Calculating Fraction of Kinetic Energy Transferred in Head-on Collision

[stunner5000pt]

Determine from classical mechanics (using a head-on collision with recoil at 180 degrees) what fraction of an electron’s kinetic energy can be transferred to a mercury atom in an elastic collision. Derive an approximate value of the fraction.

well ok let the mass of the electron be m_e
initial velocity of electron v_1
fina lvelocity v_2
mercury atom mass m_Hg
mercury atom fina lvelocity v_Hg

$$\frac{1}{2} m_{e} v_{1}^2 = \frac{1}{2} m_{Hg} v_{Hg}^2 + \frac{1}{2} m_{e} v_{2}^2$$

the fraction of kinetic energy is Kf/Ki right

$$\frac{K_{f}}{K_{i}} = \frac{v_{2}^2}{v_{1}^2}$$

do i use momentum concepts to find v2 in terms of v1

still it doesn't give me a numerical value... if that's what the question is asking...

 

[stunner5000pt]

bump :D
for help :)

 

[kyle8921]

To calculate the fraction of kinetic energy transferred in a head-on collision between an electron and a mercury atom, we can use the principles of classical mechanics. In an elastic collision, both kinetic energy and momentum are conserved. Therefore, we can use the conservation of energy equation to determine the final velocity of the mercury atom in terms of the initial velocity of the electron.

Using the equation provided, we can rearrange it to solve for v2:

v2 = sqrt((m_e/m_Hg)*(v1^2-v_Hg^2))

Substituting this into the fraction of kinetic energy equation, we get:

(Kf/Ki) = ((m_e/m_Hg)*(v1^2-v_Hg^2))/v1^2

Simplifying this further, we get:

(Kf/Ki) = (m_e/m_Hg)*(1-(v_Hg/v1)^2)

Since we know that the final velocity of the mercury atom is equal to the negative of the initial velocity of the electron (due to the head-on collision with recoil at 180 degrees), we can substitute this into the equation:

(Kf/Ki) = (m_e/m_Hg)*(1-(v_Hg/v1)^2)

= (m_e/m_Hg)*(1-(-v1/v1)^2)

= (m_e/m_Hg)*(1-1)

= (m_e/m_Hg)*0

= 0

This means that in an elastic head-on collision between an electron and a mercury atom, no kinetic energy is transferred from the electron to the mercury atom. All of the electron's kinetic energy is conserved and remains with the electron after the collision.

Therefore, the approximate value of the fraction of kinetic energy transferred in this scenario is 0. This result is due to the fact that the mercury atom is significantly larger and more massive than the electron, and thus it is not able to absorb any of the electron's kinetic energy in the collision.