Ball 1, with a mass of 100 g and traveling at 10 m/s, collides head-on with ball 2, which has a mass of 300 g and is initially at rest. What are the final velocities of each ball if the collision is (a) perfectly elastic? (b) perfectly inelastic?

Perfectly Elastic (after collision):

Velocity of Ball 1 = v₁ = -5m/s

Velocity of Ball 2 = v₂ = 5m/s

In the case of a collision, assuming perfectly elasticity, ball 2 would move at a speed of 5m/s and ball 1 would switch direction moving at a now lower speed of 5m/s.

Perfectly Inelastic (after collision):

Velocity of Ball 1 = v₁ = 0m/s

Velocity of Ball 2 = v₂ = 0m/s

Explanation:

Perfectly Elastic:

No kinetic energy is lost via transfer into any other kind, e.g. heat, so we just use the conservation of momentum formula:

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.1)(10) + (0.3)(0) = (0.1)v₁ + (0.3)v₂  (×10)

v₁ + 3v₂ = 10 → v₁ = 10 - 3v₂

Now, we have to be a bit innovative and use the kinetic energy formula for both balls:

KE = ¹/₂mv²

Firstly, before the collision:

Ball 1:

(KE)₁ = ¹/₂(0.1)(10)²

(KE)₁ = 5

Ball 2:

(KE)₂ = 0 ← it was at rest

Now after the collision:

Ball 1:

(KE)₁ = ¹/₂(0.1)(v₁)²

(KE)₁ = ¹/₂₀(v₁)²

Ball 2:

(KE)₂ = ¹/₂(0.3)(v₂)²

(KE)₂ = ³/₂₀(v₂)²

Since we are considering a perfectly elastic collision, no energy is lost in the collision so the sum of the kinetic energy of the two balls before and after the collisons has to be equal, hence we can formulate the following:

5 + 0 = ¹/₂₀(v₁)² + ³/₂₀(v₂)²

100 = (v₁)² + 3(v₂)²

Now, we can plug in the equation we found for v₁ from earlier and solve:

100 = (10 - 3(v₂))² + 3(v₂)²

100 = 100 - 60(v₂) + 9(v₂)² + 3(v₂)²

12(v₂)² - 60(v₂) = 0

(v₂)² - 5(v₂) = 0

(v₂)((v₂) - 5) = 0

v₂ = 0 OR v₂ - 5 = 0 → v₂ = 5

v₁ = 10 OR v₁ = 10 - 3(5) = -5 m/s

Perfectly Inelastic:

Final velocities would be 0;

All kinetic energy would be transferred in the collision, the balls would join together to become one in such a case.

trig 15.3 cos