A simple pendulum is made from a 2 kg. block of wood suspended from a light cord of length 1 m. When the pendulum is hanging in such a way that it is stationary and vertical, a bullet is shot horizontally into the block of wood where it sticks. The bullet has a mass of 1.00 x 10^2 kg. and has a speed of 500 m/s just before its head-on collision with the block. What is the pendulum's maximum vertical angle as it swings?

I assume a typo and bullet is 10 grams

initial momentum of bullet and mass before collision

= 500 * 10^-2 kg + 0 * 2 kg = 5 kg m/s

the momentum just after collision will be the same with a mass now of 2.010 kg
and speed v
so
2.010 v = 5

calculate v from that

now the pendulum starts with speed v at the bottom
it has kinetic energy of
(1/2)(2.01) v^2

it slows to a stop as it rises
the rise distance is
1 meter (1 - cos T)
where T is the angle you want
the potential energy increase is
m g h = 2.01 * 9.81 * (1-cos T)
so in the end

1-cos T = (1/2) v^2/ [9.81]

To find the pendulum's maximum vertical angle as it swings after the bullet collision, we need to consider the conservation of energy and the principles of rotational motion.

1. Initial energy of the pendulum: The initial energy of the pendulum is purely gravitational potential energy. Since the pendulum is stationary and vertical, this is given by mgh, where m is the mass of the block (2 kg) and h is the height of the block (1 m). Therefore, the initial potential energy (U_initial) is U_initial = mgh.

2. Final energy of the pendulum: After the collision, the pendulum starts swinging. At the maximum vertical angle, all the initial potential energy will be converted into kinetic energy. The equation for kinetic energy is K = (1/2) Iω², where I is the moment of inertia and ω is the angular velocity.

3. Conservation of energy: According to the law of conservation of energy, the initial potential energy (U_initial) will be equal to the final kinetic energy (K). Therefore, we have U_initial = K.

4. Moment of inertia (I): The moment of inertia for a simple pendulum is given by I = mL², where L is the length of the pendulum (1 m) and m is the mass of the block (2 kg). Therefore, I = mL².

5. Angular velocity (ω): The angular velocity is related to the maximum angle (θ) that the pendulum swings. It can be expressed as ω = √(2g/L) * sin(θ), where g is the acceleration due to gravity (approximated as 9.8 m/s²). Note that sin(θ) represents the vertical displacement of the pendulum.

6. Solving for the maximum angle (θ): Substituting the expressions in steps 1, 4, and 5 into the conservation of energy equation (U_initial = K), we get mgh = (1/2) Iω². Rearranging this equation and substituting the expressions for I and ω, we have mgh = (1/2) mL²[(√(2g/L) * sin(θ))²]. Canceling out common terms, we get h = (1/2) L * (sin(θ))². Rearranging this equation, we have sin(θ) = √(2h/L). Plugging in the values for h (1 m) and L (1 m), we can solve for sin(θ).

7. Calculating the maximum angle (θ): Taking the inverse sine (arcsin) of √(2h/L), we can calculate the maximum angle (θ). Thus, θ = arcsin(√(2h/L)).

Performing the calculations with the provided values, we find:

θ = arcsin(√(2(1 m) / 1 m)) ≈ 0.7854 radians ≈ 45 degrees

Therefore, the maximum vertical angle that the pendulum swings is approximately 45 degrees.