Phys1D03ExamSupplementSolutionsDec2011REVISED

Phys1D03ExamSupplementSolutionsDec2011REVISED - Prep 101(C...

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Unformatted text preview: Prep 101 (C) Phys 1D03 Exam Solutions Dec 2003 Part A DEC 2003 PART A Q1 C) The position of the first y 1 = -4.9 t 2 The position of the second, dropped a bit later, time t bit y 2 = -4.9 (t-t bit ) 2 The distance between them y 2 – y 1 = -4.9 (t 2 – 2t t bit + t bit 2 ) – (-4.9t 2 ) = 8.9t t bit – 4.9t bit 2 As time increases, the distance between them will increase. Q2 A) At first θ is dropping, so ω negative, but the slope is getting less negative, so α is positive Then θ is increasing, with ω positive, and the slope is increasing, so α is positive. Q3 C) Diagram 1, F p from the left First look at the whole system Σ F = ma F p = 3Ma a = F p /3M Now look at M Σ F = ma F large on small = Ma F l = MF P /3M = F p /3 Diagram 2, F p from the right Look at the whole thing Σ F = ma - F p = 3Ma a = -F p /3M Look at M Σ F = ma - F p + F l = M(-F p /3M) F l = F p- F p /3 F l = (2/3) F p The force is greater in diagram 2 Prep 101 (C) Phys 1D03 Exam Solutions Dec 2003 Part A Q4 C) Constant speed means acceleration is zero Σ F x = 0 100 cos 37 – F s = 0 79.86 – μ s N = 0 Get N from y Σ F y = 0 N – W + 100sin37 = 0 N = 200 – 100sin37 = 139.82 79.86 – μ s 139.82 = 0 μ s = 0.57 The closest is 0.6, C) Q5 C) You get the angular acceleration from τ = I α α = τ /I The torque is due to the weight, with τ = mg(L/2)sin θ The moment of inertia for a rod about one end is I = ML 2 /3 So we have α = mg(L/2)sin θ / (mL 2 /3) = (3/2) gsin θ /L and is therefore lower for the rod with greater L Q6 B) Equilibrium. Look at the forces Σ F x =0 -T s cos60 + T l cos30 = 0 T l = 0.58 T s Then the y component Σ F y = 0 - mg + T s sin60 + T l sin30 = 0 - mg + T s 0.87 + (0.58T s )0.5 = 0 1.16T s = mg T s = 0.86 mg The minor difference is likely due to rounding. Prep 101 (C) Phys 1D03 Exam Solutions Dec 2003 Part A Q7 A) Take the torques around the pivot, set the mass of the block to be M Σ τ = 0 +Mg(L/3) – mg(L/2 – L/3) = 0 M(L/3) – m(L/6) = 0 M = m/2 Q8 D) Power = Fv = τ ω In a case like this, we have a clear maximum of τ at around 2400 rpm, but a region where it’s almost as big for clearly larger ω , so the maximum is between 2400 and 3600. Q9 A) He walks in a closed path, and the force is always in the same direction – anywhere it does work it will be cancelled out. You can do the math W = F.d = 100(100)cos0 + 100(100)cos90 + 100(100)cos180+100(100)cos270 = 10,000 + 0 – 10,000 + 0 = 0 Q10 E) Need to work out the value first KE = ½ mv 2 = ½ 1.85 (1.8) 2 = 3.0 Then the uncertainty is Δ KE/KE = ( Δ m/m) + ( Δ v 2 /v 2 ) We find Δ v 2 /v 2 = 2 Δ v/v = 2 (0.1/1.8) = 0.11 Then Δ KE/3 = (0.09/1.85) + (0.11) Δ KE = 0.48 round to 1 decimal place (the least accurate number we were given) = 0.4 J Prep 101 (C) Phys 1D03 Exam Solutions Dec 2003 Part A Q11 B) Change in height and velocity is conservation of energy, then the collision is conservation of momentum, then rising to a height is conservation of energy again....
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Phys1D03ExamSupplementSolutionsDec2011REVISED - Prep 101(C...

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