Answer:
In an elastic collision, both momentum and kinetic energy are conserved. The total momentum before the collision is equal to the total momentum after the collision.
Let's use the principle of conservation of momentum:
\(m_1 \cdot v_1 + m_2 \cdot v_2 = m_1 \cdot v_1' + m_2 \cdot v_2'\)
Where:
- \(m_1\) and \(m_2\) are the masses of the two objects,
- \(v_1\) and \(v_2\) are their velocities before the collision,
- \(v_1'\) and \(v_2'\) are their velocities after the collision.
Given:
- \(m_1 = 130 \, \text{kg}\), \(v_1 = -3.5 \, \text{m/s}\) (negative because it's in the opposite direction),
- \(m_2 = 110 \, \text{kg}\), \(v_2 = 3 \, \text{m/s}\),
- \(v_1' = -2.4 \, \text{m/s}\) (negative because it's bouncing backward).
Now, we can solve for \(v_2'\), which is Venessa's speed after the collision.
\[130 \cdot (-3.5) + 110 \cdot 3 = 130 \cdot (-2.4) + 110 \cdot v_2'\]
Solving this equation will give you the value of \(v_2'\), which is Venessa's speed after the collision.
