test3_09Sol - LAST NAME STUDENT NR PHYS 1010 6.0 CLASS TEST...

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Unformatted text preview: LAST NAME: STUDENT NR: PHYS 1010 6.0: CLASS TEST 3 Time: 50 minutes; Calculators & formulae provided at the end = only aid; Total = 20 points. 1) [5] Alice (m = 50 kg) is on a skateboard (M = 2 kg) moving with v = 5 m/s on a flat stretch of road. She jumps off backwards such that the board is propelled forward to a speed of 8 m/s. What is her velocity with respect to the ground as she lands? Ignore friction and drag, and note that magnitude and direction are required (the board is moving in the +x direction). 2) [5] A squash ball (m = 100 g) flies very fast horizontally towards a wall (vx = 40 m/s), and makes a perfect rebound (no mechanical energy loss). Use the impulse-momentum theorem to calculate the impulse provided by the wall in SI units (be careful with the sign). Then sketch a possible Fx (t) curve together with an average force estimate assuming the collision took ∆t = 10 ms. 1 ... 3) [5] A glider (m = 200 g) is held at the top end of a tilted air track (angle with horizontal α = 30◦ ) of length L = 2m. At the other end is a short spring with constant k = 100 N/m. By how much will this spring be maximally compressed when the glider is released and reaches the bottom? Ignore gravity during the spring compression. Start with a drawing of the situation and state the laws used to solve the problem. 2 4) [5] A simple yo-yo is made from a (thin) disk of radius R and mass M (ICM = 0.5M R2 ) with a thread wound around it. The downward motion can be understood as a superposition of a fall of the CM and a rotation of the disk about the point where the string connects with the disk (IO = 1.5M R2 ). Formulate the laws of motion for the fall (don’t forget to specify the net force on the CM and the torque about the pivot point O). Then, calculate the acceleration of the CM, and the tension in the string. 3 FORMULA SHEET ￿ v (tf ) = v (ti ) + ti a(t) dt s(tf ) = s(ti ) + ttif v (t) dt vf = vi + a∆t sf = si + vi ∆t + 1 a∆t2 vf2 = vi2 + 2a∆s 2 ￿ tf g = 9.8 m/s2 f (t) = t f (t) = a = 0 F (t) = f (t) dt = at + C F (t) = anti-derivative = indefinite integral area under the curve f (t) between limits t1 and t2 : F (t2 ) − F (t1 ) x2 + px + q = 0 factored by: x1,2 = − p ± p4 − q 2 uniform circular m. ￿(t) = R(cos ω t ˆ + sin ω t ˆ); ￿ (t) = r i jv exp￿ = exp; sin￿ = cos; cos￿ = − sin. m￿ = Fnet ; a￿ FG = Gm1 m2 ; r2 d [f (g (x))] dx df dt df dt = 1 F (t) = ￿ ￿ f (t) dt = t2 2 +C ￿ 2 = d￿ r dt df dg ; dg dx = ...; ￿ (t) = a d￿ v dt = .... (f g )￿ = f ￿ g + f g ￿ 2 g= GME 2; RE ∆p 1 + ∆ p 2 = 0 ; ￿ ￿ + = + for elastic collisions. ￿ CM = a ￿ 2 ˆ ￿ = ￿ × F ; τz = rF sin(α) for ￿, F in xy plane. I = i mi ri ; I αz = τz ; (k = rot. axis) τr￿ r￿ ￿r￿ ￿τ Krot = I ω 2 ; Lz = I ωz ; d Lz = τz ; L = ￿ × p; d L = ￿ 2 dt dt fs ≤ µs n; fk = µk n; fr = µr n; µr << µk < µs . FH = −k ∆x = −k (x − x0 ). ￿ Fd ∼ −￿ ; linear: Fd = dv ; quadratic: Fd = 0.5ρAv 2 ; A = cross sectional area v W = F ∆x = F (∆r) cos θ For F (x) the work is given as area under the Fx vs x curve. ￿ ￿ ∆p = J = F (t)dt; ∆px = Jx = area under Fx (t) = F avg ∆t ; p = m￿ ; K = m v 2 ￿￿ ￿ v x in K1 in K2 fin K1 fin K2 2 m1￿ 1 +m2￿ 2 a a m1 +m2 RE = 6370 km; G = 6.67 × 10−11 Nm2 ; ME = 6.0 × 1024 kg kg 4 ...
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This note was uploaded on 09/28/2010 for the course MATH 1025 taught by Professor Tammie during the Fall '10 term at York University.

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