Calculating Relativistic Density Change in a Moving Cube | Physics Homework Help

AI Thread Summary
The discussion revolves around calculating the relativistic density of water in a moving cube, initially at rest with a density of 1000.0 kg/m3. When the cube accelerates to a speed of 0.791c, the mass increases due to relativistic effects, leading to a calculated mass of 1630 kg. However, the length contraction must be considered, as the cube's dimensions change in the observer's frame, resulting in a new volume of 0.229 m^3. The final density, after accounting for both mass and volume changes, is determined to be approximately 7130 kg/m3. The conversation emphasizes the importance of solving problems symbolically and understanding the implications of relativistic effects on physical dimensions.
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Homework Statement



A cube, one cm on an edge, is filled with pure water. As measured in a reference frame at rest relative to the cube, the density of the water is exactly 1000.0 kg/m3. The cube is now accelerated to a constant speed of 0.791c. What is the density of the water in the moving cube as measured in the same reference frame?


Homework Equations



D=m/v

mrel=m/(sqrt(1-(v^2/c^2))

The Attempt at a Solution



I found m relative to be 1630 kg by using the second equation and if I divide it my 1m^3 the density should equal the new mass but that is incorrect. Can someone help please
 
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I don't know much about this, but I believe the length would contract as well.
 
rock.freak667 said:
I don't know much about this, but I believe the length would contract as well.
Yes, that's the key.
 
Agreed. The 1 cm length that's measured in the cube's frame is not the length measured in the observer's frame.
 
So the relativistic density is 1634 kg ( at least that's what I calcuated)

Then I converted 1cm to meters and found the relativistic length which was 6.1182e-3 m
then i cubed it to find volume and got 2.29e-1m^3

Next I found the density by diving the new mass by the new volume and got 7.13e3 kg/m^3

Is that correct?
 
The relativistic length contraction is:

L' = \dfrac{L}{\gamma}

where \gamma = \dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}}

The easy way to remember this formula is to remember that \gamma is always greater than or equal to 1. Anyway, when calculating the new volume, remember that the cube is only contracted along the direction of motion. Therefore, in another reference frame it won't look like a cube, but rather a rectangular prism. You need to take this into account when finding the volume.

As a sidenote, my recommendation (both as a student and a former TA) is to solve all problems symbolically before plugging in numbers. It makes it easier for you to find mistakes, and it makes it easier for the grader to give you partial credit. Not to mention the physics insight you tend to get...
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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