👤
Answered

A 7.00 kg ball hits a 75.0 kg man standing at rest on ice. The man catches the ball. How fast does the ball need to be moving in order to send the man off at a speed of 3.00 m/s?

Answer :

Supernovago

Answer:

35.14 m/s

Explanation:

The Law of Conservation of Momentum states that the momentum before and after a collision is the same.

  • m₁ v₁ + m₂ v₂ = m₁ v₁ + m₂ v₂

Let's set the ball to have the subscript of 1 and the man to have the subscript of 2.

The initial and final mass of the ball is the same, so m₁ = 7.00 kg on both sides of the equation.

The initial velocity of the ball, v₁ on the left side of the equation, is the unknown variable we are trying to find.

The initial and final mass of the man is the same, so m₂ = 75.0 kg on both sides of the equation.

The man starts at rest, meaning that his initial velocity is v₂ = 0 m/s on the left side of the equation.

The final velocity of both the ball and the man is 3.00 m/s, so we can set v₁ and v₂ on the right side of the equation to equal 3.00 m/s.

Left side of the equation:

  • m₁ = 7.00 kg
  • v₁ = ?
  • m₂ = 75.0  kg
  • v₂ = 0 m/s

Right side of the equation:

  • m₁ = 7.00 kg
  • v₁ = 3.00 m/s
  • m₂ = 75.0  kg
  • v₂ = 3.00 m/s  

Substitute these values into the Law of Conversation of Momentum formula.

  • (7.00) v₁ + (75.0)(0) = (7.00)(3.00) + (75.0)(3.00)

Multiply and simplify.

  • 7.00 v₁ = 21 + 225
  • 7 v₁ = 246

Divide both sides of the equation by 7.

  • v₁ = 35.14 m/s

The ball needs to be moving at a speed of 35.14 m/s in order to send the man off at a speed of 3.00 m/s.

Go Teaching: Other Questions