Question: A ball of mass 0.2 kg is initially at rest on a height of 5 meters above the ground. The ball is then released and falls freely. When it hits the ground, it rebounds and reaches a height of 3 meters. Assume air resistance to be negligible throughout the motion of the ball.

a) Calculate the initial potential energy of the ball.

b) Determine the velocity of the ball just before it hits the ground.

c) Calculate the kinetic energy of the ball just before it hits the ground.

d) Determine the magnitude of the impulse experienced by the ball upon hitting the ground.

Answer:

a) The initial potential energy (U_i) of the ball can be calculated using the formula:

U_i = m * g * h

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

Given: mass (m) = 0.2 kg, height (h) = 5 m, acceleration due to gravity (g) = 9.8 m/s^2

Substituting the values into the formula:

U_i = 0.2 kg * 9.8 m/s^2 * 5 m U_i = 9.8 J

Therefore, the initial potential energy of the ball is 9.8 Joules.

b) The velocity of the ball just before it hits the ground can be determined using the conservation of energy principle. The sum of initial potential energy and initial kinetic energy is equal to the sum of final potential energy and final kinetic energy.

Initial potential energy + Initial kinetic energy = Final potential energy + Final kinetic energy

U_i + K_i = U_f + K_f

Since the ball starts from rest, the initial kinetic energy (K_i) is 0 Joules.

Therefore, we have:

U_i = U_f + K_f

Substituting the values:

9.8 J = 0 J + K_f

K_f = 9.8 J

The final kinetic energy is also equal to the initial kinetic energy since velocity changes instantaneously the moment it strikes the ground.

Therefore, the velocity of the ball just before it hits the ground is equivalent to the square root of twice its kinetic energy divided by its mass:

K_f = (1/2) * m * v^2

Solving for v:

9.8 J = (1/2) * 0.2 kg * v^2

v^2 = (9.8 J * 2) / 0.2 kg

v^2 = 98 J/kg

v = √(98 J/kg) v ≈ 9.90 m/s

So, the velocity of the ball just before it hits the ground is approximately 9.90 m/s.

c) The kinetic energy (K_f) of the ball just before it hits the ground can be calculated using the formula:

K_f = (1/2) * m * v^2

Given: mass (m) = 0.2 kg, velocity (v) = 9.90 m/s

Substituting the values into the formula:

K_f = (1/2) * 0.2 kg * (9.90 m/s)^2 K_f = 9.80 J

Therefore, the kinetic energy of the ball just before it hits the ground is 9.80 Joules.

d) The impulse experienced by the ball upon hitting the ground can be determined using the impulse-momentum principle. Impulse is the change in momentum, and since the ball changes direction upon hitting the ground, the impulse can be expressed as the change in momentum multiplied by (-1) because the ball rebounds.

Impulse (J) = (-1) * change in momentum

The change in momentum can be calculated using the formula:

Change in momentum = m * (final velocity - initial velocity)

Given: mass (m) = 0.2 kg, initial velocity (vi) = 9.90 m/s (downward direction), final velocity (vf) = 9.90 m/s (upward direction)

Substituting the values into the formula:

Change in momentum = 0.2 kg * (9.90 m/s - (-9.90 m/s)) Change in momentum = 0.2 kg * 19.80 m/s Change in momentum = 3.96 kg·m/s

The impulse experienced by the ball is:

Impulse (J) = (-1) * 3.96 kg·m/s Impulse (J) = -3.96 kg·m/s

Therefore, the magnitude of the impulse experienced by the ball upon hitting the ground is 3.96 kg·m/s.