A superball of mass m, starting at rest, is dropped from a height hi above the ground and bounces back up to a height of hf. The collision with the ground occurs over a total time tc. You may ignore air resistance.
(a) What is the magnitude of the momentum of the ball immediately before the collision? Express your answer in terms of m, hi, and g as needed (enter h_i for hi, h_f for hf, and t_c for tc).
(b) What is the magnitude of the momentum of the ball immediately after the collision? Express your answer in terms of m, hf, and g as needed (enter h_i for hi, h_f for hf, and t_c for tc).
(c) What is the magnitude of the impulse imparted to the ball? Express your answer in terms of m, hi, hf, tc, and g as needed (enter h_i for hi, h_f for hf, and t_c for tc).
(d) What is the magnitude of the average force of the ground on the ball? Express your answer in terms of m, hi, hf, tc, and g as needed (enter h_i for hi, h_f for hf, and t_c for tc).
there is a problem in the last answer, at the site from where this question is from.
m*sqrt(2*g*h_i)
m*sqrt(2*g*h_f)
(m*sqrt(2*g*h_i)+m*sqrt(2*g*h_f))
To solve these problems, we need to apply the principles of conservation of momentum and impulse. Let's break down each part step by step:
(a) What is the magnitude of the momentum of the ball immediately before the collision?
Momentum (p) is defined as the product of an object's mass (m) and its velocity (v). Since the ball is starting at rest, its initial velocity is 0. Therefore, the momentum before the collision is:
p_before = m * v_before = m * 0 = 0
So, the magnitude of the momentum of the ball immediately before the collision is 0.
(b) What is the magnitude of the momentum of the ball immediately after the collision?
Since this is an elastic collision, the momentum is conserved. After the collision, the ball bounces back up to height hf. We can calculate the final velocity (v_after) using the equation:
v_after = √(2 * g * hf)
where g is the acceleration due to gravity.
Now, we can calculate the momentum after the collision:
p_after = m * v_after = m * √(2 * g * hf)
So, the magnitude of the momentum of the ball immediately after the collision is m * √(2 * g * hf).
(c) What is the magnitude of the impulse imparted to the ball?
Impulse (J) is defined as the change in momentum of an object and can be calculated using the equation:
J = Δp = p_after - p_before
Substituting the values we found earlier:
J = (m * √(2 * g * hf)) - 0 = m * √(2 * g * hf)
So, the magnitude of the impulse imparted to the ball is m * √(2 * g * hf).
(d) What is the magnitude of the average force of the ground on the ball?
The average force (F_average) can be calculated by dividing the impulse by the time of collision (tc):
F_average = J / tc = (m * √(2 * g * hf)) / tc
So, the magnitude of the average force of the ground on the ball is (m * √(2 * g * hf)) / tc.