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

Answer:

6.25 m/s

Explanation:

mass of man (m1) = 80 kg

mass of boy (m2) = 20 kg

mass of man and boy after collision (m12)= 20 + 80 = 100 kg

velocity of man and boy after collision (v) = 2.5 m/s

angle θ = 60 °

How fast was the boy moving just before the collision ?

• From the diagram attached, the first image shows the man and the boys motion while the second diagram shows their motion rearranged to form a triangle. With the momentum of the man and the boy forming the sides of the triangle.
• M₁₂ =  total momentum after collision = m12 x v = 100 x 2.5 = 250
• Mboy = momentum of the boy before collision = m2 x Velocity of boy
• Mman = momentum of the man before collision = m1 x velocity of man  
• from the triangle, cos θ = $\frac{Mboy}{M₁₂}$

        cos 60 = $\frac{Mboy}{250}$

        Mboy = 250 x cos 60 = 125

• recall that momentum of the boy (Mboy) also = m2 x Velocity of boy

        therefore

        125 = 20 x velocity of boy

         velocity of boy = 125 / 20 = 6.25 m/s