PHYSIC 問題用黃色HIGHLIGHT了

Bungee Jump (笨豬跳) is a very exciting sport game. A player of mass 80 kg is tied to his ankle a tough elastic band of natural length 25 m. He jumps down a bridge (assuming starts from a speed of zero), falls for a certain distance and then is pulled to stop by the elastic band. Assume the player moves in a vertical line.

(a) Find the vertical speed of the payer just before the band is taut (扯直). (2 marks)




What is the time required, starting from the speed of zero, for the player to reach the speed in (a)?(2 marks)




When the band extends, it exerts an upward pulling force on the player and eventually brings him to stop momentarily. The braking time is 2.5 s.


Find his change in momentum in the process of extension of the elastic band.

What is the average force acting on the player by the elastic band during the extension?

ANS




The player undergoes free fall before the band becomes taut.Take downward as positive
v2 – u2 = 2gs
v2 – 0 = 2 x 10 x 25
v = 22.4 ms-1 (downward)



Alt : s= ut + ½gt2
25 = 0 + ½ x 10 t2
t = 2.24 s



(b) v = u + gt
22.4 = 0 + 10 t
t = 2.24 s






(i) Take downward direction as positive change in momentum = m (v – u)
= 80 ( 0 – 22.4)
= - 1792 kgms-1 (upward)


(ii) F = Δ mv / t = - 1792 / 2.5
= - 716.8 N (upward)
average force = -716.8 - 80 x 10
= - 1516.8 N (upward)

我想問點解要計Δ mv / t之後,仲要把他再減mg??