Find the magnitude and direction of the velocity

[kuruman]

So assume fully inelastic.

That's what I assumed. I got an answer for the angle which is probably correct.

 

[inigooo]

That's what I assumed. I got an answer for the angle which is probably correct.

What would be the final answer?

 

[inigooo]

That's what I assumed. I got an answer for the angle which is probably correct.

Pls. Help me with this one, what would be the final answers?

 

[inigooo]

That's what I assumed. I got an answer for the angle which is probably correct.

Can you explain me how did you arrive with your answer? With complte solution so that i can understand how, pls

 

[inigooo]

Look i know you only want the correct answer and move on. But that is not how we operate on this forum

no, i’m asking for your help so that i can understand completely… however, i can Only comprehend it further if you can provide solutions in order to arrive with the correct answers. By providing the solutions in getting the answers will help me to understand more quickly

 

[inigooo]

That's what I assumed. I got an answer for the angle which is probably correct.

So pls, provide your solutions, even without the correct answers so that i can practice computing it on my own

 

[kuruman]

So far you have only calculated the speed of ball 1 just before the collision. After the collision, ball 1 will have a velocity that in general can be written as $\mathbf{v'}_1=v'_{1x}\mathbf{\hat x}+v'_{1y}\mathbf{\hat y}$. The secondary ball can only move in the horizontal direction and will have velocity $\mathbf{v'}_2=v'_{2x}\mathbf{\hat x}$. Momentum is conserved in the horizontal direction and you can write an expression for that. That's equation 1.

Now here comes the tricky part. The primary ball delivers an impulse only in the center-to-center direction. You need to write an expression for the impulse as the change in momentum of the primary ball and set its component tangent to the surface equal to zero. This will give you an expression relating the three primed components of the velocities. That's equation 2. For this step, it would help if you wrote the normal to the surface $\mathbf{\hat n}$ and the tangent to the surface $\mathbf{\hat t}$ at the point of contact in terms of $\mathbf{\hat x}$ and $\mathbf{\hat y}.$

Finally, you need to write the energy conservation equation which is equation 3. You have 3 equations and three unknowns which you can solve.

That's what I did and I got an answer. I am not sure that this is what your teacher wants you to do.

 

[Delta2]

So far you have only calculated the speed of ball 1 just before the collision. After the collision, ball 1 will have a velocity that in general can be written as $\mathbf{v'}_1=v'_{1x}\mathbf{\hat x}+v'_{1y}\mathbf{\hat y}$. The secondary ball can only move in the horizontal direction and will have velocity $\mathbf{v'}_2=v'_{2x}\mathbf{\hat x}$. Momentum is conserved in the horizontal direction and you can write an expression for that. That's equation 1.

Now here comes the tricky part. The primary ball delivers an impulse only in the center-to-center direction. You need to write an expression for the impulse as the change in momentum of the primary ball and set its component tangent to the surface equal to zero. This will give you an expression relating the three primed components of the velocities. That's equation 2. For this step, it would help if you wrote the normal to the surface $\mathbf{\hat n}$ and the tangent to the surface $\mathbf{\hat t}$ at the point of contact in terms of $\mathbf{\hat x}$ and $\mathbf{\hat y}.$

Finally, you need to write the energy conservation equation which is equation 3. You have 3 equations and three unknowns which you can solve.

That's what I did and I got an answer. I am not sure that this is what you teacher wants you to do.

That's all very nice but I think there is something we misunderstood in the problem statement, I refuse to believe that an introductory high school problem (I suspect that the OP is a high school student) has such a complex treatment.

 

[haruspex]

So far you have only calculated the speed of ball 1 just before the collision. After the collision, ball 1 will have a velocity that in general can be written as $\mathbf{v'}_1=v'_{1x}\mathbf{\hat x}+v'_{1y}\mathbf{\hat y}$. The secondary ball can only move in the horizontal direction and will have velocity $\mathbf{v'}_2=v'_{2x}\mathbf{\hat x}$. Momentum is conserved in the horizontal direction and you can write an expression for that. That's equation 1.

Now here comes the tricky part. The primary ball delivers an impulse only in the center-to-center direction. You need to write an expression for the impulse as the change in momentum of the primary ball and set its component tangent to the surface equal to zero. This will give you an expression relating the three primed components of the velocities. That's equation 2. For this step, it would help if you wrote the normal to the surface $\mathbf{\hat n}$ and the tangent to the surface $\mathbf{\hat t}$ at the point of contact in terms of $\mathbf{\hat x}$ and $\mathbf{\hat y}.$

Finally, you need to write the energy conservation equation which is equation 3. You have 3 equations and three unknowns which you can solve.

That's what I did and I got an answer. I am not sure that this is what you teacher wants you to do.

I did it a slightly different way. I let J be the impulse between the balls and resolved that into horizontal and vertical. The horizontal component relates to both the subsequent horizontal velocity of each and to the change in vertical velocity of the free ball. Then I used those equations to eliminate J.

 

[kuruman]

That's all very nice but I think there is something we misunderstood in the problem statement, I refuse to believe that an introductory high school problem (I suspect that the OP is a high school student) has such a complex treatment.

I agree with you, hence my last sentence in post #57.

 

[malawi_glenn]

That's all very nice but I think there is something we misunderstood in the problem statement, I refuse to believe that an introductory high school problem (I suspect that the OP is a high school student) has such a complex treatment.

There is a smilar problem in one of my high school books (but not the one I use for class) where it says nothing about the radii of the balls and is indeed treated as a 1-dimensional problem in my "teachers guide". But, something regarding the chord is mentioned in the problem statement. That is why I was so inclined on that track at first.

 

[Delta2]

There is a smilar problem in one of my high school books (but not the one I use for class) where it says nothing about the radii of the balls and is indeed treated as a 1-dimensional problem in my "teachers guide". But, something regarding the chord is mentioned in the problem statement. That is why I was so inclined on that track at first.

Hmm, I wonder what exactly is mentioned about the cord that allow us to treat this as a 1-d problem.

 

[neilparker62]

Why point masses?

Here is the diagram at the time of the collision assuming that the balls have the same radius $R$. It seems to me that the momentum of the two-ball system is not conserved in the vertical direction because the string constrains part of the system from moving in that direction. Assuming that there is no friction while the balls are in contact, the top ball delivers an impulse directed along the center-to-center direction which can be calculated from the geometry.

View attachment 305408

Whatever component of velocity the secondary ball had along the collision line (line of centres per Kuruman's diagram above) is lost. The ball retains the other component perpendicular to that.

 

[haruspex]

Whatever component of velocity the secondary ball had along the collision line (line of centres per Kuruman's diagram above) is lost.

Why?

 

[neilparker62]

Why?

In essence because when equal masses collide elastically (assume one is initially stationary), momentum along the line of collision is completely transferred. That is why the post collision trajectories of two equal masses are at right angles. (although in this case the motion of the primary mass is constrained by the string).

In detail I would need to blow own trumpet and refer you to my Insights article:

https://www.physicsforums.com/insig...-to-solving-2-dimensional-elastic-collisions/

 

[haruspex]

In essence because when equal masses collide elastically (assume one is initially stationary), momentum along the line of collision is completely transferred. That is why the post collision trajectories of two equal masses are at right angles. (although in this case the motion of the primary mass is constrained by the string).

In detail I would need to blow own trumpet

and refer you to my Insights article:

https://www.physicsforums.com/insig...-to-solving-2-dimensional-elastic-collisions/

You can't treat it as two separate events: the collision of the balls and the impulse from the string. It all happens at once, and I would expect some of the string's impulse to feed back into the rebound of the falling ball. In effect, the hanging ball has greater inertia.

 

[neilparker62]

You can't treat it as two separate events: the collision of the balls and the impulse from the string. It all happens at once, and I would expect some of the string's impulse to feed back into the rebound of the falling ball. In effect, the hanging ball has greater inertia.

Well I thought about exactly that. And came to the conclusion that if the string has "negligible" mass and is perfectly elastic, the implication is that the instantaneous collision is only between the two balls. The primary ball moves 'out of the way' before any of the string's impulse can 'feed back' to the secondary ball. Otherwise as you've pointed out, we start needing information about the string's spring constant and its tensile strength.

Edit 1: Put another way if the chord is something like a bungee chord which can extend (for practical purposes) indefinitely, then there will be negligible energy transfer to it over the very short collision $\Delta t$.

Edit 2: $k\Delta x << \frac{\Delta P}{\Delta t}$ where $\Delta P$ is the collision impulse , k is the spring constant and $\Delta x$ the extension over the brief collision period $\Delta t$.

 

[haruspex]

if the string has "negligible" mass and is perfectly elastic, the implication is that the instantaneous collision is only between the two balls

Don't confuse elasticity and extensibility.
In reality, all strings and solid bodies are partly elastic and at least a bit extensible/compressible. That means they have two spring 'constants', $k_{stress}>k_{relax}>0$. Common idealisations are:

• Perfectly elastic: $k_{relax}\rightarrow k_{stress}$
• Perfectly inelastic: $k_{relax}\rightarrow 0$
• Inextensible/incompressible: $k_{stress}\rightarrow\infty$

Idealised solutions are only valid as the limits of realistic solutions. (Feel free to quote this as Haruspex' Principle, or Principia Haruspicis.)

When more than one such idealisation is applied, ambiguities can arise wrt the relative rates of the limits. In the current problem we have the spring constants for the ball collision and for the string. You have assumed $k_{stress,string}/k_{stress,balls}\rightarrow 0$.

Consider a much simpler problem: a light elastic spring sits on a rigid mass, which sits on another light elastic spring, which sits on firm ground. A second equal rigid mass falls on top at speed v.

If $k_{upper}<<k_{lower}$ then most of the energy gets stored in the upper spring and the lower mass barely moves. Having a much higher frequency of oscillation, it rebounds immediately. The lower spring might as well be rigid.

If $k_{upper}>>k_{lower}$ then we have your bungee cord model. The upper mass comes almost to a stop and the lower mass moves down at speed v, then bounces back up. In the meantime, the upper mass is in free fall, and the value of $\frac{g^2m}{v^2k_{lower}}$ will be critical to the subsequent motion. If it is very small we can ignore gravity; the upper mass is stationary while the lower rebounds and, on second collision, transfers all the KE back to the upper mass, producing the same result as above.

In between, we could consider $k_{upper}=k_{lower}$, which produces a compound oscillation involving the Golden Ratio. To figure out what happens, we need to find the first time at which one of the springs returns to its relaxed length, thereby losing contact with the mass above it. E.g. for the lower spring I got $(\lambda-1)\sin(\omega't)=\sin(\omega t)$, where $\lambda=\frac{\sqrt 5-1}2$ and $\omega'=-\frac {\omega}{\lambda^2}$. That could be solved numerically for $\alpha=\omega t$.

In terms of the post #1 problem, we have to make some assumption about the ratio of the two spring constants.

Taking the string to have a much higher constant seems reasonable ($k_{string}/k_{balls}\rightarrow \infty$), but it still not clear to me what result that produces. Most likely it does allow the combination of KE conservation with a resulting purely horizontal velocity for the suspended mass, but I am uncertain.

If instead we take $k_{string}/k_{balls}\rightarrow 0$ ("bungee" model), we have to worry about a second/concurrent collision. For a given angle of incidence, and again ignoring gravity, that will depend on the value of $\frac{v^2m}{r^2k_{string}}$, where r is the radius of the ball. If sufficiently large, your original claim applies, and since we are taking $k_{string}\rightarrow 0$, that will be true.

Finally, if we take $k_{string}=k_{balls}=k$ and ignore gravity, simulation would seem to be the only reasonable prospect. Varying $\frac{v^2m}{r^2k}$ would need to be tried.

There's a dissertation in this.

 

[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.

 

[neilparker62]

Case 1 limiting value would be $30^{\circ}$ below the horizontal as per physics of an oblique elastic collision between 2 equal masses. Perhaps we can also find a limiting value for case 3 ?

Edit: In your simulation , could you perhaps set the string's spring constant to 100 times that of ball pair system ?

 

[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.