Elastic Collision of Hydrogen and Carbon Atoms

In summary, the total momentum before a collision is equal to the total momentum after the collision.
  • #1
AN630078
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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!
 
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  • #2
AN630078 said:
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.
 
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  • #3
Careful with units. The time of the collision is 3 ms, not 3 s.
 
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  • #4
TSny said:
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 ?
 
  • #5
AN630078 said:
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?
 
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  • #6
TSny said:
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.
 
  • #7
AN630078 said:
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
 
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  • #8
TSny said:
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! 👍
 
  • #9
All your work looks good and the magnitude of the force would be positive.
 
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  • #10
TSny said:
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 ! 👍
 
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1. What is an elastic collision?

An elastic collision is a type of collision between two objects where there is no loss of kinetic energy. This means that the total kinetic energy of the two objects before the collision is equal to the total kinetic energy after the collision.

2. How do hydrogen and carbon atoms undergo an elastic collision?

Hydrogen and carbon atoms undergo an elastic collision when they collide with each other without any external forces acting on them. This means that there is no friction or other external forces that could cause a loss of kinetic energy during the collision.

3. What happens during an elastic collision between hydrogen and carbon atoms?

During an elastic collision between hydrogen and carbon atoms, the atoms will exchange energy and momentum. This means that the hydrogen atom will transfer some of its kinetic energy and momentum to the carbon atom, and vice versa.

4. How is the momentum conserved during an elastic collision between hydrogen and carbon atoms?

The momentum is conserved during an elastic collision between hydrogen and carbon atoms because the total momentum of the two atoms before the collision is equal to the total momentum after the collision. This is due to the fact that there are no external forces acting on the atoms during the collision.

5. What are some real-world applications of elastic collisions between hydrogen and carbon atoms?

Elastic collisions between hydrogen and carbon atoms are important in many industrial processes, such as the production of synthetic fuels and the creation of new materials. They are also studied in physics and chemistry research to better understand the behavior of atoms and molecules.

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