Spheres hanging by strings - Finding impulse of tension

[Saitama]

Homework Statement

Two equal spheres of mass m are suspended by vertical strings so that they are in contact with their centres at the same level. A third equal sphere of mass m falls vertically and strikes elastically the other two simultaneously so that their centres at the instant of impact form an equilateral triangle in a vertical plane. If u is the velocity of m just before impact, find the velocities just after impact and the impulse of tension of the strings.

Homework Equations

The Attempt at a Solution

After collision, the hanging spheres will gain velocity along the direction of common normal of spheres. (Common normal is the line joining the centres of sphere)

See attachment 2 (I haven't shown the strings and rotated the figure).
Conserving momentum in x-direction,
$mu=mv'+2mv''\cos(\pi/6)$
From the equation of coefficient of restitution,
$$e=\frac{\text{Relative speed after collision}}{\text{Relative speed after collision}}$$
The relative speed before collision is $u\cos(\pi/6)$ and after collision is $v''-v'\cos(\pi/6)$ and for elastic collision, coefficient of restitution is 1. Plugging and rearranging,
$$v''-v'\cos(\pi/6)=u\cos(\pi/6)$$
Solving the two equations I have doesn't give me the right answer. :(

Any help is appreciated. Thanks!

 

[voko]

Clearly the hanging spheres are constrained to move horizontally immediately after the collision. So their strings must have impulses preventing any vertical motion, just like the problem indicates.

 

[Saitama]

Clearly the hanging spheres are constrained to move horizontally immediately after the collision. So their strings must have impulses preventing any vertical motion, just like the problem indicates.

So do I assume that after collision, the hanging spheres have velocity in x direction and conserve momentum again?

 

[voko]

On your diagram, the x-axis is vertical. As I said, the hanging spheres have no motion vertically: it is prevented by the strings.

 

[Saitama]

On your diagram, the x-axis is vertical. As I said, the hanging spheres have no motion vertically: it is prevented by the strings.

Sorry about that but what should I do now? I don't really know how to begin.

 

[voko]

One thing to observe is that momentum is not conserved in vertical direction because of the impulsive tension in the strings; this yields an equation. Momentum is conserved in the horizontal direction, which has an immediate consequence (which is also obvious from symmetry). Energy is conserved, too.

 

[Saitama]

One thing to observe is that momentum is not conserved in vertical direction because of the impulsive tension in the strings; this yields an equation.

And which equation is that?

Conserving energy,
$$\frac{1}{2}mu^2=\frac{1}{2}mv'^2+2 \times \frac{1}{2}mv''^2$$
where v'' is the horizontal velocity of hanging balls after collision.

 

[voko]

And which equation is that?

Vertical momenta and impulses.

 

[Saitama]

Vertical momenta and impulses.

Change in vertical momentum i.e impulse is $mv'-mu$ but how does it help? Impulse is force multiplied by time but I don't have the time of collision here. I am sorry if I miss something obvious.

 

[voko]

Change in vertical momentum i.e impulse is $mv'-mu$ but how does it help? Impulse is force multiplied by time but I don't have the time of collision here. I am sorry if I miss something obvious.

You are not required to find the force, you are required to find the impulse, so not knowing the time is OK.

Knowing the impulse by one string should yield another equation that relates the momentum of the striking sphere, the impulse of the string, and the momentum of the hanging ball.

 

[Saitama]

Knowing the impulse by one string should yield another equation that relates the momentum of the striking sphere, the impulse of the string, and the momentum of the hanging ball.

I still don't understand what to do.

Let me clear something first. I found the impulse to be $mv'-mv$. Is this the net impulse by "two" strings? How do I find the impulse by one string? Divide the above by 2? And how do I relate it to the momentum of hanging ball? The momentum of hanging ball is always zero in the vertical direction so I don't understand what am I supposed to do here.

 

[voko]

The falling ball's momentum is purely vertical. Yet the impact is via two off-vertical directions. So the exchange of the momenta happens via these two directions. Can you decompose the incoming momentum into those two directions? Can you decompose the reaction from the hanging balls into those two directions? The hanging balls acquire strictly horizontal momenta. Add to that the impulsive tensions delivered vertically. What do you get?

 

[Saitama]

Can you decompose the incoming momentum into those two directions?

Is it $mv\cos(\pi/6)$? Just to be sure, you are talking about the line joining the centres of the spheres. Right?

Can you decompose the reaction from the hanging balls into those two directions?

Why you ask me to decompose the reaction force? I mean the reaction force is already along the line joining the centres of the spheres.

 

[voko]

Is it $mv\cos(\pi/6)$? Just to be sure, you are talking about the line joining the centres of the spheres. Right?

The direction is correct, but the magnitude is not. The sum of these two momenta must be $mv$, vertically downward.

Why you ask me to decompose the reaction force? I mean the reaction force is already along the line joining the centres of the spheres.

I did not say "force". But I did say "reaction", which was wrong. I should have said "the recoil" of the falling ball after the collision.

 

[Saitama]

The direction is correct, but the magnitude is not. The sum of these two momenta must be $mv$, vertically downward.

$mv/\sqrt{3}$?

I did not say "force". But I did say "reaction", which was wrong. I should have said "the recoil" of the falling ball after the collision.

I still don't get this. Can you please elaborate?

 

[voko]

I still don't get this. Can you please elaborate?

The ball falls down. You decomposed its initial momentum into two directions.

After the collision, it goes up. So its has some upward momentum now. Decompose it into those same directions.

Then for each direction you can set up a vector equation relating the incoming momentum, the recoil momentum, the momentum of the hanging ball, and the impulse by the string.

 

[Saitama]

Then for each direction you can set up a vector equation relating the incoming momentum, the recoil momentum, the momentum of the hanging ball, and the impulse by the string.

Let the sphere has velocity v' in the vertical direction after collision.

Are you talking about direction AB? Change in momentum of striking sphere in that direction is $mv'/\sqrt{3}+mu/\sqrt{3}$. The tension is in vertical direction and momentum of hanging balls in the horizontal direction. How do I relate them?

 

[voko]

I = incoming momentum (AB direction)
R = recoil momentum (AB direction)
S = hanging ball momentum (horizontal)
T = impulse of tension (vertical)

Total change of momentum: R + S - I.
Total impulse: T.

So?

 

[Tanya Sharma]

Some observations-

1)The hanging masses moves in horizontal direction after collision (say v).
2)The falling mass moves vertically (say w).
3)The net impulse in vertical direction on hanging mass is zero .

Now these are the steps you need to take

1) Apply impulse momentum theorem in horizontal direction on left(or right) hanging mass .
2) Apply impulse momentum theorem in vertical direction on left(or right) hanging mass .
3) Apply restitution equation along the normal on falling mass and left(or right) hanging mass .
4) Apply impulse momentum theorem on either falling mass or left(or right) hanging mass or both .

Be careful with the signs.

What do you get with step 1 ?

 

[Saitama]

I = incoming momentum (AB direction)
R = recoil momentum (AB direction)
S = hanging ball momentum (horizontal)
T = impulse of tension (vertical)

Total change of momentum: R + S - I.
Total impulse: T.

So?

Okay, I am redefining the coordinate system to make things easier. The vertical direction is y-axis and horizontal as x-axis. Sorry for the inconvenience, I should have done this from the beginning.

I have drawn a vector diagram showing R,S and I. Change in momentum is:
$$\vec{T}=\left(mv''-\frac{mv'}{2\sqrt{3}}-\frac{mu}{2\sqrt{3}}\right) \hat{i}+\left(\frac{mv'}{2}+\frac{mu}{2}\right) \hat{j}$$
What should I do now?

 

[Saitama]

3)The net impulse in vertical direction on hanging mass is zero .

How? Isn't that what the question asks? Can you please define the directions? Are you referring to the second attachment I posted in the first post?

 

[voko]

T is purely vertical, use that.

P.S. I am not completely sure about the signs, but I trust you know what you are doing.

 

[Tanya Sharma]

How? Isn't that what the question asks? Can you please define the directions? Are you referring to the second attachment I posted in the first post?

The net impulse is zero ,not impulse due to tension .The net impulse is due to Normal force(by falling mass) and the tension due to string.

Consider vertical and horizontal directions.

 

[Saitama]

T is purely vertical, use that.

P.S. I am not completely sure about the signs, but I trust you know what you are doing.

Thanks a lot voko! I have reached the correct answer.

The net impulse is zero ,not impulse due to tension .The net impulse is due to Normal force(by falling mass) and the tension due to string.

I am still confused on this. If the net impulse in the vertical direction (I hope you know I redefined my coordinate system in the previous post) is zero, then I should be able to conserve momentum in that direction.

 

[Tanya Sharma]

I am still confused on this. If the net impulse in the vertical direction (I hope you know I redefined my coordinate system in the previous post) is zero, then I should be able to conserve momentum in that direction.

What is the initial and final momentum of the hanging mass(left mass) in the vertical direction ?

 

[voko]

Perhaps a more systematic solution would first consider the falling ball alone. At impact, it experiences two impulses from the hanging balls, so U - D = N1 + N2, where D is the downward momentum, U the upward momentum, and N1 and N2 are impulses of the normal forces from the hanging balls. Then for each hanging ball S = -N + T.

 

[Saitama]

What is the initial and final momentum of the hanging mass(left mass) in the vertical direction ?

Zero.

Perhaps a more systematic solution would first consider the falling ball alone. At impact, it experiences two impulses from the hanging balls, so U - D = N1 + N2, where D is the downward momentum, U the upward momentum, and N1 and N2 are impulses of the normal forces from the hanging balls. Then for each hanging ball S = -N + T.

Let $N_1$ be from the left ball and $N_2$ be from the right ball. Then
$$\vec{N_1}+\vec{N_2}=(mv'+mu)\hat{j}$$
$$\vec{T}-\vec{N_1}=-mv''\hat{i}$$
$$\vec{T}-\vec{N_2}=mv''\hat{i}$$
Are the above equations correct?

 

[voko]

Assuming you follow the sign convention for u and v' consistently, those equations should be fine (I hope). That's a lot more unknowns than before, though.

 

[Tanya Sharma]

Zero.

So ,if the initial and final momentum of the hanging mass in vertical direction are zero, the net impulse on the hanging mass has to be zero.

 

[Saitama]

Assuming you follow the sign convention for u and v' consistently, those equations should be fine (I hope). That's a lot more unknowns than before, though.

Impulse due to tension can be found using the equations above, add the second and third. It comes out to be the same expression as #20. But I still need one more equation.

The equation in post #20 reduces to T=(mv'+mu)/2 using the fact that tension is purely vertical. So I got one more equation from there by equating the x-component to zero. How do I obtain the same using this method?

 

[Saitama]

So ,if the initial and final momentum of the hanging mass in vertical direction are zero, the net impulse on the hanging mass has to be zero.

Lets begin with step 1, that would give me that the horizontal velocities of the left and right masses are equal. But you ask me to apply it on a single sphere. Consider the right sphere, the impulse is due to reaction force, right?

 

[voko]

I would say that we have to assume in both cases that tension is purely vertical, exactly like we assume the hanging balls can only move horizontally; this is given by the constraints on the system.

Or do I misunderstand your inquiry?

 

[Tanya Sharma]

Lets begin with step 1, that would give me that the horizontal velocities of the left and right masses are equal. But you ask me to apply it on a single sphere. Consider the right sphere, the impulse is due to reaction force, right?

Right .Lets call it N(impulse due to falling mass on hanging left mass). What will be the equation ?

 

[Saitama]

I would say that we have to assume in both cases that tension is purely vertical, exactly like we assume the hanging balls can only move horizontally; this is given by the constraints on the system.

Or do I misunderstand your inquiry?

I phrased it badly, sorry.

From post #20,
$$\vec{T}=\left(mv''-\frac{mv'}{2\sqrt{3}}-\frac{mu}{2\sqrt{3}}\right) \hat{i}+\left(\frac{mv'}{2}+\frac{mu}{2}\right) \hat{j}$$
Since the tension is purely vertical, I can equate the x-component to zero. So
$$\left(mv''-\frac{mv'}{2\sqrt{3}}-\frac{mu}{2\sqrt{3}}\right)=0$$ and $$\vec{T}=\frac{mv'+mu}{2}\hat{j}$$
I already have an equation from conservation of energy. I can solve these two equations to reach the answer.

Now using the equations I posted in #27, add the second and third, that gives
$$2\vec{T}-(\vec{N_1}+\vec{N_2})=0 \Rightarrow \vec{T}=\frac{\vec{N_1}+\vec{N_2}}{2}$$
From the first equation, $\vec{N_1}+\vec{N_2}=(mv'+mu)\hat{j}$
$$\Rightarrow \vec{T}=\frac{mv'+mu}{2}\hat{j}$$
This is the same result I got before but how do I find the other equation I got by equating the x-component to zero. I hope you understand what I ask.

Right .Lets call it N(impulse due to falling mass on hanging left mass). What will be the equation ?

N=mv''?

 

[Tanya Sharma]

N=mv''?

No...

Ncos60° =mv where v is the horizontal velocity of the hanging left mass .