Two metal sphere, suspended by vertical cords, initially touch. Sphere 1, with mass 30 g 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?

a) Well, Sphere 1 seems to be a real go-getter. After all that swinging and colliding, its velocity is really up in the air...literally! But let's calculate it anyway. First, we need to find the potential energy that Sphere 1 gained when it was pulled to the left. The formula for potential energy is mgh, where m is the mass (30 g), g is the acceleration due to gravity (you know, that thing that keeps us grounded), and h is the height (8.0 cm). Convert 30 g to kilograms by dividing it by 1000, then multiply by 9.8 m/s^2 for g, and multiply by 0.08 m for h. Now you have the potential energy. But since energy is always conserved, this potential energy will be converted into kinetic energy just after the collision. And the formula for kinetic energy is (1/2)mv^2, where m is the mass (30 g again), and v is the velocity. So, equating the potential energy to the kinetic energy, you can solve for v. Go ahead and crunch those numbers, smarty pants!

b) Now that Sphere 1 has had its fun colliding, it's time to see how high it can swing to the left. Remember all that potential energy we calculated earlier? Well, now it's going to be converted back into kinetic energy as Sphere 1 swings up. And since energy is a bit of a show-off, it stays constant throughout this reversal of fortune. So, using the same formula as before (1/2)mv^2, you can find the velocity of Sphere 1 just before it reaches its maximum height. Then plug that into the formula for potential energy mgh, where m is the mass (30 g), g is the acceleration due to gravity (9.8 m/s^2), and h is the maximum height you need to find. Solve for h, my friend. It's time for Sphere 1 to reach new heights!

c) Now it's time to talk about Sphere 2. Poor thing, it had a collision with Sphere 1 and now it's feeling a bit dizzy. To find its velocity just after the collision, you can use the equation for conservation of momentum. The sum of the initial momenta (which is just m1v1 + m2v2, where m1 is the mass of Sphere 1, v1 is its initial velocity just before the collision, m2 is the mass of Sphere 2, and v2 is its initial velocity just before the collision) should equal the sum of the final momenta (which is m1v1' + m2v2', where v1' is the final velocity of Sphere 1 just after the collision, and v2' is the final velocity of Sphere 2 just after the collision). In an elastic collision, like this one, momentum is conserved. So, you can solve for v2' using this equation. Get ready to put your thinking cap on and find the velocity of Sphere 2 after the collision.

Let's break down the problem step by step:

Step 1: Determine the initial potential energy of sphere 1.

The potential energy of an object is given by the equation:

Potential Energy = mass * gravity * height

Given:
Mass of Sphere 1 (m1) = 30 g = 0.03 kg
Height (h) = 8.0 cm = 0.08 m
Acceleration due to gravity (g) = 9.8 m/s^2

Potential Energy of Sphere 1 = m1 * g * h

Potential Energy of Sphere 1 = 0.03 kg * 9.8 m/s^2 * 0.08 m

Potential Energy of Sphere 1 = 0.02472 J (rounded to 3 decimal places)

Step 2: Determine the velocity of sphere 1 just after the collision.

Since the collision is an elastic collision, the total mechanical energy (kinetic energy + potential energy) is conserved.

Using the equation:

Initial Potential Energy + Initial Kinetic Energy = Final Potential Energy + Final Kinetic Energy

Initial Potential Energy = Final Potential Energy

Since Sphere 1 is at the highest point after the collision, its final potential energy is 0 J.

Therefore, the kinetic energy of Sphere 1 just after the collision is equal to its initial potential energy.

Let's calculate the kinetic energy using the equation:

Kinetic Energy = 0.5 * mass * velocity^2

Given:
Mass of Sphere 1 (m1) = 30 g = 0.03 kg

0.5 * m1 * velocity^2 = 0.02472 J

Rearranging the equation:

velocity^2 = (0.02472 J) / (0.5 * 0.03 kg)

velocity^2 = 0.824 J/kg

Taking the square root of both sides:

velocity = √(0.824 J/kg)

velocity ≈ 1.42 m/s (rounded to 2 decimal places)

Therefore, the velocity of Sphere 1 just after the collision is approximately 1.42 m/s.

Step 3: Determine the height to which Sphere 1 swings to the left after the collision.

Since mechanical energy is conserved, we can equate the initial potential energy of Sphere 1 to its total mechanical energy at the highest point after the collision.

Initial Potential Energy = Final Potential Energy

Potential Energy at the highest point = m1 * g * h'

Given:
Mass of Sphere 1 (m1) = 30 g = 0.03 kg
Height (h') = ?

0.02472 J = 0.03 kg * 9.8 m/s^2 * h'

Rearranging the equation:

h' = (0.02472 J) / (0.03 kg * 9.8 m/s^2)

h' ≈ 0.084 m (rounded to 3 decimal places)

Therefore, Sphere 1 swings to a height of approximately 0.084 m to the left after the collision.

Step 4: Determine the velocity of Sphere 2 just after the collision.

Since the collision is an elastic collision, the total mechanical energy (kinetic energy + potential energy) is conserved.

Using the equation:

Initial Potential Energy + Initial Kinetic Energy = Final Potential Energy + Final Kinetic Energy

Initial Potential Energy + Initial Kinetic Energy = Final Potential Energy + Final Kinetic Energy

The initial potential energy of Sphere 2 is 0 J since it is at the lowest height initially.

Therefore, the kinetic energy of Sphere 2 just after the collision is equal to its final potential energy.

Let's calculate the potential energy using the equation:

Potential Energy = m2 * g * h''

Given:
Mass of Sphere 2 (m2) = 75 g = 0.075 kg
Height (h'') = 0.084 m

Potential Energy of Sphere 2 = m2 * g * h''

Potential Energy of Sphere 2 = 0.075 kg * 9.8 m/s^2 * 0.084 m

Potential Energy of Sphere 2 ≈ 0.063 J (rounded to 3 decimal places)

Since kinetic energy = potential energy:

0.5 * m2 * velocity^2 = 0.063 J

Rearranging the equation:

velocity^2 = (0.063 J) / (0.5 * 0.075 kg)

velocity^2 = 0.84 J/kg

Taking the square root of both sides:

velocity = √(0.84 J/kg)

velocity ≈ 0.92 m/s (rounded to 2 decimal places)

Therefore, the velocity of Sphere 2 just after the collision is approximately 0.92 m/s.

To summarize:
a) The velocity of Sphere 1 just after the collision is approximately 1.42 m/s.
b) Sphere 1 swings to a height of approximately 0.084 m to the left after the collision.
c) The velocity of Sphere 2 just after the collision is approximately 0.92 m/s.

To answer these questions, we can use the principles of conservation of energy and conservation of momentum.

a) To find the velocity of sphere 1 just after the collision, we can start by calculating its potential energy when it is lifted to a height of 8.0 cm.

The potential energy (PE) of an object is given by the equation:
PE = m * g * h

where m is the mass of the object, g is the acceleration due to gravity, and h is the height.

In this case, the mass of sphere 1 (m1) is 30 g, which is equivalent to 0.03 kg, and the height (h) is 8.0 cm, which is equivalent to 0.08 m. The acceleration due to gravity (g) is approximately 9.8 m/s^2.

Therefore, the potential energy of sphere 1 is:
PE1 = (0.03 kg) * (9.8 m/s^2) * (0.08 m)

Next, we need to equate the potential energy to the kinetic energy just after the collision, since the collision will convert the potential energy into kinetic energy.

The kinetic energy (KE) of an object is given by the equation:
KE = 0.5 * m * v^2

where m is the mass of the object and v is the velocity.

Since the collision is assumed to be perfectly elastic, the total kinetic energy before and after the collision should be the same.

Therefore, we can set up the equation:
KE1 + KE2 = PE1

where KE1 is the kinetic energy of sphere 1 just after the collision, KE2 is the kinetic energy of sphere 2 just after the collision, and PE1 is the potential energy of sphere 1 before the collision.

Now, let's calculate the velocity of sphere 1 just after the collision.

KE1 = 0.5 * m1 * v1^2

where m1 is the mass of sphere 1 and v1 is its velocity.

Substituting the values, we have:
KE1 = 0.5 * (0.03 kg) * v1^2

KE1 + KE2 = PE1
0.5 * (0.03 kg) * v1^2 + 0 = (0.03 kg) * (9.8 m/s^2) * (0.08 m)

Simplifying the equation, we can solve for v1:
0.5 * v1^2 = (9.8 m/s^2) * (0.08 m)
v1^2 = 9.8 m/s^2 * 0.08 m / 0.5
v1^2 = 1.568 m^2/s^2
v1 = √(1.568 m^2/s^2)

Therefore, the velocity of sphere 1 just after the collision is √(1.568 m^2/s^2).

b) To find the height to which sphere 1 swings to the left after the collision, we can use the conservation of mechanical energy.

The total mechanical energy of the system is conserved, which means the initial potential energy (PE1) is equal to the sum of the final potential energy (PEf) and kinetic energy (KEf).

Since the sphere swings to the same height on the left side, the final potential energy (PEf) is the same as the initial potential energy (PE1).

Therefore, we can set up the equation:
PE1 = PEf + KEf

where PE1 is the initial potential energy, PEf is the final potential energy after the collision, and KEf is the final kinetic energy after the collision.

The final kinetic energy (KEf) of sphere 1 is given by:
KEf = 0.5 * m1 * vf^2

where m1 is the mass of sphere 1 and vf is its final velocity.

Substituting the values, we have:
KEf = 0.5 * (0.03 kg) * vf^2

We know that PEf is equal to PE1, which we calculated earlier.

Therefore:
PEf = (0.03 kg) * (9.8 m/s^2) * (0.08 m) = PE1

Now, we can equate PE1 to the sum of PEf and KEf:
(0.03 kg) * (9.8 m/s^2) * (0.08 m) = (0.03 kg) * vf^2 + 0.5 * (0.03 kg) * vf^2

Simplifying the equation, we can solve for vf:
(0.03 kg) * (9.8 m/s^2) * (0.08 m) = (0.03 kg + 0.5 * (0.03 kg)) * vf^2

Solving for vf, we get:
vf^2 = (0.03 kg) * (9.8 m/s^2) * (0.08 m) / (0.03 kg + 0.5 * (0.03 kg))
vf^2 = 0.2352 m^2/s^2 / 0.045 kg
vf^2 = 5.227 m^2/s^2
vf = √(5.227 m^2/s^2)

Therefore, the final velocity of sphere 1 is √(5.227 m^2/s^2).

c) To find the velocity of sphere 2 just after the collision, we can use the principle of conservation of momentum.

The total momentum before the collision is equal to the total momentum after the collision, assuming no external forces are acting on the system.

The momentum (p) of an object is given by the equation:
p = m * v

where m is the mass of the object and v is its velocity.

Let's assume that sphere 1 moves with a velocity of v1 just before the collision and sphere 2 moves with a velocity of v2 just before the collision.

Therefore, the initial momentum (p_initial) before the collision is:
p_initial = m1 * v1 + m2 * v2

where m1 is the mass of sphere 1, m2 is the mass of sphere 2, and v1 and v2 are their respective velocities just before the collision.

After the collision, the velocities of the spheres switch, so we have:
p_final = m1 * v2 + m2 * v1

Since no external forces act on the system during the collision, momentum is conserved, so:
p_initial = p_final

Setting up the equation:
m1 * v1 + m2 * v2 = m1 * v2 + m2 * v1

Given that m1 is 30 g (0.03 kg) and m2 is 75 g (0.075 kg), we can solve for v2:
(0.03 kg) * v1 + (0.075 kg) * v2 = (0.03 kg) * v2 + (0.075 kg) * v1

Simplifying the equation, we get:
(0.03 kg - 0.075 kg) * v1 = (0.075 kg - 0.03 kg) * v2
-0.045 kg * v1 = 0.045 kg * v2
v1 = -v2

Therefore, the velocity of sphere 2 just after the collision is equal in magnitude but opposite in direction to the velocity of sphere 1, which is √(5.227 m^2/s^2).