Problem on conservation of momentum and collision

[Titan97]

Homework Statement

Three identical balls (of equal masses) are connected by light inextensible strings and kept on a smooth horizontal surface. The middle ball B is given a velocity $v_0$ at $t=0$. Find the velocity of ball A when A collides with C.

Homework Equations

Impulse $J=\Delta P$

The Attempt at a Solution

Since there is no net external horizontal force on the system, acceleration of Centre of Mass along $x$ is zero. Also, $v_{cm,x}=0$.

Velocity of A and B along the string should be equal to $v\sin\theta$ so that the string does not extend.

Along $x$, velocity of A is $v\sin\theta\cos\theta$. If I can assume that all three balls have small radius, the final velocity of A will be equal to velocity of B which varies due to tension.

I tried using work-energy theorem but initially Tension is impulsive and the velocity of A and B increase suddenly. (I am not sure about this but I learned I can't use work-energy theorem when there is a sudden increase in velocity after reading one of the insight articles)

That leaves me with conservation of momentum of the system. Momentum of ball A or C cannot be conserved due to tension. But I am finding it difficult to get the equation.

 

[Nathanael]

If I can assume that all three balls have small radius, the final velocity of A will be equal to velocity of B which varies due to tension.

I do not understand this part.

Along $x$, velocity of A is $v\sin\theta\cos\theta$.

I don't believe this is correct. You are correct that the components of V_A and V_B must be the same in the direction of the length of string, but there can be a nonzero component in the perpendicular direction which would contribute to the x-component of V_A.

I tried using work-energy theorem but initially Tension is impulsive and the velocity of A and B increase suddenly. (I am not sure about this but I learned I can't use work-energy theorem when there is a sudden increase in velocity after reading one of the insight articles)

Yes, using conservation of energy would be an assumption. If you're unsure (as am I in this case) you should solve it by other means and then check if energy was conserved.

(Edited:)

But I am finding it difficult to get the equation.

You could choose coordinates and try to write and solve equations for the motion in time, but there may be a simpler way.

 

[PeroK]

Homework Statement

Three identical balls (of equal masses) are connected by light inextensible strings and kept on a smooth horizontal surface. The middle ball B is given a velocity $v_0$ at $t=0$. Find the velocity of ball A when A collides with C.

You may want to forget about equations to begin with and try to figure out an important constraint that must apply when ball A collides with ball C.

 

[Titan97]

Since length of string does not change, $x^2+y^2=L^2$. then $x\frac{dx}{dt}+y\frac{dy}{dt}=0$ Is this the equation @PeroK ?
When they collide, x=0.

And,

I do not understand this part.

What I meant is, just when the ball are about to collide, the velocity of the two lower balls will be only along $y$ and will be equal to that of velocity of B.

 

[PeroK]

Since length of string does not change, $x^2+y^2=L^2$. then $x\frac{dx}{dt}+y\frac{dy}{dt}=0$ Is this the equation @PeroK ?
When they collide, x=0.

And,

What I meant is, just when the ball are about to collide, the velocity of the two lower balls will be only along $y$ and will be equal to that of velocity of B.

Your last statement is the key. The velocity of the balls (in the y direction) when they collide must be equal to the velocity of ball B.

 

[Titan97]

What kind of equation are you suggesting @PeroK ?
You mean equation of trajectory of ball A and C?

 

[haruspex]

What I meant is, just when the ball are about to collide, the velocity of the two lower balls will be only along $y$

Why?

What kind of equation are you suggesting @PeroK ?
You mean equation of trajectory of ball A and C?

More likely, conservation laws.

 

[Titan97]

It will have a component along x. But its velocity along y should be equal to velocity of ball B

 

[haruspex]

It will have a component along x. But its velocity along y should be equal to velocity of ball B

Right.
What conservation laws can you apply?

 

[Titan97]

Conservation of momentum in a direction perpendicular to string?

 

[haruspex]

Conservation of momentum in a direction perpendicular to string?

Do you mean perpendicular to the direction the string takes at a particular time, or generally?
The benefit of conservation laws is that they allow you to relate the starting state to the final state directly, without having to worry about what goes on in between.
Yes, momentum will be conserved, but remember that momentum is a vector, so in a 2D set-up that is effectively two equations.
What else will be conserved?

 

[Titan97]

Momentum of the system can be conserved. But for ball A, tension is an external force. So how can momentum of ball A or C can be conserved individually?

 

[haruspex]

Momentum of the system can be conserved. But for ball A, tension is an external force. So how can momentum of ball A or C can be conserved individually?

I'm not suggesting their momenta are conserved individually. In terms of the unknown final velocities, write out the initial and final momenta for the total system.

 

[Titan97]

Initially, its $3mv_0$ along y-axis.

Finally, its $3mv$ along y-axis and 0 along x-axis?

 

[haruspex]

Initially, its $3mv_0$ along y-axis.

how do you get that?

 

[Titan97]

Initially, a total mass of 3m has a velocity $v_0$ upwards.

 

[Orodruin]

Initially, a total mass of 3m has a velocity $v_0$ upwards.

This is not correct, only the middle ball has an initial velocity.

 

[Titan97]

$v_{cm}=\frac{mv_0}{3m}$
So initial momentum is mv₀/3 ?

 

[Orodruin]

$v_{cm}=\frac{mv_0}{3m}$
So initial momentum is mv₀/3 ?

No, initially you have one moving mass of mass $m$ which is moving with velocity $v_0$. What does that mean for the initial momentum?

 

[Titan97]

Its $mv_0$. Finally, along y-axis, its $3mv$ where $v$ is the final velocity of any ball along y-axis.

 

[haruspex]

Its $mv_0$. Finally, along y-axis, its $3mv$ where $v$ is the final velocity of any ball along y-axis.

Right. You need another equation. What else is conserved?

 

[Titan97]

The velocity of the ball A and C suddenly increases. So Energy can't be conserved (just like the chain problem). But no external force acts on the center of mass. So energy of the system can be conserved.

 

[Orodruin]

The velocity of the ball A and C suddenly increases.

Why do you think so? The initial velocities of A and C are zero.

 

[Titan97]

Tension is impulsive.

 

[Orodruin]

Tension is impulsive.

There is no tension at the initial moment.

 

[Titan97]

I understand now.
$$mv_0=3mv$$
$$v=\frac{v_0}{3}$$
$$\frac{1}{2}mv_0^2=\frac{1}{2}mv^2+2\cdot\frac{1}{2}m(v+v_x)^2$$
$$v_x=\frac{v_0}{\sqrt{3}}$$
$$|v_A|=|v_C|=\sqrt{v^2+v_x^2}=\frac{2v_0}{3}$$

 

[haruspex]

$$\frac{1}{2}mv_0^2=\frac{1}{2}mv^2+2\cdot\frac{1}{2}m(v+v_x)^2$$

That's wrong, but the next line onwards was right, so I guess that was just a transcription error.

 

[haruspex]

There is no tension at the initial moment.

That isn't the critical fact, is it? Work is conserved because there's no sudden change in speed.

 

[Orodruin]

That isn't the critical fact, is it? Work is conserved because there's no sudden change in speed.

Indeed, either argument works.

 

[haruspex]

Indeed, either argument works.

I disagree.
Suppose there is a mass on a frictionless table, attached to a taut string that passes over a pulley to a suspended mass. Initially, the first mass is held in place, then released. There is nonzero tension right from the start, but acceleration is smooth, no sudden jumps in speed, so work is conserved.

 

[Orodruin]

I disagree.
Suppose there is a mass on a frictionless table, attached to a taut string that passes over a pulley to a suspended mass. Initially, the first mass is held in place, then released. There is nonzero tension right from the start, but acceleration is smooth, no sudden jumps in speed, so work is conserved.

This in no way invalidates my argument. It is just an argument for why your argument works. In this case, there is no tension before the middle ball is hit and there is no tension immedeately after. There is no impulse change of the outer balls and therefore no violation of work conservation. The balls would have to change their speed at the instant of the hit for this to occur.

 

[haruspex]

This in no way invalidates my argument. It is just an argument for why your argument works. In this case, there is no tension before the middle ball is hit and there is no tension immedeately after. There is no impulse change of the outer balls and therefore no violation of work conservation. The balls would have to change their speed at the instant of the hit for this to occur.

I think you are saying that because the tension increases gradually from zero, it follows that there is no sudden change in speed. That is true, but my quibble with your wording in post #25 is that it might give the reader the impression that such a profile for the tension is the key requisite for work conservation here. I contend that it is not. The tension could have been nonzero right from the start yet work be conserved. Moreover, tension could suddenly change from zero to nonzero yet work be conserved. The key consideration is a sudden change in speed.

 

[Titan97]

So in this case, work is conserved since there is no sudden change in speed (acceleration is finite).

 

[haruspex]

So in this case, work is conserved since there is no sudden change in speed (acceleration is finite).

Yes.