Collision involving blocks and pulley

In summary, the conversation is discussing a problem involving a particle of mass M/2 colliding with a pan and the resulting tension and acceleration in the string connecting the pan to another block. Multiple possible solutions are proposed, including treating the two blocks as a system and considering the centripetal force on block A, but the final answer is not agreed upon. The conversation ends with a question about how the leftwards impulse on block A can result in a downwards acceleration.
  • #1
Tanya Sharma
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Homework Statement



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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 .
 

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  • #2
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.
 
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  • #3
andrevdh said:
Since A and B is connected by a taught string shouldn't the magnitude of their acceleration be the same?
Yes .
andrevdh said:
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 ?
 
  • #4
Tanya Sharma said:
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?
 
  • #5
Tanya Sharma said:
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.
 
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  • #6
Hello TSny ,

Thanks for replying .

TSny said:
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 ?
 
  • #7
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.
 
  • #8
Do you think that B will move downward?
 
  • #9
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 ?
 
  • #10
Tanya Sharma said:
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.
 
  • #11
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##.
 
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  • #12
TSny said:
I think so. I also get that B accelerates upward at g/18.

The answer given is option b) i.e g/9 :rolleyes:
 
  • #13
Tanya Sharma said:
The answer given is option b) i.e g/9 :rolleyes:
Good. We must still have something to learn! I don't see a mistake, so we can hope that someone will set us straight.
 
  • #14
TSny said:
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 .
 
  • #15
Weight of A tries to make the system move counterclockwise while weight of B acts clockwise.
 
  • #16
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 .
 
  • #17
Tanya Sharma said:
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.
 
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  • #18
TSny said:
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 .
 
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  • #19
Tanya Sharma said:
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.
 
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  • #20
Sorry...but some confusion has crept in :confused:.

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

TSny said:
The string on the left side is accelerating downward

How ? I can't visualize it :oldeek: .

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.
 
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  • #21
Tanya Sharma said:
Sorry...but some confusion has crept in :confused:.

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 :oldeek: .

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.
 
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  • #22
TSny said:
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 ?
 
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  • #23
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
 
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  • #24
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 .
 
  • #25
Doesn't ##\ddot{r}>0## imply that left string length increases in vertical direction at time just after the collision ?
 
  • #26
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.
 
  • #27
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## .
 
  • #28
Tanya Sharma said:
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 ##.
 
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  • #29
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 .

TSny said:
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 ?

TSny said:
And we have ## \ddot{r} = a_B ##.

Sign issues :oldfrown:.

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 :oldshy: .
 
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  • #30
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.
 
  • #31
TSny said:
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 :oldtongue: .

Thank you very very much :oldsmile: . I really appreciate your help .
 
  • #32
Tanya Sharma said:
Any further questions and I guess you would surely kill me :oldtongue: .

Thank you very very much :oldsmile: . 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.
 
  • #33
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.
 
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  • #34
ehild said:
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##
 
  • #35
TSny said:
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.
 
<h2>1. What is a collision involving blocks and pulley?</h2><p>A collision involving blocks and pulley is a type of physical interaction between two or more objects that are connected by a pulley system. The objects may collide with each other or with the pulley, resulting in a change in their motion and energy.</p><h2>2. How does a collision involving blocks and pulley occur?</h2><p>A collision involving blocks and pulley can occur when one or more objects connected by a pulley system are set in motion and come into contact with each other or with the pulley. This can be caused by an external force acting on one or more of the objects or by the objects' own inertia.</p><h2>3. What factors affect the outcome of a collision involving blocks and pulley?</h2><p>The outcome of a collision involving blocks and pulley can be affected by various factors such as the mass and velocity of the objects, the type of material they are made of, and the angle at which they collide. The elasticity of the objects and the friction between them and the pulley can also play a role in the outcome.</p><h2>4. What is the conservation of energy and momentum in a collision involving blocks and pulley?</h2><p>The conservation of energy and momentum states that in a closed system, the total energy and momentum remain constant before and after a collision. In a collision involving blocks and pulley, the total energy and momentum of the system will be conserved as long as there are no external forces acting on the objects.</p><h2>5. How can a collision involving blocks and pulley be calculated or analyzed?</h2><p>A collision involving blocks and pulley can be calculated or analyzed using principles of physics such as Newton's laws of motion and the conservation of energy and momentum. The mass, velocity, and angle of collision can be measured or estimated, and equations can be used to determine the final velocities and energy of the objects after the collision.</p>

1. What is a collision involving blocks and pulley?

A collision involving blocks and pulley is a type of physical interaction between two or more objects that are connected by a pulley system. The objects may collide with each other or with the pulley, resulting in a change in their motion and energy.

2. How does a collision involving blocks and pulley occur?

A collision involving blocks and pulley can occur when one or more objects connected by a pulley system are set in motion and come into contact with each other or with the pulley. This can be caused by an external force acting on one or more of the objects or by the objects' own inertia.

3. What factors affect the outcome of a collision involving blocks and pulley?

The outcome of a collision involving blocks and pulley can be affected by various factors such as the mass and velocity of the objects, the type of material they are made of, and the angle at which they collide. The elasticity of the objects and the friction between them and the pulley can also play a role in the outcome.

4. What is the conservation of energy and momentum in a collision involving blocks and pulley?

The conservation of energy and momentum states that in a closed system, the total energy and momentum remain constant before and after a collision. In a collision involving blocks and pulley, the total energy and momentum of the system will be conserved as long as there are no external forces acting on the objects.

5. How can a collision involving blocks and pulley be calculated or analyzed?

A collision involving blocks and pulley can be calculated or analyzed using principles of physics such as Newton's laws of motion and the conservation of energy and momentum. The mass, velocity, and angle of collision can be measured or estimated, and equations can be used to determine the final velocities and energy of the objects after the collision.

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