Exam info •  Exam is 7:30–10:00am on Tuesday, May 5 in Coors Event

Center. There is assigned seating. Please check D2L for your seat location. There will also be an ID check.

•  Enter at NW entrance (closest to engineering and physics) and grab lap board. Proceed down to floor level to get exam. Then take assigned seat.

•  All seats are in Section 7-13 (there will be another class on the other side of court).

•  We will be there starting around 7:10am.

•  Turn in just the bubble sheet when done.

Exam info •  There are 32 multiple choice questions on the exam. At least 6

of these will be from the 3 midterms (2 from each). There will also be questions that are similar but not identical to old questions. 8 questions are from the last three weeks (statics, simple harmonic motion, fluids).

•  Same rules as before: 1 handwritten formula sheet (front and back), calculator, #2 pencils and erasers, no cell phones.

•  Exam will have same information on the front page as in midterm #3.

More study advice •  Take the old exams (midterms and finals) using just your cheat

sheet and calculator. –  For questions you get wrong, think about what you were

thinking and how to avoid the error in the future •  Make sure you understand the CAPA, tutorial homework, and

clicker questions. •  Clicker questions since last midterm (including today) will be

added to D2L this afternoon.

Problem solving: Identify what type of problem it is: kinematics (linear or rotational), Newton’s 2nd law (linear F=ma or rotational ), conservation of momentum or angular momentum, conservation of energy (including rotational kinetic energy), simple harmonic motion, universal gravitation, fluids

τ = Iα

If you need or use time, then kinematics and/or Newton’s 2nd law are probably needed; otherwise, conservation laws are easier.

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Things to remember when solving problems

Collisions conserve momentum (as long as nothing is attached to the earth) but only conserve energy if they are elastic

For 2D problems deal with each dimension separately and combine at the end with Pythagorean theorem and trigonometry. Constant velocity means no acceleration

Impulse (change in momentum) is Force times Time while work (change in kinetic energy) is Force times Displacement

p=p0+ρgd is the hydrostatic pressure (for static liquids). For moving liquids need Bernoulli’s equation (and continuity equation).

Buoyant force is the weight of displaced fluid; for floating objects (with no other forces) it is also equal to the weight of the object.

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Final words •  Final clicker scores (after removing the 5 lowest days) will

be posted on D2L late today. •  Final tutorial scores should appear by Tuesday. •  We will be dropping the lowest CAPA, tutorial participation,

and tutorial homework grade. •  Please check all D2L grades and send email if there is a

problem (email Prof. Kinney for CAPA/tutorial/exams and Prof. Stenson for clicker).

•  I should be in my office (F317) from 3-4:30pm today and most of Monday if you have questions

•  I will also try to respond to emailed questions. •  Thank you all for being such a great class. •  Good luck on the final and have a great summer.

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Clicker question 1 An object’s position vs time graph is shown on the right. What best describes the car’s velocity vs time?

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E. None of these

A.  0 B.  5 m/s2 C.  9.8 m/s2 D.  11 m/s2

E.  None of the above

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Clicker question 2 A ball is thrown with an initial velocity of v0 = 10 m/s at an angle of θ = 60° with respect to the horizontal. At the highest point on its trajectory, the magnitude of its acceleration is…

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In free fall, the only acceleration is that of gravity: 9.8 m/s2 in the downward direction.

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Clicker question 3 Which diagram below correctly depicts the equation ?

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!v2 =!v1 +Δ

!v

Clicker question 4 Two projectiles were launched with the same speed but at different angles. Which projectile was in the air longer?

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A.  A B.  B C.  A and B were in the air the same time D.  Not enough information given

The time in the air is determined by the motion in the y direction.

The time to land is twice the time to reach the maximum height.

The time to reach the maximum height is the time to reach a vertical velocity of 0. Use with vy = 0 to get: or so the projectile with the largest initial vertical velocity will be in the air the longest.

vy = v0 y − gttpeak = v0 y / g ttotal = 2v0 y / g

A.  Earth car B.  Space car C.  Both cars have the same

acceleration D.  Not enough information

given

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Clicker question 5 Consider the following two situations: Situation I: A car on Earth rides over the top of a round hill, with radius of curvature = 100 m, at constant speed v = 35 mph. Situation II: A monorail car in intergalactic space (no gravity) moves along a round monorail, with radius of curvature = 100 m, at constant speed v = 35 mph. Which car experiences the larger acceleration?

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II

I

a = v2/r in both cases.

A.  A B.  B C.  Both bikes have the same total kinetic energy D.  Not enough information given

Clicker question 6 Two bicycles have the same mass, are traveling at the same speed, and have the same radius tires. In bike A, more mass is in the tires than the frame compared to bike B. Which bike has more total kinetic energy?

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Both bikes have the same translational kinetic energy (K = ½mv2) because they have the same mass and speed.

Both tires have the same angular velocity ω = v/r but since bike A tires are heavier, they have a higher moment of inertia (MR2) so the rotational kinetic energy (K = ½Iω2) is greater.

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Clicker question 7 Set frequency to BA A disk rolls without slipping down the left side of a valley-shaped container. The right side of the container is completely frictionless. Compared to its initial height on the left, the disk’s maximum height on the right is…

A.  lower B. higher C. the same

friction no friction

From top to bottom, gravitational potential energy is converted to rotational kinetic energy and translational kinetic energy.

Going up the other side, without friction, it is impossible for the disk to lose rotational kinetic energy so only the translational kinetic energy is converted to gravitational potential energy. So it will still be spinning when it reaches its maximum height.

mgh1 = K trans +Krot =mgh2 +Krot so h2 < h1

A. 

B. 

C. 

D.  E. 

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Clicker question 8 A pendulum of length L is swinging back and forth. At the point where θ is largest, what is the direction of the net force on the bob?

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When it is at the maximum θ, it is at rest and so there is no radial acceleration. The only possible acceleration is tangential acceleration, which is in the direction shown. This is also the direction of the net force (the string constrains motion in that direction).

L θ

A.  mg B.  mg sinθ C.  T sinθ D.  mg – T sinθ E.  Need coefficient of friction

Clicker question 9 A block is being pulled by a rope with tension T at an angle of θ from the horizontal. What is the magnitude of the normal force?

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Free body diagram:

m

Tθ

Newton’s 2nd law in x and y dimensions: Fx∑ = T cosθ −Ff =max

T

mg

n

Ff

Fy∑ = T sinθ + n−mg =may

In y direction acceleration is 0, can solve for: n =mg−T sinθ

A. The rubber ball B. The putty ball C. It makes no difference which ball D. Need more information

Clicker question 10 Set frequency to BA An unhappy student works out his aggression by attempting to knock down a large wooden bowling pin by throwing balls at it. The student has two balls of equal size and mass, one made of rubber and the other of putty. The rubber ball bounces back, while the ball of putty sticks to the pin. Which ball is most likely to topple the bowling pin?

Each ball starts with momentum of mv. The change in momentum of the ball after the collision is for the putty ball and for the rubber ball. So more momentum gets transferred to the pin with the rubber ball.

Δp =mvΔp = 2mv