Answer:
The velocity of the second ball is 15.88 m/s East
Explanation:
The given parameters are;
The type of collision = Perfectly elastic collision
The mass of the ball moving East, m₁ = 0.400 kg
The velocity of the ball moving East, v₁ = 3.7 m/s
The mass of the ball at rest, m₂ = 0.200 kg
The initial velocity of the ball at rest, v₂ = 0 m/s
The velocity of the first ball, 'm₁', after the collision, v₃ = 4.24 m/s (The direction is assumed as being towards the West)
The velocity, 'v₄', of the second ball, 'm₂', after the collision is given by the law of conservation of linear momentum as follows;
m₁·v₁ + m₂·v₂ = m₁·v₃ + m₂·v₄
Substituting the known values gives;
0.400 kg × 3.7 m/s + 0.200 kg × 0 m/s = 0.400 kg × (-4.24 m/s) + 0.200 kg × v₄
∴ 0.200 kg × v₄ = 0.400 kg × 3.7 m/s - 0.400 kg × (-4.24 m/s)
Therefore;
0.200 kg × v₄ = 0.400 kg × 3.7 m/s + 0.400 kg × 4.24 m/s = 3.176 kg·m/s
v₄ = 3.176 kg·m/s/(0.200 kg) = 15.88 m/s
The velocity of the second ball = v₄ = 15.88 m/s East.
