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Two players hit an air hockey puck simultaneously. One player hits the puck with a force of 4.15 N at an angle of 32.5 degrees. The other player hits the puck with a force of 8.85 N at an angle of 66.9 degrees. If the puck has a mass of 52.0 g, what is the acceleration of the puck and the angle of the puck's motion? Assume the friction between the puck and the table's surface is negligible.

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.

  1. 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]

  2. 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]

  3. 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]

  4. 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]

  5. 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]

  6. 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.

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