Two metal sphere, suspended by vertical cords, initially touch. Sphere 1, with mass 30g is pulled to the left to a height 8.0 cm and then released. After swinging down,

it undergoes and elastic collision with sphere 2 whose mass is 75 g.

a) What is the velocity of the sphere 1 just after the collision?
b) To what height does the sphere 1 swing to the left after the collision?
c) What is the velocity of the sphere 2 just after the collision?

To answer these questions, you will need to apply the principles of conservation of mechanical energy and conservation of momentum.

Let's break down the steps to find the answers:

a) Velocity of sphere 1 just after the collision:
1. Determine the initial potential energy of sphere 1 before it is released.
- Mass of sphere 1: 30g (0.03kg)
- Height of the release point: 8.0 cm (0.08m)
- Potential energy = mass * gravity * height
- Potential energy before release = 0.03kg * 9.8m/s^2 * 0.08m

2. Calculate the final kinetic energy of sphere 1 after the collision.
- Assuming an elastic collision, kinetic energy is conserved.
- Kinetic energy after collision = Potential energy before release

3. Determine the final velocity of sphere 1 after the collision.
- Kinetic energy = 1/2 * mass * (velocity)^2
- Solve the equation for velocity.

b) Height reached by sphere 1 after the collision:
1. Use the final velocity obtained in the previous question to determine the potential energy at the highest point.
- Kinetic energy of sphere 1 = Potential energy at maximum height
- Solve the equation for height.

c) Velocity of sphere 2 just after the collision:
1. Apply conservation of momentum.
- Momentum before collision = Momentum after collision
- Momentum before collision = mass of sphere 1 * velocity of sphere 1
- Momentum after collision = mass of sphere 1 * final velocity of sphere 1 + mass of sphere 2 * velocity of sphere 2
- Solve the equation for velocity of sphere 2.

Using these steps, you should be able to find the answers to all three questions.