Physics C – Mechanics 1991
Physics C – Mechanics 1991
1. Notice: No friction and also the inelastic collision...
(a) Use conservation of momentum:
p0 = pf
mv0 + 0 = 3mvf [pic]
(b) Ki = [pic] , here we use conservation of E (mechanical energy).
Ei = Ef
[pic] = K + (3m)gh or K + (3m)gr [pic]
(c) To complete the loop we need to have a minimum or critical speed, [pic] of...
This occurs when the Normal force of the track at the top is zero. Then only gravity will make up the centripetal force. Now back to Fnet = ma ...
[pic] Now we're asked to find is the minimum speed of the bullet which will get the block-bullet combo to that speed at the top!
Back to Conservation of Energy:
Ei = Ef (starting after the collision and ending at the top of the loop)
Recall from our work above...
[pic] = 3mg(2r) + ½ 3m[pic] = 3mg(2r) + ½ 3m(rg) = [pic] Now cancel out the m's!
[pic]
2. (a) Rotational Equilibrium requires: [pic]
r1m1g = r2m2g
½ (200) = 1.5m2(10)
[pic]
(b) Now for some 'F=ma' action! [pic]
r1(200) = 45[pic]
[pic]
(c) Note the Free-Body Diagram at the right.
200 – T = m1a Here we use [pic] to get a = .5(20/9) = 10/9 m/s2
200 – T = 20 (10/9) [pic]
(d) from rest, v0 = 0 m/s (back to constant acceleration kinematics!)
[pic]
Physics C – Mechanics 1991 p.2
3. (a) Looking at the small mass (m) in equilibrium: Fs = mg or kx = mg
With displacement, D, we get: kD = mg [pic]
(b) From Block II's point of view, maximum compression occurs when Block I comes to rest (w.r.t. Block II). From the Earth's point of view, this means both blocks have the same velocity.
Or... In terms of energy, maximum compression occurs when the relative velocity of the one mass w.r.t. the other is 0. Hence K.E. = 0J, hence we have a maximum U.
(c) Maximum compression occurs when Block I and Block II move together (as in an inelastic collision), so we can use conservation of momentum:
p0 = pf
mv0 + 2m(0) = 3m vf [pic]vf = [pic] (at the instant of maximum compression)
Hmm, wait... This isn't what they're asking for. They want the maximum compression, x?
Well, with no energy loss we can use conservation of mechanical energy (E):
E0 = Ef
[pic] Now cancel the ½ 's and substitute for k = mg/D
[pic] [pic]
(d) Now consider the elastic collision as they separate...
Using the Elastic Head-on Collision Thm (The relative velocity before the collision is the opposite or negative of the relative velocity after the collision.)
v0 – 0 = - (v1f – v2f) and also use... p0 = pf
v0 = - v1f + v2f mv0 = mv1f + 2mv2f
v0 = v1f + 2v2f
v0 = -v1f + v2f
[pic] [pic]
-----------------------
T
200 = m1g
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