Review for Test #1

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# Review for Test #1 - PowerPoint PPT Presentation

Review for Test #1. Responsible for: - Chapters 1, 2, 3, and 4 - Notes from class - Problems worked in class - Homework assignments Test format: - 15 problems: 10 simple, 4 intermediate, 1 advanced, 6 2/3 pts each - Time: 50 minutes

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Presentation Transcript
Review for Test #1
• Responsible for: - Chapters 1, 2, 3, and 4 - Notes from class - Problems worked in class - Homework assignments
• Test format: - 15 problems: 10 simple, 4 intermediate, 1 advanced, 6 2/3 pts each - Time: 50 minutes
• Test materials: - Pencil, eraser, paper, and calculator - No formulae sheet - Closed textbook and notes
Material Covered
• Chapter 1: Introduction - Units, significant figures, dimensions - Order-of-magnitude estimates
• Chapters 2 and 4: 1D and 2D Kinematics - Displacement, velocity and speed, acceleration - Equations of kinematics -> solve problems - Horizontal and free-fall (1D motion) - Projectile motion (2D)
• Chapter 3: Scalars and Vectors - Components of a vector, unit vectors - Vector addition/subtraction - Resultant vector magnitude and direction - Relative velocity

Scores returned by next Wednesday

Example Problem (intermediate)

A ball is thrown straight upward and rises to a maximum height of 16 m above its launch point. At which height above its launch point has the speed of the ball decreased to one-half of its initial value?

Solution:

Given: ymax = 16 m

Infer: vmax = 0, y0 = 0

Find: y1 when v1 = v0/2

Also, need v0

y

ymax, vmax

y1, v1

y0, v0

To maximum height:

v2max = v20 –2g(ymax-y0)

Solve for v0

v20 = v2max +2g(ymax-y0) = 2gymax

To intermediate point:

v21 = v20 –2g(y1-y0)

Solve for y1

y1 = (v20 – v21)/(2g) = [v20 – (v0/2)2]/(2g)

= v02(1-1/4)/(2g) = v02 (3/4)/(2g) = 3v02/(8g)

= 3(2gymax)/(8g) = 3ymax/4 = 3(16m)/4 = 12 m