1. **State the problem:** Hockey Player A has a mass of 60 kg and an initial velocity of +5 m/s. Player B applies a force of -500 N for 0.8 seconds. We need to find Player A's velocity after the force is applied. 2. **Relevant formula:** Impulse-momentum theorem states that impulse equals change in momentum: $$ F \Delta t = m \Delta v $$ where $F$ is force, $\Delta t$ is time interval, $m$ is mass, and $\Delta v$ is change in velocity. 3. **Calculate impulse:** $$ \text{Impulse} = F \times \Delta t = -500 \times 0.8 = -400 \text{ Ns} $$ 4. **Calculate change in velocity:** $$ \Delta v = \frac{\text{Impulse}}{m} = \frac{-400}{60} = -6.67 \text{ m/s} $$ 5. **Calculate final velocity:** $$ v_{final} = v_{initial} + \Delta v = 5 + (-6.67) = -1.67 \text{ m/s} $$ 6. **Interpretation:** The negative sign means Player A is now moving in the opposite direction at 1.67 m/s after the body check. **Final answer:** $$ v_{final} = -1.67 \text{ m/s} $$