Question

An 80-kg man is skating northward and happens to suddenly collide with a 20-kg boy who is ice skating toward the east. Immediately after the collision, the man and boy are seen to be moving together at 2.5 m/s in a direction 60° north of east. How fast was the boy moving just before the collision

Answer 1

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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Answer 2

The initial speed of the boy before the collision is 6.25 m/s.

The given parameters;

• mass of the man, m₁ = 80 kg
• mass of the boy, m₂ = 20 kg
• final speed of the man and the boy, v = 2.5 m/s north east

Apply the principle of conservation of linear momentum to determined the speed of the boy before the collision;

$m_1 u_1_x + m_2 u_2_x = v_x(m_1 + m_2)$

Since the boy was moving in x - direction, we will consider x - direction for the initial momentum and final momentum.

$20(u_1 cos(0)) + 80(u_2 cos(90)) = 2.5\times cos(60)\times (20 + 80)\\\\20u_1 = 125\\\\u_1 = \frac{125}{20} \\\\u_1 = 6.25 \ m/s$

Thus, the initial speed of the boy before the collision is 6.25 m/s.

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