Problem Solving Steps 1. Geometry & drawing: trajectory, vectors, coordinate axes free-body diagram, … 2. Data: a table of known and unknown quantities, including “implied data”. 3. Equations ( with reasoning comments ! ), their solution in algebraic form, and the final answers in algebraic form !!! 4. Numerical calculations and answers. 5. Check: dimensional, functional, scale, sign, … analysis of the answers and solution.

Exam Example 1 : Coin Toss V y =0 y 0 y0y0 v 0y yvyvy a y t 0+6m/s??-g=-9.8m/s 2 ? Questions: (a) How high does the coin go? (b) What is the total time the coin is in the air? Total time T= 2 t = 1.2 s (c) What is its velocity when it comes back at y=0 ? for y=0 and v y <0 yields v y 2 = v 0 2 → v y = -v 0 = - 6m/s (problem 2.85)

Exam Example 2: Accelerated Car (problems 2.7 and 2.17) Data: x(t)= αt+βt 2 +γt 3, α=6m/s, β=1m/s 2, γ = -2 m/s 3, t=1s Find: (a) average and instantaneous velocities; (b) average and instantaneous accelerations; (c) a moment of time t s when the car stops. Solution: (a) v(t)=dx/dt= α+2βt +3γt 2 ; v 0 =α; (b) a(t)=dv/dt= 2β +6γt; a 0 =2β; (c) v(t s )=0 → α+2βt s +3γt s 2 =0 x 0 V(t) t 0 tsts α a(t) 2β2β

Exam Example 3: Truck vs. Car (problem 2.34) x 0 Data: Truck v=+20 m/s Car v 0 =0, a c =+3.2 m/s 2 Questions: (a) x where car overtakes the truck; (b) velocity of the car V c at that x; (c) x(t) graphs for both vehicles; (d) v(t) graphs for both vehicles. Solution: truck’s position x=vt, car’s position x c =a c t 2 /2 (a) x=x c when vt=a c t 2 /2 → t=2v/a c → x=2v 2 /a c (b) v c =v 0 +a c t → v c =2v x t 0 truck car truck car t V(t) v 0 t=2v/a c v/a c v c =2v t=2v/a c

Exam Example 4: Free fall past window (problem 2.84) Data: Δt=0.42 s ↔ h=y 1 -y 2 =1.9 m, v 0y =0, a y = - g Find: (a) y 1 ; (b) v 1y ; (c) v 2y y 0 y1y1 y2y2 V 0y =0 V 1y V 2y ayay h 1 st solution: (b) Eq.(3) y 2 =y 1 +v 1y Δt – gΔt 2 /2 → v 1y = -h/Δt + gΔt/2 (a) Eq.(4) → v 1y 2 = -2gy 1 → y 1 = - v 1y 2 /2g = -h 2 /[2g(Δt)] 2 +h/2 – g(Δt) 2 /8 (c) Eq.(4) v 2y 2 = v 1y 2 +2gh = (h/Δt + gΔt/2) 2 2 nd solution: (a)Free fall time from Eq.(3): t 1 =(2|y 1 |/g) 1/2, t 2 =(2|y 2 |/g) 1/2 → Δt+t 1 =t 2 (b) Eq.(4) → (c) Eq.(4) →

Exam Example 5: Relative motion of free falling balls (problem 2.94) y 0 H Data: v 0 =1 m/s, H= 10 m, a y = - g Find: (a) Time of collision t; (b) Position of collision y; (c) What should be H in order v 1 (t)= Solution: (a) Relative velocity of the balls is v 0 for they have the same acceleration a y = –g → t = H/v 0 (b) Eq.(3) for 2 nd ball yields y = H – (1/2)gt 2 = H – gH 2 /(2v 0 2 ) (c) Eq.(1) for 1 st ball yields v 1 = v 0 – gt = v 0 – gH/v 0, hence, for v 1 =0 we find H = v 0 2 /g

Projectile Motion a x =0 → v x =v 0x =const a y = -g → v oy = v oy - gt x = x 0 + v ox t y = y o + v oy t – gt 2 /2 v 0x = v 0 cos α 0 v 0y = v 0 sin α 0 tan α = v y / v x Exam Example 6: Baseball Projectile Data: v 0 =22m/s, α 0 =40 o x0x0 y0y0 v 0x v 0y axax a y xyvxvx vyvy t 00 ? ?0-9.8m/s 2 ????? Find: (a) Maximum height h; (b) Time of flight T; (c) Horizontal range R; (d) Velocity when ball hits the ground Solution: v 0x =22m/s·cos 40 o =+17m/s; v 0y =22m/s·sin40 o =+14m/s (a)v y =0 → h = (v y 2 -v 0y 2 ) / (2a y )= - (14m/s) 2 / (- 2 · 9.8m/s 2 ) = +10 m (b)y = (v 0y +v y )t / 2 → t = 2y / v 0y = 2 · 10m / 14m/s = 1.45 s; T = 2t =2.9 s (c)R = x = v 0x T = 17 m/s · 2.9 s = + 49 m (d)v x = v 0x, v y = - v 0y (examples , problem 3.12)

Exam Example 7: Ferris Wheel (problem 3.29) Data: R=14 m, v 0 =3 m/s, a || =0.5 m/s 2 Find: (a) Centripetal acceleration (b) Total acceleration vector (c) Time of one revolution T Solution: (a) Magnitude: a c =a ┴ = v 2 / r Direction to center: (b) θ (c)

Exam Example 8: Relative motion of a projectile and a target (problem 3.56) Data: h=8.75 m, α=60 o, v p0 =15 m/s, v tx =-0.45 m/s 0 y x Find: (a) distance D to the target at the moment of shot, (b) time of flight t, (c) relative velocity at contact. Solution: relative velocity (c) Final relative velocity: (b) Time of flight (a) Initial distance