Find the magnitude and direction of the velocity

[inigooo]

Since it says that ‘in terms of its velocity before contact’? Then the magnitude of its velocity is 0m/s or unchanged, then the direction is upward or opposite to v1?

Is this wrong?

 

[malawi_glenn]

Is this wrong?

Why do you think your answer is correct?

 

[inigooo]

Why do you think your answer is correct?

Since it says that ‘in terms of its velocity before contact’. Then the magnitude of its velocity is 0m/s or unchanged, then the direction is upward or opposite to v1.

 

[inigooo]

Since it says that ‘in terms of its velocity before contact’. Then the magnitude of its velocity is 0m/s or unchanged, then the direction is upward or opposite to v1.

Pls. Help me point out what is wrong, if it is wrong…

 

[malawi_glenn]

Then the magnitude of its velocity is 0m/s or unchanged, then the direction is upward or opposite to v1.

Why is it so?

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

 

[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.