• Support PF! Buy your school textbooks, materials and every day products Here!

Galilean Relativity (Invariance) Problem

  • #1
175
9

Homework Statement


Imagine two inertial frames, S and S'. Inertial frame S' moves with velocity v0 = 5 m = s in the upward (positive y) direction as seen by an observer in frame S. Now imagine that a person at rest in frame S throws a ball with mass m straight up into the air with initial velocity v 0 (with respect to frame S) from an initial height of y=y0 = 0 at time t= 0. Use Galilean relativity, and take gravity into account. Use a reasonable number of signi cant digits and appropriate units.

Question 1: Does the ball appear to be moving in S'?
Q2: Acceleration in both S and S'
Q3: Max y height in S?
Q4: What is the position of the ball y'ymax according to frame S' when it reaches its maximum height ymax in frame S?

Homework Equations


Kinematic Equations

The Attempt at a Solution


So for Q1 I say that the ball does not appear to be moving. To better understand this I thought about a person in an elevator and a person in front stationary. The stationary person throws a ball up just as I pass in the elevator. During that time t>0 the ball will appear stationary since I will be essentially moving with the ball. However, since the ball has velocity v0 and I have a velocity of 5m/s this is where I thought that this may be wrong. Instead it would appear to be moving, but moving much more slowly. To understand this better I imagined the train scenario. A train is moving at 5m/s and someone throws a ball in the same direction of the train, it is moving slower since I am TRAVELLING WITH the ball essentially. Is that the correct reasoning? How would one write this in mathematics?

For Q2 I just used the kinematics equations from physics 1 and applied them to each reference frame. In the case of S it would be accelerating at 9.8 m/s with the upward direction being positive so for S it is -9.8 m/s. Now things get confusing for S'. This frame of reference IS moving WITH the ball. This means that there is a different look to it initially. I then realized that since the laws of physics match for all frames of reference wouldn't that mean the acceleration is still -9.8m/s? I am lost here.

Q3 should be h = √(1/(4.9m)) as I used (1/2)mgh and solved for h, but I think this is wrong.

Q4 is completely confusing and I have no idea where to begin or how to think about it.
 

Answers and Replies

  • #2
Merlin3189
Homework Helper
Gold Member
1,576
717
So for Q1 I say that the ball does not appear to be moving.
... I thought about a person in an elevator and a person in front stationary. The stationary person throws a ball up just as I pass in the elevator.
During that time t>0 the ball will appear stationary since I will be essentially moving with the ball. We are told that the ball is moving with velocity v0 at t=0, but it will not be moving at v0 at t>0

However, since the ball has velocity v0 and I have a velocity of 5m/s this is where I thought that this may be wrong. But v0 IS 5m/s !???
...
I agree with your idea that the observer in S' is moving at the same velocity as the ball at time t=0 according to the person in S. So at that time, the ball and the person in S' have the same velocity for all observers, including the observer in S' - he thinks he is not moving, so he thinks the ball is not moving, at t=0.

According to the person in S the ball does not continue at 5m/sec upwards, so he will see the ball moving at a different velocity to S' when t>0. So other observers will also see the ball and S' moving with different velocities when t>0.

If you simply ignore S' for the moment, you can work out how the ball moves according to the person in S.
When you know that, you can simply apply Galilean relativity to determine what the observer in S' sees.

Alternatively, now you know the starting condition for the ball in S', you can calculate the path of the ball in S' without reference to S at all.

Q2 I agree with you. Both reference frames are moving at constant velocity with respect to each other, so experience the same acceleration due to gravity. Why are you lost?

Q3. I'm not sure what you did here? Even if it were correct, (1/2)mgh seems to be a problem, since we don't know m. (Perhaps you used 1/2 mv^2 = mgh and cancelled m ?)
Just ignore S' . You are standing still in S, you throw the ball up at 5m/sec, what is its max height?

Q4. I think it would help to know when the ball reaches its max height.
 
  • #3
175
9
I agree with your idea that the observer in S' is moving at the same velocity as the ball at time t=0 according to the person in S. So at that time, the ball and the person in S' have the same velocity for all observers, including the observer in S' - he thinks he is not moving, so he thinks the ball is not moving, at t=0.

According to the person in S the ball does not continue at 5m/sec upwards, so he will see the ball moving at a different velocity to S' when t>0. So other observers will also see the ball and S' moving with different velocities when t>0.

If you simply ignore S' for the moment, you can work out how the ball moves according to the person in S.
When you know that, you can simply apply Galilean relativity to determine what the observer in S' sees.

Alternatively, now you know the starting condition for the ball in S', you can calculate the path of the ball in S' without reference to S at all.

Q2 I agree with you. Both reference frames are moving at constant velocity with respect to each other, so experience the same acceleration due to gravity. Why are you lost?

Q3. I'm not sure what you did here? Even if it were correct, (1/2)mgh seems to be a problem, since we don't know m. (Perhaps you used 1/2 mv^2 = mgh and cancelled m ?)
Just ignore S' . You are standing still in S, you throw the ball up at 5m/sec, what is its max height?

Q4. I think it would help to know when the ball reaches its max height.
Alright I think I managed to figure this problem out as a whole. So please bear with me;

Q1: This piece makes much more senes that the ball is moving. As gravity acts on the ball when t>0 we as an observer in S' would notice the ball is very slow in movement by some fraction due to gravity acting on the ball. So this means we as the observer do see the ball moving.

Q2: I was initially confused if there was additional acceleration in frame S' which is clearly errored thinking. It makes much more sense that both frames hold an acceleration equal to -9.8 m/s^2.

Q3: So this problem I simply did KEf+PEf=KEi+PEi meaning that PEf and KEi are both 0. From there we know what these are by writing it as (1/2)mv^2=mgh and solving for h gives h=((1/2)v^2)/g and pluging in v=5m/s and g=-9.8m/s^2 I get -1.28m or 1.28m.

Q4: This is the only part I am stuck on as I am confused on what I need to do or use. I know that Ymax is 1.28m so I am assuming that there will be some addition to the 5m/s variable to find the Y'max value. In other words, I am going to have to somehow use galilean relativity in the sense of u' = u, but that is where I am confused. This equation is for velocitys and accelerations, not position. How do I use this for position? Or am I simply thinking wrong about this scenario and it should be 1.28m in both frames?

Thanks for the help!!!
 
  • #4
Merlin3189
Homework Helper
Gold Member
1,576
717
Q1 - This actually seems a strange question to me. Once the ball is released, its motion is not constant velocity because it is being accelerated by gravity. So in any frame of reference which is moving with constant velocity, including 0, the ball must move and move with non-constant velocity.
The only way this question makes any sense to me is if it was, "Is the ball moving in S' at the moment it is released?"

Q3 - Ok I agree. In giving your answer, bear in mind the comment about significant figures: if you use g=9.8 m/sec, can you calculate the height as 1.28m?

Q4 - Again I'm a bit puzzled by the question here. " a person at rest in frame S throws a ball...straight up into the air with initial velocity v 0 (with respect to frame S) from an initial height of y=y0 = 0 at time t= 0" I'm not sure what frame of reference they are using for the height. I presume the y simply means the vertical direction, y0 means the y position at time t=0 and that this is 0 in the S frame. What height it is in the S' frame does not appear to be mentioned. Maybe we suppose the observer in S' uses the same labels, so that the height 0 is at the same place as the ball is released - ie y0 = y'0 = 0 for both people and their vertical rulers match at the instant of release. Then their rulers slide past each other at 5m/sec, so that after 2 sec, say, the 0 of S' ruler is level with the +10m mark on the S ruler.
There are two approaches to this now.
A- You've worked out where the ball is on the S ruler when it peaks. If you know how much the rulers have slid, you can say what position that matches on the S' ruler.
B- (more interesting I think, but not easier) You now know what happens in S - the ball is thrown up at 5m/sec at t=0 , it decelerates at g and after some time it stops rising at height 1.28m and starts to fall accelerating at g.
So how does this look from S' ? At t=0 the ball appears how? (Relative to S, you in S' are moving up at 5m/sec and the ball is moving up at 5m/sec.)
Then what does the ball do ?
If you work out the time the ball peaks in S, then you can calculate where it will be in S' at the same time.

Edit: thinking a bit more about it, you can do without the time, because you know the relative speeds. When the ball has v(0)=+5 in the S frame, it has v'(0)=0 in the S' frame. The frames have a constant relative speed of 5m/sec. When the ball peaks for S, you know v(peak) = 0 in S, so you know what v'(peak) is in S'. So you can calculate y'(peak) from v'(0), v'(peak) and g
 
Last edited:
  • #5
175
9
Q1 - This actually seems a strange question to me. Once the ball is released, its motion is not constant velocity because it is being accelerated by gravity. So in any frame of reference which is moving with constant velocity, including 0, the ball must move and move with non-constant velocity.
The only way this question makes any sense to me is if it was, "Is the ball moving in S' at the moment it is released?"

Q3 - Ok I agree. In giving your answer, bear in mind the comment about significant figures: if you use g=9.8 m/sec, can you calculate the height as 1.28m?

Q4 - Again I'm a bit puzzled by the question here. " a person at rest in frame S throws a ball...straight up into the air with initial velocity v 0 (with respect to frame S) from an initial height of y=y0 = 0 at time t= 0" I'm not sure what frame of reference they are using for the height. I presume the y simply means the vertical direction, y0 means the y position at time t=0 and that this is 0 in the S frame. What height it is in the S' frame does not appear to be mentioned. Maybe we suppose the observer in S' uses the same labels, so that the height 0 is at the same place as the ball is released - ie y0 = y'0 = 0 for both people and their vertical rulers match at the instant of release. Then their rulers slide past each other at 5m/sec, so that after 2 sec, say, the 0 of S' ruler is level with the +10m mark on the S ruler.
There are two approaches to this now.
A- You've worked out where the ball is on the S ruler when it peaks. If you know how much the rulers have slid, you can say what position that matches on the S' ruler.
B- (more interesting I think, but not easier) You now know what happens in S - the ball is thrown up at 5m/sec at t=0 , it decelerates at g and after some time it stops rising at height 1.28m and starts to fall accelerating at g.
So how does this look from S' ? At t=0 the ball appears how? (Relative to S, you in S' are moving up at 5m/sec and the ball is moving up at 5m/sec.)
Then what does the ball do ?
If you work out the time the ball peaks in S, then you can calculate where it will be in S' at the same time.

Edit: thinking a bit more about it, you can do without the time, because you know the relative speeds. When the ball has v(0)=+5 in the S frame, it has v'(0)=0 in the S' frame. The frames have a constant relative speed of 5m/sec. When the ball peaks for S, you know v(peak) = 0 in S, so you know what v'(peak) is in S'. So you can calculate y'(peak) from v'(0), v'(peak) and g
Ok I know v(peak)=0, but how do I find v'(peak)? Im sorry, but this whole relative frames thing is really confusing me. I am mulling over some equation for galilean relativity and cannot make a connection. I know that velocities are the same and acceleration is the same, but I don't know how to correlate that to v(peak). Or wait, is v'(peak)=0 since the frame S also has v(peak)=0 meaning that the y'(peak) is 0?
 
  • #6
Merlin3189
Homework Helper
Gold Member
1,576
717
After my last post I spent some time drawing pictures to help you. But after sleeping on it, I did wonder whether a mathematical approach was more what you wanted.
The simple connection between the two frames is
v' = v -5 eg. something stationary in S is going down at 5m/sec in S'
or v = v' +5 eg. something stationary in S' is going up at 5m/sec in S
So for eg. when the ball is moving up at 3m/sec in S, then it is moving at 3-5 = -2 or 2m/sec down in S'

For distances you can integrate that:
## \int v' \, dt = \int (v - 5) \, dt ##
## y' = y -5t +k ## the k being why I wondered how the two frames were matched up for distances.
I assumed they were equal at t=0, y'(0) = y(0) so k=0 and y'(t) = y(t) -5t
So for eg. if the ball were at 0.785m in S after 0.4sec, then for S' it would be at 0.785 - 5x0.4 = -1.215m
Because S' is moving up faster than the ball, someone in S' sees the ball going down rather than up
galilean_relativity.png
 
  • #7
175
9
After my last post I spent some time drawing pictures to help you. But after sleeping on it, I did wonder whether a mathematical approach was more what you wanted.
The simple connection between the two frames is
v' = v -5 eg. something stationary in S is going down at 5m/sec in S'
or v = v' +5 eg. something stationary in S' is going up at 5m/sec in S
So for eg. when the ball is moving up at 3m/sec in S, then it is moving at 3-5 = -2 or 2m/sec down in S'

For distances you can integrate that:
## \int v' \, dt = \int (v - 5) \, dt ##
## y' = y -5t +k ## the k being why I wondered how the two frames were matched up for distances.
I assumed they were equal at t=0, y'(0) = y(0) so k=0 and y'(t) = y(t) -5t
So for eg. if the ball were at 0.785m in S after 0.4sec, then for S' it would be at 0.785 - 5x0.4 = -1.215m
Because S' is moving up faster than the ball, someone in S' sees the ball going down rather than up
View attachment 210623
Wow!!! Thank you so much for going to that length of helping me draw it out. I had a wrong picture as I thought it was an xy plane where the x-axis was stationary and y-axis was moving up and down and that explains my utter frustration in trying to figure this out. So essentially when finding the time the ball reaches max height which you say is 0.5 and I calculated about 0.52 then you use that same time and simply subtract the max height from that of S' velocity and time. So essentially I am calculating what the height would be in S' when it moves 5m/s upward at 0.5s. Then taking that height I subtract from the S frames height and get an approximate of what the height is for the S' frame. It's starting to make much more sense considereing the analogy of two rulers where one moves upward at 5m/s and the other is stationary. So would that mean the answer is supposed to be negative since we are stating that the origin point is 0? I thank you very much for the time and effort put into helping me with this.
 
  • #8
Merlin3189
Homework Helper
Gold Member
1,576
717
Yes. It sounds like you've got the right idea now.

There are different ways to come at it, but all should get you to the same place.

The interesting thing about this question was that the ball is thrown up at exactly the same speed as the other frame is moving. So my first thought was that for the guy in S' the ball never moves up at all. It starts of moving at exactly the same speed as him - so to him it starts from rest - then just falls under gravity.
Now in S, the ball goes up, reaches its peak, then falls back down. We know that the falling part takes exactly the same time as the rising to a peak, so this falling part as the ball comes back down, looks to S guy much like what the S' guy sees during the first part.
So what the S guy sees as the ball starting up at 5m/sec, rising to a peak and coming to rest, the S' guy sees as the ball starting from rest and falling exactly the same distance until it reaches -5m/sec. So whatever height S sees it rise to, that is the distance S' sees it fall in the same time.
So the simple answer to Q4 is y' max = -y max and no further calculation needed.

(The time was a bit over 0.5sec, I just got fed up with writing extra digits! Especially when they depend on what value you use for g. I decide I wanted one more digit and used 9.81, but I'm not sure what accuracy they were expecting.)
 

Related Threads on Galilean Relativity (Invariance) Problem

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
18
Views
976
  • Last Post
Replies
1
Views
2K
Replies
9
Views
734
  • Last Post
Replies
3
Views
831
Replies
1
Views
1K
Replies
4
Views
1K
Replies
7
Views
16K
Replies
2
Views
2K
Replies
2
Views
3K
Top