A 50.0 kg soccer player jumps vertically upwards and heads the 0.45 kg ball as it is descending vertically with a speed of 27.0 m/s. If the player was moving upward with a speed of 2.80 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.)

To find the speed of the ball immediately after the collision, we can use the principle of conservation of momentum. In this case, we can assume that the soccer player and the ball are the only objects involved in the system.

The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision, as long as there are no external forces acting on the system. Mathematically, this can be expressed as:

m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'

where m1 is the mass of the soccer player, v1 is the initial velocity of the soccer player, m2 is the mass of the ball, v2 is the initial velocity of the ball, v1' is the final velocity of the soccer player, and v2' is the final velocity of the ball.

In this case, the soccer player jumps vertically upwards, so his initial velocity is 2.80 m/s upwards. The ball is descending vertically, so its initial velocity is 27.0 m/s downwards.

Substituting the given values into the equation, we get:

50.0 kg * 2.80 m/s + 0.45 kg * (-27.0 m/s) = 50.0 kg * v1' + 0.45 kg * v2'

Simplifying the equation:

140 kg·m/s - 12.15 kg·m/s = 50.0 kg * v1' + 0.45 kg * v2'

Multiply both sides of the equation by -1:

12.15 kg·m/s - 140 kg·m/s = 50.0 kg * (-v1') - 0.45 kg * v2'

Rearranging the equation:

-127.85 kg·m/s = -50.0 kg * v1' - 0.45 kg * v2'

Now, since the collision is elastic, the relative velocity of separation after the collision is equal to the relative velocity of approach before the collision. In this case, the soccer player was moving upwards with a velocity of 2.80 m/s before the collision, and the ball was initially moving downwards with a velocity of 27.0 m/s.

Therefore, the relative velocity of approach is the sum of these velocities, which is:

vrel = v1 - v2 = 2.80 m/s - (-27.0 m/s) = 29.8 m/s

Now we can substitute this value into the equation:

-127.85 kg·m/s = -50.0 kg * v1' - 0.45 kg * v2'
-127.85 kg·m/s = -50.0 kg * 29.8 m/s - 0.45 kg * v2'

Solving for v2', the final velocity of the ball:

v2' = (-127.85 kg·m/s + 50.0 kg * 29.8 m/s) / 0.45 kg

v2' = (-127.85 kg·m/s + 1490 kg·m/s) / 0.45 kg

v2' = 1362.15 kg·m/s / 0.45 kg

v2' ≈ 3027.00 m/s

Therefore, the speed of the ball immediately after the collision, if the collision is elastic, is approximately 3027.00 m/s.