Unformatted text preview: Massachusetts Institute of Technology  Physics Department Exam #3
SOLUTIONS Physics  8.01 Fall 1999 Problem 1 35 points
8 pts a) The diagrams: Nx
Ny N N2 =m2g
T2
mg
2 9 pts
9 pts 9 pts T1 T2
m2
2 T1
g m1g The contact force N on the pulley must have components Nx = T2 and Ny = m2 g + T1
2
for the pulley to stay in place.
g
b) T2 = m2a = m2 2
a
c) R(T1 ; T2 ) = I
I = 1 m2 R2 = m24R2
=R
22
g
g
a
R(T1 ; m2 2 ) = I R = 2IR = m28gR
) T1 = m82 g + m22 g = 5 m2g
8
d) m1 g ; T1 = m1a = m1 g
2
g
m1 2 = T1 = 5 m2 g
8
5
) m1 = 4 m2 1 Problem 2 30 points
6 pts a) Since m2 m1 , the bullet will leave with the same relative speed with which it came in.
V1 = V1
b) The speed of the bullet after the collision is now V1 + V2 relative to m2 . Therefore, the
speed of m1 is V1 = V1 + 2V2
11
c) Momentum conservation gives the speed of both masses after the collision, V = mm+Vm2 .
1
0 8 pts 0 8 pts 0 This kinetic energy brings the system to a height given by
1
2
2 (m12+2m2 )V = (m1 + m2 )gh
1 m1 V1 = g L(1 ; cos
max)
2 (m1 +m2 )2
m1+m2 p2gL(1 ; cos max) Notice for max = 0, V1 = 0 as it should!
) V1 = m1
+
d) If max = 90 , then cos max = 0 and V1 = m1m1m2 p2gL, which is possible. For example,
m2
if L = 1 m and m1 102 , then V1 450 m=sec.
0 8 pts � 2 Problem 3 35 points 7 pts a) p = jrp F j = ;bM g sin
~
b) Ip = Ic + M b2 = 1 M R2 + M b2
2
P
c) p = Ip
1
;bM g sin = ( 2 M R2 + M b2 )
) + 1 Rbg+b2 sin = 0
2
2
d) Under the small angle approximation, sin 7 pts e) There must be a force at P or the CM would accelerate straight downwards. 7 pts
7 pts
7 pts , and the equation of motion is given
frequency given by by + 1 Rb2g+b2 = 0 which is simple harmonic with an angular
r
r2
1 R2 +b2
bg . The period is given by T = 2 = 2
2
! = 1 R2 +b2
!
bg .
2 3 ...
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 Fall '09
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 Physics, Force, Mass

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