Answer to Question #140010 in Physics for Kaycee

Question

A 0.125-kg hockey puck is sliding North across a frictionless ice surface at a constant speed of 39 m/s when a hockey stick exerts a force of 32 N [S25oE] on the puck for 0.15 s. Find the final velocity of the hockey puck

Answer 1

The puch initialy has the momentum $\mathbf{p}_0$ with the coordinates:

$$\mathbf{p}_0 = (0,mv_0)$$

where $m = 0.125kg$ is the mass of the puck and $v_0 = 39m/s$ is its initial speed. The force exerted to the puck has the following coordinates:

$$\mathbf{F} = F (\cos\theta, -\sin\theta )$$

where $F = 32N$ is its magnitude. According to the second Newton's law, this force exerts the following momentum change to the puck:

$$\Delta\mathbf{ p} = \mathbf{F}\Delta t = (F \Delta t\cos\theta, -F \Delta t\sin\theta )$$

where $\Delta t = 0.15s$ is the time of interaction.

Thus, the final momentum will be:

$$\mathbf{p} = \mathbf{p}_0+\Delta\mathbf{p}\\ \mathbf{p} = (0+F \Delta t\cos\theta,\space mv_0-F \Delta t\sin\theta) \approx(4.35, 2.85)kg\cdot m/s$$

The final speed is:

$$\mathbf{v} = \dfrac{\mathbf{p}}{m} = (34.8, 22.8)m/s$$

Answer. $(34.8, 22.8)m/s$.