Question

A 71 kg soccer player jumps vertically upwards and heads the
0.45 kg ball as it is descending vertically with a speed of 27 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/s^{2}

Answer #1

Here ,

let the final speed of the ball is v1

final speed of the person is v2

Using conservation of momentum

71 * v2 + 0.45 * v1 = 71 * 4 - 0.45 * 27

as the collision is elastic

1 = (v1 - v2)/(27 + 4)

v1 - v2 = 31

solving for v1 and v2

v1 = 34.6 m/s

v2 = 3.61 m/s

rebound vertical velocity is 34.6 m/s

b)

average acceleration = change in velocity/time taken

average acceleration = (34.6 - (-27))/0.020

average acceleration = 3080 m/s^2

the average acceleration is 3080 m/s^2

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