Problem 94 - University Physics with Modern Physics

Question

Energy Sharing in Elastic Collisions. A stationary object with mass \(m_{B}\) is struck head-on by an object with mass \(m_{A}\) that is moving initially at speed \(v_{0} .\) (a) If the collision is elastic, what percentage of the original energy does each object have after the collision? (b) What does your answer in part (a) give for the special cases (i) \(m_{A}=m_{B}\) and (ii) \(m_{A}=5 m_{B} ?(\mathrm{c})\) For what values, if any, of the mass ratio \(m_{A} / m_{B}\) is the original kinetic energy shared equally by the two objects after the collision?

Step-by-Step Solution

Answer

(a) Mass A retains \( \left( \frac{m_A - m_B}{m_A + m_B} \right)^2 \)% energy; mass B gets \( \left( \frac{2m_A}{m_A + m_B} \right)^2 \)%. (b) (i) All energy to B, (ii) 44% A, 56% B. (c) Equal energy for \( m_A = 3m_B \).

Step 1: Understanding Elastic Collisions

In an elastic collision, both energy and momentum are conserved. The kinetic energy of the system and the total momentum before and after the collision remain constant.

Step 2: Applying Conservation of Momentum

The law of conservation of momentum states \( m_A v_0 = m_A v_A' + m_B v_B' \), where \( v_A' \) and \( v_B' \) are the velocities after the collision.

Step 3: Applying Conservation of Energy

For elastic collisions, the initial kinetic energy \( \frac{1}{2}m_A v_0^2 \) equals the sum of the kinetic energies after collision: \( \frac{1}{2} m_A v_A'^2 + \frac{1}{2} m_B v_B'^2 \).

Step 4: Solving for Post-collision Velocities

Using the two conservation equations, solve for the post-collision velocities:\[ v_A' = \frac{(m_A - m_B)}{(m_A + m_B)} v_0 \]\[ v_B' = \frac{2m_A}{(m_A + m_B)} v_0 \]

Step 5: Calculating Percentages of Initial Energy

The kinetic energies can be expressed as percentages of the initial energy:For mass \( A \):\[ \frac{E_A'}{E_0} = \left( \frac{m_A - m_B}{m_A + m_B} \right)^2 \]For mass \( B \):\[ \frac{E_B'}{E_0} = \left( \frac{2m_A}{m_A + m_B} \right)^2 \]

Step 6: Special Case (i) Equal Masses

If \( m_A = m_B \), masses cancel in formulas giving:- For mass \( A \): \( \frac{E_A'}{E_0} = 0 \).- For mass \( B \): \( \frac{E_B'}{E_0} = 1 \).Mass A transfers all its energy to mass B.

Step 7: Special Case (ii) Mass Ratio 5:1

Substitute \( m_A = 5m_B \) in the energy percentages:- For mass \( A \): \( \frac{E_A'}{E_0} = \left( \frac{4}{6} \right)^2 = \frac{4}{9} \approx 44\% \).- For mass \( B \): \( \frac{E_B'}{E_0} = \left( \frac{10}{6} \right)^2 = \frac{100}{36} = \frac{25}{9} \approx 56\% \).

Step 8: Equal Energy Sharing Condition

For equal energy sharing, set \( \left( \frac{m_A - m_B}{m_A + m_B} \right)^2 = \left( \frac{2m_A}{m_A + m_B} \right)^2 \). Solving this equation leads to a quadratic form. Simplifying gives \( m_A = 3m_B \), meaning each object will have the same energy post-collision.

Key Concepts

Conservation of Momentum

In physics, the law of conservation of momentum is a cornerstone concept that is fundamental to understanding elastic collisions. When two objects collide, the total momentum before the collision must equal the total momentum after the collision. This implies that no momentum is lost, only transferred.
Consider a scenario where two objects collide: initially, one object with mass \( m_A \) is moving with a velocity \( v_0 \), while the other with mass \( m_B \) is stationary. The total momentum initially is \( m_A v_0 \). After the collision, the velocities of the objects are \( v_A' \) and \( v_B' \). Using conservation of momentum, we have:

• Initial total momentum = \( m_A v_0 \)
• Final total momentum = \( m_A v_A' + m_B v_B' \)

By setting these two expressions equal, one can solve for the velocities \( v_A' \) and \( v_B' \), which are essential for determining the kinetic energy distribution after the collision.

Conservation of Energy

In an elastic collision, not only is momentum conserved, but so is kinetic energy. This means the total kinetic energy of the system before the collision must equal the total kinetic energy after the collision. This principle allows us to further explore the behavior of the objects involved in the collision.
Initially, the moving object has a kinetic energy calculated as \( \frac{1}{2} m_A v_0^2 \). Post-collision, the energies of both objects become important:

• Post-collision kinetic energy of object A: \( \frac{1}{2} m_A v_A'^2 \)
• Post-collision kinetic energy of object B: \( \frac{1}{2} m_B v_B'^2 \)

Using the conservation of energy equation:\[ \frac{1}{2}m_A v_0^2 = \frac{1}{2} m_A v_A'^2 + \frac{1}{2} m_B v_B'^2 \]This equation ensures that energy is merely redistributed between objects, not lost, assuming perfect elastic conditions. Solving for these helps in determining the kinetic energy fractions of each object post-collision.

Kinetic Energy Distribution

The redistribution of kinetic energy in an elastic collision depends heavily on the masses of the objects involved. The mass ratio \( \frac{m_A}{m_B} \) plays a crucial role in how the kinetic energy is divided post-collision. Here is how it unfolds:

• For equal masses (\( m_A = m_B \)): Mass A transfers all its kinetic energy to mass B, ending with 0% of the initial energy, while mass B takes 100% of it.
• When \( m_A = 5m_B \): Mass A retains approximately 44% of the original energy, and mass B gains about 56% of it.
• Special equal sharing occurs when \( m_A = 3m_B \). Both masses end up with equal kinetic energies post-collision.

Through these ratios, it becomes clear how kinetic energy can shift between objects, illustrating the elegance of elastic collisions. For any mass ratio, we can use the derived formulas to predict exactly how energy splits between the two objects post-collision.