# Elastic Collision of Hydrogen and Carbon Atoms

Homework Statement:
Hello I have been revising momentum and collisions and have found the question below focusing on an elastic collision. I have answered all of the questions but I am struggling particularly with question 3 and would appreciate if anyone could offer any guidance A hydrogen atom has a mass 1.7x10^-27 kg and moves at 500 m/s. It collides elastically with a stationary carbon atom of mass 2.0x10^-26 kg and rebounds with a speed of 420 m/s in the opposite direction to its original motion.

Question 1; Use a momentum equation to find the speed of the carbon atom after collision.  Question 2; Verify that the collision is perfectly elastic, within the number of significant figures given.
Question 3; If the collision takes 3.0 ms, what is the average force acting in this time interval?
Relevant Equations:
p=mv
Ek=1/2mv^2
F=ma
1.p=mv
Before the collision:
p hydrogen = 1.7x10^-27 * 500 =8.5*10^-25 kg ms^-1
p carbon = 2.0x10^-26 * 0 = 0 kg ms^-1
p total before = 8.5*10^-25 kg ms^-1

The sum of momentum prior to the collision is equal to the total momentum after a collision, momentum is constant, therefore;
p before = p after

After:
p hydrogen = 1.7x10^-27 * -420 =-7.14*10^-25 kg ms^-1
p carbon = 8.5*10^-25 - -7.14*10^-25 kg ms^-1 = 1.564*10^-24 kgms^-1

Rearrange p=mv in terms of the speed of the carbon atom after the collision;
v=p/m
v=1.564*10^-24/2.0x10^-26
v=78.2 ms^-1

2.In an elastic collision, total kinetic energy is conserved.
KE before=KE after
Before;
Ek hydrogen = 1/2*1.7x10^-27*500^2=2.125 *10^22 J
Ek carbon = 1/2*2.0x10^-26*0^2=0J
Ek total before = 2.125 *10^-22 J~ 2.1*10^-22 J

After;
Ek hydrogen = 1/2*1.7x10^-27*420^2=1.4994*10^-22J
Ek carbon = 1/2*2.0x10^-26*78.2^2=6.11524*10^-23 J
Ek total after = 1.4994*10^-22 + 6.11524*10^-23 = 2.110924*10^-22 J ~ 2.1 *10^-22J

KE before=KE after
Thus, 2.1 *10^-22J = 2.1 *10^-22J

3. This is where I am most confused, I do not know whether the equations I have used are correct here nor am I entirely sure which force it is referring to? Would this be the force acting on the hydrogen atom in the collision? Also, would I take the velocity to be 500ms^-1 during the 3 seconds of this collision?

In yes then;
Force=ma=mv/t
Force = 1.7x10^-27 *500/3
Force = 2.83 * 10^-25 N

Or would I use;

Force=change in momentum/time

Change in momentum = mvf-mvi=(1.7x10^-27 *-420) -(1.7x10^-27 *500)= -7.14*10^-25 - 8.5*10^-25= - 1.564 *10^-24
Force = - 1.564 *10^-24 / 3
Force = - 5.21 *10^-25 N

I am rather confused on this last section, also would the negative sign be applicable here? Sorry I think I have confused myself!

## Answers and Replies

PeroK
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Or would I use;

Force=change in momentum/time

Change in momentum = mvf-mvi=(1.7x10^-27 *-420) -(1.7x10^-27 *500)= -7.14*10^-25 - 8.5*10^-25= - 1.564 *10^-24
Force = - 1.564 *10^-24 / 3
Force = - 5.21 *10^-25 N
I didn't check the numbers but this is the correct approach.

• AN630078
TSny
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Careful with units. The time of the collision is 3 ms, not 3 s.

• AN630078
Careful with units. The time of the collision is 3 ms, not 3 s.
Thank you for your reply. Sorry my mistake! So would v=3ms^-1?

Force=change in momentum
Change in momentum = mv =1.7x10^-27 *3= 5.1*10^-27 kgms^-1
Force = 5.1*10^-27 N ?

TSny
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Thank you for your reply. Sorry my mistake! So would v=3ms^-1?

Force=change in momentum

This should be as you originally wrote it: Force = (change in momentum)/time

Here "time" is the time interval over which the collision takes place. This is given to be 3 ms, which I interpret as 3 milliseconds.

If you want the average force experienced by the hydrogen atom, use the change in momentum of the hydrogen atom. For the average force on the carbon atom, use the change in momentum of the carbon atom. According to Newton's third law, what do you expect for how these forces compare?

• AN630078, PeroK and hutchphd
This should be as you originally wrote it: Force = (change in momentum)/time

Here "time" is the time interval over which the collision takes place. This is given to be 3 ms, which I interpret as 3 milliseconds.

If you want the average force experienced by the hydrogen atom, use the change in momentum of the hydrogen atom. For the average force on the carbon atom, use the change in momentum of the carbon atom. According to Newton's third law, what do you expect for how these forces compare?
Thank you for your reply. Sorry, I misread it and was wrong again.

Force=change in momentum/time
Change in momentum = mvf-mvi=(1.7x10^-27 *-420) -(1.7x10^-27 *500)= -7.14*10^-25 - 8.5*10^-25= - 1.564 *10^-24
Force = - 1.564 *10^-24 / 3 *10^-3
Force = - 5.21 *10^-22 N

Newton's 3rd Law states that whenever two objects interact, they exert equal and opposite forces on each other. So the force exerted on the hydrogen atom is equal and opposite to the force exerted on the carbon atom.

TSny
Homework Helper
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Force=change in momentum/time
Change in momentum = mvf-mvi=(1.7x10^-27 *-420) -(1.7x10^-27 *500)= -7.14*10^-25 - 8.5*10^-25= - 1.564 *10^-24
Force = - 1.564 *10^-24 / 3 *10^-3
Force = - 5.21 *10^-22 N
OK, this is the average force experienced by the hydrogen atom.

Newton's 3rd Law states that whenever two objects interact, they exert equal and opposite forces on each other. So the force exerted on the hydrogen atom is equal and opposite to the force exerted on the carbon atom.
Yes. You can check this by calculating:

Force on carbon atom = (change in momentum of carbon atom)/time

• AN630078
OK, this is the average force experienced by the hydrogen atom.

Yes. You can check this by calculating:

Force on carbon atom = (change in momentum of carbon atom)/time

Thank you for your reply.
Force on carbon atom=(change in momentum of carbon atom)/time
Change in momentum=78.2*2.0x10^-26-2.0x10^-26*0=1.564*10^-24 kgms^-1
Force on carbon atom=1.564*10^-24/3*10^-3
Force on carbon atom=5.21*10^-22 N

Therefore, the force on the hydrogen atom is of equal magnitude but opposite direction to the force on the carbon atom.

Should I give my final answer as 5.21*10^-22 N (removing the negative sign as only the magnitude of the force not the direction is asked for?)

Also, would my solutions to the earlier parts 1 and 2 of the question be correct?
Thank you for your help! TSny
Homework Helper
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All your work looks good and the magnitude of the force would be positive.

• AN630078
All your work looks good and the magnitude of the force would be positive.
Thank you very much for your reply and for all of your help, I truly appreciate it ! • TSny