Question

A 65-kg soccer player jumps vertically upwards and heads the 0.45-kg ball as it is descending vertically with a speed of 23 m/s.

a.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
b.If the ball is in contact with the player's head for 22 ms, what is the average acceleration of the ball?_______m/s

Answer 1

given,

mass of soccer player = M = 65 Kg

mass of football = m = 0.45 Kg

speed of the ball which is descending (u₂)= 23 m/s

Speed of the player vertically upward(u₁) = 4 m/s

a) Using conservation of momentum

M u₁ + m u₂ = M v₁ + m v₂

65 x 4 - 0.45 x 23 = 65 v₁ + 0.45 v₂

65 v₁ + 0.45 v₂ = 249.65................(1)

we know,

`e= (v_2-v_1)/(4 -(- 23))`

e = 1 for elastic collision

`1= (v_2-v_1)/(4 + 23)`

`v_2-v_1= 27`.....................(2)

now putting value of v₂ in equation(1)

65 v₁ + 0.45 (27 + v₁) = 249.65

65.45 v₁ = 237.5

v₁ = 3.63 m/s

v₂ = 30.63 m/s

b)

acceleration of ball =
`(v_2 - u_2)/(t)`

acceleration of ball =
`(30.63-(-23))/(0.022)`

acceleration of ball =2437.73 m/s²