Question:

A ball with a mass of 0.2 kg is released from rest at a height of 2 meters. It rolls down a frictionless inclined plane with an angle of 30 degrees. The ball reaches the bottom of the incline and collides with a small spring with a spring constant of 500 N/m. The spring compresses by 0.1 meters before the ball comes to a stop momentarily. Determine the maximum compression of the spring when the ball rebounds.

Given: Mass of the ball (m) = 0.2 kg Height of the incline (h) = 2 m Angle of the inclined plane (θ) = 30° Spring constant (k) = 500 N/m Compression (x) = 0.1 m (when the ball comes to a stop momentarily)

a) Calculate the initial potential energy of the ball at the starting height. b) Calculate the velocity of the ball just before hitting the spring. c) Calculate the maximum potential energy stored in the compressed spring. d) Calculate the maximum compression of the spring when the ball rebounds.

Answer:

a) To calculate the initial potential energy (PE_initial) of the ball at the starting height, we will use the formula: PE = mgh, where m is the mass, g is the acceleration due to gravity, and h is the height.

Given: m = 0.2 kg, h = 2 m, g = 9.8 m/s²

Substituting the given values into the formula, we have:

PE_initial = (0.2 kg) x (9.8 m/s²) x (2 m)

PE_initial = 3.92 J

b) To calculate the velocity of the ball just before hitting the spring, we will use the conservation of energy principle. The initial potential energy (PE_initial) will be converted into the final kinetic energy (KE_final) of the ball.

Since there is no friction, the total mechanical energy (E_initial) at the starting height will be conserved throughout the motion. Therefore, we have:

E_initial = E_final

PE_initial + KE_initial = PE_final + KE_final

At the starting height, the ball has no initial kinetic energy (KE_initial) since it is released from rest. Hence, KE_initial = 0.

PE_initial = PE_final + KE_final

Substituting the values, we have:

3.92 J = (0.2 kg) x g x h + KE_final

3.92 J = (0.2 kg) x (9.8 m/s²) x (2 m) + KE_final

3.92 J = 3.92 J + KE_final

KE_final = 0 J

Since the ball comes to rest momentarily just before hitting the spring, its final kinetic energy (KE_final) is equal to 0. Therefore, the velocity just before hitting the spring is also 0.

c) To calculate the maximum potential energy stored in the compressed spring (PE_max), we will use the formula: PE = (1/2)kx², where k is the spring constant and x is the compression.

Given: k = 500 N/m, x = 0.1 m

Substituting the given values into the formula, we have:

PE_max = (1/2)(500 N/m)(0.1 m)²

PE_max = 2.5 J

d) To calculate the maximum compression of the spring when the ball rebounds, we will use the conservation of mechanical energy.

At the bottom of the incline, all the potential energy of the ball is converted into the maximum potential energy stored in the compressed spring.

PE_initial = PE_max

(0.2 kg) x g x h = (1/2)kx²

(0.2 kg) x (9.8 m/s²) x (2 m) = (1/2)(500 N/m)x²

Rearranging the equation, we have:

x² = [(0.2 kg) x (9.8 m/s²) x (2 m)] / [(1/2)(500 N/m)]

x² = 0.784 m²

x = √(0.784 m²)

x ≈ 0.885 m

Therefore, the maximum compression of the spring when the ball rebounds is approximately 0.885 meters.