Final answer:
The hockey puck has an east component of momentum of 1.5 kg·m/s and a north component of momentum of 7.5 kg·m/s after a force is applied due north for 1.5 seconds.
Step-by-step explanation:
The question is asking to calculate the momentum components of a hockey puck after it experiences an impulse as a result of a force acting on it. Initially, it is given that a hockey puck with a mass of 150 g (0.15 kg) is sliding due east on a frictionless table with a speed of 10 m/s. Then, a constant force of 5 N directed due north is applied to the puck for 1.5 seconds. Using the formula for impulse (J = Ft), we can calculate the change in momentum due to this force. The impulse provided in the north direction is J = (5 N)(1.5 s) = 7.5 Ns. The east component of momentum remains unchanged since no force acts in that direction. Therefore, the east component of momentum (Pe) is the initial momentum in the east direction: 0.15 kg × 10 m/s = 1.5 kg·m/s.
The north component of momentum (Pn) is the product of the mass of the puck and the change in velocity in the north direction, which is equal to the impulse since there was no initial velocity in the north direction. As such, Pn = 7.5 kg·m/s. Finally, the puck's total momentum at the end of the 1.5-second interval has two components: Pe = 1.5 kg·m/s to the east and Pn = 7.5 kg·m/s to the north.