A 3.32 kg mud ball has a perfectly inelastic collision with a second mud ball that is initially at rest. The composite system moves with a speed equal to one-eighth the original speed of the 3.32 mud ball. What is the mass of the second mud ball?

Conservation of momentum:

(M1+M2)V1/8 = M1*V1-M2*0
Multiply both sides by 8/V1:
M1 + M2 = 8M1
3.32 + M2 = 8*3.32
M2 = 8*3.32 - 3.32 = 23.24 kg.

Ah, mud balls, the perfect subjects for some muddy calculations. Let's get down and dirty with the numbers.

We have two mud balls in action here. The first mud ball has a mass of 3.32 kg and an initial speed, which we'll call V1. The second mud ball is initially at rest, so its initial speed, V2, would be a big, fat zero.

During the collision, the two mud balls stick together and become one big, messy composite system. This composite system then moves with a speed equal to one-eighth the original speed of the first mud ball, which means its final speed, Vf, would be (1/8)V1.

To solve this muddy mystery, we need to understand the concept of conservation of momentum. According to this law, in the absence of any external forces, the total momentum before and after the collision remains constant.

So, let's write down the momentum before the collision:

Initial momentum = mass of first mud ball (m1) * initial speed of first mud ball (V1) + mass of second mud ball (m2) * initial speed of second mud ball (V2)

Now, let's write down the momentum after the collision:

Final momentum = total mass of composite system (m1 + m2) * final speed of composite system (Vf)

Since the collision is inelastic, the two mud balls stick together and move as one unit. Therefore, the final speed of the composite system after the collision is equal to one-eighth the original speed of the first mud ball, as we mentioned before.

Now, let's set up these equations and do some muddy math:

m1 * V1 + m2 * V2 = (m1 + m2) * Vf,

3.32 kg * V1 + m2 * 0 = (3.32 kg + m2) * (1/8)V1,

Expanding and simplifying:

3.32 kg * V1 = (3.32 kg + m2) * (1/8)V1,

Now, let's divide by V1:

3.32 kg = (3.32 kg + m2) * (1/8),

24.16 kg = 3.32 kg + m2,

Now, let's subtract 3.32 kg:

m2 = 24.16 kg - 3.32 kg,

m2 = 20.84 kg.

So, the mass of the second mud ball is approximately 20.84 kg. That's one heavy and muddy mud ball!

To solve this problem, we can apply the conservation of momentum.

The momentum before the collision is equal to the momentum after the collision. In this case, only one mud ball is initially in motion, so we can write:

(mass of the first mud ball) * (original velocity of the first mud ball) = ((mass of the first mud ball) + (mass of the second mud ball)) * (final velocity of the composite system)

Let's denote the mass of the first mud ball as m1 and the mass of the second mud ball as m2.

Given:
Mass of the first mud ball (m1) = 3.32 kg
Final velocity of the composite system = (1/8) * (original velocity of the first mud ball)

Substituting these values into the equation above, we have:

3.32 kg * (original velocity of the first mud ball) = (3.32 kg + m2) * ((1/8) * (original velocity of the first mud ball))

Simplifying the equation:

(original velocity of the first mud ball) = ((1/8) * (original velocity of the first mud ball)) * (3.32 kg + m2)

(original velocity of the first mud ball) = (0.125) * (original velocity of the first mud ball) * (3.32 kg + m2)

Cancelling out the original velocity of the first mud ball from both sides of the equation:

1 = 0.125 * (3.32 kg + m2)

Dividing both sides of the equation by 0.125:

8 = 3.32 kg + m2

Rearranging the equation:

m2 = 8 - 3.32 kg

m2 = 4.68 kg

Therefore, the mass of the second mud ball is 4.68 kg.

To find the mass of the second mud ball, we can use the principle of conservation of momentum. In an inelastic collision, the total momentum before the collision is equal to the total momentum after the collision.

Before the collision, the mud ball with a mass of 3.32 kg moves with a certain speed, which we'll call v_1. The second mud ball is at rest, so its initial velocity is 0.

After the collision, the composite system (the two mud balls together) moves with a speed equal to one-eighth (1/8) the original speed of the 3.32 kg mud ball. Therefore, the final speed of the composite system is (1/8)v_1.

Let's now write the equation for the conservation of momentum:

Total momentum before collision = Total momentum after collision

(mass of 1st mud ball * initial velocity of 1st mud ball) + (mass of 2nd mud ball * initial velocity of 2nd mud ball) = (mass of 1st mud ball + mass of 2nd mud ball) * final velocity of composite system

(3.32 kg * v_1) + (mass of 2nd mud ball * 0) = (3.32 kg + mass of 2nd mud ball) * (1/8)v_1

Now, let's simplify the equation:

3.32 kg * v_1 = (3.32 kg + mass of 2nd mud ball) * (1/8)v_1

Let's cancel out v_1 from both sides:

3.32 kg = (3.32 kg + mass of 2nd mud ball) * (1/8)

Now, let's solve for the mass of the second mud ball:

8 * 3.32 kg = 3.32 kg + mass of 2nd mud ball

26.56 kg = 3.32 kg + mass of 2nd mud ball

Subtracting 3.32 kg from both sides, we get:

mass of 2nd mud ball = 26.56 kg - 3.32 kg

mass of 2nd mud ball = 23.24 kg

Therefore, the mass of the second mud ball is 23.24 kg.