Spheres hanging by strings - Finding impulse of tension

[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 .

 

[Saitama]

No...

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

Yes, sorry about that. It slipped my mind while I was writing the post.

Should I move on to step 2?

 

[Tanya Sharma]

Yes...Let impulse due to tension - T.

 

[Saitama]

Yes...Let impulse due to tension - T.

$T=N\sin(\pi/3)$?

 

[Tanya Sharma]

$T=N\sin(\pi/3)$?

:thumbs:...Now step 3

Velocity of the falling mass -u
Velocity of the left mass -v
Recoil velocity of the falling mass -w

 

[Saitama]

:thumbs:...Now step 3

The striking sphere bounces back with velocity v' upwards. Applying the equation of coefficient of restitution,
$$u\cos(\pi/6)=v''\cos(\pi/3)+v'\cos(\pi/6)$$
Correct?

I am unsure about step 4. Didn't I do the same in step 1 and 2?

 

[Tanya Sharma]

Is v'' horizontal velocity of left mass after collision ?

 

[Saitama]

Is v'' horizontal velocity of left mass after collision ?

Yes.

 

[Tanya Sharma]

The striking sphere bounces back with velocity v' upwards. Applying the equation of coefficient of restitution,
$$u\cos(\pi/6)=v''\cos(\pi/3)+v'\cos(\pi/6)$$
Correct?

Well done ...Keep going ...Now the final step :)

I am unsure about step 4. Didn't I do the same in step 1 and 2?

No...Consider the falling mass .The impulse is due to the two normal forces(one from left mass,other from right) .

 

[Saitama]

No...Consider the falling mass .The impulse is due to the two normal forces(one from left ,other from right) .

The impulse on falling mass is $2N\cos(\pi/3)=mv'+mu$. Since N=mv'', $\sqrt{3}mv''=mv'+mu$.

 

[Tanya Sharma]

The impulse on falling mass is $2N\cos(\pi/3)=mv'+mu$. Since N=mv'', $\sqrt{3}mv''=mv'+mu$.

Why do you keep writing N=mv'' when it is Ncos(π/3)=mv'' ?

Recheck the component of impulse due to Normal in vertical direction.

 

[Saitama]

Why do you keep writing N=mv'' when it is Ncos(π/3)=mv'' ?

Recheck the component of impulse due to Normal in vertical direction.

Very sorry again, the correct equation is $2\sqrt{3}mv''=mv'+mu$.

 

[Tanya Sharma]

What is the component of impulse due to normal by left mass alone in vertical direction ?

 

[voko]

Take the 3rd equation from #27. Dot-multiply with T; you get $T^2 = T N_2 \cos \pi/6$, where $T$ is known, so you can find the $N_2$. Same for $N_1$. Thus you get the magnitudes of $N_1$ and $N_2$, and, because their directions are known, you get the entire vectors. Then you get $v''$ from the third equation.

 

[Saitama]

What is the component of impulse due to normal by left mass alone in vertical direction ?

$N\cos(\pi/6)$?

Take the 3rd equation from #27. Dot-multiply with T; you get $T^2 = T N_2 \cos \pi/6$, where $T$ is known, so you can find the $N_2$. Same for $N_1$. Thus you get the magnitudes of $N_1$ and $N_2$, and, because their directions are known, you get the entire vectors. Then you get $v''$ from the third equation.

Awesome. Your method of solving the problems using vectors is really cool. I will give that a try tomorrow. Its getting late here.

Thank you both for spending your precious time to help me.

 

[Tanya Sharma]

Very sorry again, the correct equation is $2\sqrt{3}mv''=mv'+mu$.

Yes...That is the correct equation

You have all four equations . Now it is just a matter of algebra .

 

[Saitama]

Yes...That is the correct equation

You have all four equations . Now it is just a matter of algebra .

Thanks a lot Tanya!