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.
