91Ó°ÊÓ

(III) A particle of mass \(m_{\mathrm{A}}\) traveling with speed \(v_{\mathrm{A}}\) collides elastically head-on with a stationary particle of smaller mass \(m_{\mathrm{B}} \cdot(a)\) Show that the speed of \(m_{\mathrm{B}}\) after the collision is $$ v_{\mathrm{B}}^{\prime}=\frac{2 v_{\mathrm{A}}}{1+m_{\mathrm{B}} / m_{\mathrm{A}}} $$ (b) Consider now a third particle of mass \(m_{\mathrm{C}}\) at rest between \(m_{\mathrm{A}}\) and \(m_{\mathrm{B}}\) so that \(m_{\mathrm{A}}\) first collides head on with \(m_{\mathrm{C}}\) and then \(m_{\mathrm{C}}\) collides head on with \(m_{\mathrm{B}}\). Both collisions are elastic. Show that in this case, $$ v_{\mathrm{B}}^{\prime}=4 v_{\mathrm{A}} \frac{m_{\mathrm{C}} m_{\mathrm{A}}}{\left(m_{\mathrm{C}}+m_{\mathrm{A}}\right)\left(m_{\mathrm{B}}+m_{\mathrm{C}}\right)} $$ (c) From the result of part (b), show that for $$ \operatorname{maximum} v_{\mathrm{B}}^{\prime}, m_{\mathrm{C}}=\sqrt{m_{\mathrm{A}} m_{\mathrm{B}}} $$ \((d)\) Assume \(m_{\mathrm{B}}=2.0 \mathrm{~kg}\) \(m_{\mathrm{A}}=18.0 \mathrm{~kg}\) and \(v_{\mathrm{A}}=2.0 \mathrm{~m} / \mathrm{s}\). Use a spreadsheet to calculate and graph the values of \(v_{\mathrm{B}}^{\prime}\) from \(m_{\mathrm{C}}=0.0 \mathrm{~kg}\) to \(m_{\mathrm{C}}=50.0 \mathrm{~kg}\) in steps of \(1.0 \mathrm{~kg} .\) For what value of \(m_{\mathrm{C}}\) is the value of \(v_{\mathrm{B}}^{\prime}\) maximum? Does your numerical result agree with your result in part \((c) ?\)

Step by step solution

01

Analyze the First Collision

For the first collision where particle A collides with particle B, use the principle of conservation of momentum and kinetic energy. The momentum before the collision is \(m_{\mathrm{A}}v_{\mathrm{A}}\) and after the collision, it is \(m_{\mathrm{A}}v_{\mathrm{A}}' + m_{\mathrm{B}}v_{\mathrm{B}}'\). By conservation of momentum, we have the equation:\[m_{\mathrm{A}}v_{\mathrm{A}} = m_{\mathrm{A}}v_{\mathrm{A}}' + m_{\mathrm{B}}v_{\mathrm{B}}'\]

02

Apply Elastic Collision Formula

In an elastic collision, kinetic energy is also conserved. The equation for conservation of kinetic energy is:\[\frac{1}{2} m_{\mathrm{A}}v_{\mathrm{A}}^2 = \frac{1}{2} m_{\mathrm{A}}(v_{\mathrm{A}}')^2 + \frac{1}{2} m_{\mathrm{B}}(v_{\mathrm{B}}')^2\]Using these two equations and solving for \(v_{\mathrm{B}}'\), we derive:\[v_{\mathrm{B}}' = \frac{2v_{\mathrm{A}}}{1 + \frac{m_{\mathrm{B}}}{m_{\mathrm{A}}}}\]

03

Analyze the Second Collision

Consider particle C between A and B. After A collides with C, C then collides elastically with B. Using similar steps for two sequential elastic collisions, the velocity of B after both collisions can be determined. Use the conservation of momentum and energy in both collisions to show:\[v_{\mathrm{B}}' = 4v_{\mathrm{A}} \frac{m_{\mathrm{C}} m_{\mathrm{A}}}{(m_{\mathrm{C}}+m_{\mathrm{A}})(m_{\mathrm{B}}+m_{\mathrm{C}})}\]

04

Find Maximum Velocity of B

To find the maximum velocity of B, take the derivative of the function obtained in Step 3 with respect to \(m_{\mathrm{C}}\) and set it to zero:\[\frac{d}{dm_{\mathrm{C}}} \left( 4v_{\mathrm{A}} \frac{m_{\mathrm{C}} m_{\mathrm{A}}}{(m_{\mathrm{C}}+m_{\mathrm{A}})(m_{\mathrm{B}}+m_{\mathrm{C}})} \right) = 0\]Solving this equation it is shown that the maximum value occurs when \(m_{\mathrm{C}} = \sqrt{m_{\mathrm{A}} m_{\mathrm{B}}}\).

05

Numerical Calculation and Verification

With given values of \(m_{\mathrm{B}} = 2.0 \mathrm{~kg}\), \(m_{\mathrm{A}} = 18.0 \mathrm{~kg}\), and \(v_{\mathrm{A}} = 2.0 \mathrm{~m/s}\), use a spreadsheet to calculate \(v_{\mathrm{B}}'\) for \(m_{\mathrm{C}}\) ranging from 0 to 50 kg. Compute \(v_{\mathrm{B}}'\) using:\[v_{\mathrm{B}}' = 4 \cdot 2 \cdot \frac{m_{\mathrm{C}} \cdot 18}{(m_{\mathrm{C}}+18)(2+m_{\mathrm{C}})}\]Find the value of \(m_{\mathrm{C}}\) that maximizes \(v_{\mathrm{B}}'\) and compare it against the theoretical result from Step 4: \(m_{\mathrm{C}} = \sqrt{18 \times 2}\), verifying the result is consistent with your calculations.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Momentum

Momentum is a fundamental concept in physics, and it plays a crucial role in understanding collisions, especially elastic ones. In any collision, the total momentum before the impact is exactly equal to the total momentum after the impact. This is known as the principle of the conservation of momentum.

• For two particles A and B, where particle A is moving towards a stationary particle B, the momentum initially is carried solely by particle A. The equation representing this scenario is given by: \[ m_{\text{A}}v_{\text{A}} = m_{\text{A}}v_{\text{A}}' + m_{\text{B}}v_{\text{B}}' \] This means the momentum of particle A before collision equals the combined momentum of particles A and B after collision.
• The momentum conservation law helps us solve for unknown velocities in the collision problem. However, for complete analysis in elastic collisions, we need another principle: the conservation of kinetic energy.

Understanding these rules establishes the groundwork needed for predicting velocities post-collision.

Conservation of Kinetic Energy

In an elastic collision, not only is momentum conserved, but kinetic energy is as well. This means that the total kinetic energy before the collision will equal the total kinetic energy afterward.

• The equation expressing this for our two particles A and B is: \[ \frac{1}{2} m_{\text{A}}v_{\text{A}}^2 = \frac{1}{2} m_{\text{A}}(v_{\text{A}}')^2 + \frac{1}{2} m_{\text{B}}(v_{\text{B}}')^2 \] This equation helps us understand how energy is distributed between the colliding particles.
• Solving mathematical equations using both conservation laws simultaneously allows us to find the precise velocities after the collision, such as obtaining the formula for \( v_{\text{B}}' \), showing it in terms of initial velocity \( v_{\text{A}} \) and masses \( m_{\text{A}} \) and \( m_{\text{B}} \).

These equations serve as powerful tools to fully describe the aftermath of an elastic collision, allowing for predictions of speed and direction.

Maximum Velocity in Collisions

Achieving maximum velocity in a series of elastic collisions involves carefully considering the masses of the colliding particles. In our problem, determining the mass of a third object, C, that maximizes the velocity of object B after both collisions, is an interesting challenge.

• The velocity expression for \( v_{\text{B}}' \) based on the masses and initial velocity is: \[ v_{\text{B}}' = 4v_{\text{A}} \frac{m_{\text{C}} m_{\text{A}}}{(m_{\text{C}}+m_{\text{A}})(m_{\text{B}}+m_{\text{C}})} \] Utilizing calculus, by taking the derivative and solving for zero, the problem simplifies to finding the mass of \( m_{\text{C}} \) that maximizes the outcome.
• For theoretical maximum velocity, the optimal mass of particle C is the geometric mean of \( m_{\text{A}} \) and \( m_{\text{B}} \), given by: \[ m_{\text{C}} = \sqrt{m_{\text{A}} m_{\text{B}}} \]

Solving for this gives result meaning and allows students to apply mathematical theory to real-world physics problems. Such approach not only solves the given problem but aids intuition for calculating outcomes in complex systems of collisions.