Elastic Collision/Kinetic Energy Problem

In summary, the fraction of the neutron's initial kinetic energy transferred to the carbon nucleus in an elastic head-on collision is 0.284, or 28.4%. Additionally, the final kinetic energy of the neutron is 1.15 x 10^-13 J and the kinetic energy of the carbon nucleus after the collision is 4.54 x 10^-14 J.
  • #1
NoPhysicsGenius
58
0
[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.
 
Last edited:
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  • #2
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?
 
  • #3
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.
 
  • #4
Hi NoPhysicsGenius,


NoPhysicsGenius said:
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.

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.
 
  • #5
alphysicist said:
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.


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

1. What is an elastic collision?

An elastic collision is a type of collision in which the total kinetic energy of the system is conserved. This means that the total energy before the collision is equal to the total energy after the collision.

2. What is the difference between an elastic collision and an inelastic collision?

In an elastic collision, the total kinetic energy of the system is conserved. In an inelastic collision, some of the kinetic energy is lost in the form of heat, sound, or deformation of the objects involved.

3. How do you calculate the kinetic energy of an object?

The kinetic energy of an object can be calculated using the formula KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

4. How does the mass and velocity of objects affect the outcome of an elastic collision?

In an elastic collision, the mass and velocity of the objects involved determine the amount of kinetic energy that is transferred between them. A heavier object with a higher velocity will transfer more kinetic energy to a lighter object with a lower velocity.

5. Can you have an elastic collision between two objects with different masses?

Yes, an elastic collision can occur between two objects with different masses. However, the final velocities of the objects will depend on their masses and initial velocities, and the total kinetic energy will still be conserved.

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