Find the magnitude and direction of the velocity

[inigooo]

Why is it so?

If it is 0m/s how can ut have a direction?

Ohhh. so it should be, no direction? Then a) 0m/s and b) no direction?

 

[malawi_glenn]

Ohhh. so it should be, no direction? Then a) 0m/s and b) no direction?

Why should it be 0m/s?

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

 

[inigooo]

Why should it be 0m/s?

Because (v’2)/(v’1)
(7.67m/s)/(0m/s)= 0m/s?
if it’s wrong, what should i do to get the magnitude of the velocity in terms of its velocity before contact?

 

[malawi_glenn]

Because (v’2)/(v’1)
(7.67m/s)/(0m/s)= 0m/s?
if it’s wrong, what should i do to get the magnitude of the velocity in terms of its velocity before contact?

Divide by 0?
How do you know that the secondary ball has velocity 0m/s after collision? Can you calculate it?

 

[inigooo]

Divide by 0?
How do you know that the secondary ball has velocity 0m/s after collision? Can you calculate it?

how can i get the v’1?

 

[inigooo]

how can i get the v’1?

pls. what should i do to get the final answers

 

[Steve4Physics]

@inigooo, Can I chip in. The question says:

“A secondary ball with the same weight as the first ball is ... just touching the cord” [my underlining]​

The wording in the question is (IMO) poor. I guess it should also say that the two balls have the same radius. So the centre of the upper ball is above the edge of the lower ball.

A good diagram is needed showing the positions of the balls at the moment of impact.

I suggest stepping-back a bit; revise/read-up how to solve simple ‘oblique collision’ problems. There is plenty of help available, e.g. a YouTube search for “elastic oblique collision in two dimensions” gives a number of videos. The first video I got (which is not necessarily the best) was:


EDIT: Can I add this, on edit. I'm not saying that the impact is a simple oblique collision because the presence of the (presumably inextensible) chord is a major constraint. However, considering the line-of-centres and the tangent plane between the balls - as we would for a simple oblique collision - is important.

 

[malawi_glenn]

Something like this:

Because if the balls have to be treated as pointmasses, one have to assume something regarding the chord.

 

[kuruman]

Something like this:
View attachment 305409

Because if the balls have to be treated as pointmasses, one have to assume something regarding the chord.

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.

 

[malawi_glenn]

Why point masses?

I mean something like this

so that it effectively becomes a 1-d problem. But then one has to assume something about the chord.

 

[kuruman]

If this were a 1-d problem, then why ask for the direction of the velocity in part (b)?

 

[haruspex]

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

I'm not sure the problem is fully defined. There are in effect two impacts, one between the balls, which we are told is elastic, but one also in the sudden tensioning of the string. We can assume the string is inextensible but is it elastic or fully inelastic?

Edit: in the case of an elastic string, we would also need to know the ratio between the two "spring constants". So assume fully inelastic.

Edit 2:
Correction: the solution @kuruman and I got independently assumes no loss of KE, so everything is assumed to be perfectly elastic.
Our other assumption is that the struck ball only moves sideways. Whether e.g. the string's spring constant being much greater than that of the balls would make that true I am not sure. More generally, it seems feasible that the struck ball would also acquire some vertically upward velocity.

But what if we take the string to be perfectly inelastic? Still puzzling over that one.

 

[malawi_glenn]

If this were a 1-d problem, then why ask for the direction of the velocity in part (b)?

Same/oppsite are directions no?

 

[inigooo]

I'm not sure the problem is fully defined. There are in effect two impacts, one between the balls, which we are told is elastic, but one also in the sudden tensioning of the string. We can assume the string is inextensible but is it elastic or fully inelastic?

Edit: in the case of an elastic string, we would also need to know the ratio between the two "spring constants". So assume fully inelastic.

Hello. Is the problem unsolvable?

 

[inigooo]

Same/oppsite are directions no?

But the direction has to be in degrees with respect to North,south, east, or west

 

[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 ?