In: Physics
A 74 kg soccer player jumps vertically upwards and heads the
0.45 kg ball as it is descending vertically with a speed of 26 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?
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.)
Let us consider the upwards direction as positive and the downwards direction as negative.
Mass of the soccer player = m1 =74 kg
Mass of the ball = m2 = 0.45 kg
Initial velocity of the soccer player = V1 = 4 m/s
Initial velocity of the ball = V2 = -26 m/s (Negative as it is directed downwards)
Velocity of the soccer player after the collision = V3
Velocity of the ball after the collision = V4
The collision is elastic.
Coefficient of restitution = e = 1
V4 - V3 = 30
V3 = V4 - 30
By conservation of linear momentum,
m1V1 + m2V2 = m1V3 + m2V4
(74)(4) + (0.45)(-26) = (74)(V4 - 30) + (0.45)V4
296 - 11.7 = 74V4 - 2220 + 0.45V4
74.45V4 = 2504.3
V4 = 33.64 m/s
Time period the ball is in contact with the player's head = T = 20 ms = 0.02 sec
Average acceleration of the ball = a
V4 = V2 + aT
33.64 = -26 + a(0.02)
a = 2982 m/s2
A) Speed of the ball immediately after the ball rebounds upwards = 33.64 m/s
B) Average acceleration of the ball = 2982 m/s2