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Sagot :
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.
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