]

This solution appears to neglect the downward acceleration of A due to the changing length of the string on the left.

How does leftwards impulse on block A gives it a downwards acceleration ? I understand that it makes A move in circular path about the top pulley , but how does it provide downward acceleration to A ?

Downwards acceleration of block A means that string length of left string increases in vertical direction and that of right string decreases in vertical direction. Isn't that so ?

I would like to understand your reasoning .

 

[TSny]

Downwards acceleration of block A means that string length of left string increases in vertical direction and that of right string decreases in vertical direction. Isn't that so ?

The string on the left side is accelerating downward, but mass A has a total acceleration upward due to the upward centripetal acceleration. For mass A, the net upward acceleration is

$a_{net, A} = v^2_A/l - a$

where $a$ is the downward contribution to the acceleration of A due to the lengthening of the string on the left. Note, $a = a_{B}$ where $a_B$ is the upward acceleration of B.

2nd law for mass A: $T-Mg = M(v_A^2/l-a_B)$

For mass B: $T-Mg = Ma_B$.

These are the essentially the same as you already set up.

 

[Tanya Sharma]

Sorry...but some confusion has crept in .

So,basically both A and B have a net acceleration g/18 upwards . Right ?

The string on the left side is accelerating downward

How ? I can't visualize it .

I still do not see how leftwards movement of A causes vertically lengthening of the left string . I can see horizontal lengthening of left string causing an overall lengthening of left string .Please help me understand how A accelerates downwards i.e how string length of left string increases in vertical direction.

 

[TSny]

Sorry...but some confusion has crept in .

So,basically both A and B have a net acceleration g/18 upwards . Right ?

Yes. Just after the collision A and B experience exactly the same vertical forces: Mg acts down on each mass while T acts up on both masses. So, A and B must have the same vertical component of acceleration at that time.

How ? I can't visualize it .

I still do not see how leftwards movement of A causes vertically lengthening of the left string . I can see horizontal lengthening of left string causing an overall lengthening of left string .Please help me understand how A accelerates downwards i.e how string length of left string increases in vertical direction.

That was poor wording on my part. If we let $r$ be the length of the string on the left side, then we agree that $\ddot{r} > 0$. We are interested in the time just after the collision when the string on the left is still vertical. The vertical component of acceleration of A, with upward taken as positive, is given by $a_y = v_A^2/r - \ddot{r}$. Thus, $\ddot{r}$ is seen to contribute in the downward direction to the acceleration of A. That’s all I should have said. I did not mean to imply that the "length of left string increases in vertical direction". That would have more to do with y-component of velocity of A rather than acceleration.

 

[Tanya Sharma]

The vertical component of acceleration of A, with upward taken as positive, is given by $a_y = v_A^2/r - \ddot{r}$.

I think i am messing up with the basics . Why can't $a_y = v_A^2/r$ ? How did you write the expression $a_y = v_A^2/r - \ddot{r}$ ? Can you relate this expression to some other simple kinematic problem ?

 

[TSny]

If $r$ is the length of the left string and $\theta$ is the angle of the left string from vertical, then $r$ and $\theta$ are just polar coordinates for A with the origin at the point where the string meets the pulley. The radial component of acceleration in polar coordinates is $a_r = \ddot{r} - r\dot{\theta}^2$. When the string is vertical, the radial direction is downward. So, the upward component of acceleration is $a_y = - a_r = r\dot{\theta}^2 - \ddot{r} = v_A^2/r - \ddot{r}$.

See equation (4) here:
http://ocw.mit.edu/courses/aeronaut...fall-2009/lecture-notes/MIT16_07F09_Lec05.pdf

 

[Tanya Sharma]

My sincere apologies for being so careless . I understood the expression but just as I was about to edit my previous post , saw your above response .

 

[Tanya Sharma]

Doesn't $\ddot{r}>0$ imply that left string length increases in vertical direction at time just after the collision ?

 

[TSny]

Just after the collision, $\dot{r} = 0$ and $\ddot{r} > 0$. This means that $r$ will increase. So, the string on the left will get longer. But it doesn't necessarily mean that the string length increases "in vertical direction". That is, block A does not necessarily move downward as it moves to the left.

 

[Tanya Sharma]

Okay .

How did you determine that $\ddot{r} > 0$ ? I am asking this because earlier I was incorrectly believing that the string length changes is the reason behind $\ddot{r} > 0$ .

 

[TSny]

How did you determine that $\ddot{r} > 0$ ? I am asking this because earlier I was incorrectly believing that the string length changes is the reason behind $\ddot{r} > 0$ .

Applying Newton's second law to each mass and solving for $a_B$ gave us a positive answer for $a_B$. And we have $\ddot{r} = a_B$.

 

[Tanya Sharma]

I regret my inability to understand trivial issues . Thanks for your time and energy in putting up the explanations .The question is a bit tricky . Even good teachers and students I know are fumbling on this question . Everyone is going by the solution I mentioned previously.I am trying hard to understand this question .

Applying Newton's second law to each mass and solving for $a_B$ gave us a positive answer for $a_B$.

You mean by solving $T-Mg = M(v_A^2/l-a_B)$ and $T-Mg = Ma_B$ , we get $a_B = v_A^2/2l$ , a positive number. Right ?

Does that mean that the component of radial acceleration $r\dot{\theta}^2$ i.e $v_A^2/l$ necessitates the presence of $\ddot{r}$ component ,else the two force equations would not be consistent ?

I mean , just as in the solution given they have assumed $a=0$ , but then it made the two force equations inconsistent.

Is that so ?

And we have $\ddot{r} = a_B$.

Sign issues .

1) If we take downward positive for both A and B , would we have written $\ddot{r} = -a_B$ ?

2) If we take downward positive for both A and B , should the force equations be $Mg-T = M(a_B-v_A^2/l)$ and $Mg-T = -M(a_B)$ ?

But then we discussed in a previous thread that we consider signs with forces , not with 'a' in ΣF = Ma . Why should we put a negative sign in front of $a_B$ ?

3) Since in polar coordinates with origin at pulley, increasing r is downwards , sign of $a_B$ should be negative irrespective of whether we choose upwards or downwards as positive or negative ?

4) How do we determine sign of $\ddot{r}$ term ?

Again, thanks for your patience . I hope it is not getting irksome for you .

 

[TSny]

Let $r$ be the length of the string on the left and $s$ be the length of the string on the right. Let’s take downward as the positive y-direction for both A and B. The 2nd law is

$\sum{F_{A,y}} = M a_{A,y}$ and $\sum{F_{B,y}} = M a_{B,y}$

Then $a_{A,y} = \ddot{r} - \frac{v_A^2}{r^2}$. No assumption is being made about whether or not $\ddot{r}$ is a positive or negative quantity. Solving the equations will determine the value of $\ddot{r}$.

$a_{B,y} = \ddot{s}$. No assumption is being made about whether or not $\ddot{s}$ is positive or negative. But $a_{B,y}$ and $\ddot{s}$ must have the same sign if positive y is downward.

We have the constraint that $r + s$ = const. So, $\ddot{r} =-\ddot{s}$. Or, $\ddot{r} =-a_{B,y}$.

So, the 2nd law equations are

$Mg-T = M \left (-a_{B,y} - \frac{v_A^2}{r^2} \right )$ and $Mg-T = Ma_{B,y}$

Solving gives $a_{B,y} = -\frac{v_A^2}{2r^2} = -\frac{g}{18}$. So, B accelerates upward at g/18.

 

[Tanya Sharma]

Let $r$ be the length of the string on the left and $s$ be the length of the string on the right. Let’s take downward as the positive y-direction for both A and B. The 2nd law is

$\sum{F_{A,y}} = M a_{A,y}$ and $\sum{F_{B,y}} = M a_{B,y}$

Then $a_{A,y} = \ddot{r} - \frac{v_A^2}{r^2}$. No assumption is being made about whether or not $\ddot{r}$ is a positive or negative quantity. Solving the equations will determine the value of $\ddot{r}$.

$a_{B,y} = \ddot{s}$. No assumption is being made about whether or not $\ddot{s}$ is positive or negative. But $a_{B,y}$ and $\ddot{s}$ must have the same sign if positive y is downward.

We have the constraint that $r + s$ = const. So, $\ddot{r} =-\ddot{s}$. Or, $\ddot{r} =-a_{B,y}$.

So, the 2nd law equations are

$Mg-T = M \left (-a_{B,y} - \frac{v_A^2}{r^2} \right )$ and $Mg-T = Ma_{B,y}$

Solving gives $a_{B,y} = -\frac{v_A^2}{2r^2} = -\frac{g}{18}$. So, B accelerates upward at g/18.

Any further questions and I guess you would surely kill me .

Thank you very very much . I really appreciate your help .

 

[TSny]

Any further questions and I guess you would surely kill me .

Thank you very very much . I really appreciate your help .

You are very welcome. Your questions are always good. If you or someone figures out why the answer should actually be g/9, then please post it.

 

[ehild]

I assume "plastic" collision means inelastic ?

I do not like this type of questions "what happens just after the collision" . One can not know when is the collision finished and what is everything at that instant.
Anyway, the process is not instantaneous, something happens earlier and implies the other things to follow.
Here the collision happens first, block A gains a horizontal velocity, and then the tension changes in the vertical rope, and block B starts to accelerate. .

We can imagine the collision with the pan happening in a very short time. The pan has no mass, so the speed of the particle is reduced as block A is set on motion by the string between the pan and the block. The tension is the same both in the vertical and the horizontal parts of the string (it is massless) and the the radial part of the forces of tension are absorbed by the normal force of the axis of the left pulley which is fixed to the support.
Block A is accelerated horizontally during the collision by the horizontal tension, and it gains a horizontal velocity of magnitude U which is the same as the vertical velocity of the particle and pan.
It is an inelastic collision. So the particle on the pan and the block A will move together after the collision with speed U, the pan downward, A to the left. The speed of the particle before collision is v. v²= gL. Conservation of momentum says that 3/2 MU = M/2 v -->U = v/3.
Before the collision happened, the tension in the other string was Mg.
At the first instant, there is no displacement yet. A hangs vertically, and has a horizontal velocity - this is circular motion at that instant. To maintain that motion, the tension had to change The tension in the vertical string is T. The centripetal force is MU²/L=T-Mg.
T=Mg+Mv²/(9L)=Mg+MgL/(9L)=Mg+Mg/9.. The acceleration of B is determined by the upward tension and the downward force of gravity.
The problem does not asks what happens afterwards.

 

[TSny]

The centripetal force is MU²/L=T-Mg.

It seems to me that the assumption $r + s$ = constant is generally made for these idealized problems. $r$ is the length of the string on the left side of the pulley and $s$ is the length on the right side. Then $\ddot{r} = -\ddot{s} = a_B$, where $a_B$ is the upward acceleration of B.

So, at any instant when B has a nonzero acceleration, $\ddot{r} \neq 0$. But, $\ddot{r}$ is part of the acceleration of A.

So, it seems to me that we must have $M \left (U^2/L - \ddot{r} \right) = T - Mg$.

That is, $M \left (U^2/L - a_B \right) = T - Mg$ instead of $M U^2/L = T - Mg$

 

[ehild]

It seems to me that the assumption $r + s$ = constant is generally made for these idealized problems. $r$ is the length of the string on the left side of the pulley and $s$ is the length on the right side. Then $\ddot{r} = -\ddot{s} = a_B$, where $a_B$ is the upward acceleration of B.

So, at any instant when B has a nonzero acceleration, $\ddot{r} \neq 0$. But, $\ddot{r}$ is part of the acceleration of A.

B starts to accelerate when the collision is completed.
A gains speed during the collision, but its position does not change yet. The length of the left piece of string is still L, and the force from the other string is perpendicular to it.
The tension in the long strings builds up because of the motion of A.

So the collision causes the velocity of A. The motion of A causes the change of tension in the long string. The change of tension causes the acceleration of B and also the radial acceleration of A, but that is not relevant for the problem. There is a lot of guessing, that is why I do not like such problems.

 

[Chestermiller]

I analyzed this problem independently, and ended up with the same analysis that TSny presented. To implement the analysis, I assumed that the radii of the two pulleys are very small compared to the lengths of the ropes, so that I didn't need to take into account the effect of the change in the wrap angle on the kinematics of the motion. I confirm TSny's result of g/18.

Chet

 

[insightful]

The speed of the particle before collision is v. v2= gL. Conservation of momentum says that 3/2 MU = M/2 v -->U = v/3.
Before the collision happened, the tension in the other string was Mg.
At the first instant, there is no displacement yet. A hangs vertically, and has a horizontal velocity - this is circular motion at that instant. To maintain that motion, the tension had to change The tension in the vertical string is T. The centripetal force is MU2/L=T-Mg.
T=Mg+Mv2/(9L)=Mg+MgL/(9L)=Mg+Mg/9.. The acceleration of B is determined by the upward tension and the downward force of gravity.

I find this to be correct, which means the answer is g/9.

 

[TSny]

Let $r$ be the length of the vertical string between the pulley and mass A. The point of disagreement seems to be with how one answers the following questions.

(1) If block B has an upward acceleration just after the collision, then does that mean that $\ddot{r} \neq 0$ at the same instant of time? If not, why not? If so, then proceed to question (2).

(2) Since $\ddot{r} \neq 0$, then doesn’t that mean that $\ddot{r}$ should contribute to the acceleration of mass A? If not, why not? If so, then proceed to question (3).

(3). [Edited to correct typo] If $\ddot{r}$ contributes to the acceleration of A, then doesn’t that mean that the acceleration of A just after the collision is $v^2/r - \ddot{r}$ rather than just $v^2/r$? If not, why not?

It seems odd to me to allow the effect of the velocity of A to be transmitted essentially instantaneously by the string to B (and causing B to accelerate) while not allowing the effect of the acceleration of B to be transmitted essentially instantaneously back to A.

If you allow any "time delays" due to elasticity of the string, then the question becomes ambiguous. I think this is ehild's point and why he doesn't like the question. You could then argue that the acceleration of B is zero immediately after the collision since there has not been enough time for the effect of the collision at A to propagate to B.

 

[insightful]

the acceleration of A just after the collision is mv2/r−r¨mv^2/r - \ddot{r} rather than just mv2/rmv^2/r? If not, why not?

Why is m in these expressions?

 

[TSny]

Why is m in these expressions?

The mass should not be there. Thanks!

 

[insightful]

So, the effect of the string accelerating is that the radius of movement of mass A is actually twice as long (2xl)? This would halve my tension to also give a=g/18 for mass B.

 

[TSny]

No, just after the collision the length of the string on the left side of the pulley is still $l$. So, the acceleration of mass A due to its circular motion is $v^2/l$. However, I believe there is an additional acceleration of A due to the acceleration of the string ($\ddot{r}$) associated with the acceleration of B.

 

[insightful]

But wouldn't that immediate additional acceleration immediately start mass A on a path different from a circle of radius l ?

 

[Chestermiller]

The equations I obtained for time zero were:

Radial Force Balance from upper pulley to mass A:
$$T-mg=m\left[r\left(\frac{dθ}{dt}\right)^2-\frac{d^2r}{dt^2}\right]$$
where r is the distance between the upper pulley and mass A, θ is the angle that the rope segment from the upper pulley to mass A makes with the vertical, and T is the tension in the upper rope.

Vertical Force Balance on mass B:
$$T = mg+m\frac{d^2r}{dt^2}$$

If we combine these equations, we get:

$$\frac{d^2r}{dt^2}=r\left(\frac{dθ}{dt}\right)^2-\frac{d^2r}{dt^2}$$

Solving for $d^2r/dt^2$, we obtain:

$$\frac{d^2r}{dt^2}=\frac{r}{2}\left(\frac{dθ}{dt}\right)^2=\frac{v^2}{2r}$$

In my judgement, this confirms TSny's g/18 result.

Chet

 

[Satvik Pandey]

Let $r$ be the length of the vertical string between the pulley and mass A. The point of disagreement seems to be with how one answers the following questions.

(1) If block B has an upward acceleration just after the collision, then does that mean that $\ddot{r} \neq 0$ at the same instant of time? If not, why not? If so, then proceed to question (2).

(2) Since $\ddot{r} \neq 0$, then doesn’t that mean that $\ddot{r}$ should contribute to the acceleration of mass A? If not, why not? If so, then proceed to question (3).

(3). [Edited to correct typo] If $\ddot{r}$ contributes to the acceleration of A, then doesn’t that mean that the acceleration of A just after the collision is $v^2/r - \ddot{r}$ rather than just $v^2/r$? If not, why not?

It seems odd to me to allow the effect of the velocity of A to be transmitted essentially instantaneously by the string to B (and causing B to accelerate) while not allowing the effect of the acceleration of B to be transmitted essentially instantaneously back to A.

If you allow any "time delays" due to elasticity of the string, then the question becomes ambiguous. I think this is ehild's point and why he doesn't like the question. You could then argue that the acceleration of B is zero immediately after the collision since there has not been enough time for the effect of the collision at A to propagate to B.

I too have same doubt. This question came in test papers of some coaching institution. And in the solution they simply wrote $a=v^(2) /l.$ I asked same question about the radial acceleration of $M_{a}$ here:(https://brilliant.org/discussions/thread/please-help-18/?ref_id=771297) but I didn't get any convincing answer.

Mass A is not pivoted about a fixed point so here mass A will not have centripetal acceleration only. So I think it is not evident to consider centripetal acceleration of block only. I think the answer should be $g/18$.

 

[ehild]

If you allow any "time delays" due to elasticity of the string, then the question becomes ambiguous. I think this is ehild's point and why he doesn't like the question. You could then argue that the acceleration of B is zero immediately after the collision since there has not been enough time for the effect of the collision at A to propagate to B.

Yes, it is my point. Collision between two bodies can be solved by applying conservation of momentum and the condition for energy but the problem is undetermined in case of three or more bodies without knowing how the mechanical deformation propagates in them.
In this problem, the pan gets an impulse. Its mass is negligible, so the impulse is transmitted to the string around the left pulley to block A.Block A hangs on the other string, and the tension in it just compensates its weight, so A starts to move horizontally. That is the outcome of the first stage of collision.
I do not know how you would solve the problem if all bodies take part in the collision instantaneously.
A nice example for collision among more bodies is Newton's cradle. What we see is that after the first ball hits the next, the last ball emerges. http://en.wikipedia.org/wiki/Newton's_cradle. That result is obtained if we assume collisions happening one after the other, between a pair of balls as if the balls do not touch each other. But they do. So the true solution would consider elastic wave traveling in the balls.

Block A hangs on the vertical string around the right pulley. If the pulley had mass the block was able to start its motion along a circle. So its motion would not change the length L₀ of the string, but changes the tension in it. Well, the pulley has negligible mass, but the situation could be the same.

Assuming A moves horizontally with speed u. With TSny notation, r he length of the left piece of the rope changes as $r=\sqrt{L^2+(ut)^2}$.
If t=0, the rate of change is zero, and the acceleration is $\ddot r = \frac{u^2}{L }$. It implies that the second derivative of the right piece is the negative of it, $\ddot s = - \frac{u^2}{L } = a_B$.

But my opinion is that such problems can not be solved and should not be given to high-school students.
For collision among more bodies, Newton's cradle is a nice example. The simple solution assumes that collisions happen between two balls, one after other, as if the balls do not touch each other. But they do, and the exact solution should include the propagation of the elastic wave inside the balls. http://en.wikipedia.org/wiki/Newton's_cradle

 

[TSny]

Assuming A moves horizontally with speed u. With TSny notation, r he length of the left piece of the rope changes as $r=\sqrt{L^2+(ut)^2}$.
If t=0, the rate of change is zero, and the acceleration is $\ddot r = \frac{u^2}{L }$. It implies that the second derivative of the right piece is the negative of it, $\ddot s = - \frac{u^2}{L } = a_B$.

It seems to me that your expression for $r$ as a function of time is equivalent to treating A as moving along a horizontal line with constant speed just after the collision. Thus, A would have zero vertical component of acceleration. Another way to see this is to use the standard expression for radial acceleration in polar coordinates to write $a_{A,y} = r \dot{\theta}^2 - \ddot{r}$ where y is taken as positive upward. According to your result, $\ddot{r} = \frac{u^2}{L }$. Therefore, $a_{A,y} = r \dot{\theta}^2 - \ddot{r} = \frac{u^2}{L } - \frac{u^2}{L } = 0$.

But if $a_{A,y} = 0$, then $T-Mg = Ma_{A,y} = 0$. So, $T-Mg = 0$ for A. But then $T-Mg = 0$ also for B if we assume the tension is the same throughout the string. Therefore the net force on B is zero and B would have zero acceleration, which is not consistent with $a_B = \ddot{r}$

 

[ehild]

It seems to me that your expression for $r$ as a function of time is equivalent to treating A as moving along a horizontal line with constant speed just after the collision. Thus, A would have zero vertical component of acceleration. Another way to see this is to use the standard expression for radial acceleration in polar coordinates to write $a_{A,y} = r \dot{\theta}^2 - \ddot{r}$ where y is taken as positive upward. According to your result, $\ddot{r} = \frac{u^2}{L }$. Therefore, $a_{A,y} = r \dot{\theta}^2 - \ddot{r} = \frac{u^2}{L } - \frac{u^2}{L } = 0$.

But if $a_{A,y} = 0$, then $T-Mg = Ma_{A,y} = 0$. So, $T-Mg = 0$ for A. But then $T-Mg = 0$ also for B if we assume the tension is the same throughout the string. Therefore the net force on B is zero and B would have zero acceleration.

That is what I assumed. B does not accelerate during the collision between the falling particle, pan and A, but starts to accelerate after as consequence of the motion of A.
Of course, it is an approximation. In reality, everything is involved but then you have to take elasticity of the strings and mass of the pulleys into account.

 

[TSny]

That is what I assumed. B does not accelerate during the collision between the falling particle, pan and A, but starts to accelerate after as consequence of the motion of A.

How do you make $a_B = 0$ consistent with $a_B = \ddot{r} \neq 0$?

 

[ehild]

a_B=0 during the collision, a_B≠0 after the collision.
There is some time delay before B starts to move. The limit of it is zero when the string is ideal. See the video when a ball collides with two others, connected by a spring. The collision happens between the touching balls, and the other ball starts to move a bit later, when it feels the tension built up in the string.