Answer:
2 m/s
Explanation:
From the law of conservation of momentum,
Total momentum before collision = total momentum after collision
mu+m'u' = V(m+m') .................................Equation 1
Where m = mass of the first player, u = initial speed of the first player, m' = mass of the second player, u' = initial speed of the second player, V = combined speed of both players.
Making V the subject of the equation,
V = (mu+m'u')/(m+m')................ Equation 2
Note: Taking the direction of the first player as positive.
Given: m = 75 kg, m' = 100 kg, u = 6 m/s, u' = -8 m/s (opposite the first player),
Substituting into equation 2
V = [(75×6)+(100×(--8))]/(75+100)
V = (450-800)/175
V = 350/175
V = - 2 m/s.
Note: The negative signs tells that the combined speed is in the direction of the second player.
Hence the combined speed of the two players = 2 m/s
