HW5Sol old

# HW5Sol old - Oregon State University Physics 201 Fall Fall...

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Oregon State University Physics 201 , Fall 2008 HW5 (due Nov. 3 at 5:00 p.m.) Page 1 Oregon State University Physics 201 Fall Term, 2008 HW5 Solutions This HW assignment, due November 3, covers topics in Chapters o 7 and 9 of the text—freely mixed 7 together (and youʼll need the basics of kinematics/projectile motion, too). Donʼt forget the extra credit point available. Follow the recommended format for solutions, with SI units unless directed otherwise and 3 signi± cant digits in all numerical answers. Use a separate page for each solution—donʼt crowd your solutions on a page! 1. (Chapter 7, Problem #10) The groundʼs force on the student changes the studentʼs momentum— from his momentum just before the collision to his momentum just after the collision: F net.ext Δ t = P P 0 In other words: F ground Δ t = mv mv 0 In this case, the velocity just after the collision is zero; the velocity just before is the impact velocity: F ground Δ t = m (0) – mv impact Simplifying: F ground Δ t = mv impact Solving for v impact : v impact = F ground Δ t / m Keep in mind that these are vector quantities here: F ground and d v impact have directions, so we need to deF ne an axis. Itʼs easiest to use convention: This problem is all about vertical motion, so deF ne the y -direction as positive upward. Therefore F net.ext is positive (and v impact will turn out to be negative). Now do a bit of kinematics. What height would be necessary to fall from (starting at rest) to produce a certain impact velocity? Use this equation: v y 2 = v 0y 2 + 2 a y y Solve this for y : y = ( v y 2 v 0y 2 )/2 a y ±or the falling motion itself (before impact), the initial velocity is zero; and the F nal velocity is the impact velocity: y = ( v impact 2 – 0 2 )/[2(– g )] Simplify: y = –( v impact 2 )/2 g Substitute: y = –[( F ground Δ t / m ) 2 ]/2 g The numbers: y = –{[–(18,000)(0.010)/63] 2 }/[2(9.80)] = 0.416 m The student fell from a height of 0.416 m (41.6 cm).

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Oregon State University Physics 201 , Fall 2008 HW5 (due Nov. 3 at 5:00 p.m.) Page 2 2. A 1.9 kg basketball is dropped from rest from a height of 2.46 m onto a bathroom scale. As it bounces, its contact with the scale lasts for 0.25 s. The ball bounces up to a height of 1.78 m. What was the average reading of the scale during the time when the ball was in contact with it? (Ignore air resistance and assume the basketball is a point mass.) F net.ext Δ t = P P 0 In other words: F scale Δ t = mv mv 0 In this case, the velocity just before the collision is the F nal velocity of the ball after being dropped from its initial height (i.e. the impact velocity). And the velocity just after the collision is the initial velocity necessary to launch the ball to its F nal height; : F scale Δ t = mv launch mv impact Simplifying: F scale Δ t = m ( v launch v impact ) Solving for F scale : F scale = m ( v launch v impact )/ Δ t Keep in mind that these are vector quantities here: F
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## This note was uploaded on 01/13/2010 for the course PH 201 taught by Professor Staff during the Fall '08 term at Oregon State.

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HW5Sol old - Oregon State University Physics 201 Fall Fall...

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