## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

We can find the speed of the heavier skater after they push off. $v = \frac{distance}{time}$ $v = \frac{30~m}{20~s}$ $v = 1.5~m/s$ After the two skaters push off, they will have the same magnitude of momentum. We can find the speed $v_L$ of the lighter skater. $(50~kg)~v_L = (75~kg)(1.5~m/s)$ $v_L = \frac{(75~kg)(1.5~m/s)}{50~kg}$ $v_L = 2.25~m/s$ We can find the time it takes the lighter skater to travel 30 meters. $t = \frac{distance}{speed}$ $t = \frac{30~m}{2.25~m/s}$ $t = 13.3~s$ It takes the lighter skater 13.3 seconds to reach the edge.