PHY2053
Summer C 2013
Exam 1 Solutions
1. The force is
F
Gm1m2
r2
Forming a ratio
Gm1m2
2

F
 r
F Gm1m2
r2
r 2 mm
 2 1 2
r  m1m2
The only combination that gives F'/F = 2 is doubling m1.
2. The speed of light is
c  186,282
miles 8 furlongs 60 s 60 min 24 hr 14 days





 1.8 1012 furlongs / fortnight
s
mile
1min
1hr
1day fortnight
3. The distance traveled for the entire trip is
x
t
x  vav , x t
vav , x 
 1 hr 
 mi 

  36 (75 min)
 hr 
 60 min 
 45 mi
The distance covered during the first part is
x
t
x1  vav,x t
vav,x 
 1 hr 
 mi 

  30 (30 min)
 hr 
 60 min 
1
x1  15 mi
The distance for the second part is
x  x1  x2
x2  x  x1
 45 mi  15 mi
 30 mi
The duration of the second part is
t  t1  t2
t2  t  t1
 75 min  30 min
 45 min
The average velocity for the second part is
x
t
 30 mi  60 min 


 
 45 min  1 hr 
vav , x 
 40 mph
4. For the first 10 s, the distance is 120 m. This gives,
x  vix t  12 a x (t ) 2
x1  vix t1  12 a x (t1 ) 2
120 m  vix (10 s)  12 a x (10 s) 2
12
m
 vix  (5 s)a x
s
Two seconds later,
x2  x1  48 m  120 m  48 m  168m
t 2  t1  2 s  10 s  2 s  12 s
x2  vix t2  12 ax (t2 ) 2
168m  vix (12 s)  12 ax (12 s) 2
2
14
m
 vix  (6 s)ax
s
There are two equations and two unknowns,
m
 vix  (5 s)a x
s
m
14  vix  (6 s)a x
s
12
Solve the first equation for ax,
12
m
 vix  (5 s)ax
s
m v
ax  2.4 2  ix
s 5s
Substitute into the second equation,
m
 vix  (6 s)a x
s

m v 
 vix  (6 s) 2.4 2  ix 
s 5s 

m
 vix  14.4  1.2vix
s
m
m
14  14.4  0.2vix
s
s
m
 0.4  0.2vix
s
m
vix  2.0
s
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5. The acceleration is
x  vix t  12 a x (t ) 2
x  0  12 ax (t ) 2
ax 
2x 2(100 m)

 2 m/s2
2
2
(t )
(10 s)
The total distance is x = 100 m + 300 m = 400 m. The final velocity is
3
v fx  vix  2a x x
2
2
v fx  vix  2a x x
2
2
 0  2a x x
v fx  2(2 m/s 2 )(400 m)  40 m/s
6. The time it takes for the first stone to hit the water is
y  viy t  12 a y (t ) 2
y  0  12 a y (t ) 2
2y

ay
t 
2(19.6 m)
 2.0 s
 9.8 m / s 2
The second stone must hit the water in t = 2.0 s – 1.0 s = 1.0 s.
y  vix t  12 a y (t ) 2
viy t  y  12 a y (t ) 2
y 1
 2 a y t
t
 19.6 m 1

 2 (9.8 m/s2 )(1s)
1 .0 s
 14.7 m/s
viy 
7. The run looks like,
B
45o
C
A
To add vectors we take components. By inspection for A
4
Ax  0
Ay  20 yards
For B,
Bx   B sin 45  (30 yards) sin 45  21.2 yards
By  B cos 45  (30 yards) sin 45  21.2 yards
The components of C
Cx  Ax  Bx  0  (21.2 yards)  21.2 yards
C y  Ay  By  20.0 yards  21.2 yards  41.2 yards
The magnitude of C
C  C x  C y  (21.2 yards) 2  (41.2 yards) 2  46.3 yards
2
2
8. The picture looks like

A
B
o
35
C
The vector relation is
  
C  AB
  
B  CA
The components of A and C are
Ax  A cos35  (30 N) cos35  24.6 N
Ay  A sin 35  (30 N) sin 35  17.2 N
Cx  C cos 0  (75 N) cos 0  75 N
C y  C sin 0  (75 N) sin 0  0
The components of B are
5
Bx  Cx  Ax  75 N  24.6 N  50.4 N
By  C y  Ay  0  17.2 N  17.2 N
The magnitude of B is
B  Bx  By  (50.4 N) 2  (17.2 N) 2  53 N
2
2
The angle is
  arctan
By
Bx
 arctan
 17.2 N
 19
50.4 N
9. The figure is
vRS
vWS
vRW

The velocity of the rowboat relative to the shore is
vRS 
x
250 m 1min


 0.99 m/s
t 4.2 min 60 s
The vector equation is



v RS  v RW  vWS
From the diagram
  arctan
vWS
 0.61m/s
 arctan
 32
vRS
 0.99 m/s
10. The time it takes for the ball to hit the ground is
y  viy t  12 a y (t ) 2
 0  12 a y (t ) 2
t 
2y

ay
2(78.4 m)
 4.0 s
 9.8 m/s 2
6
The ball falls 80 m from the bottom of the building
x  vix t  12 ax (t ) 2
 vi t  0
vi 
x 80 m

 20 m/s
t
4s
Since the ball is thrown horizontally, the ball is in the air for the same amount of time
regardless of the magnitude of the throw. So again t = 4.0 s. The new range is 160 m,
x  vix t
vi 
x 160 m

 40 m/s
t
4s
11. The range equation is
v sin 2
R i
g
2
The time of flight equation is
t F 
2vi sin
g
Solve the time of flight equation for vi and substitute into the range equation,
2vi sin
g
gt F
vi 
2 sin
t F 
The range equation becomes,
vi sin 2
g
2
R
 gt F  2 sin  cos


g
 2 sin  
g 2 (t F ) 2 2 sin  cos

4 sin 2 
g
2
7
g (t F ) 2 cos
R
2
sin 
2
sin  g (t F )

cos
2R
g (t F ) 2
tan  
2R
 9.8 m/s2 (5.85 s) 2 
 g (t F ) 2 
  35
  arctan
  arctan
2(240 m)
 2R 


Use the time of flight equation to find vi
t F 
vi 
2vi sin
g
gt F
(9.8 m/s2 )(5.85 s)

 50 m/s
2 sin
2 sin 35
Since the field is level, the ball hits the ground with the same velocity.
Easier solution: The golf ball travels 240 m in the x-direction. It takes 5.85 s to travel
that distance. The x-component of the velocity is constant for a projectile and is
vx 
x 240 m

 41.0 m/s
t 5.85 s
The ball reaches its highest point at th = tf/2 = (5.85 s)/2 =2.93 s. The y-component of
the initial velocity is
v fy  viy  a y t
0  viy   gth
viy  gth  (9.8 m/s 2 )(2.93s)  28.7 m/s
The initial velocity is
vi  vix  viy  (41.0 m/s) 2  (28.7 m/s) 2  50.0 m/s
2
2
The final velocity equals the initial velocity since the motion is symmetric.
12. The weight is the Earth pulling down on the block. The interaction partner is the block
pulling up on the Earth.
8
13. The free body diagram is
F
mg
Using Newton’s second law,
F
y
 ma y
F  mg  ma y
F  mg 49000N  (1000kg )(9.8 m/s2 )
ay 

 39 m/s2
m
1000kg
14. The free body diagram is
N
Fext

f
mg
Using Newton’s second law
F
x
 ma x
 f  Fext cos  0
f  Fext cos
F
y
 ma y
N  Fext sin   mg  0
N  mg  Fext sin 
9
The coefficient of friction is defined in
f  N
Substituting for f and N,
f  N
Fext cos   (mg  Fext sin  )
 mg  Fext sin 
Fext cos  Fext sin   mg
Fext 
mg
cos   sin 
(0.3)(4 kg )(9.8 m/s2 )
cos 25  (0.3) sin 25
 11.4 N

15. The free body diagram is
x
N
mg
o
30
30o
Using Newton’s second law,
F
x
 ma x
mg sin  ma x
ax  g sin  (9.8 m / s 2 ) sin 30  4,9 m / s 2
Since the acceleration is independent of the mass, so are all the other kinematic variables.
The final velocity is still 10 m/s when the mass of the roller coaster is doubled.
16. The free body diagram for mA is
10
T
N
37o
37o
mAg
Using Newton’s second law,
F
x
 mA a x
T  mA g sin  mAa Ax
For mB,
T
mBg
Again using Newton’s second law,
F
y
 mB aBy
T  mB g  mB aBy
Since blocks A and B are connected they move together. Suppose we call their
acceleration a. If A moves up the incline, B will descend. This gives,
aAx  a
and
aBy  a
The two equations from Newton’s second law become
T  mA g sin  mAa Ax
T  mA g sin  mAa
11
and
T  mB g  mB aBy
T  mB g  mB (a)
mB g  T  mB a
We have two equations with two unknowns (T and a)
T  mA g sin   mAa
mB g  T  mB a
Adding the two equations cancels the T,
mB g  mA g sin   mAa  mB a
(mB  mA sin  ) g  (mA  mB )a
a
(mB  mA sin  ) g (5 kg  10 kg sin 37)(9.8 m/s2 )

 0.67 m/s2
(mA  mB )
(5 kg  10 kg)
The acceleration is positive if B ascends. Since the acceleration is negative, B descends
with an acceleration of 0.67 m/s2.
17. The angular velocity of the wheel is
  36
rev 2 rad 1min
rad


 3.77
min
rev
60 s
s
The angular displacement is

t
   t  (3.77 rad/s)(45s)  170 rad

18. A diagram would be
mg
N
12
There are two forces acting on the water in the bucket, its weight and the normal force of
the bucket pushing on the water. At the top of the circle, both forces point down. When
we apply Newton’s second law,
F
r
 mar
N  mg  m
v2
r
(Since the radial direction is downward, forces downward are positive.) When the water
just barely stays in the bucket, the bucket is not really pushing on the water. That means
N = 0,
v2
r
v2
0  mg  m
r
N  mg  m
v  rg  (1 m)(9.8 m/s2 )  3.1 m/s
19. The operating speed of the Eye is

 2 rad 1min


 3.49 103 rad/s
t 30 min 60 s
The angular displacement while it accelerates is
  12 ( f  i )t  12 (3.49 103 rad/s  0)(20 s)  3.49 102 rad
20. The net acceleration is the sum of the radial and tangential accelerations,

 
a net  a r  a t
Since the radial and tangential directions are perpendicular to each other,
anet  ar  at
2
2
2
From the radius and the angular speed, the radial acceleration can be found,
ar   2r  (2 rad/s)2 (0.25 m)  1.0 m/s2
Since the net acceleration is also known, the tangential acceleration can be determined,
13
anet  ar  at
2
2
2
at  anet  ar
2
2
2
at  anet  ar  (1.6 m/s2 ) 2  (1.0 m/s2 ) 2  1.25 m/s2
2
2
The tangential acceleration is related to the angular acceleration,
at  r

at 1.25 m/s 2

 5.0 rad/s 2
r
0.25 m
14