Ball 1, with a mass of 100g and traveling at 12 m/s , collides head on with ball 2, which has a mass of 340g and is initially at rest.

What are the final velocities of each ball if the collision is perfectly elastic?
What are the final velocities of each ball if the collision is perfectly elastic?

Use conservation of momentum

m1v1=m1*v1' + m2v2'
solve that for v1'
Then use conservation energy.
1/2 m1*v1^2=1/2 m2*v1'^2 + 1/2 m2*v2'^2
put the v1' expression in that, and turn the algebra cranks, put it in quadratic form, and solve the quadratic equation.

To find the final velocities of each ball in a perfectly elastic collision, we can use the principle of conservation of momentum and conservation of kinetic energy.

1. First, let's calculate the initial momentum of each ball:

Initial momentum of Ball 1 (P1_initial) = mass * velocity
P1_initial = (100g) * (12 m/s) = 1200 g·m/s

Initial momentum of Ball 2 (P2_initial) = (mass * velocity)
P2_initial = (340g) * (0 m/s) = 0 g·m/s (as it is initially at rest)

2. Next, let's calculate the total initial momentum before the collision:

Total initial momentum (P_initial) = P1_initial + P2_initial
P_initial = P1_initial + P2_initial
P_initial = 1200 g·m/s + 0 g·m/s = 1200 g·m/s

3. Since it is a perfectly elastic collision, both momentum and kinetic energy are conserved. Hence, the total momentum after the collision (P_final) will also be 1200 g·m/s.

4. Now, let's set up the conservation of momentum equation for the final velocities of the balls:

Total final momentum (P_final) = (mass1 * velocity1) + (mass2 * velocity2)
P_final = (mass1 * velocity1) + (mass2 * velocity2)

Using the given masses of the balls (in grams) and the final velocities (in m/s) as variables:

P_final = (100g * velocity1) + (340g * velocity2)

5. Since we have two unknowns (velocity1 and velocity2), we need another equation to solve the system of equations. For a perfectly elastic collision, kinetic energy is also conserved. So we can set up the conservation of kinetic energy equation:

Initial kinetic energy = Final kinetic energy

For Ball 1:
(1/2) * mass1 * (initial velocity1)^2 = (1/2) * mass1 * (final velocity1)^2

For Ball 2:
(1/2) * mass2 * (initial velocity2)^2 = (1/2) * mass2 * (final velocity2)^2

6. Substituting the values and simplifying the equations:

(1/2) * 100g * (12 m/s)^2 = (1/2) * 100g * (velocity1)^2 ---> Equation 1

(1/2) * 340g * (0 m/s)^2 = (1/2) * 340g * (velocity2)^2 ---> Equation 2

7. Solving Equation 1 for velocity1:

(1/2) * 100g * (12 m/s)^2 = (1/2) * 100g * (velocity1)^2
(1/2) * 12 m/s = (velocity1)^2

Taking the square root of both sides:
velocity1 = √(6 m/s)

8. Substituting the value of velocity1 into Equation 2:

(1/2) * 340g * (0 m/s)^2 = (1/2) * 340g * (velocity2)^2
0 = (velocity2)^2

Since the initial velocity of Ball 2 is 0, its final velocity will also be 0.

9. Therefore, the final velocities of each ball are:
Ball 1: √(6 m/s)
Ball 2: 0 m/s