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bagirrra123 [75]
2 years ago
11

a golf ball has a mass 0.046 kkg rests on a tee it is struck by a golf club with an effectiv mass of 0.22 and speed of 44 assumi

ng the collision is elastic, find the speed of the ball when it leaves the tee
Physics
1 answer:
Fiesta28 [93]2 years ago
6 0

Answer:

210 m/s

Explanation:

We will use this equation for an elastic collision:

  • m₁v₁ + m₂v₂ = m₁v₁ + m₂v₂  
  • The left side of the equation is before the collision; the right side is after the collision.

In this case, m₁ = mass of golf ball (0.046 kg) and m₂ = mass of the golf club (0.22 kg).

<h3><u>Left side of equation:</u></h3>

v₁ is 0 m/s since the ball is at rest on the tee.

v₂ is 44 m/s, given in the problem.

<h3><u>Right side of equation:</u></h3>

v₁ is what we are trying to find.

v₂ is 0 m/s since the ball will be at rest on the ground.

<h3>⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯</h3>

Plug these values into the elastic collision equation:

  • (0.046 kg)(0 m/s) + (0.22 kg)(44 m/s) = (0.046 kg)(v₁) + (0.22 kg)(0 m/s)

Simplify this equation.

  • (0.22 kg)(44 m/s) = (0.046 kg)(v₁)  

Get rid of the units.

  • (0.22)(44) = 0.046v₁

Multiply the left side of the equation.

  • 9.68 = 0.046v₁

Divide both sides by 0.046.

  • 210.434782609 = v₁
  • 210 m/s = v₁

The speed of the ball when it leaves the tee is about 210 m/s.

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Answer:

r = 0.11 m

Explanation:

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<em>v: is the proton's velocity</em>

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<em>B: is the magnetic field = 0.040 T </em>

Solving equation (1) for r, we have:

r = \frac{mv}{qB}   (2)

By conservation of energy, we can find the velocity of the proton:

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<u>Where:</u>

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Solving equation (3) for v, we have:

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Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!  

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