A Piece of Cake Will cover all physics topics since August 10, 2016. The Physics Final Examination for Semester 1, 2016-2017 Will cover all physics topics since August 10, 2016. Work in your current teams (with some exceptions due to exempted seniors). Every team member needs to use a different color pen. No pencils. Formulae will be provided. Bring your calculator. Use books, etextbook, Sketchbooks, physics related websites/videos. Cell phones may be used strictly for physics only. No social media. You will need the full 2 hours. Arrive on time, collaborate intensely. Focus on linking physics concepts together, not fragments. Your solution must take the form of a narrative, supported by calculations and sketches. Your narrative should reveal sound reasoning. Solutions without narratives earn 0 points, regardless of the quality of sketches and calculations. See illustration on Red Panda. Your score will be based on your team's work and your individual work.

Work, Mechanical Energy, Collisions Work done by net force = Fd cos θ, KE = (1/2)MV2 Worknet = change in Kinetic Energy = ∆KE = (1/2)M(Vf2 - Vi2) Worknet = work done by Kinetic Friction = Fkdcosθ = µkmgdcosθ PEg = mgh ; gravitational PE = mass  free-fall acceleration  height PEelastic = (1/2) k x2 ; elastic PE = (1/2)  spring constant  (distance compressed or stretched)2 W ext + PEg + PEelastic + KE = PEg + PEelastic + KE + Fkdcosθ Fk = friction force Power = Work / ∆t F ∆t = mvf – mvi ; force  time interval = Change in Momentum MVi (object A) + MVi (object B) = MVf (object A) + MVf (object B) ; Conserv of Momentum Elastic Collisions: Show that both Momentum and Kinetic Energy are Conserved Perfectly Inelastic Collisions: Show that Momentum is Conserved, but Kinetic Energy is Not

Kinematics v = Δx / Δt Δx = xf - xi a = Δv / Δt Δv = vf - vi x = Vi t + 1/2 (a) t2 Vf2 = Vi2 + 2 (a) x Newton’s 1st Law, Net External Force, Mass v. Inertia Newton’s 2nd Law, Force, F = mass x acceleration Newton’s 3rd Law, Action and Reaction Mass v. weight, the Normal Force, Force of Friction (static v kinetic), coefficient of friction Net External Force on Free Body Diagrams Action – Reaction pairs in different situations

Circular Motion F = ma Fcentripetal = m acentripetal = m v2tangential / r Fcentripetal can be due to gravity, string tension, weight, friction, normal force Rotational Motion s = θ r θ, ω, α are in rads. There are 2 π rads per revolution. v = ω r a = α r Ƭ = Torque = F d (perpendicular) Ƭ = Torque = Force x Lever Arm) Ƭ = Torque = I α (Moment of Inertia x Angular Accleration) KE (translational) is 1/2 (m) V2 KE (rotational) is 1/2 (I ) ω 2 I = moment of inertia, depends on object's shape, M, R

Distance, Velocity, Acceleration v Time

F net = 58.8 – 24.5 N = 34.3 N toward right and up the ramp F gx = (5 kg) (9.8 m/s2) sin 30 = 24.5 N F bx = (6 kg) (9.8 m/s2) = 58.8 N F net = 58.8 – 24.5 N = 34.3 N toward right and up the ramp F = ma, so F net will impact the acceleration of both the 5 kg and the 6 kg blocks as a system. Thus, F net = (5 kg + 6 kg) a, where a is acceleration of both blocks acting as one. 34.3 N = (11 kg) a a = 34.3 N / 11 kg = 3.1 m/s2 Thus, the 5 kg block accelerates up the ramp and the 6 kg block accelerates down the cliff at 3.1 m/s2 F bx F gx

30 degrees KEi + PEi = KEf + PEf + Ffriction d Ffriction = μ FN Ffriction = μ mg cos 30 degrees 30 degrees 10 kg 4 kg Friction on ramp Frictionless ramp

velocity r m1 A Fc m2 Illustration of Well Written Solution (Your solution must take the form of a narrative, supported by calculations and sketches. Your narrative should reveal sound reasoning. Solutions without narratives earn 0 points, regardless of the quality of sketches and calculations.) Problem: What is the tangential velocity of moon A as it revolves around the planet? Moon A’s mass is 10,000 kg and the Planet’s mass is 100,000 kg. The distance between them is 20,000 m. Solution: Because moon A moves in a circular orbit, it must experience a centripetal force, Fc. Because the gravitational force, Fg, between moon A and the planet is the source of Fc, Fc = Fg. I can determine the specific magnitude of Fg by using this equation for gravitational force: Subsequently, I can determine the effect of this Fg on the motion of moon A by using this equation for centripetal force in which moon A is m: To solve for V, I set Fc = Fg and note that m is m1 and then solve for V: G m1m2 = mv2 G m2 = v2 r2 r r Inserting 100,000 kg for m2, 20,000 m for r and (6.7 x 10 -11) for G, I calculate V2 to be 3.4 x 10-10 and thus V = 1.8 x 10-5 m/s. A r Fc m2 m1 velocity