
#109
Jan412, 06:46 AM

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P: 16,481

Is this a correct understanding of your position? If I am misunderstanding then can you please provide a clear yes/no answer to whether or not you agree with the proof of post 59. 



#110
Jan412, 07:00 AM

P: 60





#111
Jan412, 07:08 AM

Mentor
P: 16,481

Btw, it would really help if you were more careful about distinguishing your quantities. Ie. Is your p the change in momentum or is it the initial momentum? I follow the usual convention where [itex]\Delta p[/itex] is the change in momentum, [itex]p_i[/itex] is the initial momentum, and [itex]p_f=p_i+\Delta p[/itex] is the final momentum. 



#112
Jan412, 07:09 AM

P: 60

In the first case, the car's 50kJ, is transferred to the battery. So, the second case, which is a mirror, should see the energy come from the battery to the car. The total remains in each case at 50kJ. In the example you gave, and in the first case, the energy to the ground is a result of a change of Earth's velocity, 1e21m/s. That results in 5e18J In the other case, the change in Earth's velocity is Vcar*(Mearth  Mcar)/Mearth Vcar =10m/s Mearth = 1e25kg Mcar =1e3kg so the change in KE is related to (10  1e21)m/s. That , I think, accounts for the difference of relative velocities and energy. The energy "transferred back to the car" should be Mearth/Mcar*5e18J =50kJ 



#113
Jan412, 07:19 AM

Mentor
P: 16,481





#114
Jan412, 07:29 AM

P: 60

p^2/2mcar is the KE of the car with momentum p p^2/2mearth is the change in the Earth's KE when p is transferred to it. 



#115
Jan412, 07:37 AM

Mentor
P: 16,481

Also, clearly [itex]p_i[/itex] depends on [itex]p_i[/itex] even though momentum adds linearly, so adding linearly is not a justification for a lack of dependency in general. If you want to show that x doesn't depend on y because y adds linearly then you need to take the linearity property, use it in the expression for x, and show that y drops out algebraically. Btw, I note that you still avoid the question. 



#116
Jan412, 07:50 AM

P: 60

The change in KE for the car is p_{i,c}^2/2m_{c} and for the Earth p_{i,c}^2/2m_{e} Works both ways. 



#117
Jan412, 07:57 AM

Mentor
P: 16,481





#118
Jan412, 08:05 AM

P: 60

Makes no sense. 



#119
Jan412, 08:09 AM

P: 60





#120
Jan412, 08:17 AM

PF Gold
P: 3,072





#121
Jan412, 08:26 AM

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P: 16,481

http://physics.ucr.edu/~wudka/Physic...ww/node47.html You are literally centuries out of date in your thinking. These are both correct descriptions of the same physical situation. Momentum and energy are conserved in both cases, and Newton's laws are obeyed in both cases. 



#122
Jan412, 08:30 AM

Mentor
P: 16,481





#123
Jan412, 08:32 AM

P: 60

p= mv ( I assume you know that) p^{2} = m^{2}v^{2} p^{2}/2m = m^{2}v^{2}/2m = 1/2mv^{2} = KE. 



#124
Jan412, 08:34 AM

Mentor
P: 16,481

Btw, I think that Huber's continued avoidance of the question is because he recognizes that the derivation is correct, but he feels that he would lose face to admit it.




#125
Jan412, 08:43 AM

Mentor
P: 14,459

This nonsense has gone on long enough. Thread locked.



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