Understanding Velocity Addition Laws for People on Train & Ground

[rgtr]

Homework Statement

q1
Bob is standing in the middle of a train moving at velocity v at constant velocity in the right direction. Bob throws 2 identical baseballs in opposite direction at the same speed at a constant velocity on the train. From Alice perceptive on the ground what is the velocity of the train?

q2
Now imagine I have 2 people running at a constant velocity towards each other in the same direction lets call the person on the left Alice and the person on the right Bob. Calculate from Alice's frame and Bob's frame the constant velocity?

q3
Now imagine I have 2 people running to the right in the same direction located at different locations let's call the person on the left Alice and the person on the right Bob. Calculate from Alice's frame and Bob's frame the constant velocity?

I am also assuming the earth is an inertial reference frame in all the questions

Relevant Equations

velocity addition vectors.

I kind of just made up the questions. I realize this is a basic question but my knowledge of physics is very limited.q1 answer

v_left_ball = v_left_ball - v_train
v_right_ball = v_right_ball + v_train

q2 answer

To get the speed from Bob's frame I would use v_Bob = v_Bob + v_Alice
To get the speed from v_Alice = v_Bob + v_Alice


q3 answer

To get the speed from Bob's frame I would use v_Bob = v_Bob - v_Alice
To get the speed from Alice's frame I would use v_Alice = v_Alice - V_Bob
Assuming I am correct

Why are the people on the train and the ground different velocity addition laws?
I can't seem to find a discernible pattern.

Let's find the velocity_Left_ball for on the train.

IOW's q1 answer can be drawn as from Alice's frame.

This is a shorter distance
v_train-------><-------v_left_ball
vs
q1 answer from Alice's frame I don't understand how to draw it or whyIn Alice's frame for the people on the ground running, I intuitively understand it but can't think of a vector diagram. Can someone help ?If I made any mistakes then please correct me.

 

[Drakkith]

q1
Bob is standing in the middle of a train moving at velocity v at constant velocity in the right direction. Bob throws 2 identical baseballs in opposite direction at the same speed at a constant velocity on the train. From Alice perceptive on the ground what is the velocity of the train?

Given your answer to this question I assume you meant to ask what's the velocity of each ball with respect to the ground.

q1 answer

v_left_ball = v_left_ball - v_train
v_right_ball = v_right_ball + v_train

The first answer is not correct. Let's say the train is moving at a speed of 5A along the x-axis in the positive direction (towards the right). Let's also say that Bob throws each baseball at a speed of A in opposite directions along the x-axis with respect to the train.

The velocity of each ball is the sum of the train's velocity with respect to the ground and the balls' velocities with respect to the train.
Left ball with respect to ground: V_lbg = V_train + V_lbt = (5A,0,0)+(-A,0,0) = (4A,0,0). The left ball is moving to the right at a speed of 4A with respect to the ground.
Right ball with respect to ground: V_rbg = V_train + V_rbt = (5A,0,0)+(A,0,0) = (6A,0,0). The right ball is moving to the right at a speed of 6A with respect to the ground.

The velocity of each ball with respect to the ground is in this case just simple vector addition. This is NOT the case for your next two questions. Questions 2 and 3 are about Galilean transformations, not just about finding the speed of two objects with respect to the same inertial frame.

q2
Now imagine I have 2 people running at a constant velocity towards each other in the same direction let's call the person on the left Alice and the person on the right Bob. Calculate from Alice's frame and Bob's frame the constant velocity?

Ignoring that two people can't run towards each other AND run in the same direction, it's unclear which velocity you are trying to find just by looking at the question. Given your stated answer I assume you're looking for the velocity of Bob or Alice with respect to the other, but as written it almost looks like you're asking for the velocity of Alice and Bob with respect to the ground. Also, your answer is giving the speed, not the velocity, which might seem pedantic but I think mistaking speed for velocity is part of why you're confused.

Moving on, if you try to add the two velocities, as viewed from the frame of the ground, both will sum to zero since they are equal in magnitude but opposite in direction. Instead you subtract. Let's say the velocity of Alice and bob with respect to the ground is V and V' along the X-axis respectively and that the magnitude of each velocity is A. To find the velocity of Bob as viewed from Alice's frame, V_ba, you would subtract V from V'. So V_ba = V'-V. Decomposing into vectors we get V_ba = V'-V = (-A,0,0)-(A,0,0) = (-2A,0,0).

If Alice was moving at 2A wrt the ground then Bob's velocity in her frame is V_ba = V'-V = (-A,0,0)-(2A,0,0)=(-3A,0,0) instead.

q3
Now imagine I have 2 people running to the right in the same direction located at different locations let's call the person on the left Alice and the person on the right Bob. Calculate from Alice's frame and Bob's frame the constant velocity?

The same method from above applies here. Let's say Alice is moving at a speed of A and Bob at 2A wrt to the ground. V_ba=V'-V=(2A,0,0)-(A,0,0)=(A,0,0). Bob appears to be moving away from Alice at a speed of A to the right along the X-axis.

To reiterate what I said above, questions 2 and 3 are about Galilean transformations. You are finding the velocity of a single object as viewed from multiple reference frames. Alice's and the Ground's frames in these examples since our initial velocities are given with respect to the ground and then we move into Alice's frame.

 

[kuruman]

All your equations are incorrect. That's because you use the same symbol to denote two different things. This is no different from calling both observers Alice. To illustrate, your first equation says

v_left_ball = v_left_ball - v_train

The correct way to proceed algebraically is to subtract v_left_ball from both sides of the equation and end up with 0 = -v_train which is nonsense. The way to deal with relative velocity problems is to use double subscripts to denote what the velocity of what is relative to what. In this case, one would define

$V_{\text{TA}}=$ velocity of Train relative to Alice. (Bob is at rest w.r.t. the train so he doesn't count.)
$V_{\text{1T}}=$ velocity of ball 1 relative to Train
$V_{\text{1A}}=$ velocity of ball 1 relative to Alice.

The pattern that you are looking for is to say that one of these terms is the sum of the other two. To know which two are added, look at the subscripts and find the subscript that appears both on the left and on the right. Add these two and set them equal to the third velocity. In the above example you see that "A" is only on the right, "1" is only on the left but "T" is on the left and on the right. Therefore,
$V_{\text{1A}}=V_{\text{TA}}+V_{\text{1T}}.$

An additional rule is $V_{\text{BA}}=-V_{\text{AB}}.$ This comes in very handy in case you end up with $V_{\text{AB}},~V_{\text{BC}},~V_{\text{CA}}$ where all the three subscripts appear on the left and on the right. You swap one pair and have $-V_{\text{BA}},~V_{\text{BC}},~V_{\text{CA}}$ to break the symmetry. Then you have "C" only on the left and the right so that you can write $-V_{\text{BA}}=V_{\text{BC}}+V_{\text{CA}}.$
This sets up the equation. With a little algebra, you can find anyone of the three velocities if you know the other two. It doesn't matter which pair you swap; the equation that you get will not be new.

 

[robphy]

Therefore,
$V_{\text{1A}}=V_{\text{TA}}+V_{\text{1T}}.$

It might be better to write this as
$V_{\text{1A}}=V_{\text{1T}}+V_{\text{TA}}$
so that the train-frame is an intermediate frame of reference for velocity composition.
(And one can easily generalize this form to a chain of intermediate frames.)

In addition, (using $V_{\text{BA}}=-V_{\text{AB}}$)
one can write relative velocities
$V_{\text{1A}}=V_{\text{1T}}-V_{\text{AT}}$
so that again the train-frame is an intermediate frame of reference.

 

[kuruman]

I think an "intermediate" frame of reference adds an extra layer of complexity to worry about without being really useful. Once one writes the equation placing the velocities to be added correctly, one can always swap indices and move terms around depending on what is unknown. The crucial step is to put the velocities with the correct signs in their proper place. Besides, what makes a frame intermediate? Suppose we are given the question shown below. How does the use of an intermediate frame simplify the solution shown in the spoiler?

Three ships are sailing in a straight line. The velocity of the Nina relative to the Pinta is 3 knots and the velocity of the Pinta relative to the Santa Maria is 2 knots. What is the velocity of the Santa Maria relative to the Nina?

Spoiler

$V_{\text{NP}}=3$ kn; $V_{\text{PS}}=2$ kn; $V_{\text{SN}}=?$
All 3 subscripts appear on the left and on the right.
Swap: $V_{\text{PN}}=-3$ kn which keeps only S on the left and on the right.
$V_{\text{PN}}=V_{\text{SN}}+V_{\text{PS}}$
$V_{\text{SN}}=V_{\text{PN}}-V_{\text{PS}}=-3$ kn $-2$ kn $=-5$ kn.

 

[robphy]

The intermediate frame (a third frame) is already in the problem.
I am not introducing one.
I am giving it a name.

I am merely suggesting a notation that clarifies the calculation, which can be generalized for more complicated situations.

 

[kuruman]

The intermediate frame (a third frame) is already in the problem.
I am not introducing one.
I am giving it a name.

I am merely suggesting a notation that clarifies the calculation, which can be generalized for more complicated situations.

I understand that you are not adding a frame. I don't understand (a) the rule that you use for identifying which of the three frames is "intermediate" and (b) how this identification simplifies/clarifies the rule to be used to write the equation relating the three frames. Perhaps you can illustrate using the example of the three ships in post #5.

 

[robphy]

Again, this is an issue of notation... there is no new physics being introduced.

Therefore,
$V_{\text{1A}}=V_{\text{TA}}+V_{\text{1T}}.$

It might be better to write this as
$V_{\text{1A}}=V_{\text{1T}}+V_{\text{TA}}$
with "1" and "A" on the outside (in that order) and "T" on the inside
which is
akin to writing the chain rule for $f(T(x))$ as
$\frac{df}{dx}=\frac{dT}{dx}\frac{df}{dT}$ suggesting $\frac{d}{dx}f=\frac{dT}{dx}\frac{d}{dT}f$
instead of
$\frac{df}{dx}=\frac{df}{dT}\frac{dT}{dx}$ (which does not have a similar re-write).

By extension, the chain rule for $f(S(T(x)))$ should be written as
$\frac{df}{dx}=\frac{dT}{dx}\frac{dS}{dT}\frac{df}{dS}$ suggesting $\frac{d}{dx}f=\frac{dT}{dx}\frac{dS}{dT}\frac{d}{dS}f$
instead of
$\frac{df}{dx}=\frac{df}{dS}\frac{dS}{dT}\frac{dT}{dx}$.

---

The quantity $V_{\text{1A}}$ primarily concerns frame-$1$ and frame-$A$.

• If we have a third frame frame-$T$ and we know $V_{\text{1T}}$ and $V_{\text{TA}}$,
  we have $V_{\text{1A}}=V_{\text{1T}}+V_{\text{TA}}$.
• If instead we had a fourth frame frame-$S$ and we know $V_{\text{1S}}$ and $V_{\text{SA}}$,
  we have $V_{\text{1A}}=V_{\text{1S}}+V_{\text{SA}}$.
• If instead we have both frames and we know $V_{\text{1S}}$ , $V_{\text{ST}}$ and $V_{\text{TA}}$,
  we have $V_{\text{1A}}=V_{\text{1S}}+V_{\text{ST}}+V_{\text{TA}}$.
• It seems appropriate to refer to "the frames other than the primary frames $1$ and $A$
  for the quantity $V_{\text{1A}}$" as "intermediate frames" (but one may choose other names).

My suggestion of this notation makes (in my opinion)
the bookkeeping and, possibly, the interpretation easier.

 

[kuruman]

My suggestion of this notation makes (in my opinion)
the bookkeeping and, possibly, the interpretation easier.

Yes, I now understand what you are saying and I agree.

 

[rgtr]

Thanks everyone.

 

[rgtr]

Just a follow up question does these rules work for someone running toward or away from a waves? Assume this doesn't involve light waves.

 

[Drakkith]

The same principles apply, but there is some additional math for finding frequency, wavelength, etc.

 

[David Lewis]

The velocity of each ball with respect to the ground is in this case just simple vector addition. This is NOT the case for your next two questions. Questions 2 and 3 are about Galilean transformations, not just about finding the speed of two objects with respect to the same inertial frame.

I assume the speed of the baseball is given with respect to train (Bob), but the question asks for the speed of the baseball with respect to ground (Alice).

Aren't Bob and Alice in two different frames?

 

[Drakkith]

Aren't Bob and Alice in two different frames?

Certainly.

 

[rgtr]

All your equations are incorrect. That's because you use the same symbol to denote two different things. This is no different from calling both observers Alice. To illustrate, your first equation says

The correct way to proceed algebraically is to subtract v_left_ball from both sides of the equation and end up with 0 = -v_train which is nonsense. The way to deal with relative velocity problems is to use double subscripts to denote what the velocity of what is relative to what. In this case, one would define

$V_{\text{TA}}=$ velocity of Train relative to Alice. (Bob is at rest w.r.t. the train so he doesn't count.)
$V_{\text{1T}}=$ velocity of ball 1 relative to Train
$V_{\text{1A}}=$ velocity of ball 1 relative to Alice.

The pattern that you are looking for is to say that one of these terms is the sum of the other two. To know which two are added, look at the subscripts and find the subscript that appears both on the left and on the right. Add these two and set them equal to the third velocity. In the above example you see that "A" is only on the right, "1" is only on the left but "T" is on the left and on the right. Therefore,
$V_{\text{1A}}=V_{\text{TA}}+V_{\text{1T}}.$

An additional rule is $V_{\text{BA}}=-V_{\text{AB}}.$ This comes in very handy in case you end up with $V_{\text{AB}},~V_{\text{BC}},~V_{\text{CA}}$ where all the three subscripts appear on the left and on the right. You swap one pair and have $-V_{\text{BA}},~V_{\text{BC}},~V_{\text{CA}}$ to break the symmetry. Then you have "C" only on the left and the right so that you can write $-V_{\text{BA}}=V_{\text{BC}}+V_{\text{CA}}.$
This sets up the equation. With a little algebra, you can find anyone of the three velocities if you know the other two. It doesn't matter which pair you swap; the equation that you get will not be new.

Sorry I took a break from physics for a while I got busy and just have a followup/ confirmation question.
Where ball 1 or B_1 is the left ball. Also obviously Alice is on the ground seeing the train move at velocity v in the right direction.

$VPA = VTA -(VB_1A)$ , when using velocity addition on the right hand of the equal sign does the right hand of the equal sign always need to equal the same frame unless performing algebra and changing it to plus? That kind of trips me up? What I mean is $Vxy = Viy - Vty$? Just assume I used random variables for $Vxy = Viy - Vty$.

 

[kuruman]

$VPA = VTA -(VB_1A)$ , when using velocity addition on the right hand of the equal sign does the right hand of the equal sign always need to equal the same frame unless performing algebra and changing it to plus? That kind of trips me up? What I mean is $Vxy = Viy - Vty$? Just assume I used random variables for $Vxy = Viy - Vty$.

Please explain clearly what you are asking. What is $VPA$ and what are you trying to say with $Vxy = Viy - Vty$? The subscripts denote reference frames fixed on objects moving relative to each other. For example, $V_{\text{TA}}$ is the velocity of a frame fixed on the Train relative to a frame fixed on Alice. In your equation you have x, y, i and t which means that you have 4 moving objects. What are they and how are they relevant to the original question?

 

[rgtr]

Sorry I was multitasking while posting the first time.$VBLA = VBLT + VTA$ =
$VBLT = VBLA - VTA$

When using the example $VBLT = VBLA - VTA$ velocity addition on the right hand of the equal sign of the equal sign, does the right hand of the equal sign always need to equal the same frame unless performing algebra and changing it to plus? I guess it could be a minus also but you get what I am saying.

 

[kuruman]

I would advise you to be consistent and use only two subscripts only and also not to string letters one after the other. For all I know $VBLT$ could stand for "vegan BLT sandwich." I get what you are trying to say, but it should be self-explanatory. This is how I would it.

1. Start with clear definitions of what stands for what.
$V_{\text{LT}}~$= velocity of the Left ball relative to the Train.
$V_{\text{RT}}~$= velocity of the Right ball relative to the Train.
$V_{\text{TA}}~$= velocity of Train ball relative to Alice.

2. Decide what you wish to find and write it down. Say you want the velocity of the left ball relative to Alice.
$V_{\text{LA}}~=~?$ The velocity of the right ball is irrelevant to this. You see that the letter "T" appears on the left and on the right in the subscripts which means that you connect the corresponding two velocities with a "+" sign and set the sum equal to the other velocity:
$V_{\text{LA}}=V_{\text{LT}}+V_{\text{TA}}.$
This equation says that the velocity of the left ball relative to Alice can be obtained by adding the velocity of the left ball relative to the train to the velocity of the train relative to Alice.

It is correct to algebraically move one term to the other side, change sign and write
$V_{\text{LA}}-V_{\text{LT}}=V_{\text{TA}}.$
What does this say? Well, you can swap left and right hand sides in the equation to have only one velocity on the left.
$V_{\text{TA}}=V_{\text{LA}}-V_{\text{LT}}.$
You can also eliminate the negative sign by swiping subscripts in the second term. That's because $V_{\text{LT}}=-V_{\text{TL}}$ which says the obvious: if the ball sees the train move in a given direction with a given speed, the train sees the ball move in the opposite direction with the same speed.
So finally the equation is transformed to
$V_{\text{TA}}=V_{\text{LA}}+V_{\text{TL}}.$
Note that this equation obeys the rule that the same subscript appears both on the left and on the right in the velocities that are connected with the "+" sign. It says that the velocity of the train relative to Alice is equal to the velocity of the left ball relative to Alice plus the velocity of the train relative to the left ball.

Let's see how this works for a specific problem with numbers.
Bob, at rest on a train, throws a ball to the left with speed 5 m/s. Alice, at rest on the ground, sees the ball moving to the right with speed 15 m/s. Find the velocity of Alice as seen by Bob on the train.
Solution
1. Write down in double-subscript notation what you are given and what you are looking for.
$V_{\text{LT}}=-5~$m/s; $V_{\text{LA}}=+15~$m/s; $V_{\text{AT}}=?$
2. See which subscript appears both on the left and on the right, combine the corresponding to velocities with a "+" sign and set them equal to the third. Note that "A" appear both on the left and on the right. Then
$V_{\text{LT}}=V_{\text{LA}}+V_{\text{AT}}.$
Put in the known numbers and solve
$-5~\rm{m/s}=15~\rm{m/s}+V_{\text{AT}} \implies V_{\text{AT}}=-25~\rm{m/s}.$
Answer: Bob sees Alice moving to the left with a speed of 25 m/s.

Finally, if it so happens that you are given and what you are looking for have double subscripts which are symmetric, i.e. all three letters appear both on the left an on the right, then swap subscripts in one velocity (doesn't matter which) and change sign to break the symmetry.

Do you see now how it all works?