The speed at which the boy was moving just before the collision is;
6.25 m/s
We are given;
Mass of man; m₁ = 80 kg
Mass of boy; m₂ = 20 kg
Initial velocity of man; u₁ = 0 m/s
Initial velocity of the boy; u₂ = ?
Speed at which the boy and the man move together after collision; v_f = 2.5 m/s
Direction of movement after collision; θ = 60° North East
Now, in momentum when two bodies stick together after collision and move with a common velocity, the equation is represented as;
m₁u₁ + m₂u₂ = (m₁ + m₂)v
Since they move with same velocity at an angle of 60°, then;
v = 2.5cos 60
Thus;
(80 × 0) + (20 × u₂) = (80 + 20) × 2.5 cos 60
20u₂ = 125
u₂ = 6.25 m/s
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The initial speed of the boy before the collision is 6.25 m/s.
The given parameters;
Apply the principle of conservation of linear momentum to determined the speed of the boy before the collision;
[tex]m_1 u_1_x + m_2 u_2_x = v_x(m_1 + m_2)[/tex]
Since the boy was moving in x - direction, we will consider x - direction for the initial momentum and final momentum.
[tex]20(u_1 cos(0)) + 80(u_2 cos(90)) = 2.5\times cos(60)\times (20 + 80)\\\\20u_1 = 125\\\\u_1 = \frac{125}{20} \\\\u_1 = 6.25 \ m/s[/tex]
Thus, the initial speed of the boy before the collision is 6.25 m/s.
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