A 58.0 kg soccer player jumps vertically upwards and heads the 0.45 kg ball as it is descending vertically with(b) If the ball is in contact with the players head for 19 ms, what is the average acceleration of the ball? a speed of 26.0 m/s. If the player was moving upward with a speed of 4.20 m/s just before impact. (a) What will be the speed of the ball immediately after the collision if the ball rebounds vertically upwards and the collision is elastic? (Note that the force of gravity may be ignored during the brief collision time.)

Question

A 58.0 kg soccer player jumps vertically upwards and heads the 0.45 kg ball as it is descending vertically with(b) If the ball is in contact with the players head for 19 ms, what is the average acceleration of the ball? a speed of 26.0 m/s. If the player was moving upward with a speed of 4.20 m/s just before impact. (a) What will be the speed of the ball immediately after the collision if the ball rebounds vertically upwards and the collision is elastic? (Note that the force of gravity may be ignored during the brief collision time.)

Answer 1

Calculate the initial velocity of the tennis ball the instant it is released

Answer 2

To find the answer to this question, we can use the principle of conservation of momentum.

Let's break down the problem into two parts: before and after the collision.

Before the collision:
The soccer player has a mass of 58.0 kg and is moving upward with a speed of 4.20 m/s. The ball has a mass of 0.45 kg and is descending with a speed of 26.0 m/s.
We can calculate the initial momentum (p_initial) of the system before the collision as follows:

p_initial = (58.0 kg * 4.20 m/s) + (-0.45 kg * 26.0 m/s)
p_initial = 243.6 kg·m/s + (-11.7 kg·m/s)
p_initial = 231.9 kg·m/s

After the collision:
The collision is described as elastic, which means that both objects will rebound without losing any kinetic energy.
To find the final speed of the ball (v_final), we use the conservation of momentum principle:

p_initial = p_final

Let's assume the speed of the ball immediately after the collision is v_ball.

p_initial = (58.0 kg * 4.20 m/s) + (-0.45 kg * 26.0 m/s)
p_final = (58.0 kg * 4.20 m/s) + (-0.45 kg * v_ball)

Since the collision is elastic, we can assume that the rebound speed of the ball is equal in magnitude but opposite in direction to the initial speed of the ball:

v_ball = -26.0 m/s

Now, we can calculate the final momentum (p_final) of the system after the collision:

p_final = (58.0 kg * 4.20 m/s) + (-0.45 kg * -26.0 m/s)
p_final = 243.6 kg·m/s + 11.7 kg·m/s
p_final = 255.3 kg·m/s

Using the conservation of momentum principle (p_initial = p_final), we can solve for the final speed of the ball (v_final):

p_initial = p_final
231.9 kg·m/s = (58.0 kg * 4.20 m/s) + (-0.45 kg * v_final)

Rearrange the equation to solve for v_final:

v_final = (231.9 kg·m/s - (58.0 kg * 4.20 m/s)) / -0.45 kg
v_final = (231.9 kg·m/s - 243.6 kg·m/s) / -0.45 kg
v_final = -11.7 kg·m/s / -0.45 kg
v_final = 26.0 m/s

Therefore, the speed of the ball immediately after the collision (v_final) is 26.0 m/s.

Now let's calculate the average acceleration of the ball during the collision.

Using the definition of acceleration (a = Δv/Δt), where Δv is the change in velocity and Δt is the time duration, we can find the average acceleration of the ball during the 19 ms (0.019 s) collision.

Since the ball rebounds vertically upwards, we consider the change in velocity to be from the initial downward velocity (-26.0 m/s) to the final upward velocity (v_final = 26.0 m/s).

Δv = v_final - initial_velocity
Δv = 26.0 m/s - (-26.0 m/s)
Δv = 52.0 m/s

Now, we can calculate the average acceleration (a_avg) during the collision:

a_avg = Δv/Δt
a_avg = 52.0 m/s / 0.019 s

a_avg ≈ 2737 m/s²

Therefore, the average acceleration of the ball during the collision is approximately 2737 m/s².