phy7a_midterm2_review_solutions

# phy7a_midterm2_review_solutions - PHY-7A Review for Midterm...

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: PHY-7A Review for Midterm 2: Solutions Solution to Problem 1. To solve this problem, we will adopt the following procedure: 1. Use conservation of energy for the bouncy ball and basketball individually to find their speeds v 1B and v 2B just before the basketball hits the ground. 2. Use conservation of energy of just the basketball to find the speed of the basketball immedi- ately after it hits the ground, but before it collides with the bouncy ball. 3. Use conservation of energy and momentum for the system to find the speeds v 1C and v 2C of the balls just after the collision. 4. Use conservation of energy for just the bouncy ball to find its final height. The points of interest in this problem are (A) the initial position of the two balls, (B) the point at which the basketball hits the ground, (B’) the point immediately after which the basketball bounces upwards but before it has collided with the bouncy ball, (C) the point immediately after the basketball has collided with the bouncy ball, and finally (D) the point at which the bouncy ball reaches its maximum height. To find the velocity of the bouncy ball at (B), use conservation of energy: mgh = 1 2 m v 2 1B ⇒ v 1B =- p 2 gh. (1.1) Take the negative root since the bouncy ball is falling down . Likewise, we can find the velocity of the basketball at (B): Mgh = 1 2 Mv 2 2B ⇒ v 2B = v 1B =- p 2 gh. (1.2) The bouncy ball and the basketball both have the same velocity at point (B), as expected. (Falling objects have the same acceleration.) Next, we need to find the velocity of the basketball immediately after it hits the ground. Since energy is conserved by hypothesis (the problem statement tells us to treat all collisions as elastic), we must have v 2B =- v 2B =- v 1B = p 2 gh. (1.3) The velocity vector switches directions but keeps the same magnitude. Note that the momentum of the basketball is not conserved during the bounce because there is a net external force present: the normal force of the ground on the basketball. Now we can examine the collision of the basketball with the bouncy ball. Because this is a 1D head-on collision, we can use the following result: v 2B- v 1B =- ( v 2C- v 1C ) ⇒ v 2C = v 1C + 2 v 1B . (1.4) This is Equation (9-8) in Giancoli. Conservation of momentum can be used to solve for the velocity of the bouncy ball just after the collision: m v 1B + Mv 2B = m v 1C + Mv 2C ⇒ v 1C =- 3 M- m M + m v 1B . (1.5) Here we used the fact that v 2B =- v 1B to eliminate v 2B from the above equations. 1 Finally, we use conservation of energy for the bouncy ball to find the height H at point (D): mgH = 1 2 m v 2 1C = 1 2 m 3 M- m M + m v 1B 2 = 3 M- m M + m 2 mgh ⇒ H = 3 M- m M + m 2 h....
View Full Document

## This test prep was uploaded on 04/03/2008 for the course PHYSICS 7A taught by Professor Lanzara during the Spring '08 term at University of California, Berkeley.

### Page1 / 9

phy7a_midterm2_review_solutions - PHY-7A Review for Midterm...

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

View Full Document
Ask a homework question - tutors are online