Question

A 60 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/s²

Answer 1

Let us consider the upwards direction as positive and the downwards direction as negative.

Mass of the soccer player = m₁ = 60 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 = -27 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 = 31

V₃ = V₄ - 31

By conservation of linear momentum,

m₁V₁ + m₂V₂ = m₁V₃ + m₂V₄

(60)(4) + (0.45)(-27) = (60)(V₄ - 31) + (0.45)V₄

240 - 12.15 = 60V₄ - 1860 + 0.45V₄

60.45V₄ = 2087.85

V₄ = 34.54 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

34.54 = -27 + a(0.02)

a = 3077 m/s²

A) Speed of the ball immediately after the ball rebounds upwards = 34.54 m/s

B) Average acceleration of the ball = 3077 m/s²