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Chapter 9: Problem 69

Two objects, one initially at rest, undergo a one-dimensional elastic collision. If half the kinetic energy of the initially moving object is transferred to the other object, what is the ratio of their masses?

Short Answer

Expert verified
The ratio of their masses is 1:1.

Step by step solution

01

Define the given variables

Let's denote the masses of the two objects as \(m_1\) and \(m_2\). Let \(v_1\) represent the initial velocity of the moving object (\(m_1\)), and \(v_1'\) and \(v_2'\) be the final velocities of \(m_1\) and \(m_2\). The given problem states that half the kinetic energy is transferred from \(m_1\) to \(m_2\), expressed as: \(\frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_2 {v_2'}^2.\)
02

Apply the principle of conservation of momentum

The conservation of momentum states that the sum of the initial momenta of the two objects is equal to the sum of the final momenta, expressed by: \(m_1 v_1 = m_1 v_1' + m_2 v_2'\). Using the data from Step 1, \(v_2'\) can be expressed as: \(v_2' = v_1 \sqrt{\frac{m_1}{m_2}}\). Substitute this equation into the momentum conservation equation to get a new equation for \(v_1'\).
03

Solve for \(v_1'\)

After substituting the expression of \(v_2'\) into the momentum conservation equation, we find that \(v_1' = v_1 (1 - \sqrt{\frac{m_1}{m_2}})\).
04

Apply the principle of conservation of kinetic energy

The conservation of kinetic energy states that the sum of the initial kinetic energy of the two objects is equal to the sum of their final kinetic energies, expressed as: \(\frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_1 {(v_1')}^2 + \frac{1}{2} m_2 {(v_2')}^2\). Substitute \(v_1'\) and \(v_2'\) in this equation and simplify the equation.
05

Solve for the ratio of masses

After substituting for \(v_1'\) and \(v_2'\) in the kinetic energy conservation equation and simplifying, we obtain \(m_1:m_2 = 1:1\).

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