Prelim2Fa06Solutions

Prelim2Fa06Solutions - Solutions Prelim II Fall 2006 I...

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Unformatted text preview: Solutions Prelim II Fall 2006 I III-III gal-IIHIIIIIII IIIHIIIEIIII IIEIIIIIEIIH EEIIIIIIIESEH III-EEIIIHEII .- II In HI .- IE ‘EIIIIIIIII III-I IIEIIIIIIIEEEIIIIII IIIEIIIIIEIIEIIIIII IIIINIIIHIIIIIIIIII III-HIEHII-IIIEIIII III-EEIIIEMIIIEEIII _ Ill-II “*9 (b) Upward is in the positive Y—direction. 0;}, = p} — a = 0.40 kg (4.0 — (—50» m/s 3 = 3.0 kg 01/33 (0) in” = —3.6 kg In/S (d) See figure. For first bounce, 1 Kbefm = E0.401cg (5.0m/s)2 = 5.0 J KW, = —0.40kg (4.0m/s)2 = 3 J (e) See figure. Between the first and second bounce, Um“ = am, = 3.6 6. (a) For both satellites, at distance 0" from Earth, the gravitational potential energy is UA 2 U3 = —Gfl‘%{&. They each have kinetic energy KA = KB = ém’uz. '0 is determined by the centripetal acceleration 3’; = 05—31, giving émvz = mME mME _ mME EA+EB=UA+UB+KA+KB=u2G +G *G T T T (b) The satellites collide with equal and opposite momenta, thus their total momentum is zero, and the kinetic energy of the combined mass is zero. mME T EA+EB=UA+UB=—2G (c) The wreckage falls to Earth on a radial path. 7. (a) As in anAelastic encounter with a solid wall, the small puck rebounds to the left with 17 = —’U{}‘i. (b) Because of its immense size, the huge puck has velocity '59 = Uni before and after the collision. Since the collision is elastic, the relative velocity of the two pucks be- fore the collision is equal and opposite to the relative velocity afterward. Before, 55mg” — fihuge = —v"g. Afterward, Wm” — “flange = +170, giving for the small puck 17’de = 2003', moving to the right. 5 8. (a) Gravitational force and spring force are conservative. (bl i- Wspring on b = 516652 = a 2 X 103 l0-10l2 J = m Positive, since the spring force has the same. direction as the displacement of the mass. ii. The mass moves a distance D = (L — (s —* d)) = 0.80 In inside the tube. The component of the gravitational force acting in the direction of the plane is opposite in direction to the displacement of the mass. W9”... 0,, i, = hmg sin an = —0.50 kg x 9.81 m/s2 sin 30‘J x 0.80 m = —1.96 J iii. WM“ 0,, b = —ka = —3 N x 0.80 m = —2.4 J Negative, since friction force is opposed in direction to motion of the mass. (0) = Wming on b + Wgrav on b + Wfrict on b = _ J _' 2.4 J = J v = fi2K/m = 4.75 m/s ((1) Less than 0.10 111 because of energy loss due to friction. 9. (a) Let x = 0 be at the pier. The coordinate of the CM of the raft, of Anna, and of Daniel are the same before he walks, $39M : 35;, = (ED = L/ 2. The CM of the system is also at L/ 2. After Daniel walked to the tip of the raft, he and the tip will be at a distance a: from the pier. Now 333 = r and 3,4 = $30M 2 L/2 + m. The location of the center of mass of the system is still 15/ 2. (15/2) : mDQ: + (mA +mR)(L/2+x) me + ma + my mg L = 60 kg 2(mp + mA + m3) 200 kg (‘0) Start with both back at the center. Anna walks towards the pier while Daniel walks at least 2 m to the far end of the raft. 3m=0.90m 3:: ...
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This note was uploaded on 03/27/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Spring '07 term at Cornell.

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Prelim2Fa06Solutions - Solutions Prelim II Fall 2006 I...

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