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

Answer

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,

 $e= \dfrac{v_2-v_1}{4 -(- 23)}$

   e = 1  for elastic collision

 $1= \dfrac{v_2-v_1}{4 + 23}$

 $v_2-v_1= 27$.....................(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 =$\dfrac{v_2 - u_2}{t}$

     acceleration of ball =$\dfrac{30.63-(-23)}{0.022}$

     acceleration of ball =2437.73 m/s²