Answer :
To find the acceleration of the puck and the angle of its motion, we need to consider the forces exerted by both players and their directions.
Resolve Forces into Components:
For the first player with a force of 4.15 N at an angle of 32.5 degrees:
[tex]F_{1x} = 4.15 \cos(32.5^\circ)[/tex]
[tex]F_{1y} = 4.15 \sin(32.5^\circ)[/tex]For the second player with a force of 8.85 N at an angle of 66.9 degrees:
[tex]F_{2x} = 8.85 \cos(66.9^\circ)[/tex]
[tex]F_{2y} = 8.85 \sin(66.9^\circ)[/tex]
Calculate the Component Forces:
Calculate numeric values for each component:
- [tex]F_{1x} = 4.15 \times \cos(32.5^\circ) \approx 3.50 \text{ N}[/tex]
- [tex]F_{1y} = 4.15 \times \sin(32.5^\circ) \approx 2.24 \text{ N}[/tex]
- [tex]F_{2x} = 8.85 \times \cos(66.9^\circ) \approx 3.52 \text{ N}[/tex]
- [tex]F_{2y} = 8.85 \times \sin(66.9^\circ) \approx 8.14 \text{ N}[/tex]
Find the Net Force Components:
Add the x-components:
[tex]F_{net x} = F_{1x} + F_{2x} = 3.50 + 3.52 = 7.02 \text{ N}[/tex]Add the y-components:
[tex]F_{net y} = F_{1y} + F_{2y} = 2.24 + 8.14 = 10.38 \text{ N}[/tex]
Calculate the Net Force Magnitude:
- Using Pythagorean theorem:
[tex]F_{net} = \sqrt{(F_{net x})^2 + (F_{net y})^2} = \sqrt{7.02^2 + 10.38^2} \approx 12.56 \text{ N}[/tex]
- Using Pythagorean theorem:
Find the Angle of Motion:
- The angle [tex]\theta[/tex] from the positive x-axis can be found using:
[tex]\theta = \tan^{-1} \left( \frac{F_{net y}}{F_{net x}} \right) = \tan^{-1} \left( \frac{10.38}{7.02} \right) \approx 56.7^\circ[/tex]
- The angle [tex]\theta[/tex] from the positive x-axis can be found using:
Calculate the Acceleration:
The mass of the puck is 52.0 g, which is 0.052 kg.
Using Newton's second law, [tex]F = ma[/tex]:
[tex]a = \frac{F_{net}}{m} = \frac{12.56}{0.052} \approx 241.54 \text{ m/s}^2[/tex]
Therefore, the acceleration of the puck is approximately 241.54 m/s² and the angle of the puck's motion is approximately 56.7 degrees from the positive x-axis.
