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.
Multiply the left side of the equation.
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.
