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 increasingl tense. Assume that you have cords that are 15 m long, and that the cords stretch in the jump an additional 24 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?). At the top, when the person has fallen 0 m. O When the person has fallen 15 m. O When the person has fallen between 15 m and the bottom. O When the person has fallen between 0 m and 15 m. O At the bottom, when the person has fallen 39 m. (c) What is the jumper's speed at this instant, when the tension is greatest in the cords? v= m/s (d) Is the jumper's momentum changing at this instant or not? (That is, is dp,/dt nonzero or zero?) O No, the jumper's momentum is not changing. Yes, the jumper's momentum is changing. (e) Which of the following statements is a valid basis for answering part (d) correctly? If the momentum weren't changing, the momentum would remain zero forever. A very short time ago the momentum was downward (and nonzero). Since the momentum is zero, the momentum isn't changing. ✔After a very short time the momentum will be upward (and nonzero). Two core slack ini mg

icon
Related questions
Question

Please answer all. Parts

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 15 m long, and that the cords stretch in the jump an additional 24 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?)
At the top, when the person has fallen 0 m.
O When the person has fallen 15 m.
O When the person has fallen between 15 m and the bottom.
O When the person has fallen between 0 m and 15 m.
O At the bottom, when the person has fallen 39 m.
(c) What is the jumper's speed at this instant, when the tension is greatest in the cords?
V =
m/s
(d) Is the jumper's momentum changing at this instant or not? (That is, is dp/dt nonzero or zero?)
O No, the jumper's momentum is not changing.
Yes, the jumper's momentum is changing.
(e) Which of the following statements is a valid basis for answering part (d) correctly?
If the momentum weren't changing, the momentum would remain zero forever.
A very short time ago the momentum was downward (and nonzero).
Since the momentum is zero, the momentum isn't changing.
✔ 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 dp/dt is equal to the net force, dp,/dt must be zero.
Two cords
slack initially
mg
Transcribed Image Text: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 15 m long, and that the cords stretch in the jump an additional 24 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?) At the top, when the person has fallen 0 m. O When the person has fallen 15 m. O When the person has fallen between 15 m and the bottom. O When the person has fallen between 0 m and 15 m. O At the bottom, when the person has fallen 39 m. (c) What is the jumper's speed at this instant, when the tension is greatest in the cords? V = m/s (d) Is the jumper's momentum changing at this instant or not? (That is, is dp/dt nonzero or zero?) O No, the jumper's momentum is not changing. Yes, the jumper's momentum is changing. (e) Which of the following statements is a valid basis for answering part (d) correctly? If the momentum weren't changing, the momentum would remain zero forever. A very short time ago the momentum was downward (and nonzero). Since the momentum is zero, the momentum isn't changing. ✔ 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 dp/dt is equal to the net force, dp,/dt must be zero. Two cords slack initially mg
Expert Solution
steps

Step by step

Solved in 3 steps with 4 images

Blurred answer