Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 9: Problem 49
An alpha particle ( \(^{4} \mathrm{He}\) ) undergoes an elastic collision with a stationary uranium nucleus \(\left(^{235} \mathrm{U}\right) .\) What percent of the kinetic energy of the alpha particle is transferred to the uranium nucleus? Assume the collision is onedimensional.
Short Answer
Expert verified
Approximately 6.49% of the initial kinetic energy of the alpha particle is transferred to the stationary uranium nucleus during the elastic collision.
Step by step solution
01
Define the initial known conditions
We are given the following information:
- The alpha particle has mass \(m_\alpha = 4\) amu (atomic mass units)
- The uranium nucleus has mass \(m_U = 235\) amu
- The initial velocity of the alpha particle is \(v_\alpha\)
- The initial velocity of the uranium nucleus is \(v_U = 0\), since it is stationary.
Since we are only interested in the percentage of the kinetic energy transferred, we don't need the actual numerical value of the initial velocities. We can work symbolically and express the final result in terms of \(v_\alpha\).
02
Apply the conservation of momentum
In a one-dimensional elastic collision, the conservation of momentum states that the total initial momentum of the system equals the total final momentum of the system.
Mathematically,
\(m_\alpha v_\alpha + m_U v_U = m_\alpha v_\alpha' + m_U v_U'\)
where,
- \(v_\alpha'\) is the final velocity of the alpha particle,
- \(v_U'\) is the final velocity of the uranium nucleus.
Since the uranium nucleus is initially stationary, \(v_U = 0\), and the equation simplifies to:
\(m_\alpha v_\alpha = m_\alpha v_\alpha' + m_U v_U'\)
03
Apply the conservation of kinetic energy
In an elastic collision, the total kinetic energy is conserved as well.
Mathematically,
\(\frac{1}{2} m_\alpha v_\alpha^2 = \frac{1}{2} m_\alpha v_\alpha'^2 + \frac{1}{2} m_U v_U'^2\)
04
Solve simultaneously for the final velocities
We have two equations now:
1. \(m_\alpha v_\alpha = m_\alpha v_\alpha' + m_U v_U'\)
2. \(\frac{1}{2} m_\alpha v_\alpha^2 = \frac{1}{2} m_\alpha v_\alpha'^2 + \frac{1}{2} m_U v_U'^2\)
First, we solve equation (1) for \(v_U'\):
\(v_U' = \frac{m_\alpha (v_\alpha-v_\alpha')}{m_U}\)
Next, we substitute equation (3) into equation (2) and solve for \(v_\alpha'\):
\(v_\alpha' = \frac{m_\alpha - m_U}{m_\alpha + m_U}v_\alpha\)
05
Calculate the percentage of kinetic energy transferred
The initial kinetic energy of the alpha particle is:
\(KE_\alpha^{initial} = \frac{1}{2} m_\alpha v_\alpha^2\)
The final kinetic energy of the alpha particle is:
\(KE_\alpha^{final} = \frac{1}{2} m_\alpha (v_\alpha')^2\)
After replacing \(v_\alpha'\) from step 4 in the final kinetic energy equation and simplifying, we get:
\(KE_\alpha^{final} = \frac{(m_\alpha - m_U)^2}{(m_\alpha + m_U)^2} \cdot \frac{1}{2} m_\alpha v_\alpha^2\)
Now, we can find the percentage of the kinetic energy transferred:
\(\text{% kinetic energy transferred} = \frac{KE_\alpha^{initial} - KE_\alpha^{final}}{KE_\alpha^{initial}} \cdot 100\%\)
After substituting the initial and final kinetic energies equations, the expression simplifies to:
\(\text{% kinetic energy transferred} = \frac{4m_\alpha m_U}{(m_\alpha + m_U)^2} \cdot 100\%\)
Finally, we substitute the masses of the alpha particle and uranium nucleus, and evaluate the percentage:
\(\text{% kinetic energy transferred} = \frac{4(4)(235)}{(4 + 235)^2} \cdot 100\% \approx 6.49 \% \)
Hence, approximately 6.49% of the initial kinetic energy of the alpha particle is transferred to the stationary uranium nucleus during the elastic collision.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Conservation of Momentum
When discussing elastic collisions, understanding the conservation of momentum is crucial. It's a fundamental principle stating that, in a closed system, the total momentum remains constant provided no external forces are acting on it.
The momentum for a given object is calculated using the equation:
\[ p = m \cdot v \]
where \( p \) represents momentum, \( m \) is the mass, and \( v \) is the velocity of the object. In the context of our exercise, the alpha particle and uranium nucleus make up the system. Initially, the alpha particle is in motion, while the uranium nucleus is stationary. The conservation of momentum dictates that the motion of our alpha particle will be transferred to the uranium nucleus in such a way that the total momentum before and after the collision is equal. This reflects in our problem where we use
\[ m_\alpha v_\alpha = m_\alpha v_\alpha' + m_U v_U' \]
to show that the sum of the momenta before collision is equal to the sum after the collision, with the primes indicating the final velocities post-collision.
The momentum for a given object is calculated using the equation:
\[ p = m \cdot v \]
where \( p \) represents momentum, \( m \) is the mass, and \( v \) is the velocity of the object. In the context of our exercise, the alpha particle and uranium nucleus make up the system. Initially, the alpha particle is in motion, while the uranium nucleus is stationary. The conservation of momentum dictates that the motion of our alpha particle will be transferred to the uranium nucleus in such a way that the total momentum before and after the collision is equal. This reflects in our problem where we use
\[ m_\alpha v_\alpha = m_\alpha v_\alpha' + m_U v_U' \]
to show that the sum of the momenta before collision is equal to the sum after the collision, with the primes indicating the final velocities post-collision.
Conservation of Kinetic Energy
In one-dimensional elastic collisions, we also talk about the conservation of kinetic energy. 'Elastic' indicates that no kinetic energy is lost as heat or sound, unlike what would occur in an inelastic collision. Basically, the energy you start with is the energy you end with.
The kinetic energy of an object is given by the equation:
\[ KE = \frac{1}{2} m v^2 \]
For our alpha particle and uranium nucleus before and after the collision, their collective kinetic energy should be constant. Thus, we set up an equation similar to that of momentum conservation, but instead for kinetic energies, which in our exercise is:
\[ \frac{1}{2} m_\alpha v_\alpha^2 = \frac{1}{2} m_\alpha v_\alpha'^2 + \frac{1}{2} m_U v_U'^2 \]
Through this, we can understand that the initial kinetic energy of the moving alpha particle is redistributed between it and the uranium nucleus following the collision.
The kinetic energy of an object is given by the equation:
\[ KE = \frac{1}{2} m v^2 \]
For our alpha particle and uranium nucleus before and after the collision, their collective kinetic energy should be constant. Thus, we set up an equation similar to that of momentum conservation, but instead for kinetic energies, which in our exercise is:
\[ \frac{1}{2} m_\alpha v_\alpha^2 = \frac{1}{2} m_\alpha v_\alpha'^2 + \frac{1}{2} m_U v_U'^2 \]
Through this, we can understand that the initial kinetic energy of the moving alpha particle is redistributed between it and the uranium nucleus following the collision.
Kinetic Energy Transfer
We've seen that during an elastic collision, kinetic energy is conserved. But how is this energy shared between the colliding bodies? The answer lies in the term called kinetic energy transfer.
Essentially, kinetic energy transfer is the process by which energy moves from one body to another as a result of the collision. In our case, because the uranium nucleus starts at rest, all the kinetic energy it has after the collision comes from the alpha particle, representing the transferred energy. We capture this transfer quantitatively by looking at the difference in the alpha particle's kinetic energy before and after the collision. As expressed by
\[ \text{% kinetic energy transferred} = \frac{KE_\alpha^{initial} - KE_\alpha^{final}}{KE_\alpha^{initial}} \cdot 100\% \]
This gives us the percentage which, after applying the conservation principles and plugging in the known values, tells us precisely how much energy the alpha particle gives up to the uranium nucleus.
Essentially, kinetic energy transfer is the process by which energy moves from one body to another as a result of the collision. In our case, because the uranium nucleus starts at rest, all the kinetic energy it has after the collision comes from the alpha particle, representing the transferred energy. We capture this transfer quantitatively by looking at the difference in the alpha particle's kinetic energy before and after the collision. As expressed by
\[ \text{% kinetic energy transferred} = \frac{KE_\alpha^{initial} - KE_\alpha^{final}}{KE_\alpha^{initial}} \cdot 100\% \]
This gives us the percentage which, after applying the conservation principles and plugging in the known values, tells us precisely how much energy the alpha particle gives up to the uranium nucleus.
One-Dimensional Collision
All the principles we've talked about converge in the context of a one-dimensional collision. A one-dimensional collision refers to a scenario where two objects collide and move along a single straight line.
For such collisions, analyzing the physics becomes more straightforward because all motion is restricted to one dimension 鈥� there are no perpendicular or secondary forces to consider. Consequently, the mathematics that governs the collision can be greatly simplified, illustrated by the linear equations used to solve our problem. The fact that our objects鈥攁n alpha particle and a uranium nucleus鈥攐nly move along the direction of the initial alpha particle's movement, simplifies the analysis enabling us to use conservation laws directly to find final velocities and the percentage of kinetic energy transferred.
For such collisions, analyzing the physics becomes more straightforward because all motion is restricted to one dimension 鈥� there are no perpendicular or secondary forces to consider. Consequently, the mathematics that governs the collision can be greatly simplified, illustrated by the linear equations used to solve our problem. The fact that our objects鈥攁n alpha particle and a uranium nucleus鈥攐nly move along the direction of the initial alpha particle's movement, simplifies the analysis enabling us to use conservation laws directly to find final velocities and the percentage of kinetic energy transferred.
