Relativity problems


by F.B
Tags: relativity
F.B
F.B is offline
#1
Dec24-05, 09:02 AM
P: 87
I need help with two of these questions.

1. An electron is travelling at 0.866c, with respect to the face of a television picture tube. What is the value of its relativistic momentum to the tube?

Im not goin to post every single step but i'll post my answer.

p=mv/sqrt(1-v^2/c^2)
p=9.11 x 10^-31 x (0.866 x 3.00 x 10^8)/sqrt(0.250044)
p=4.74 x 10^-22

Thats the answer i get but in my book, they have 1.58 x 10^-30.

2. The electron and positron each have a rest mass of 9.11 x 10^-31 kg. In a certain experiment, an electron and positron collide and vanish, leaving only electromagnetic radiation after the interaction. Each particle is moving at a speed of 0.20c relative to the laboratory before the collision. Determine the energy of the electromagnetic radiation.

First of all do i have to solve this question like a collision problem, if i do there is no after situation.

Anyways i think i have to solve for Ek.

So Et=Ek + Erest

Et=mc^2/sqrt(1-v^2/c^2)
Et=9.11 x 10^-31 x (3.00 x 10^8)^2/sqrt(0.96)
Et=8.37 x 10^-14
Erest = 8.2 x 10^-14
Ek= 8.37 x 10^-14 - 8.2 x 10^-14
Ek= 1.7 x 10^-15

But this answer isnt right. In my book they have 0.615 MeV and when i convert my answer to that i dont get that answer. So what did i do wrong ,cant anyone help me please.
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Tide
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#2
Dec24-05, 10:45 AM
Sci Advisor
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P: 3,149
1. Did your textbook provide an answer for p/c? It looks like they dropped a factor of c.

2. Since the rest energy of an electron is 0.511 MeV then the annihilation of an electron positron pair must yield a photon of at least twice that energy (1.022 MeV). For the parameters you've given the result should be 1.043 MeV with the extra energy being provided by the kinetic energy of the electron and positron.

What textbook are you using?
F.B
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#3
Dec24-05, 11:12 AM
P: 87
I'm using the Grade 12 nelson physics book.

So is my answer for number 1 correct?

For number 2 if i do:

Ek=2(8.37 x 10^-14) - 2(8.2 x 10^-14) would this be right that way im taking into account both the positron and electron or does one of these energies cancel out because they have to be going in opposite directions to cancel out.

Tide
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#4
Dec24-05, 03:08 PM
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P: 3,149

Relativity problems


#1 looks good to me.

In #2 all of the energy initially present (kinetic plus rest) is transformed into electromagnetic radiation so [itex]E = 2 \gamma m_0 c^2[/itex].
krab
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#5
Dec24-05, 03:24 PM
Sci Advisor
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P: 905
Actually, your answer to #1 is incorrect, since it lacks units. I say this not to be a smartass, but because units are one of the ways of checking and verifying. For example, your answer should be in kg m/sec. If the book's answer is simply in kg, you know they've divided by a velocity (for whatever reason). But if the book states 1.58 x 10^-30 kg m /sec, then the book is in error.
F.B
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#6
Dec24-05, 04:52 PM
P: 87
Hey Tide i've never seen that formula before. It's not in any of the sections that relate to that question i asked. I'll check my whole book but i dont know what that gamma symbol is.
Gamma
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#7
Dec24-05, 05:40 PM
PF Gold
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P: 330
gamma is the common symbol used in relativity:


gamma = 1 / sqrt(1-v^2/c^2)


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