# Project 4 Solution: Part A: Escape from Korath

The problem makes use of three models, two old ones: constant velocity and constant force, and a new one: spring-ball model of materials. In the first part, students have to recognize they can drop the apple to determine distance. This happens fairly quickly once a couple groups figure it out. You should give students time to think about why they'd drop the apple and what it buys them. Saying they can't see the platform well enough is fine so they can use the reflected sound. The extension of the different metal wires is a big challenge, so having students focus on a single wire can help them make quicker work. Sometimes, students try to do all of them at once and that can be arduous. It's a good point to push them to use symbols and not numbers because you can ask them about the other metal wires after they finish the first one.

The students need to use the apple to determine the height $H$ of the cliff. After dropping the apple, it takes $t_{\rm tot}=4.9\,{\rm s}$ for the group to hear it hit the ground. So, the total time $$t_{\rm tot}=t_{\rm drop}+t_{\rm air}$$ For the time of the drop, we will use the kinematic equation $$x_{f} = v_{xi} \Delta t + \dfrac{1}{2}\dfrac{F_{net,x}}{m} \Delta t^2$$ Where the initial velocity is zero, making that term go to zero. Then we can solve the equation for the time it takes the apple to fall to the platform. $$t_{\rm drop}=\sqrt{\frac{2H}{g}}$$

Then we can use the same equation to determine the time it take for the sound to travel back to the drop point, where the acceleration is zero making that term go to zero. Then we can solve that equation for time, finding the time it takes for the sound of the apple's impact to return.

$$t_{\rm air}=\frac{H}{v_{\rm air}}$$

We can then plug these two equations into $$t_{\rm tot}=t_{\rm drop}+t_{\rm air}$$ $$t_{\rm tot}=\sqrt{\frac{2H}{g}}+\frac{H}{v_{\rm air}}$$ The students can solve this by hand or on Wolfram.

$v_{\rm air}\approx 343\,{\rm m/s}$ (The speed of sound). Solving this for the height gives $H\approx 103\,{\rm m}$.

##### Tutor Question

**Question:**Did you take into account the time it takes for the sound to travel?**Expected Answer:**If they didn't, they need to.

**Question:**Why are convinced dropping the apple will work for you? What if you had some other object?**Expected Answer:**In the absence of air resistance, things take the same time to fall.

**Question:**Is there an inconsistency between these two models (i.e., free fall and air speed)?**Expected Answer:**Absolutely! There's no air in the first model!

Now, given the diameter $D$ of the cables and the interatomic bond length $d$, they can figure out the number of atoms in a cross sectional area $N_{\rm A}$ (Think “a” for across). From the Young's Modulus notes: $$N_{chains\:in\:wire} = \dfrac{A_{wire}}{A_{bond}}$$ Then plug the formulas for the bond and wire cross sectional area $$N_{\rm A}=\frac{\pi(D/2)^{2}}{d^{2}}.$$ Similarly, for a length of wire $L$, the number of bonds in a single chain is given from the same notes on the Young's Modulus, $$N_{bonds\:in\:chain} = \dfrac{L_{wire}}{L_{bond}}$$ Or (Think “l” for length) $$N_{\rm L}=L/d.$$ So, given the interatomic spring constant $k_{i}$ (formula in the same section of the notes), we find the effective spring constant $$k_{\rm eff}=\frac{N_{\rm A}}{N_{\rm L}}k_{i}=\frac{\pi D^{2}}{4dL}k_{i}.$$

##### Tutor Question

**Question:**Can you draw how you are counting the number of atoms inside the cross sectional area?**Expected Answer:**…

**Question:**Can you draw how you are counting the number of atoms inside a single chain?**Expected Answer:**…

**Question:**When determining the number of atoms inside the cross sectional area, why is okay to fit squares (simple cubic structure) into the circular area?**Expected Answer:**The squares are super small.

Suggest that the students use Wolfram Alpha to handle any tedious algebra, and keep everything in variables until the very end of the problem. It is important that they recognize that they can solve two equations in two unknowns, but in this limited time situation their efforts are better used elsewhere.

Finally, the limiting scenario is when (the distance to the platform should be equal to the length of the rope plus the stretch of the rope, so that the metal of the rope doesn't touch the platform and Korath doesn't come to kill us all) $$H=L+s,$$

From the notes Modeling a Solid Wire: Springs in series and parallel we find the equation $$k_{s,eff} = \dfrac{mg}{s}$$ Which can be solved for the stretch $$s=\frac{Mg}{k_{\rm eff}}$$ We can then plug the equation derived above for $k_{\rm eff}$ $$k_{\rm eff}=\frac{\pi D^{2}}{4dL}k_{i}.$$ And we get the following formula for the stretch $$s=\frac{4dgLM}{k_{i}\pi D^{2}}.$$ Then plug stretch into $H=L+s$ giving $$H=L+\frac{4dgLM}{k_{i}\pi D^{2}}.$$

Solving for the limiting length, we find $$L=\frac{H\pi D^{2}k_{i}}{4dMg+k_{i}\pi D^{2}}.$$

##### Tutor Question

**Question:**Where are the group members holding onto the wire as you consider the stretch? Does this make a difference?**Expected Answer:**We assumed they were holding onto the end of the wire, but the stretch increases as each person lowers down the wire.

**Question:**Can you draw a free body diagram of a person on the wire? What about a cross sectional area of the wire itself?**Expected Answer:**…

Suggesting that the group choose an initial length and type of wire to give them an idea of the associated stretch will help to direct their solution. They should at least have one calculation of the stretch of one type of wire as a solid stopping point.

If we choose a total mass $M=4m$ for a group of four ($m\approx 80\,{\rm kg}$), we find the following limiting lengths for the various materials:

Element | Limiting Length $L$ (${\rm m}$) |
---|---|

Adamantium | $98.9$ |

Unobtainium | $94.9$ |

Vibranium | $76.3$ |

##### Tutor Question

**Question:**Does the mass of the wire play a role here? Would it increase or decrease the stretch?**Expected Answer:**Yes, a massive wire would pull itself down more, but we are assuming a massless wire.

##### Common Difficulties:

- Students will struggle to decide how to measure the distance to the platform. Let them work on this, but you can also ask them about previous models that might help them (i.e., constant velocity and constant force). Give them time to choose this, don't tell them exactly how to do it.
- How the macro/micro analysis of materials is used can be tough for them to make sense of. You may point them to think about the connections from the notes.
- The algebra in this problem can be hairy, but students can make use of WolframAlpha to symbolic manipulation and calculations. Let them (or guide them) to use it.

##### Main Points

- Students should be able to talk thorough the plan and calculation for determining the distance to the platform.
- Students should be able to talk through the connection between macro and micro properties of materials and how they were used in this problem.

**Conceptual questions:**

- What assumptions did you make when dropping the apple? Are any of the assumptions you made conflicting and create an inconsistent model?
- Draw a free-body diagram of the forces acting on the person hanging from the wire.
- When determining the number of atoms inside the cross-sectional area, why is it okay to fit squares (simple cubic structure) into the circular area?
- Can you draw how you are counting the number of atoms inside the cross-sectional area?
- Can you draw how you are counting the number of atoms in a chain?