A 61 kg soccer player jumps vertically upwards and heads the 0.45 kg ball as it is descending vertically with a speed of 29 m/s. 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
If the ball is in contact with the player's head for 20 ms, what is the average acceleration of the ball? (Note that the force of gravity may be ignored during the brief collision time.)
m/s2
In an elastic collision, kinetic energy is conserved. The kinetic energy of the player and the ball just before collision is equal to that after collision. Assume that just after heading the ball, the velocity of the player is zero.
You can also use the conservation of vertical momentum.
To find the speed of the ball immediately after the collision, we can use the law of conservation of momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is given by the product of its mass and velocity. We can calculate the initial momentum of the player and ball using the following formula:
Initial momentum = (mass of player * velocity of player) + (mass of ball * velocity of ball)
Given:
Mass of the player (m1) = 61 kg
Velocity of the player (v1) = 4.0 m/s
Mass of the ball (m2) = 0.45 kg
Velocity of the ball (v2) = -29 m/s (negative sign indicates downward direction)
Initial momentum = (61 kg * 4.0 m/s) + (0.45 kg * -29 m/s)
Initial momentum = 244 kg⋅m/s - 13.05 kg⋅m/s
Initial momentum = 230.95 kg⋅m/s
Since the collision is elastic, the total momentum before the collision is equal to the total momentum after the collision:
Initial momentum = Final momentum
Final momentum = (mass of player * final velocity of player) + (mass of ball * final velocity of ball)
We need to solve for the final velocity of the ball.
Final velocity of the ball (v2f) = (Initial momentum - mass of player * final velocity of player) / mass of ball
Plugging in the values:
230.95 kg⋅m/s = (61 kg * 4.0 m/s) + (0.45 kg * v2f)
Rearranging the equation:
(0.45 kg * v2f) = 230.95 kg⋅m/s - (61 kg * 4.0 m/s)
(0.45 kg * v2f) = 230.95 kg⋅m/s - 244 kg⋅m/s
(0.45 kg * v2f) = -13.05 kg⋅m/s
Finally, divide both sides by 0.45 kg to solve for the final velocity of the ball:
v2f = (-13.05 kg⋅m/s) / 0.45 kg
v2f = -29 kg⋅m/s (approximately)
The speed of the ball immediately after the collision is approximately 29 m/s.
Now, let's move on to the average acceleration of the ball during the collision.
Average acceleration can be calculated using the equation:
Average acceleration = change in velocity / time
Given:
The time of contact (t) = 20 ms = 0.02 s
Change in velocity (Δv) = final velocity - initial velocity
Since the ball rebounds vertically upwards, the initial velocity is -29 m/s and the final velocity is 29 m/s.
Δv = 29 m/s - (-29 m/s)
Δv = 58 m/s
Average acceleration = (58 m/s) / (0.02 s)
Average acceleration = 2900 m/s² (approximately)
The average acceleration of the ball during the collision is approximately 2900 m/s².
To find the speed of the ball immediately after the collision, we can use the principle of conservation of momentum.
The momentum before the collision is equal to the momentum after the collision in an elastic collision:
(mass of player * velocity of player before collision) + (mass of ball * velocity of ball before collision) = (mass of player * velocity of player after collision) + (mass of ball * velocity of ball after collision)
Let's break down the given information:
Mass of player (m1) = 61 kg
Velocity of player before collision (v1) = 4.0 m/s
Mass of ball (m2) = 0.45 kg
Velocity of ball before collision (v2) = -29 m/s (negative because it is descending)
After the collision, the player will continue moving upwards and the ball will rebound vertically upwards. Let's assume the player's velocity after the collision is v1' and the ball's velocity after the collision is v2'.
Now we can substitute these values into the equation:
(61 kg * 4.0 m/s) + (0.45 kg * -29 m/s) = (61 kg * v1') + (0.45 kg * v2')
Solving for v2', we get:
v2' = [(61 kg * 4.0 m/s) + (0.45 kg * -29 m/s) - (61 kg * v1')] / 0.45 kg
Calculating this expression, the speed of the ball immediately after the collision is:
v2' = [244 kg*m/s - 13.05 kg*m/s - (61 kg * 4.0 m/s)] / 0.45 kg
v2' ≈ -303 m/s
Note that the negative sign indicates that the ball rebounds vertically upwards after the collision.
To find the average acceleration of the ball during the collision, we can use the formula:
Average acceleration = (change in velocity) / (time interval)
Given information:
Time interval (Δt) = 20 ms = 0.02 s
Change in velocity is the final velocity minus the initial velocity:
Change in velocity = v2' - v2 (velocity after collision - velocity before collision)
Using the values we already have:
Change in velocity = (-303 m/s) - (-29 m/s) = -274 m/s
Substituting these values into the formula, we get:
Average acceleration = (-274 m/s) / (0.02 s)
Average acceleration ≈ -13,700 m/s^2
Therefore, the average acceleration of the ball during the collision is approximately -13,700 m/s^2.