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AP Physics – Mechanics – Chapter 3 – Vectors and Motion in 2 dimensions Text chapter 3 - Reading pp. 51-68 textbook HW -- #2,3,5,6,8,9,12,18,20,17,19,22,42,44,45,48 3.1-3.2 Vectors and Scalars Revisited List the TEN vector quantities you learned last year: Displacement, velocity, acceleration, force, weight, impulse, momentum, Gravitational, electric, and magnetic field strengths I. Graphically, how are vectors combined? A. Parallelogram Method – Good for two concurrent vectors. B. Chain Method- Good for two or more vectors Vectors at right angles #1) A car travels 100 m south, then 30 m east, and finally 80m north. Find the resultant displacement. [36 m]

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3.3 Mathematically combining vectors at any angle by summation of x and y components Remember: Ax = AcosѲ Ay = AsinѲ #2) Find the resultant displacement [661 m, Ѳ=45.7°]

Now go home and try HW -- #2,3,5,6,8,9,12

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#3) A man walk from A-B-C-D as shown. He ends up at pt D (-40 m, -60 m). What is the distance from C to D? [205 m]

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#4) Find T1 and T2 Hmmmm… What little fact is REALLY important here? [T1=407 N, T2=307 N]

(Review Equilibrium)

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Using equations to analyze the relationship between variables #5)

ϴ1 ϴ2

a) If ϴ1=ϴ2, what is the relationship between T1 and T2?

b) If ϴ1=ϴ2, what is the relationship between T and ϴ?

c) If ϴ1>ϴ2, what is the relationship between T1 and T2?

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#6) Two charged spheres, each with mass m and charge q repel each other with force Fe . a. Write two equations, one representing the net horizontal force, and the other representing the net vertical force, acting on the sphere on the left in the diagram below.

∑Fx = Tcos60°-Fe ∑Fy = Tsin60°-W b. Derive an equation for Fe in terms of W. Fe =Wcot60° Funny, though, the MC answer is Fe =Wtan30°… why?

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AP Physics – Mechanics – Chapter 3 – Motion in 2 dimensions - a.k.a. Projectile Motion 3.5 Projectile Motion: The motion in the y-direction is INDEPENDENT of the motion in the x-direction. Thus simultaneous calculations must be done vertically and horizontally. I. Draw vectors representing vx, vy, ax, and ay at each point below. Be sure the magnitude of each vector is relative to all others.

Equations: y-component: think x-component: think constant acceleration of -g constant velocity of a=0 1) vy = v0y - gt

vy = (v0sinѲ) - gt 2) y = v0yt – ½ gt2

y = (v0sinѲ)t – ½ gt2 3) vy

2 = v0y2 - 2gy

vy2 = (v0sinѲ)2 - 2gy

1) vx = v0x + at vx = (v0cosѲ) + at

2) x = v0xt + ½ at2 x = (v0cosѲ)t + ½ at2

3) vx2 = v0x

2 + 2ax vy

2 = (v0cosѲ)2 + 2ax

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#7) A daredevil is fired from a cannon 30o to the horizontal with an initial velocity of 40 m/s. A) Find v0x and v0y [34.6 m/s, 20 m/s] B) Find the time to reach the maximum Height [2 s]

C) Find the maximum height D) Find the total time the daredevil is

that the daredevil rises. [20 m] in the air. [4 s] E) Find the horizontal distance F) Find the velocity at time t = 3.6 s that the daredevil travels. [138 m/s] [38 m/s]

Now go home and try HW -- #18,20,17,19

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#8) A potato is fired from the roof of the high school, which is 35 m above the level ground. He fires it at 36o above the horizontal at a rate of 55 m/s. See note below.

A) Determine the time the potato is in the air. [7.4 s] B) Determine the potato’s speed as it hits the level ground. [61 m/s] C) At what height above the ground is the speed 63 m/s?

D) At what height above the ground is the speed 59 m/s? [12.1 m]

Note: After the Anthony Cascione - EHS class of 2011 incident, this demo is no longer done, unless… someone builds a compressor-based potato launcher.

Now go home and try HW – #22, 42, 44, 45

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#9) A Projectile is fired at an initial velocity v0 and at an angle θ to the horizontal. a) Determine the range R in terms of θ, v0 , and g.

You may need the trig identity sin2θ = 2sinθcosθ

b) Determine the height h in terms of θ, v0 , and g.

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c) Use the equation from part b to support the statement: maximum horizontal distance occurs at an angle of 45⁰.

R is maximized when: sin2θ = 1 thus: 2 θ = sin-1(1) 2 θ = 90⁰ θ = 45⁰

** Hand out – Extra Credit problem (for baseball fans)

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Horizontal launches: #10) A block of mass M is pushed off a table of height h with velocity vo. For parts a-d, express all in terms of vo, h, g, and M.

a) Find the time t to hit the ground.

b) Find the horizontal distance x.

c) Find the speed when it hits the ground. d) Find the height above the ground in which the speed is 2vo

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#11 (HW Prob.) a) A plane, carrying a relief package to drop to the survivors of a small nuclear accident at the US Navy Nuclear Power School in Charleston, SC, is moving horizontally at 68 m/s and is at an altitude of 250 m above the small drop zone. At what horizontal distance

does the plane have to release the package so that it hits the drop zone? [486 m] 485.7 m

b) If the operator of the plane from part a releases the package too late, and releases it at a horizontal distance of only 125 m from the drop zone, what initial vertical velocity would the package have to have in order to still hit the target? Note: vertical thrusters are standard issue on all relief packages. [126 m/s down]

-131.5 m/s Final Extra Credit Problem: #48 Road Runner Wile E. Coyote text problem.