Question

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 assuming the collision is elastic, find the speed of the ball when it leaves the tee

Answer 1

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).

Left side of equation:

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

v₂ is 44 m/s, given in the problem.

Right side of equation:

v₁ is what we are trying to find.

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

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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.