Question

Here's a basketball problem: A 87.2 kg basketball player is running in the positive direction at 7.0 m/s. She is met head-on by a 102.0 kg player traveling at 5.2 m/s toward her. If the 102.0 kg player is knocked backwards at 2.9 m/s, what is the resulting velocity of the 87.2 kg player?

Answer 1

Answer:

2.47 m/s backwards

Explanation:

From the law of conservation of momentum,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂...................... Equation 1

Where m₁ and m₂ = mass of the first basketball player and second basket ball player respectively, u₁ and u₂ = initial velocity of the first basket player and the second basketball player respectively, v₁ and v₂ = The final velocity of the first basket ball player and second basket ball player respectively.

Making v₁ the subject of the equation,

v₁ = (m₁u₁ + m₂u₂ - m₂v₂)/m₁.......................... Equation 2.

Given: m₁ = 87.2 kg, m₂ = 102.0 kg, u₁ = 7.0 m/s, u₂ = -5.2 m/s, v₂ = 2.9 m/s

Note: u₂ is negative because it moves towards the first basket ball player.

Substitute into equation 2

v₁ = [87.2(7.0)+102(-5.2) - (102×2.9)]/87.2

v₁ = (610.4-530-295.8)/87.2

v₁ = -215.4/87.2

v₁ = -2.47 m/s.

Thus the velocity of the 87.2 kg player = 2.47 m/s backwards.