A 80-kg soccer player jumps vertically upwards and heads the 0.45-kg ball as it

Question

A 80-kg soccer player jumps vertically upwards and heads the 0.45-kg ball as it is descending vertically with a speed of 24 m/s. (Note that the force of gravity may be ignored during the brief collision time.)

(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?
  


(b) If the ball is in contact with the player's head for 15 ms, what is the average acceleration of the ball?
  

Explanation / Answer

a.
speed of ball after collision be V and that of player be u
Ball will not effect the player's speed.
Final speed of player will still be 4 m/s upward
Initial momentum = final momentum
80*4 + 0.45*(-24) = 80*u + 0.45*(v)
80u + 0.45v = 309.2 ...eqn 1

Since collision is elastic, value of e = 1
1 = - ( u - v) / ( -24 - 4)
28 = v - u
u = v-28 ...eqn 2
use this in eqn 1
80u + 0.45v = 309.2
80*(v-28) + 0.45v = 309.2
80v - 2240 +0.45v = 309.2
v= 31.69 m/s
Answer: 31.69 m/s
b)
Force acting on ball = change in momentum / time
= m*(31.69 - (-24)) / time
a = Force/mass
= (31.69 - (-24)) / time
=(31.69+24)/ (15*10^-3)
=3713 m/s^2
Answer: 3713 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.