A 6.60 kg ball moving initially at 10.9 m/s hits a 5.24 kg ball at rest.

Question

In a linear elastic collision. What is the speed of the first ball after the collision? (the one initially moving)

What is the speed of the second ball after the collision? (the one initially at rest)

Answer 1

Because the collision is elastic, both momentum AND kinetic energy are conserved.

Let the 6.6 kg ball be ball #1 and the 5.24 kg ball be ball #2. Let the initial time be "time A" and the post collision be "time B".

Conservation of Momentum:

m₁V_1A + m₂V_2A = m₁V_1B + m₂V_2B

6.6(10.9) + 5.24(0) = 6.6V_1B + 5.24V_2B

71.94 = 6.6V_1B + 5.24V_2B

V_1A = (71.94 - 5.24V_2B)/6.6

V_1A = 10.9 - 0.7839V_2B

Conservation of Kinetic Energy:

(1/2)m₁V_1A² + (1/2)m₂V_2A² = (1/2)m₁V_1B² + (1/2)m₂V_2B²

(1/2)(6.6)(10.9)² + 0 = (1/2)(6.6)V_1B² + (1/2)(5.24)V_2B²

392.073 = 3.3V_1B² + 2.62V_2B²

Substituting the V_1A = 10.9 - 0.7839V_2B into the Conservation of Kinetic Energy equation we get:

392.073 = 3.3(10.9 - 0.7839V_2B)² + 2.62V_2B²

Simplifying the equation, we get:

4.7V_2B² - 57.116V_2B = 0

Solving, we get V_2B = 0 and V_2B = 12.152

Plugging V_2B = 12.152 into the first equation we get V_1B = 1.252

So the speed of the first ball after the collision is 1.25 m/s

The speed of the second ball after the collision is 12.2 m/s