Spheres hanging by strings - Finding impulse of tension

[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!