Linear Conservation of Momentum of electron

[Gannon]

An electron collides elastically with a stationary hydrogen atom. The mass of the hydrogen atom is 1837 times that of the electron. Assume that all motion, before and after the collision, occurs along the same straight line. What is the ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision? (hydrogen / electron)

Momentum is conserved, so

m1v1i +m2v2i = m1v1f + m2v2f

I can rearrange this to get

v2f = [ 2m1 / (m1 + m2) ] v1i

So this pretty much gives me the ratio I need (final hydrogen in comparison to initial electron). Since m2 = 1837(m1), I get

v2f = [ 2m1 / m1 + 1837m1 ] v1i

Canceling out mass, I get

v2f = [ 2 / 1 + 1837 ] v1i
v2f = [ 2 / 1838 ] v1i

So the hydrogen has more than 900 times the velocity as the electron does? How does this go into KE? This is where I fall apart. Any help is appreciated.

 

[gills]

you state that momentum is conserved. What else is conserved in an elastic collision?

 

[ogb p]

To calculate the kinetic energy, we can use the formula KE = 1/2 * m * v^2, where m is the mass and v is the velocity.

For the electron before the collision, the kinetic energy is KE1 = 1/2 * me * v1i^2.

For the hydrogen atom after the collision, the kinetic energy is KE2 = 1/2 * mh * v2f^2.

Substituting in the value for v2f that we calculated above, we get:

KE2 = 1/2 * mh * [2 / 1838]^2 * v1i^2

Since we know that mh = 1837*me, we can simplify this to:

KE2 = 1/2 * [1837*me] * [2 / 1838]^2 * v1i^2

Simplifying further, we get:

KE2 = 1/2 * me * v1i^2 * [2 / 1838]^2 * [1837 / 1837]

KE2 = 1/2 * me * v1i^2 * [2 / 1838]^2 * [1 / 1]

KE2 = KE1 * [2 / 1838]^2

Therefore, the ratio of the kinetic energy of the hydrogen atom after the collision to that of the electron before the collision is:

KE2 / KE1 = [2 / 1838]^2 = 0.000002

This means that the hydrogen atom has a significantly lower kinetic energy compared to the electron before the collision. This is because the mass of the hydrogen atom is much higher than that of the electron, so even though it has a higher velocity, its kinetic energy is still lower.

I hope this helps clarify the concept of linear conservation of momentum and how it relates to kinetic energy in this scenario.