[SOLVED] Elastic Collision/Kinetic Energy Problem

NoPhysicsGenius

1. The problem statement, all variables and given/known data


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.)


2. Relevant 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]


3. 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.

dynamicsolo

For an elastic collision, you need two equations to cover the two unknowns you'll have (often, those are the final speeds of each of the objects which collided). So you have the consequence of kinetic energy being conserved (which defines an "elastic collision").

What happens when you apply conservation of linear momentum?

NoPhysicsGenius

O.K. Applying conservation of linear momentum gives:

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

Noting that:

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

and

[tex]m_2=12m_1[/tex]

we get:

[tex]m_1v_{1i}=m_1v_{1f}+12m_1v_{2f}[/tex]
[tex]\Rightarrow v_{1i}=v_{1f}+12v_{2f}[/tex]


Now we can take the conservation of linear momentum equation, square it, and substitute it into the equation we derived from the conservation of kinetic energy equation:

[tex]{v_{1f}}^2 + 144{v_{2f}}^2 + 24v_{1f}v_{2f} = {v_{1f}}^2 + 12{v_{2f}}^2[/tex]
[tex]\Rightarrow 132{v_{2f}}^2 + 24v_{1f}v_{2f} = 0[/tex]
[tex]\Rightarrow v_{2f}(132v_{2f}+24v_{1f}) = 0[/tex]
[tex]\Rightarrow v_{2f}= 0[/tex]

or

[tex]\Rightarrow v_{2f} = \frac{-24v_{1f}}{132} = \frac{2v_{1f}}{11} = -0.182v_{1f}[/tex]


The answer in the back of the book is 0.284, or 28.4%. What have I done wrong? Thank you for your help.

alphysicist

Hi NoPhysicsGenius,


They want to find the fraction of the neutron's initial energy transferred to the carbon nucleus. So you want to eliminate [itex]v_{1f}[/itex], not [itex]v_{1i}[/itex].

Once you find the ratio of the speeds, you can then use that to find the ratio of the kinetic energies that they ask for.

NoPhysicsGenius

Thank you for your help ... I am able to get the book's answer now!