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**[SOLVED] Elastic Collision/Kinetic Energy Problem**

## Homework Statement

A neutron in a reactor makes an elastic head-on collision with the nucleus of a carbon atom initially at rest. (a) What fraction of the neutron's kinetic energy is transferred to the carbon nucleus? (b) If the initial kinetic energy of the neutron is 1 MeV = 1.6 x 10^-13 J, find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision. (The mass of the carbon nucleus is about 12 times the mass of the neutron.)

## Homework Equations

[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]

[tex]\frac{1}{2}m_1{v_{1i}}^2 + \frac{1}{2}m_2{v_{2i}}^2 = \frac{1}{2}m_1{v_{1f}}^2 + \frac{1}{2}m_2{v_{2f}}^2[/tex]

## The Attempt at a Solution

First of all, the answers in the back of the book are as follows:

(a) 0.284, or 28.4%

(b)

[tex]K_n = 1.15 x 10^{-13} J[/tex]

[tex]K_c = 4.54 x 10^{-14} J[/tex]

Using the answer from part (a), I can easily solve part (b) as follows ...

[tex]K_n = (1.00 - 0.284)(1.6 x 10^{-13}J) = 1.15 x 10^{-13} J[/tex]

[tex]K_c = (0.284)(1.6 x 10^{-13}J) = 4.54 x 10^{-14} J[/tex]

However, I haven't the slightest clue how to solve part (a). Here is my attempt ...

First note the following:

[tex]m_2 = 12m_1[/tex]

[tex]v_{2i} = 0m/s[/tex]

[tex]\frac{1}{2}m_1{v_{1i}}^2 + \frac{1}{2}m_2{v_{2i}}^2 = \frac{1}{2}m_1{v_{1f}}^2 + \frac{1}{2}m_2{v_{2f}}^2[/tex]

[tex]\Rightarrow \frac{1}{2}m_1{v_{1i}}^2 + \frac{1}{2}(12m_1)(0 m/s)_^2 = \frac{1}{2}m_1{v_{1f}}^2 + \frac{1}{2}(12m_1){v_{2f}}^2[/tex]

[tex]\Rightarrow \frac{1}{2}m_1{v_{1i}}^2 = \frac{1}{2}m_1{v_{1f}}^2 + \frac{1}{2}(12m_1){v_{2f}}^2[/tex]

[tex]\Rightarrow {v_{1i}}^2 = {v_{1f}}^2 + 12{v_{2f}}^2[/tex]

... Then what?

Please help. Thank you.

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