Answer:
First, let's calculate the total momentum before the push:
\[
\text{Total momentum before} = (50 \, \text{kg} \times 2 \, \text{m/s}) + (75 \, \text{kg} \times 0 \, \text{m/s}) = 100 \, \text{kg m/s}
\]
Now, using the principle of conservation of momentum, we can set the total momentum before the push equal to the total momentum after the push:
\[
100 = 50 \times v_1 + 75 \times v_2
\]
Given that one skater moves at 2 m/s after the push, we have \( v_1 = 2 \, \text{m/s} \). Plugging this into the equation:
\[
100 = 50 \times 2 + 75 \times v_2
\]
\[
100 = 100 + 75 \times v_2
\]
\[
75 \times v_2 = 0
\]
Therefore, the velocity of the second skater after the push is 0 m/s. This makes sense intuitively, as the first skater pushes off the second skater and moves away, leaving the second skater stationary.
