To prove the conservation of momentum, we can use the principle that the total momentum before the push must equal the total momentum after the push.
Before they push off each other, both skaters are at rest, so the initial momentum is:
\[ p_{\text{initial}} = 0 \]
After they push off, the momentum of each skater can be expressed as:
\[ p_A = m_A \cdot v_A \] \[ p_B = m_B \cdot v_B \]
Where:
- \(m_A = 72 , \text{kg}\) (mass of Skater A)
- \(m_B = 55 , \text{kg}\) (mass of Skater B)
- \(v_B = 3.0 , \text{m/s}\) (velocity of Skater B)
- \(v_A\) is the unknown velocity of Skater A in the opposite direction, so it will be negative in our equation.
The total momentum after they push off is:
\[ p_{\text{final}} = m_A \cdot v_A + m_B \cdot v_B \]
Setting the initial momentum equal to the final momentum:
\[ 0 = m_A \cdot v_A + m_B \cdot v_B \]
Substituting the known values:
\[ 0 = (72 , \text{kg}) \cdot v_A + (55 , \text{kg}) \cdot (3.0 , \text{m/s}) \]
This simplifies to:
\[ 0 = 72 \cdot v_A + 165 \]
Rearranging gives:
\[ 72 \cdot v_A = -165 \]
Now, solving for \(v_A\):
\[ v_A = \frac{-165}{72} \approx -2.29 , \text{m/s} \]
Since we are interested in the magnitude in the opposite direction, we can consider it as approximately 2.3 m/s.
Thus, the answer is:
2.3 m/s