HW #6-6-solutions

# HW #6-6-solutions - pasha(sep635 – HW 6-6 – Antoniewicz...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: pasha (sep635) – HW 6-6 – Antoniewicz – (56445) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 15.0 points Two protons are hurled straight at each other, each with a kinetic energy of 0 . 35 MeV, where 1 mega electron volt is equal to 1 . 6 × 10- 13 J. What is the separation from the protons from each other when they momentarily come to a stop (a turning point)? Use 1 4 π = 9 × 10 9 N · m 2 / C 2 , e = 1 . 6 × 10- 19 C . Correct answer: 2 . 05714 × 10- 15 m. Explanation: Convert the kinetic energy of a proton to joules to obtain KE i, proton = 5 . 6 × 10- 14 J . Define the system to be the two protons, so KE i, total = 2(5 . 6 × 10- 14 J) = 1 . 12 × 10- 13 J . Assume that they are initially very far apart, so U i ≈ 0. At the final state, their speeds are zero at the turning point, which is the point of closest approach. The work done on the system is zero and their rest energies are constant. Thus, the total kinetic energy is completely converted to the final electric potential energy. KE i, total = U f K i = 1 4 π q 1 q 2 r Using q 1 = q 2 = 1 . 6 × 10- 19 C , we obtain r = 2 . 05714 × 10- 15 m . 002 (part 1 of 2) 8.0 points A pendulum consists of a very light but sti ff rod of length L hanging from a nearly fric- tionless axle, with a mass m at the end of the rod. Calculate the gravitational potential en- ergy as a function of the angle, θ , measured from the vertical. Set U = 0 at the location of the mass when the pendulum is hanging straight down. 1. U = mg L (1- sin( θ )) 2. U = mg L θ 3. U = mg L cos( θ ) 4. U =- mg L sin( θ ) 5. U = mg L (1- cos( θ )) correct 6. U =- mg L cos( θ ) 7. U = mg L sin( θ ) Explanation: We must find the vertical distance the mass is from the bottom so we can use the potential energy formula for gravity on Earth. We can see that the height y can be found by subtracting L cos( θ ) from L . The first value is the vertical distance of the mass from the pivot, while hanging at an angle of θ , and the second is the distance of the U = 0 location from the pivot. Subtracting these two values gives us the height of the mass from the U = 0 location. Therefore, the potential energy is given by: U = mg y = mg ( L- L cos( θ )) U = mg L (1- cos( θ )) . 003 (part 2 of 2) 8.0 points Suppose that you hit the stationary hanging mass so it has an initial speed v i . What is the minimum initial speed needed for the pendulum to go over the top ( θ = 180 ◦ )? 1. v i = 2 p g L correct 2. v i = 8 p g L 3. v i = 1 2 p g L pasha (sep635) – HW 6-6 – Antoniewicz – (56445) 2 4. v i = s g L 2 5. v i = p g L 6. v i = p 2 g L 7. v i = 4 p g L Explanation: At the initial state, right after the instant that the mass is given the initial speed, we have zero potential energy and nonzero ki- netic energy. At the final state, if we had given the pendulum the absolute minimum speed for the pendulum to go over the top,...
View Full Document

## This note was uploaded on 11/17/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.

### Page1 / 6

HW #6-6-solutions - pasha(sep635 – HW 6-6 – Antoniewicz...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online