This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 (tt 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....
View
Full Document
 Spring '08
 N. MCKAY
 Force, Ω, θ, Ebefore

Click to edit the document details