To prove the conservation of momentum, we can use the principle that the total momentum before an event must equal the total momentum after the event.
Before the skaters push off each other, they are both at rest, so their initial momentum is:
\[ \text{Initial Momentum} = 0 , \text{kg m/s} \]
After they push off each other, skater A (mass \(m_A = 72 , \text{kg}\)) moves with a velocity \(v_A = 5 , \text{m/s}\) in one direction, and skater B (mass \(m_B = 48 , \text{kg}\)) must move in the opposite direction with some velocity \(v_B\).
The momentum after they push off will be:
\[ \text{Final Momentum} = m_A \cdot v_A + m_B \cdot v_B \]
Since skater B moves in the opposite direction, we will consider the velocity of skater B as negative. Thus, we have:
\[ \text{Final Momentum} = (72 , \text{kg}) \cdot (5 , \text{m/s}) + (48 , \text{kg}) \cdot (-v_B) \]
Setting the initial momentum equal to the final momentum gives:
\[ 0 = (72 \cdot 5) + (48 \cdot -v_B) \]
Calculating \(72 \cdot 5\):
\[ 0 = 360 - 48v_B \]
Now, solving for \(v_B\):
\[ 48v_B = 360 \]
\[ v_B = \frac{360}{48} \]
\[ v_B = 7.5 , \text{m/s} \]
Thus, skater B must move with a velocity of 7.5 m/s in the opposite direction after the push to prove conservation of momentum.