An 8.0-kg object moving 4.0 m/s in the positive x direction has a one-dimensional collision with a 2.0-kg object moving 3.0 m/s in the opposite direction. The final velocity of the 8.0-kg object is 2.0 m/s in the positive x direction. What is the total kinetic energy of the two-mass system after the collision?

Kf = 41 J

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:

p=m*v

Where:

p : Linear momentum

m: mass

v:velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For the three cases the total linear momentum quantity is conserved:

P₀ = Pf Formula (1)

P₀ : Initial linear momentum quantity

Pf : Final linear momentum quantity

Data

m₁ = 8.0-kg : mass of object₁

m₂ = 2.0-kg : mass of object₂

v₀₁ = 4.0 m/s, to the right : Initial velocity of m₁

v₀₂ = 3.0 m/s, to the left : Initial velocity of m₂

vf₁ = 2.0 m/s, to the right : Final velocity of m₁

Problem development

We assume that the 2kg object move to the right at the end of the collision, so, the sign of the final speeds (vf₂), we assume vf₂ positive:

We appy the formula (1):

P₀ = Pf

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂

(8) * (4) + (2) * (-3) = (8) * (2) + (2) * vf₂

32-6 = 16 + (2) * vf₂

32-6-16 = (2) * vf₂

10 = (2) * vf₂

vf₂ = 10 / (2)

vf₂ = 5 m/s, to the right

Kinetic energy

Kinetic energy is that which is due to the movement of bodies and is calculted like this:

K = (1/2) mv²

Where:

K : Kinetic energy (J)

m: mass (kg)

v : speed (m/s)

Total kinetic energy of the two-mass system after the collision (Kf)

Kf = Kf₁ + Kf₂

Kf₁ = (1/2) m₁vf₁² = (1/2) (8) (2) ² = 16 J

Kf₂ = (1/2) m₂vf₂² = (1/2) (2) (5) ² = 25 J

Kf = 16 J + 25 J

Kf = 41 J