Finding recoil velocity of an object during a collision

[jigsaw21]

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 ?

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.

 

[Raihan amin]

$\frac v c =\frac 1 5$ ,not so negligible,so you have to perform relativistic calculations

 

[Raihan amin]

I wonder how a photon could have a mass.

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

 

[SammyS]

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 ?

I suspect that OP meant to say cosmic ray proton .

 

[jigsaw21]

Yes, sorry about that. That should've been proton!

 

[jigsaw21]

$\frac v c =\frac 1 {20}$ ,not so negligible,so you have to perform relativistic calculations

In this course, we haven't gotten to relativistic calculations at all yet. The topic we're on is conservation of momentum, impulse, force and center of mass.

 

[jigsaw21]

Can anyone else tell me if I'm on the right track with this question, or if I'm not?

 

[SammyS]

Can anyone else tell me if I'm on the right track with this question, or if I'm not?

Yes, you are on the right track.

You can simplify your calculations somewhat by noticing that M = 16⋅m .

To check your answer:
Initial momentum = $\ m \cdot v_{p\text{-initial}} = m\cdot (6.0\times10^{7} ) \$

Final momentum = $\ m \cdot v_{p\text{-final}} + M \cdot v_{o\text{-final}}= m\cdot (-5.3\times10^{7} ) + 16\cdot m\cdot (0.7\times 10^{7}) \$

$= m \cdot (5.9 \times 10^{7} )$​

That's good to plus or minus 1 for the 2nd significant digit.

Using 2 significant figures in your final answer should give the velocity of the Oxygen nucleus to be 0.71×10⁷ m/s, which is rounded from 0.706×10⁷ m/s .

 

[haruspex]

After converting atomic mass units to kg's,

Why bother? All masses given use the same units, so any units conversion will cancel out.

0.7 * 10^7 m/s.

Yes, but I would give it to two significant figures to match the precisions of the given velocities.

 

[jigsaw21]

I didn't even think to try going backwards and checking, but thanks for that!
I also didn't think to simplify the mass to be 16m

Yes, you are on the right track.

You can simplify your calculations somewhat by noticing that M = 16⋅m .

To check your answer:
Initial momentum = $\ m \cdot v_{p\text{-initial}} = m\cdot (6.0\times10^{7} ) \$

Final momentum = $\ m \cdot v_{p\text{-final}} + M \cdot v_{o\text{-final}}= m\cdot (-5.3\times10^{7} ) + 16\cdot m\cdot (0.7\times 10^{7}) \$

$= m \cdot (5.9 \times 10^{7} )$​

That's good to plus or minus 1 for the 2nd significant digit.

Using 2 significant figures in your final answer should give the velocity of the Oxygen nucleus to be 0.71×10⁷ m/s, which is rounded from 0.706×10⁷ m/s .