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;

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards