Question

problem two billiard balls of identical mass move toward each other as in active figure 6.13, with the positive \\(x\\)-axis to the right (steps 1 and 2). assume that the collision between them is perfectly elastic. if the initial velocities of the balls are \\(+30.0\text{ cm/s}\\) and \\(-20.0\text{ cm/s}\\), what are the velocities of the balls after the collision? assume friction and rotation are unimportant.

Explanation:

🆕 New Concept Discovered: Elastic Collisions
Objects bounce off each other, conserving momentum and kinetic energy.

Step 1: Identify given values and conservation laws

We have two billiard balls of identical mass, so \( m_1 = m_2 = m \).

The initial velocities are:
\[ v_{1i} = +30.0\text{ cm/s} \]
\[ v_{2i} = -20.0\text{ cm/s} \]

For any perfectly elastic collision, both total momentum and total kinetic energy are conserved.

Step 2: Apply momentum conservation

Because the masses are equal, the conservation of momentum equation simplifies significantly:
\[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \]

Since \( m_1 = m_2 = m \), we can divide the entire equation by \( m \):
\[ v_{1i} + v_{2i} = v_{1f} + v_{2f} \]

Substitute the known initial values:
\[ 30.0 + (-20.0) = v_{1f} + v_{2f} \]
\[ v_{1f} + v_{2f} = 10.0\text{ cm/s} \quad \text{--- (Equation 1)} \]

Step 3: Apply the relative velocity relation for elastic collisions

For a perfectly elastic head-on collision, the relative velocity of approach equals the relative velocity of separation:
\[ v_{1i} - v_{2i} = -(v_{1f} - v_{2f}) \]
\[ v_{1i} - v_{2i} = v_{2f} - v_{1f} \]

Substitute the known initial values:
\[ 30.0 - (-20.0) = v_{2f} - v_{1f} \]
\[ v_{2f} - v_{1f} = 50.0\text{ cm/s} \quad \text{--- (Equation 2)} \]

Step 4: Solve the system of equations

We have two linear equations:

1. \( v_{1f} + v_{2f} = 10.0 \)
2. \( -v_{1f} + v_{2f} = 50.0 \)

Add Equation 1 and Equation 2 together:
\[ (v_{1f} + v_{2f}) + (-v_{1f} + v_{2f}) = 10.0 + 50.0 \]
\[ 2v_{2f} = 60.0 \]
\[ v_{2f} = 30.0\text{ cm/s} \]

Now, substitute \( v_{2f} \) back into Equation 1 to find \( v_{1f} \):
\[ v_{1f} + 30.0 = 10.0 \]
\[ v_{1f} = -20.0\text{ cm/s} \]

Note: When two identical masses undergo a head-on elastic collision, they simply exchange their velocities.

Answer:

The velocities of the balls after the collision are:

• Ball 1: \(-20.0\text{ cm/s}\) (moving to the left)
• Ball 2: \(+30.0\text{ cm/s}\) (moving to the right)