Question

A 0.160-kg {\rm kg} hockey puck is moving on an icy, frictionless, horizontal surface. At t=0 t=0 the puck is moving to the right at 2.91m/s m/s .
1)Calculate the magnitude of the velocity of the puck after a force of 25.9N N directed to the right has been applied for 6.0×10−2s s .
2)If instead, a force of 12.7N N directed to the left is applied from t=0 t=0 to t= t= 6.0×10−2s s , what is the magnitude of the final velocity of the puck?

Answer 1

A 0.160-kg $\{\mathrm{rm~kg}\}$ hockey puck is moving on an icy, frictionless, horizontal surface. At $t = 0$ to $t = 0$ the puck is moving to the right at $2.91\mathrm{m / s~m / s}$.

1) Calculate the magnitude of the velocity of the puck after a force of 25.9N N directed to the right has been applied for $6.0 \times 10^{-2} \, \text{s} \, \text{s}$.

2) If instead, a force of 12.7N N directed to the left is applied from $t = 0$ to $t = 0$ to $t = 6.0 \times 10^{-2} \, \text{s} \, \text{s}$, what is the magnitude of the final velocity of the puck?

Solution:

Let

$$v_{0} = 2.91 \, \mathrm{m/s} - \text{magnitude of hockey puck velocity before force action}$$ $$v_{f} - \text{magnitude of hockey puck velocity after force action}$$

1) Calculate the magnitude of the velocity of the puck after a force of 25.9N N directed to the right has been applied for $6.0 \times 10^{-2} \, \text{s} \, \text{s}$.

Since the force and velocity have the same direction:

$$F = m \frac{v - v_{0}}{t} \rightarrow v = v_{0} + \frac{Ft}{m} = 2.91 + \frac{25.9 * 6 * 10^{-2}}{0.16} = 7.26 \, \mathrm{m/s}$$

2) If instead, a force of 12.7N N directed to the left is applied from $t = 0$ to $t = t = 6.0 \times 10^{-2} \, \text{s} \, \text{s}$, what is the magnitude of the final velocity of the puck?

Since the force and velocity have the opposite direction:

$$- F = m \frac{v - v_{0}}{t} \rightarrow v = v_{0} - \frac{Ft}{m} = 2.91 - \frac{12.7 * 6 * 10^{-2}}{0.16} = -1.853 \, \mathrm{m/s}$$

a minus sign indicates that the puck will fly in the opposite direction

Answer: 1. $7.26\mathrm{m / s}$ $\nu = 1.85\mathrm{m / s}$