Conservation of Momentum: Relativity

In summary, Ann (mass 50kg) is standing at the left end of a 15m long 500kg cart that has frictionless wheels and rolls on a frictionless track. Initially Ann and the cart are at rest. Suddenly, Ann starts running along the cart at a speed of 5.0m/s relative to the cart. When she reaches the right end of the cart, she will have run 13.6m relative to the ground. The velocity of the cart in its own reference frame is always zero, and the momentum of the system is conserved in any inertial reference frame. Therefore, the momentum conservation equation simplifies to Vf(Ma + Mc) = MaVai, where Vf
  • #1
Buri
273
0
Ann (mass 50kg) is standing at the left end of a 15m long 500kg cart that has frictionless wheels and rolls on a frictionless track. Initially Ann and the cart are at rest. Suddenly, Ann starts running along the cart at a speed of 5.0m/s relative to the cart.
How far will Ann have have run relative to the ground when she reached the right end of the cart?Solution

When doing this problem I got the right answer only if I added in a negative sign at the end-hardly the right way do to physics. (First year honors).

This is the way I defined my variables:
Ma=mass of Ann
Mc= mass of cart
Vaf = final velocity of Ann
Vai = initial velocity of Ann
Vcf, Vci defined similarly for the cart
and the equation for relativity x = x' + Vxt
where x = Earth's reference frame
x' = carts reference frame
Vx = relative velocity
t = time

I begin with the equation for the conservation of momentum:
MaVaf + McVcf = MaVai + McVci
However I know once Ann reaches the other side her velocity will be zero, however since she's on the cart she will actually have the velocity of the cart in Earth's reference frame. Therefore the equation above simplifies:
Vf(Ma + Mc) = MaVai
Where we call Vaf = Vcf = Vf and Vci = 0
and I find Vf = 0.45m/s
Here is where my problem starts; Vf is positive! I think it should be negative - but I don't see a way to make it "come out" negative.

ADDING the negative sign and realizing it will take Ann 3s to get across, relativity gives me:
x = x' + Vxt
= 15 - (0.45)(3)
= 13.6m

Which is the correct answer! But I had to ADD the negative, something that I think I shouldn't have to do.

Any help?
 
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  • #2
Hi Buri! :wink:
Buri said:
… However I know once Ann reaches the other side her velocity will be zero

No, her velocity in the cart frame will still be 5

Try again! :smile:
 
  • #3
Buri said:
Vaf = final velocity of Ann
Vai = initial velocity of Ann
Vcf, Vci defined similarly for the cart
These velocities are relative to what? I think that if you write the momentum conservation equation with all velocities relative to the ground, you will see what is going on. Don't forget that once Ann starts moving the center of mass of Ann + cart remains at rest.
 
  • #4
Wow, what a lot of work! Why not take advantage of the fact that a system not influenced by external forces will have a center of mass that moves inertially?

Draw a diagram of a cart with Ann at one end. Mark the center of gravity. Now draw another picture of the cart showing Ann at the other end. Again mark the center of gravity. Line up the two diagrams so that their centers of gravity overlap. What's the distance Between 'Anns'?
 
  • #5
tiny-tim said:
Hi Buri! :wink:


No, her velocity in the cart frame will still be 5

Try again! :smile:

Her velocity in the carts frame CANNOT be 5m/s. This would imply she runs off the cart!
 
  • #6
Buri said:
Her velocity in the carts frame CANNOT be 5m/s. This would imply she runs off the cart!

Why would that matter? You're only interested in her position at the instant she reaches the end of the cart.
 
  • #7
Hi Buri! :smile:

(just got up :zzz: …)
Buri said:
Ann (mass 50kg) is standing at the left end of a 15m long 500kg cart that has frictionless wheels and rolls on a frictionless track. Initially Ann and the cart are at rest. Suddenly, Ann starts running along the cart at a speed of 5.0m/s relative to the cart.
How far will Ann have have run relative to the ground when she reached the right end of the cart?
Buri said:
Her velocity in the carts frame CANNOT be 5m/s. This would imply she runs off the cart!

You're reading too much into the question! :rolleyes:

Maybe she does run off the end …

it's none of your business! …

who are you to condemn her, or to tell her what to do?

has she ever told you how to do physics?

if she wants to run off the end of the cart, your job is simply to advise her what will happen when she hit the ground running! not to advise her whether to do it!

she will listen to your advice on the physics, and then make up her own mind!

know your place! :smile:
 
  • #8
tiny-tim said:
Hi Buri! :smile:

(just got up :zzz: …)



You're reading too much into the question! :rolleyes:

Maybe she does run off the end …

it's none of your business! …

who are you to condemn her, or to tell her what to do?

has she ever told you how to do physics?

if she wants to run off the end of the cart, your job is simply to advise her what will happen when she hit the ground running! not to advise her whether to do it!

she will listen to your advice on the physics, and then make up her own mind!

know your place! :smile:
If I analyze the momentum equation with the reference frame attached to the cart I indeed find her velocity to be 5 m/s, but I also gather no information about the cart. As the carts velocity in its own reference frame is always zero. I would need the velocity of the cart in the Earth's reference frame in order to find Anns velocity relative to the earth. I don't see how I can proceed by knowing Anns velocity in the carts frame.

I was just thinking...wouldn't momentum not be conserved in the carts reference frame as the system would only be Ann and not both Ann and the cart?
 
  • #9
Buri said:
I was just thinking...wouldn't momentum not be conserved in the carts reference frame as the system would only be Ann and not both Ann and the cart?
The momentum of a system is always conserved in any inertial reference frame. Remember, the CM of the Ann+cart system is initially at rest and remains at rest no matter where Ann is. So (a) find the initial position of the Ann+cart CM before she starts moving, (b) find how far in the opposite direction to Ann's motion the cart has to move to keep the CM in the same position (c) use this information to find how far with respect to the ground Ann has moved.
 
  • #10
kuruman said:
The momentum of a system is always conserved in any inertial reference frame. Remember, the CM of the Ann+cart system is initially at rest and remains at rest no matter where Ann is. So (a) find the initial position of the Ann+cart CM before she starts moving, (b) find how far in the opposite direction to Ann's motion the cart has to move to keep the CM in the same position (c) use this information to find how far with respect to the ground Ann has moved.

I don't know how to use center of mass in this problem - I haven't learned that yet. All I can use is conservation of momentum and relativity. The problem is a challenge problem in the text maybe that's why its harder only using conservation of momentum.
 
  • #11
OK. We'll do this in steps.

Step 1. What is the total momentum of the Ann+cart system before she starts moving? Give me a number.

Step 2. Can you write an expression for the total momentum of the Ann+cart system after she starts running? Express all velocities relative to the ground and never mind if you don't have numbers for them.

Step 3. Conserve momentum.

Step 4. Write an expression relating vAG (Ann relative to ground) to vAC (Ann relative to cart) and vCG (cart relative to ground)

Step 5. Use the relativity equation to eliminate vCG in the momentum conservation equation.

At this point you should be able to finish this on your own.
 
  • #12
kuruman said:
OK. We'll do this in steps.

Step 1. What is the total momentum of the Ann+cart system before she starts moving? Give me a number.

Step 2. Can you write an expression for the total momentum of the Ann+cart system after she starts running? Express all velocities relative to the ground and never mind if you don't have numbers for them.

Step 3. Conserve momentum.

Step 4. Write an expression relating vAG (Ann relative to ground) to vAC (Ann relative to cart) and vCG (cart relative to ground)

Step 5. Use the relativity equation to eliminate vCG in the momentum conservation equation.

At this point you should be able to finish this on your own.
Thank-you I was able to solve it. I guess my problem was that I was confusing velocities from different reference frames...I expected this while I was doing the problem, however I didn't see a way to separate the velocities-the answer was a system of equations. Thank-you.
 

What is the conservation of momentum?

The conservation of momentum is a fundamental principle in physics that states that the total momentum of a closed system remains constant over time, regardless of any internal changes or interactions within the system. This means that the total amount of momentum before and after a collision or interaction must be the same.

What is the relationship between conservation of momentum and relativity?

The conservation of momentum is a fundamental law of classical mechanics, but it has also been shown to hold true in the theory of relativity. In fact, momentum is considered a conserved quantity in both classical and relativistic physics. However, the concept of momentum in relativity is more complex, as it is tied to the concept of energy and mass.

How does the conservation of momentum apply to the theory of relativity?

In the theory of relativity, the conservation of momentum is extended to include the concept of relativistic momentum, which takes into account the effects of mass and energy on an object's momentum. This means that the total relativistic momentum of a closed system must also remain constant over time, just like in classical mechanics.

Why is the conservation of momentum important in the theory of relativity?

The conservation of momentum is important in the theory of relativity because it is a fundamental law that helps us understand the behavior of objects in motion. It allows us to predict the outcome of interactions and collisions between objects, and it also helps to explain the behavior of objects at high speeds and in extreme gravitational fields.

What are some real-world applications of the conservation of momentum in relativity?

The conservation of momentum in relativity has many practical applications in fields such as astrophysics, nuclear physics, and particle physics. It is also used in engineering and technology, such as in the design of spacecraft and high-speed vehicles. Additionally, the principles of conservation of momentum and relativity are applied in medical imaging techniques, such as positron emission tomography (PET) scans.

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