Question

g Two masses are involved in a collision on an axis (one dimensional). One mass is six times the mass of the second. Both masses are moving to the right before they collide and the heavier mass is further right then the lighter mass. The lighter mass has an initial speed of 4m/s and the heavier mass has an initial speed of 2m/s to the right. What is the final speed and direction of each mass mass? Was the collision elastic or inelastic and why?

Answer 1

v₁f = 0.5714 m/s (→)

v₂f = 2.5714 m/s (→)

e = 1

It was a perfectly elastic collision.

Step-by-step explanation:

m₁ = m

m₂ = 6m₁ = 6m

v₁i = 4 m/s

v₂i = 2 m/s

v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i + ((2m₂) / (m₁ + m₂)) v₂i

v₁f = ((m – 6m) / (m + 6m)) * (4) + ((2*6m) / (m + 6m)) * (2)

v₁f = 0.5714 m/s (→)

v₂f = ((2m₁) / (m₁ + m₂)) v₁i + ((m₂ – m₁) / (m₁ + m₂)) v₂i

v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)

v₂f = 2.5714 m/s (→)

e = - (v₁f - v₂f) / (v₁i - v₂i) ⇒ e = - (0.5714 - 2.5714) / (4 - 2) = 1

It was a perfectly elastic collision.