Two balls undergoing an elastic collision

Question

Two balls of masses m1=0.3 kg and m2=0.2 kg are hanging on the same kind of thin threads that are attached to the same point. Keeping the thread tense, the first ball is raised to the height of H=7 cm and released. What are the heights h1 and h2 the first and the second ball will reach after an elastic collision?

Answer 1

Given the units used in the question, I will also use kg and cm.

Let

v₁ be the velocity with which the first ball strikes the second,

w₁ be the velocity of the first ball after the collision,

w₂ be the velocity of the second ball after the collision,

g be gravitational acceleration.

This is the strategy:

Use conservation of energy to calculate v₁.

Use conservation of momentum and energy to calculate w₁, w₂.

Use conservation of energy to calculate h₁, h₂.

Throughout I will use the balls' resting position as reference point for potential energy.

CALCULATE v₁

Before the first ball is released its kinetic energy is 0 and potential energy is m₁gH.

As it strikes the second ball its kinetic energy is m₁v₁²/2 and potential energy is 0.

By conservation of energy

m₁v₁²/2 = m₁gH

So

v₁² = 2gH (1)

CALCULATE w₁ and w₂

By conservation of momentum

m₁w₁ + m₂w₂ = m₁v₁ (2)

As the collision is elastic, we also have conservation of energy

m₁w₁²/2 + m₂w₂²/2 = m₁v₁²/2 (3)

So we have two equations and two unknowns w₁ and w₂.

I will not go through the detailed calculations, only the important equations.

From (2)

w₂ = (v₁ - w₁)m₁/m₂

After plugging that into (3) and simplifying we get the quadratic equation

(m₁ + m₂)w₁² - 2m₁v₁w₁ + (m₁ - m₂)v₁² = 0

That gives the solution

w1 = v₁(m₁ ± m₂)/(m₁ + m₂)

So we have two solutions:

w₁ = v₁, w₂ = 0

w₁ = v₁(m₁ - m₂)/(m₁ + m₂), w₂ = 2v₁m₁/(m₁ + m₂) (4)

The first solution lets the first ball simply pass through the second ball without disturbing it.

While philosophically interesting, we reject that solution on practical grounds.

So we have only the second solution.

CALCULATE h₁

Right after the collision, the first ball has potential energy 0 and kinetic energy m₁w₁²/2.

When it reaches its maximal height h₁ it has potential energy m₁gh₁ and kinetic energy 0.

By conservation of energy

m₁gh₁ = m₁w₁²/2

So

h₁ = w₁²/2g

Substituting from (4)

h₁ = v₁²((m₁ - m₂)/(m₁ + m₂))²/2g

Substituting from (1)

h1 = 2gH((m₁ - m₂)/(m₁ + m₂))²/2g = H((m₁ - m₂)/(m₁ + m₂))²

Substituting actual values

h₁ = 7×(0.1/0.5)² = 0.28 cm

CALCULATE h₂

Right after the collision, the second ball has potential energy 0 and kinetic energy m₂w₂²/2.

When it reaches its maximal height h₂ it has potential energy m₂gh₂ and kinetic energy 0.

By conservation of energy

m₂gh₂ = m₂w₂²/2

So

h₂ = w₂²/2g

Substituting from (4)

h₂ = v₁²(2m₁/(m₁ + m₂))²/2g

Substituting from (1)

h₂ = 2gH(2m1/(m₁ + m₂))²/2g = H(2m₁/(m₁ + m₂))²

Substituting actual values

h₂ = 7×(0.6/0.5)² = 10.08 cm

NOTE:

You can double check the results by comparing the very original potential energy

with the total final potential energy.