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Unformatted text preview: Solutions Prelim II Fall 2006 I IIIIII
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IIIEEIIIEMIIIEEIII _ IllII “*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 l010l2 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.
 Spring '07
 LECLAIR,A
 mechanics

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