PHY2053
Summer 2013
Final Exam Solutions
1. We do deal with vectors. We deal with their components
  
A B  C
  
BCA
Taking the x-components
B x  C x  Ax  C cos 90  A cos 40  0  (50 m) cos 40  38.3 m
The y-components
B x  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  2 B x  A cos 40  2 B x  (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  1500m
For the second car,
1
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(1500m)
 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
 ma r
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
 ma t
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

g

L cos 30
9.8 m/s 2
 1.9 rad/s
(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. riction does the nonconservative work. Recall W =
F x cos  = f x cos 180º = −f x
( K f  K i )  0   fx
0  12 mv i   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 (1500kg )(6 m/s)  (1000kg )(5 m/s)

 5.6 m/s
m1  m2
1500kg  1000kg
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/s2 ) 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
 ma y
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 MR 2  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  mR 2 )

mgR
(2 kg )(9.8 m/s 2 )(0.20 m)

 5.76 rad/s 2
I  mR 2 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
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 )  (1000kg/m 3 )(9.8 m/s2 )(10 m)  9.8 104 Pa
10. The period of a simple harmonic oscillator is given by
T  2
m
k
Forming a ratio,
T2

T1
m2
k 
m1
2
k
2
6
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
Formula
y  A cos((40 rad/s)t  (20 rad/m) x)
y  A cos((30 rad/s)t  (20 rad/m) x)
y  A cos((20 rad/s)t  (20 rad/m) x)
y  A cos((40 rad/s)t  (30 rad/m) x)
y  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
v = /k (m/s)
2.0
1.5
1.0
1.3
0.67
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)
For a string resonating at it fundamental frequency,  = 2L = 2m. The speed of the wave in the
string is
v  f  (2 m)(170Hz)  340m/s
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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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