Question

A person of mass 60 kg sliding along level ice at a speed of 3.0 m/s collides with a second person who is initially at rest. Assume that friction is negli-gible. If the two people hold onto each other and their common final speed after colliding is 2.0 m/s, the mass of the second person is

Answer 1

Answer:

Explanation:

Law of conservation of momentum : States that if two bodies collide in a closed system, the total momentum before collision is equal to the total momentum after collision

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = V(m₁+m₂).............. Equation 1

Where m₁ = mass of the first person, m₂ = mass of the second person, u₁ = initial velocity of the first person, u₂ = initial velocity of the second person, V = common velocity.

Making m₂ the subject of the equation above

m₂ = (m₁u₁ - m₁V)/(V-u₂).............. Equation 2

Given: m₁ = 60 kg, u₁ = 3.0 m/s, u₂ = 0 m/s (at rest). V = 2.0 m/s

Substituting these values into equation 2

m₂ = (60×3 - 60×2)/(2-0)

m₂ = (180 - 120)/2

m₂ = 60/2

m₂ = 30 kg

The mass of the second person = 30 kg