- #1
jigsaw21
- 20
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Hello. I recently began this problem, but the site I'm working this problem on isn't showing me whether the answer is correct or not. Can someone please check my work and just verify that it's correct or state if it isn't and why?
1. Homework Statement
A cosmic-ray photon (with mass m = 1 at. mass unit) from outer space arrives with velocity 6.0 * 10^7 m/s, when it makes a head-oncollision with an oxygen nucleus (with mass M = 16 amu), part of an oxygen molecule in the atmosphere. Subsequently the proton bounces backward with velocity 5.3 * 10^7 m/s. What is the recoil velocity of the oxygen nucleus in m/s ?
The relevant equations for this question are with momentum conservation, p(initial) = p(final)
p(initial) = m*v(photon) + M*v(oxygen nucleus)
p(final) = m*v(photon-final) + M*v(recoil velocity)
First thing I did was calculate P(initial).
After converting atomic mass units to kg's, the mass of m (1 amu) in kg's would be 1.66*10^-27kg. Multiplying that by the initial velocity of the photon (6.0*10^7 m/s) would give me 9.96 * 10^-20 kg m/s. Since it doesn't indicate, I'm assuming that the oxygen molecule is at rest, and thus it's M*v would be 0. So the entire P(initial) would just be 9.96*10^-20 kg m/s
I then calculated P(final), which would be m*v(photon-final) + M * v(oxygen molecule-final), which was (1.66*10^-27 kg)(-5.3 * 10^7 m/s) + (2.66 * 10^-26 kg)* v(oxygen-final). And this simplified to -8.798 * 10^-20 kg m/s + (2.66 * 10^-26 kg)*v(oxygen-final)
After setting P(initial) = P(final), combining like terms and solving algebraically, I ended up with a final answer for v(oxygen-final) of 0.7 * 10^7 m/s.
There's no way for me to check if this was correct.
Is this the correct answer? Intuitively, it looked like it to me since the mass of the oxygen molecule was 16 times larger, so thus it would be more difficult for it to move anywhere near as fast as the photon initially moved or even to match the negative velocity the photon traveled with after the collision.
Thanks for any help anyone can provide.
1. Homework Statement
A cosmic-ray photon (with mass m = 1 at. mass unit) from outer space arrives with velocity 6.0 * 10^7 m/s, when it makes a head-oncollision with an oxygen nucleus (with mass M = 16 amu), part of an oxygen molecule in the atmosphere. Subsequently the proton bounces backward with velocity 5.3 * 10^7 m/s. What is the recoil velocity of the oxygen nucleus in m/s ?
Homework Equations
The relevant equations for this question are with momentum conservation, p(initial) = p(final)
p(initial) = m*v(photon) + M*v(oxygen nucleus)
p(final) = m*v(photon-final) + M*v(recoil velocity)
The Attempt at a Solution
First thing I did was calculate P(initial).
After converting atomic mass units to kg's, the mass of m (1 amu) in kg's would be 1.66*10^-27kg. Multiplying that by the initial velocity of the photon (6.0*10^7 m/s) would give me 9.96 * 10^-20 kg m/s. Since it doesn't indicate, I'm assuming that the oxygen molecule is at rest, and thus it's M*v would be 0. So the entire P(initial) would just be 9.96*10^-20 kg m/s
I then calculated P(final), which would be m*v(photon-final) + M * v(oxygen molecule-final), which was (1.66*10^-27 kg)(-5.3 * 10^7 m/s) + (2.66 * 10^-26 kg)* v(oxygen-final). And this simplified to -8.798 * 10^-20 kg m/s + (2.66 * 10^-26 kg)*v(oxygen-final)
After setting P(initial) = P(final), combining like terms and solving algebraically, I ended up with a final answer for v(oxygen-final) of 0.7 * 10^7 m/s.
There's no way for me to check if this was correct.
Is this the correct answer? Intuitively, it looked like it to me since the mass of the oxygen molecule was 16 times larger, so thus it would be more difficult for it to move anywhere near as fast as the photon initially moved or even to match the negative velocity the photon traveled with after the collision.
Thanks for any help anyone can provide.