- #1

AN630078

- 242

- 25

- 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!

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!