A ball with a mass of 100 g, moving at a speed of 6 m / s, catches up with another ball with a mass of 200 g

A ball with a mass of 100 g, moving at a speed of 6 m / s, catches up with another ball with a mass of 200 g, moving in the same direction at a speed of 2 m / s. Find the speed of the balls after an elastic impact if the speed of the second ball has doubled.

Let’s use the law of conservation of momentum, write down the formula:
m₁ * v₁ + m₂ * v₂ = m₁ * v₁ ‘+ m₂ * v₂’;
Let’s transform:
m₁ * v₁ + m₂ * v₂ = m₁ * v₁ ‘+ m₂ * 2 * v₂;
We get:
m₁ * v₁ ‘= m₁ * v₁ – m₂ * v₂;
Let’s transform:
v₁ ‘= (m₁ * v₁ – m₂ * v₂) / m₁;
Let’s substitute the values into the formula:
v₁ ‘= (0.1 * 6 – 0.2 * 2) / 0.1;
v₁ ‘= (0.6 – 0.4) / 0.1;
v₁ ‘= 0.2 / 0.1;
v₁ ‘= 2 m / s;
Find v₂:
v₂ ‘= 2 * 2;
v₂ ‘= 4 m / s;
Answer: 2 m / s and 4 m / s.