Question

9.In a perfectly elastic collision, a 0.400 kg ball moving toward the east at 3.7 m/s suddenly

Answer 1

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.