A 50-kg soccer player jumps vertically upwards and heads the 0.45-kg ball as it is descending vertically with a speed of 30 m/s.
(a) If the player was moving upward with a speed of 4.0 m/s just before impact, what will be the speed of the ball immediately after the collision if the ball rebounds vertically upwards and the collision is elastic?
m/s
(b) If the ball is in contact with the player's head for 21 ms, what is the average acceleration of the ball? (Note that the force of gravity may be ignored during the brief collision time.)
a. Given:
M1 = 50kg, V1 = 4 m/s.
M2 = 0.45 kg, V2 = -30 m/s.
V3 = Velocity of M1 after colliding.
V4 = Velocity of M2 after colliding.
Momentum before = Momentum after
M1*V1 + M2*V2 = M1*V3 + M2*V4.
50*4 + 0.45*(-30) = 50*V3 + 0.45*V4.
Eq1: 50V3 + 0.45V4 = 186.5.
V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (4(50-0.45) - 13.5)/50.45),
V3 = (198.2 - 13.5)/50.45 = 3.66 m/s.
In Eq1, replace V3 with 3.66 and solve for V4.
Correction:
V3 = (4(50-0.45) - 27)/(50.45),
V3 = (198.2 - 27)/50.45 = 3.39 m/s.
b. a = (V4-V2)/0.021
(a) The speed of the ball immediately after the collision, if the collision is elastic, can be calculated using the law of conservation of momentum.
Let's assume the initial velocity of the player and the ball in the upward direction as positive, and downward direction as negative.
The momentum before the collision is given by:
Momentum(player) = mass(player) * velocity(player) = (50 kg) * (4.0 m/s) = 200 kg·m/s
Momentum(ball) = mass(ball) * velocity(ball) = (0.45 kg) * (-30 m/s) = -13.5 kg·m/s
Since momentum is conserved in an elastic collision, the total momentum before the collision should be equal to the total momentum after the collision.
Total momentum before collision = Total momentum after collision
Momentum(player) + Momentum(ball) = Momentum(player) + Momentum(ball)
200 kg·m/s + (-13.5 kg·m/s) = Momentum(player) + Momentum(ball)
Simplifying the equation, we find:
186.5 kg·m/s = Momentum(player) + Momentum(ball)
Since the ball rebounds vertically upwards, the velocity of the ball after the collision will be positive.
Let's assume the speed of the ball immediately after the collision as V.
Momentum(player) + Momentum(ball) = (50 kg) * (V m/s) + (0.45 kg) * (V m/s)
186.5 kg·m/s = 50V + 0.45V
Simplifying and solving for V:
186.5 kg·m/s = 50.45V
V = 186.5 kg·m/s / 50.45
V ≈ 3.69 m/s
Therefore, the speed of the ball immediately after the collision, if the ball rebounds vertically upwards and the collision is elastic, would be approximately 3.69 m/s.
(b) The average acceleration of the ball can be calculated using the formula:
Acceleration = Change in velocity / Time
Given that the ball is in contact with the player's head for 21 ms (0.021 seconds), we can calculate the change in velocity.
Change in velocity = Final velocity - Initial velocity
Since the ball rebounds vertically upwards, the final velocity after the collision (V) is given by:
V = 3.69 m/s (from part a)
The initial velocity of the ball is -30 m/s.
Change in velocity = 3.69 m/s - (-30 m/s) = 33.69 m/s
Using the formula for average acceleration:
Acceleration = Change in velocity / Time
Acceleration = 33.69 m/s / 0.021 s
Acceleration ≈ 1604.28 m/s²
Therefore, the average acceleration of the ball during the collision is approximately 1604.28 m/s².
To find the speed of the ball immediately after collision, we can use the principle of conservation of momentum. The conservation of momentum states that the total momentum before a collision is equal to the total momentum after the collision, assuming no external forces act on the system.
(a) Let's first calculate the initial momentum and final momentum before and after the collision.
Initial momentum before collision:
The initial momentum of the player is given by:
momentum_player_initial = mass_player * velocity_player_initial
Final momentum after collision:
The final momentum of the player is given by:
momentum_player_final = mass_player * velocity_player_final
Initial momentum of the ball before collision:
momentum_ball_initial = mass_ball * velocity_ball_initial
Final momentum of the ball after collision:
The final momentum of the ball is given by:
momentum_ball_final = mass_ball * velocity_ball_final
Using the conservation of momentum, we can write the equation:
momentum_player_initial + momentum_ball_initial = momentum_player_final + momentum_ball_final
Since the collision is elastic, the initial and final velocities of the player are given. We can now substitute the given values into the equation and solve for the final velocity of the ball:
mass_player * velocity_player_initial + mass_ball * velocity_ball_initial = mass_player * velocity_player_final + mass_ball * velocity_ball_final
(50 kg) * (4.0 m/s) + (0.45 kg) * (-30 m/s) = (50 kg) * ? + (0.45 kg) * ?
Now we have two unknowns: velocity_player_final and velocity_ball_final. We need one more equation to solve for both of them.
(b) To find the average acceleration of the ball during contact, we can use the equation:
average acceleration = change in velocity / time
We are given the time of contact (21 ms) and need to find the change in velocity of the ball during that time.
The change in velocity can be calculated using the formula:
change in velocity = final velocity - initial velocity
Since we know the initial velocity of the ball is 30 m/s downwards, the final velocity will be the value we calculated in part (a). Now we can substitute the values into the equation to find the average acceleration:
average acceleration = (velocity_ball_final - velocity_ball_initial) / time
Substituting the values will give us the average acceleration.
momentum is conserved, and energy in an elastic collision
up is positive
(50 * 4.0) - (.45 * 30) = (50 p) + (.45 b)
... 186.5 = 50 p + .45 b
1/2 [(50 * 4.0^2)+(.45 * 30^2)] =
... 1/2 [(50 * p^2)+(.45 * b^2)]
... 1205 = 50 p^2 + .45 b^2
solve the system of equations for b
... using substitution
(b) a = (b - -30) / .021 m/s^2