PHY2053
Summer 2013
Final Exam Solutions
1. We do not deal with vectors. We deal with their components
  
A B  C
  
BCA
Taking the x-components
Bx  C x  Ax  C cos 90  A cos 40  0  (50 m) cos 40  38.3 m
The y-components
Bx  C x  Ax  C sin 90  A sin 40  (50 m) sin 90  (50 m) sin 40  17.9 m
The new resultant is



C   A  2B
The x-component
C x  Ax  2Bx  A cos 40  2Bx  (50 m) cos 40  2(38.3 m)  38.3 m
The y-component
C y  Ay  2B y  A sin 40  2B y  (50 m) sin 40  2(17.9 m)  67.9 m
The magnitude is
C   (C x ) 2  (C y ) 2  (38.3 m) 2  (67.9 m) 2  78.0 m
2. The distance covered by the first car is
x1  v1t1  12 a1 (t1 ) 2  (25 m/s )(60 s)  12 (0)(60 s) 2  1500 m
For the second car,
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x 2  v 2 t 2  12 a 2 (t 2 ) 2
 (0)t 2  12 a 2 (t 2 ) 2
2x 2

a2
t 2 
2(1500 m)
 32 s
3 m/s 2
3. The free body diagram
36.9o
36.9o
Use Newton’s second law along the inclined plane (x-axis)
F
x
 ma x
mg sin 36.9  ma
a  g sin 36.9  5.9 m/s 2
This answer is None of these.
4. Use the free body diagram and Newton’s second law.
T
30o
mg
For the radial component,
2
F
r
 mar
T sin 30  m 2 r
The radius from the pole is r = L sin30º. Substituting,
T sin 30  m 2 L sin 30
T  m 2 L
For the tangential component,
F
t
 mat
T cos 30  mg  0
T cos 30  mg
Substituting for T from the radial equation,
T cos 30  mg
(m 2 L) cos 30  mg

9.8 m/s 2
g

 1.9 rad/s
L cos 30
(3 m) cos 30
5. The free body diagram for the rock is
N
f
mg
Using Newton’s second law for the y-component,
F
y
 ma y
N  mg  0
N  mg
The frictional force is
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f  N  mg  (0.26)(0.5 kg)(9.8 m/s 2 )  1.27 N
Use the work-energy theorem.
K  U  Wnc
Since the parking lot is level U = 0. Friction does the nonconservative work. Recall W
= F x cos  = f x cos 180º = −f x
( K f  K i )  0   fx
0  12 mvi   fx
2
vi 
2 fx

m
2(1.27 N)(16 m)
 9.0 m/s
(0.5 kg)
6. Linear momentum is conserved in a collision.
pix  p fx
m1v1i  m2 v2i  m1v1 f  m1v2 f
After the collision the cars stick together. This means, v1f = v2f = vf.
m1v1i  m2v2i  m1v f  m1v f
m1v1i  m2v2i  (m1  m2 )v f
vf 
m1v1i  m2v2i (1500 kg)(6 m/s )  (1000 kg)(5 m/s )

 5.6 m/s
m1  m2
1500 kg  1000 kg
7. The ladder is equilibrium. Take torques about the foot of the ladder.
W
mg
N
65o
f
4
  0
 mg 23 L cos 65  WL sin 65  0
2mg cos 65 2(25 kg)(9.8 m/s 2 ) cos 65
W

 76 N
3 sin 65
3 sin 65
8. There are two systems to work with: the mass hanging from the rope and the rope
exerting a torque on the cylinder. The free body diagram for the hanging mass is
T
mg
Use Newton’s second law
F
y
 may
T  mg  m(a)
T  m( g  a )
The mass accelerates downward so, ay = −a. For the cylinder, use Newton’s second law
for rotation,
  I
The torque is due to the tension in the rope wrapped around the cylinder
  Tr  (mg  ma) R
The moment of inertia for a cylinder is
I  12 MR2  12 (30 kg)(0.20 m)2  0.60 kg  m2
Newton’s second law for rotation becomes
  I
(mg  ma) R  I
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The mass accelerates as it falls, causing the cylinder to spin faster. The relationship
between the acceleration and the angular acceleration is
a  R
Substituting into the rotational Newton’s second law equation,
(mg  ma) R  I
(mg  mR) R  I
mgR  mR 2  I
mgR  ( I  mR2 )

mgR
(2 kg)(9.8 m/s 2 )(0.20 m)

 5.76 rad/s 2
I  mR2 0.60 kg  m 2  (2 kg)(0.20 m) 2
The angular velocity is found
 f  i  t
 f  i  t  0  (5.76 rad/s 2 )(10 s)  58 rad/s
This answer is None of these.
9. Since the tube has constant area, the continuity equation implies that the speed of the
water is constant throughout the pipe. Using Bernoulli’s equation,
P1  gy1  12 v1  P2  gy2  12 v2
2
2
The kinetic energy terms cancel since v1 = v2.
P1  gy1  P2  gy2
P2  P1  gy1  gy2  g ( y1  y2 )  (1000 kg/m 3 )(9.8 m/s 2 )(10 m)  9.8 104 Pa
10. The period of a simple harmonic oscillator is given by
T  2
Forming a ratio,
6
m
k
m2
T2
k 

T1
m1
2
k
2
m2
m1
Here T1 = 5 s and T2 = 10 s. Substituting
10 s

5s
4
m2
m1
m2
m1
m2  4m1
Quadruple the mass.
11. The standard form for a traveling wave is
y  A cos(t  kx)
The speed of the wave is v = /k. For the choices available
y
y
y
y
y
Formula
A cos((40 rad/s )t  (20 rad/m) x)
A cos((30 rad/s )t  (20 rad/m) x)
A cos((20 rad/s )t  (20 rad/m) x)
A cos((40 rad/s )t  (30 rad/m) x)
A cos((20 rad/s )t  (30 rad/m) x)
(rad/s)
40
30
20
40
20
k (rad/m)
20
20
20
30
30
The combination that gives the largest speed is
y  A cos((40 rad/s )t  (20 rad/m) x)
12. The frequencies for a tube of length L open at both ends are given by
fn  n
v
2L
For a one meter long tube, the fundamental frequency is
f1  1
340 m/s
v

 170 Hz
2L
2(1 m)
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v = /k (m/s)
2.0
1.5
1.0
1.3
0.67
For a string resonating at it fundamental frequency,  = 2L = 2m. The speed of the wave
in the string is
v  f  (2 m)(170 Hz)  340 m/s
The speed of a wave on a string is related to the tension in the string and its linear
density,
F
v

Solving for the tension,
v2 
F

F  v 2  (4.5  10 4 kg/m )(340 m/s ) 2  52.0 N
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