A puck of mass 80.0 g and radius 4.10 cm slides along an air table at a speed of 1.50 m/s as shown in the figure below. It makes a glancing collision with a second puck of radius 6.00 cm and mass 130.0 g (initially at rest) such that their rims just touch. Because their rims are coated with instant-acting glue, the pucks stick together and spin after the collision.

To solve this problem, we need to use the principles of conservation of momentum and conservation of angular momentum.

Let's start by finding the initial momentum of each puck:

The momentum (p) of an object is given by the product of its mass (m) and velocity (v): p = m * v

For the first puck:
Mass (m1) = 80.0 g = 0.08 kg
Velocity (v1) = 1.50 m/s

p1 = m1 * v1 = 0.08 kg * 1.50 m/s

For the second puck:
Mass (m2) = 130.0 g = 0.13 kg (The mass should be converted to kilograms)
Velocity (v2) = 0 m/s (since it is initially at rest)

p2 = m2 * v2 = 0.13 kg * 0 m/s

Now, let's find the final momentum of the combined system after the collision. Since the pucks stick together, we can simply add their individual momenta:

pf = p1 + p2

Next, we apply the principle of conservation of angular momentum. The total angular momentum before and after the collision must be the same.

The angular momentum (L) of an object is given by the product of its moment of inertia (I) and angular velocity (ω): L = I * ω

For each puck, the moment of inertia (I) is given by: I = (2/5) * m * r^2, where r is the radius of the puck.

For the first puck:
Moment of Inertia (I1) = (2/5) * m1 * r1^2 = (2/5) * 0.08 kg * (0.041 m)^2

For the second puck:
Moment of Inertia (I2) = (2/5) * m2 * r2^2 = (2/5) * 0.13 kg * (0.06 m)^2

Let's assume the final angular velocity of the combined system is ωf.

According to the conservation of angular momentum, the initial angular momentum (L1) of the first puck is equal to the final angular momentum (Lf) of the combined system:

L1 = Lf

I1 * ω1 = (I1 + I2) * ωf

Substituting the values of I1, I2, and ω1:

((2/5) * 0.08 kg * (0.041 m)^2) * ω1 = ((2/5) * 0.08 kg * (0.041 m)^2 + (2/5) * 0.13 kg * (0.06 m)^2) * ωf

Now, we have two equations to solve for ω1 and ωf.

To analyze this collision problem between the two pucks, we need to apply the law of conservation of linear momentum and angular momentum.

1. Conservation of Linear Momentum:
The law of conservation of linear momentum states that the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces act on the system.
Mathematically, this can be written as:
(m1 * v1) + (m2 * v2) = (m1 + m2) * vf

Where:
m1 and v1 are the mass and initial velocity of the first puck (80.0 g and 1.50 m/s, respectively).
m2 is the mass of the second puck (130.0 g).
v2 is the initial velocity of the second puck (0 m/s since it is initially at rest).
vf is the final velocity of the combined system.

2. Conservation of Angular Momentum:
The law of conservation of angular momentum states that the total angular momentum before the collision is equal to the total angular momentum after the collision, assuming no external torques act on the system.
Mathematically, this can be written as:
I1 * ω1 + I2 * ω2 = (I1 + I2) * ωf

Where:
I1 and ω1 are the moment of inertia and initial angular velocity of the first puck. Since it only slides along the table and does not rotate, I1 = 0 and ω1 = 0.
I2 is the moment of inertia of the second puck, which can be calculated as I2 = (2/5) * m2 * R^2, where R is the radius of the second puck.
ω2 is the initial angular velocity of the second puck (also zero since it is initially at rest).
ωf is the final angular velocity of the combined system.

By applying these two conservation principles and solving the equations simultaneously, we can determine the final linear velocity (vf) and final angular velocity (ωf) of the combined puck system after the collision.