Find the magnitude and direction of the velocity

[haruspex]

Case 1 limiting value would be 30∘ below the horizontal as per physics of an oblique elastic collision between 2 equal masses.

Good point. And that does validate my simulation, more or less.

could you perhaps set the string's spring constant to 100 times that of ball pair system ?

Unfortunately the simulation breaks down there. In a simple simulation of SHM, there's a tendency for the energy to grow exponentially. There are probably some standard tricks for taming this, but I haven't managed to find one.

 

[neilparker62]

Similar problem discussed in this thread:

https://www.physicsforums.com/threa...ings-finding-impulse-of-tension.712709/page-1

 

[haruspex]

Fwiw, I have now simulated it, ignoring gravity.
For some arbitrary choices of initial ball speed, mass, etc.:
- With the string having a tenth the spring constant of the ball pair, the left hand ball loses contact with a trajectory of 30.6^o below the horizontal.
- With the ball pair and string having the same spring constant, the trajectory is 21.5^o below the horizontal.
- With the string having ten times the constant, the trajectory is 13.1^o above the horizontal.

So as I surmised, if all is elastic, the relative spring constants are crucial.

Edit: although I intended to ignore gravity, I made the mistake of taking the string as always in tension. I need to allow it to go slack. So ignore the above results.

Having managed to tame the simulation somewhat (by taking any three consecutive values of a position or velocity coordinate to follow a quadratic) I am satisfied that when the string's spring constant is much the greater the suspended ball does not leave with any significant vertical velocity.

 

[neilparker62]

In the case where the string's spring constant is very high compared with that of the ball pair, I found this solution. (from page 11 of the referenced pdf). Am still trying to figure out exactly what they are doing but it looks like the limiting angle is about $16^{\circ}$ above the horizontal.

 

[neilparker62]

Just wondering if this situation is the same as if a ball were dropped onto a 60 degree wedge on a smooth frictionless table ? The wedge has the same mass as the ball and there is no 'slippage' on impact. Collision is elastic.

 

[haruspex]

Just wondering if this situation is the same as if a ball were dropped onto a 60 degree wedge on a smooth frictionless table ? The wedge has the same mass as the ball and there is no 'slippage' on impact. Collision is elastic.

I would say it’s identical. In principle, the table surface has some elasticity and that would behave just the same as the string in the original.

 

[neilparker62]

In the following analysis it should be noted that with respect to the Cartesian plane, $V_0$ is a negative quantity. Arbitrarily but conveniently readers may consider its value to be $-10ms^{-1}$.

Conservation of kinetic energy: $${V_0}^2={V_B}^2+{V_{AX}}^2+{V_{AY}}^2$$
Zero net horizontal momentum: $$V_{AX}=-V_B\implies {V_0}^2=2{V_{AX}}^2+{V_{AY}}^2$$
Collision impulse(s) acting along the normal are equal and opposite $\implies$ no change in relative velocity along the normal assuming $m_A=m_B$: $$V_0\cos\theta=V_B\sin\theta-V_{AY}\cos\theta-V_{AX}\sin\theta=-V_{AY}\cos\theta-2V_{AX}\sin\theta$$ $$\implies V_{AY}=-V_0-2V_{AX}\tan\theta$$ $$\implies {V_{AY}}^2={V_0}^2 + 4{V_{AX}}^2\tan^2\theta+4V_0V_{AX}\tan\theta$$ Substituting for $V_B$ and $V_{AY}$ in the kinetic energy equation, we obtain: $$2{V_{AX}}^2+4{V_{AX}}^2\tan^2\theta+4V_0V_{AX}\tan\theta=0$$ $$\implies V_{AX}=\frac{-2V_0\tan\theta}{1+2\tan^2\theta};V_{AY}=-V_0\frac{1-2\tan^2\theta}{1+2\tan^2\theta}$$ $$\tan\beta=\frac{1-2\tan^2\theta}{2\tan\theta}$$ where $\beta$ is the angle between the post collision trajectory of ball A and the horizontal axis.