# 3 - homework 03 – KERR KELSEY – Due 11:00 pm 1 Question...

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: homework 03 – KERR, KELSEY – Due: Jan 25 2008, 11:00 pm 1 Question 1, chap 2, sect 5. part 1 of 1 10 points An electron, starting from rest and moving with a constant acceleration, travels 5 . 9 cm in 7 ms. What is the magnitude of this acceleration in km/s 2 ? Correct answer: 2 . 40816 km / s 2 (tolerance ± 1 %). Explanation: An electron starting from rest with con- stant acceleration will travel a distance of d = 1 2 at 2 in t seconds. Given the time and distance traveled we solve for a : a = 2 d t 2 making any necessary unit conversions along the way. Question 2, chap 2, sect 7. part 1 of 1 10 points The acceleration of a marble in a certain fluid is proportional to the speed of the marble squared, and is given by a = α v 2 , where v > 0 m/s and α =- 4 . 8 m − 1 . If the marble enters this fluid with a speed of 1 . 31 m / s, how long will it take before the marble’s speed is reduced to half of its initial value? Correct answer: 0 . 159033 s (tolerance ± 1 %). Explanation: Basic Concept: a = d v dt Solution: a = d v dt = α v 2 Separating variables, dv v 2 = α dt . Integrating this, we have integraldisplay dv v 2 = α integraldisplay dt. With our limits of integration, this becomes integraldisplay v v = v u − 2 du = α integraldisplay t t =0 dt ′- 1 v + 1 v = α t. We want to know t for which v = v / 2: t = 1 α parenleftbigg 1 v- 2 v parenrightbigg =- 1 α v =- 1 (- 4 . 8 m − 1 )(1 . 31 m / s) = 0 . 159033 s . Question 3, chap 2, sect 6. part 1 of 2 10 points A ball is thrown upward. After reaching a maximum height, it continues falling back towards Earth. On the way down, the ball is caught at the same height at which it was thrown upward. Neglect: Air resistance. Its initial vertical speed v , acceleration of gravity is 9 . 8 m / s 2 , and maximum height h max are shown in the figure below. b b b b b b b b b bb b b b b b b b b b v 9 . 8m / s 2 h max homework 03 – KERR, KELSEY – Due: Jan 25 2008, 11:00 pm 2 If the time ( up and down ) the ball remains in the air is 2 . 04 s, calculate its speed when it caught. Correct answer: 9 . 996 m / s (tolerance ± 1 %). Explanation: Basic Concept: For constant accelera- tion, we have v = g t (1) y = y + v t + 1 2 a t 2 . (2) Solution: The velocity at the top is zero. Since we know velocities and acceleration, Eq. 1 containing h max = y- y , a , and t . Choose the positive direction to be up, then a =- g and t up = v g ....
View Full Document

## This note was uploaded on 04/18/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

### Page1 / 6

3 - homework 03 – KERR KELSEY – Due 11:00 pm 1 Question...

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

View Full Document
Ask a homework question - tutors are online