Question

Design a "bungee jump" apparatus for adults. A bungee jumper falls from a high platform with two elastic cords tied to the ankles. The jumper falls freely for a while, with the cords slack. Then the jumper falls an additional distance with the cords increasingly tense. Assume that you have cords that are 11 m long, and that the cords stretch in the jump an additional 23 m for a jumper whose mass is 140 kg, the heaviest adult you will allow to use your bungee jump (heavier customers would hit the ground). (a) It will help you a great deal in your analysis to make a series of 5 simple diagrams, like a comic strip, showing the platform, the jumper, and the two cords at the following times in the fall and the rebound: 1 while cords are slack (shown here as an example to get you started) 2 when the two cords are just starting to stretch 3 when the two cords are half stretched 4 when the two cords are fully stretched 5 when the two cords are again half stretched, on the way up On each diagram, draw and label vectors representing the forces acting on the jumper, and the jumper's velocity. Make the relative lengths of the vectors reflect their relative magnitudes. (b) At what instant is there the greatest tension in the cords? (How do you know?) When the person has fallen 11 m. At the bottom, when the person has fallen 34 m. When the person has fallen between 0 m and 11 m. When the person has fallen between 11 m and the bottom. At the top, when the person has fallen 0 m. (c) What is the jumper's speed at this instant, when the tension is greatest in the cords? m/s (d) Is the jumper's momentum changing at this instant or not? (That is, is nonzero or zero?) (e) Which of the following statements is a valid basis for answering part (d) correctly? Since the momentum is zero, the momentum isn't changing. If the momentum weren't changing, the momentum would remain zero forever. After a very short time the momentum will be upward (and nonzero). Since the net force must be zero when the momentum is zero, and since is equal to the net force, must be zero. A very short time ago the momentum was downward (and nonzero). The number of significant digits is set to 3; the tolerance is +/-2% (f) Focus on this instant of greatest tension and, starting from a fundamental principle, determine the spring stiffness for each of the two cords. N/m (g) What is the maximum tension that each one of the two cords must support without breaking? (This tells you what kind of cords you need to buy.) N (h) What is the maximum acceleration (in "g's") that the jumper experiences? (Note that | if is small compared to .) g's (acceleration in m/s2 divided by 9.8 m/s2) (i) What is the direction of this maximum acceleration? (j) What approximations or simplifying assumptions did you have to make in your analysis which might not be adequately valid? (Don't check any approximations or simplifying assumptions which in fact have negligible effects on your numerical results.) Assume that the gravitational force hardly changes from the top of the jump to the bottom. Neglect air resistance, despite fairly high speeds. Assume the speeds are very small compared to the speed of light. Assume tension in cord proportional to stretch, even for the very large stretch occurring here.

Answer 1

a)

b)

let s be the spring stiffness of each of the cord,

so, tension T = s*x here x is the elongation of the cord from mormal position,

since tension is directly proportional to the elomgation x, so maximum tension will act on the cord, when it is fullu stretched i.e. x = 23 m as in case iv of above diagram i.e. at the bottom when person has fallen (11 + 23 = 34 m)

c)

When the tension is greatest, the jumper has touched to the ground, so velocity V will be zero a this instant,

d)

Yes, at this instant instant i.e. when tension is greatest, there is change in the momentum i.e. non zero, as the velocity of the jumper just before touching the ground and just after touching the ground will be in opposite direction.

e) on the basis of above, After a very short time, the momentum will be in upward direction, and a very short time ago the momentum was downward

f)

let s be the spring stffness of each cord,

Assuming jumper and cord as a single system , Apply the law of conservation of energy when the jumper is at the top, and when he reaches at the groung assuming ground as a zero potential line,

Energy at the top = Energy at the ground

PE_{jumper at the top}   + PE_{cord at the top}  =    PE_{jumper at the bottom} + PE_{cord at the bottom of both}

mg* 34 + 0 = 0 + 2 *1/2 * s*23²

140 * 9.81 * 34 = s* 23²

so, s = 88.27 N/m

g) maxim tension in each of the cord, T = s* x_max

= 88.27 * 23 = 2030.21 N

h)

Maximum acceleration of the jumper will be at bottom as,

2* T - mg = ma

2 * 2030.21 - 140 * 9.8 = 140 * a

so, acceleration, a = 19.203 m/s² = 1.96 g  

i) maximum acceleration will act in upward direction

j) Assuming no air resistance and  tension in cord proportional to stretch even for the very large stretch may not be adequately valid