What are the speeds of P and M while riding Ents and playing catch with rings?

[Macy]

Homework Statement

P and M are each riding Ents while playing catch with two rings. P throws his gold ring with a speed relative himself that is twice the speed that M throws his silver ring relative to himself. P and M start off going with the same speed and direction, 30m apart with M in front of P. Both P and M throw and catch their rings at the same time. Ans P throws his ring with a speed that is 5 m/s faster relative to himself than the speed that he is traveling. What are the speeds that P and M riding?

I set Vp= Velocity of P with respect to earth
T = time
Vpb = Velocity of P ball with respect to earth
Vmb= Velocity of M ball with respect to earth

Homework Equations

Since it said P throws his ring 5 m/s faster I determined that in the Earth's reference frame the velocity Vpb = 2Vp +5 by simple Galilean Relativity. Then we know the speed of the other ball is half of what P perceives in M's reference frame so according to M Vmb = (V+5)/2, and according to the earth, this Vmb = Vp - (V+5)/2 because the ball is going in the opposite direction as Vp.

The Attempt at a Solution

So using the above work, we know that
30- Vmb(t)=Vp*(t) and then 30+Vp= Vpb(t).
I rearranged the equations and got Vp(t) on both sides and subtracted but I am still left with Vp and T.. Specifically, I got 30 = t(2Vp-10)..What am I doing wrong and how do I get the velocity.

 

[Macy]

Can anyone help me with this?

 

[TSny]

Hi, Macy. I'm having trouble understanding the problem statement. The problems states that

(1) "P and M start off going with the same speed and direction".

(2) "Both P and M throw and catch their rings at the same time".

(3) " P throws his gold ring with a speed relative [to] himself that is twice the speed that M throws his silver ring relative to himself."Statement (1) implies that if you go to the frame of reference moving with P, then in this frame both P and M will be at rest. Then, in order for statement (2) to be true, wouldn't P and M need to throw their rings with the same speed relative to themselves? But this contradicts statement (3).

Am I misinterpreting the problem?

EDIT: I just realized that there is not necessarily a contradiction since P and M can throw their rings at different angles relative to the ground. So, I will go to the drawing board and think some.

EDIT 2: I'm finding that allowing for different angles of toss doesn't help.

 

[Merlin3189]

Perhaps a diagram will help?

 

[Macy]

I don't think the angles have anything to do with it..?

Attached is my work:

https://imgur.com/a/g8xjy

 

[TSny]

I set Vp= Velocity of P with respect to earth
T = time
Vpb = Velocity of P ball with respect to earth
Vmb= Velocity of M ball with respect to earth

I determined that in the Earth's reference frame the velocity Vpb = 2Vp +5 by simple Galilean Relativity.

I agree with this equation. (We've switched from rings to balls and we're neglecting gravity.)

Then we know the speed of the other ball is half of what P perceives in M's reference frame so according to M Vmb = (V+5)/2

Does the V on the right side represent the same thing as Vp (the velocity of P relative to the earth)? If so, wouldn't the right side represent the speed of M's ball relative to M? But you defined the symbol Vmb to be the velocity of M's ball relative to the earth.

and according to the earth, this Vmb = Vp - (V+5)/2 because the ball is going in the opposite direction as Vp.

I agree with the equation Vmb = Vp - (V+5)/2 if V is the same as Vp. If this is true, then you can simplify this equation.

30- Vmb(t)=Vp*(t)

It would help if you motivated your equation with some explanation. It appears that this equation is trying to express the fact that at the time t (when P catches M's ball), the position of P relative to the Earth must be the same as the position of M's ball relative to the earth. If so, I don't think the minus sign is correct on the left side of the equation.

30+Vp= Vpb(t)

There's a typo on the left side where Vp should be multiplied by t. Otherwise, I agree with this equation. It says that the position of M relative to the Earth equals the position of P's ball relative to the Earth at the instant t when M catches P's ball.

I rearranged the equations and got Vp(t) on both sides and subtracted but I am still left with Vp and T.. Specifically, I got 30 = t(2Vp-10)..What am I doing wrong and how do I get the velocity.

I'm not following this.

But, I still think the problem has no solution since it is impossible for P and M to throw the balls with different speeds relative to themselves in such a way that they throw the balls simultaneously and catch the balls simultaneously.

 

[Macy]

So is the answer that it is impossible?

 

[TSny]

So is the answer that it is impossible?

It's impossible according to the way I interpret the problem. Is the problem from a textbook? Was a diagram provided?

 

[Macy]

https://imgur.com/a/g8xjy

Here is the diagram I drew and my work at the top of the page. Its not from a textbook. Also, I do think that you can solve for the velocities but I am stuck because I keep ending up with Just Vp and T.

Thanks so much for your help TSny..!

 

[TSny]

I agree with the following equations except for the sign circled in red

If you change that minus sign to a plus sign, you will find that the two equations are inconsistent.

 

[Macy]

Shouldn't it be negative though, because the velocity of the ball is positive and we need to subtract it from 30? I am sorry the negative signs are a bit confusing in Galilean for me..

EDIT: I see my mistake

 

[Merlin3189]

faster than the speed that he is traveling relative to a stationary observer
Your diagram has made my misunderstanding clear. I thought they were on parallel tracks! Now they are in line, it works ok.

Whether it can be solved depends on how much information we have.

I don't understand

Ans P throws his ring with a speed that is 5 m/s faster relative to himself than the speed that he is travelling.

Is this the answer we are supposed to find? Or is it more information to enable us to find the answer to

What are the speeds that P and M riding?

What I don't understand about the Äns is: Say he were traveling at 100m/sec would that mean he threw it at 105m/sec relative to himself, or at 5m/sec relative to himself but we (stationary observer) would think it was 105 m/sec because he is already moving at 100 m/sec relative to us?

Edit: Just to make that clearer. Is the Ans saying,
P throws his ring with a speed relative to himself that is 5 m/s faster than the speed that he is traveling relative to a stationary observer
or
P throws his ring with a speed that is 5 m/s faster relative to himself than the speed that he is traveling relative to a stationary observer
or
P throws his ring with a speed that is 5 m/s faster relative to himself than the speed that he is traveling relative to himself.

It's the "faster relative to" that puzzles me. Whatever we measure speeds relative to, their difference remains the same. So faster is never relative.

Edit: Ah! I've just noticed S is next to D on the kbd. Is Ans a typo for And ?

 

[Macy]

Yeah sorry Ans is a typo sorry!

 

[Macy]

To clarify, Ans is supposed to be And, and the ball is moving 5 m/s faster relative to his speed relative to the Earth in his frame of reference.

 

[Merlin3189]

Ok. So it's solvable now.
I've used a different type of diagram, so I'm still trying to match my calculations up with yours.

All my equations are in a stationary frame of reference.

 

[Merlin3189]

Now. I have a problem! Trying to solve it in the P reference frame, I've got a different result! So I've done something wrong.

It's late. I'll have to look again in the morning.

Edit: quick last thought before I go. I think TSny is right. My mistake was in assuming it is solvable and ignoring an inconsistency. that way you maybe can get multiple answers.

 

[Macy]

I think I may have an answer but I do not know if it is correct.

We know P's ball is moving at 2x, because it is twice the speed of M's ball in M's reference frame which is X.

So since P and M are both moving with velocity V, I just shifted it so the system is not moving meaning the velocity of P's ball is now 2x-V ( Because you subtract v so the system is at rest) and X+v (since the ball was going the opposite direction we add v).

Since they move the same distance and the time is equal I found 2x-v=x+v

And since we know v+5 =2x, I got v =5/3

Does this work look ok?

 

[TSny]

We know P's ball is moving at 2x, because it is twice the speed of M's ball in M's reference frame which is X.

To be clear, these are speeds as measured in the frame moving with P and M. Right?

So since P and M are both moving with velocity V, I just shifted it so the system is not moving

Is this non-moving system the same as the reference frame of the ground?

meaning the velocity of P's ball is now 2x-V ( Because you subtract v so the system is at rest)

If P's ball has speed 2x relative to P, then the velocity of P's ball relative to the ground would be 2x + V.

and X+v (since the ball was going the opposite direction we add v).

Relative to the ground, the velocity of M's ball would be V - x. This could come out positive or negative depending on the relative magnitudes of V and x. If V - x is positive, then relative to the ground M's ball is traveling in the same direction that P is moving relative to the ground. If V - x is negative, then M's ball is moving relative to the ground in a direction opposite to the direction that P is moving relative to the ground.

Since they move the same distance and the time...

Relative to the ground, the balls do not move equal distances in equal times.

 

[Merlin3189]

I've now looked at this from both points of view and with tentative assumptions about the size of their common velocity, but I am not only able to get different results, but to get equations from a single scenario that are blatantly contradictory. TSny was right in the first place.
Looking at my assumptions I conclude that the fly in the ointment is the assumption that, they can throw the rings at different speeds, but both throw and catch them at the same times. If that is impossible, then any calculation based on that premise must be flawed.

When you look at it for the point of view of P and M themselves, it seems obviously impossible.
P throws a ring to M 30m away at the same time as M throws a ring to him. They both catch the rings at the same (absolute) time. Both think they are not moving and that the distance between them is fixed. So they MUST calculate the same speed for each ring. To claim that one throws his ring faster than the other is to claim an impossibility. They are lying!

TSny introduces the idea that they may throw the rings in a direction not towards the other person and that an external agency such as gravity also affects them. I had not considered this as I assumed they were riding on starship Enterprises in space. So what is an ent?
Since I've only just noticed this, I haven't tried any maths on it yet. If they are moving horizontally, it still seems that the horizontal velocity in the direction PM must be the same and that the vertical motion must be the same, because the same field is operating over the same time. If they are moving not perpendicular to the field, then I need to give it some more thought. We can't then use their frame of reference, since they are being accelerated in the direction of motion, but the contradictions in the world frame I think will still exist.

I wonder, since this question specifically mentions Galilean relativity, whether it is simply a question to see whether you can spot the impossibility of the question's premises?