A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed. (a) What is the mass of the other body?
(b) What is the speed of the two-body center of mass if the initial speed of the 2.0 kg body was 4.0 m/s?

(a) the mass of the second body is 1.2 kg

(b)  the speed of the two-body center of mass 2.5 m/s

Explanation:

Given;

mass of the body, m₁ = 2 kg

let the mass of the second body = m₂

let the initial speed of the first body, = u₁ = 4 m/s

then, the final speed of the first body, v₁= ¹/₄u₁ = 0.25u₁

initial speed of the second body, u₂ = 0

let the final speed of the second body = v₂

Apply principle of conservation of linear momentum to determine the mass of the second body;

m₁u₁  +  m₂u₂ = m₁v₁  +  m₂v₂

2u₁  +  0(m₂)  =  2(0.25u₁)  +  m₂v₂

2u₁   =  0.5u₁  +  m₂v₂

1.5(4) = m₂v₂

6 = m₂v₂

Apply one-directional velocity

u₁ + v₁ = u₂ + v₂

u₁ + (0.25u₁) = 0 + v₂

1.25u₁ = v₂

1.25(4) = v₂

5 = v₂

Then, the mass of the second body is calculated as;

m₂v₂ = 6

5m₂ = 6

m₂ = 6/5

m₂ = 1.2 kg

(b)  the speed of the two-body center of mass after collision;

if they have a finite resultant