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AP Physics
Semester Review
26 is torque
32, 45-52, 58, 59 deal with
momentum

1. The position of a particle
moving along the x axis is given
2
by x  21 22t  6t m, where t is in seconds. What is
the average velocity during the time interval t = 1.0 s to
t = 3.0 s?


x  x 3  x1


x  21 223  63  21 221  61
2
2
m
x  33  37m
x  4m
x 4m
m
v 
 2 sec
t 2sec
2
2. A bullet is fired through a board, 50.0 cm thick, with its line
of motion perpendicular
to the face of the board. If it enters
m
m
with a speed of 500 sec and emerges with a speed of 200 sec ,
what is the bullet's acceleration as it passes through the
board?
 2

v f  v  2ax
2
i
v v
200   500

a
2x
20.5m
2
f
2
i
m 2
sec

m 2
sec
a  210000 secm 2

3
3. The position of a particle moving along the x axis is given
by x = 6.0t2 - 1.0t3, where x is in meters and t in seconds.
What is the position of the particle when it achieves its
maximum speed in the positive x direction?
dx
v
 12t  3t 2
dt
dv
a
 12  6t
dt
0  12  6t
t  2sec
x  6.0t 2 1.0t 3
x  6.02 1.02
2
3
x  24  8m
x  16m

4
4.
A particle moving along the x axis has a position given by
x = (54t - 2t3)m, where t is in seconds. How far is the
particle from the origin (x = 0) when the particle is not
moving?
dx
v
 54  6t 2
dt
2
0  54  6t
t  3sec
x  54t  2.0t 3
x  54 3  2.03
3
x  162  54 m
x  108m
The particle is not
moving at the
instand the velocity

equals zero.
5
5. An object is thrown vertically and has an upward velocity of
v when it reaches one fourth of its maximum height above its
launch point. What is the initial (launch) speed of the object?
vTop  0
3H
4
 v  v
1
4
v  v 1  2g43 H 
v 2f  v o2  2gH
0  v 21  2g43 H 
v o  2gH
0  v  g H 
2v 2 
v o  2g 
 3g 
2
f
2
4
4
2
2v
2
3g
vo
 H
3
2
vo 
4 v2
3
vo 
4v
3
6

6.
A ball thrown vertically from ground level is caught 3.0
seconds later by a person on a balcony which is 30 meters
above the ground. Determine the initial speed of the ball.
y  v iy t  12 gt 2


2
2
1
m
1
30  2 10 sec2 3sec
y  2 gt
 v iy 
t
3sec
m
v iy  25 sec
7
4
7. Vx is the velocity of a particle moving along the x axis as
shown. If x = 2.0 m at t = 1.0 s, what is the position of the
particle at t = 6.0 s?
vx secm 
-2
-1
0
1
2
3
x  3 2 12  3m
1
2
3
4
5
6
tsec
2m  3m  1m

∆x is the area under the
 t =
line. The area between
1 sec and t = 3 sec cancels
out. So we’ll find the area
from t = 3 sec to t = 6 sec.
8
8. An object ism thrown vertically upward such that it has a
speed of 25 sec when it reaches two thirds of its maximum
height above the launch point. Determine this maximum
height.

2
vTop  0
vTop
 v 22  2g 13 H
3
 
1H
3
v 2  25
3
3
  H  325
2
3 v2
m
sec
3

2g
2H
3

3v 22  2gH
vo


m 2
sec
m
2 10 sec
2

H  93.75m
9
9. Objectm A is thrown downward at t = 0 with an initial speed
of 10 sec from a height of 60m above the ground. At the same
instant, a object B is propelledm vertically upward from
sec
ground level with a speed of 40 . At what height above the

ground will the two objects pass each other?
H  60 v A t  12 gt 2 
H  v B t  12 gt 2
60  v A t  12 gt 2  v B t  12 gt 2
60  v A t  12 gt 2  v B t  12 gt 2
60 10t  40t
60  50t
1.2sec  t
10
m
sec
10. A car travelsm north at 30 for one half hour. It then travels
south at 40 sec for 15 minutes. The total distance the car has
traveled and its displacement are:

60sec 
60sec 
m
x total 30 30min 
 40 sec 15min 
 90km
 min 
 min 
m
sec


x  54 ˆjkm 36 ˆjkm 18km, north.
11
11. A particle starts from the origin at t = 0 with a velocity of
8.0i
and moves in the xy plane with a constant
acceleration of (-2.0i + 4.0j) . At the instant the particle
achieves its maximum positive x coordinate, how far is it
from the origin?
x  8t  t
2
x  8t  t
2
x  84  4
dx
v
 8  2t
dt
x  16m
0  8  2t
y  v yi t  12 a ˆj t 2
t  4 sec
2
y 

1
2

4 ˆj
y  2t 2
y  32m
m
sec 2
r i  j
2
2
r  35.8m
t 
2

12
12. A rock is projected from the edge of the top of a 100-ft tall
building at some unknown angle above the horizontal. The rock
strikes the ground a horizontal distance of 160 ft from the base of
the building 5.0 seconds after being projected. Assume that the
ground is level and that the side of the building is vertical.
Determine the speed with which the rock was projected.
y  v yi t  12 gt 2
1
2
gt 2  y  v yi t
1
2
gt 2  y
 v yi
t
16 sec2 5sec 100 ft
 v yi
5sec
ft
60 sec
 v yi
ft
2
x 160 ft
vx  
 32 secft
t 5sec
v  v v
2
xi
v  68
2
yi
ft
sec
13
13. A football is thrown upward at a 30° angle to the
horizontal. To throw a 40.0-meter pass, what must be the
initial speed of the ball?
v o2 sin 2
R
g
Rg
 v o2
sin 2
Rg

vo 
sin 2
m
v o  21.5 sec

40m 10 secm 2

sin 60
14
14. The horizontal surface on which the objects slide is
frictionless. If M = 2.0 Kg and the Tension in string 1 is 18
Newtons, Determine F.
FNet  ma
T1  3Ma
T1
a
3M
18N
m
 a  3 sec
2
6Kg
F  5Ma
 
F  52Kg 3 secm 2
F  30N
15
15. A particle
starts from the origin at t = 0 with a velocity of
m
8j sec and moves in the xy plane with a constant
acceleration of a  4i  2 j  secm 2 . At the instant the x
coordinate of the particle is 32 meters, what is the
magnitude of the y coordinate?
x  ai t
1
2
2
 
32m  12 4 secm 2 t 2
t  4 sec
y  v yi t  a j t
1
2
y 8
m
sec
2
4sec 
1
2
2 4sec
m
sec2
2
y  48m
16
16. The surface of an inclined plane is frictionless. If F = 30N,
what is the magnitude of the force exerted on the 3kg block
by the 2kg block
FNet  ma
FP
F  FP  m1  m2 a
F  m1  m2 gsin   m1  m2 a


N
30N  5Kg 9.8 Kg
0.5
5Kg
30N  24.5N
 a  1.1 secm 2
5Kg
a
17
FNet  ma
P  FP  ma
FP
P  ma  FP
Push

P  ma  mgsin  




3Kg


P  3Kg 1.1 secm 2  3Kg 9.8 KgN 0.5
P  3.3N 14.7N
P 18N
18
17. If P = 5.0 N, what is the magnitude of the force exerted
on block 1 by block 2?
3
2
1
P
5 Kg
2 Kg
First, find the
acceleration of
the system!
3 Kg
FNet P
a

m
m
5N
a
 0.5 secm 2
10Kg
19

P
Push
2Kg
Push
P  Push  ma
P 
ma  Push

5N  2Kg 0.5
Push  5N 1N
m
sec2
 P

ush
3Kg
Push  ma


5Kg

Push  8Kg 0.5 secm 2

Push  4N
Push  4N
For every action Force there is an
equal and opposite reaction

Force!
20
18. If a = (20i + 2.7j), and b is as shown, what is the
magnitude of the Resultant of the two vectors?
20cos120  10i
y
b
20
20sin120  17.3 j
30
°
b  10i  17.3 j 
x

a  b  10i  20 j 
a  b  10 2  20 2
a  b  22.4
21
19. The graph below shows the force on an object of mass m
as a function of time. For the time interval t = 0 to t = 3
seconds, the total change in the momentum of the object is
p 
 Fdt
p  20N  sec 10N  sec
p 10N  sec

22
20. The position of a particle moving along the x axis is given
by x  2t 3  6t 2  4 , where x is in meters and t is in
seconds. What is the average acceleration between 2 sec <
t < 4 sec ?
dx
2
v   6t 12t
dt
dv
a  12t 12
dt
ai  a f
a
2
m
12

36

 sec2
a
2
m
a  24 sec
2
23
21. If M = 5.0 Kg, what is the Tension in the string between
the two objects of equal mass?
All surfaces are
frictionless.
3M  2M
a
g
3M  2M
M
a
g  15 g
5M
M
3M
T

M



24

FNet  ma
T
M
Mg


T  Mg  Ma
T  Ma  Mg
T  M a  g


T  565 g
T  5 15 g  55 g
T  60N
25
22. If the tension T = 15 N and the acceleration a  5 sec2 , what
is the mass m of the suspended object? All surfaces are
frictionless!
FNet  ma
a
m
mg 
T  ma
mg  ma  T
T
T
m
mg
m
mg  a  T
T
m
ga
15N
m  m  3Kg
5 sec2
26
23. If the Tension in string 1 is 40N and the Tension in string 2
is 25N, what is mass m of the object?
60
°
30
25N
40N

sin 30 sin Z
1
2
mg

Z  53
180  83  97  D
D
Z
2
m
25N
W

W  49.6N
sin 30 sin 97
W 49.6N
m 
 4.96kg
N
g 10 kg
27
24.
A roller-coaster car has a mass of 500kg when fully
loaded with passengers. The car passes over a hill
of radius 20
m
m. At the top of the hill, the car has a speed ofsec8.0 . What is
the force of the track on the car at the top of the hill?


N
FC  mg  N
mg
mv 2
N  mg 
R
500kg8 


 
N  500kg 10
N
kg
m 2
sec
20m
N  3400N, up
28
25. A uniform ladder, 15 ft long and weighing 60 lbs., is
leaning against a frictionless wall at an angle of 53° above
the horizontal. A 120 lb boy climbs 6.0 ft up the ladder.
What is the magnitude of the friction force exerted on the
f cos   f sin 
ladder by the floor?
  53

Nwall




N ground





 120lbs
 f

60lbscos
60lbs
29
 
CW
CCW
60lbscos537.5 ft  120lbscos536 ft  f cos3715 ft



60lbs 35 7.5  120lbs 35 6  f 45 15
270lbs  432lbs  f 12
702 lbs
 f  58.5 lbs
12
30
26. A uniform, horizontal meter stick, supported at the 50 cm
mark, has a mass of 500 grams hanging at the 20 cm mark
and a 300 gram mass hanging at the 60 cm mark.
Determine the position on the meter stick at which one
would hang a third mass of 600 grams to keep the meter
stick balanced.

CCW



300g
500g
600g
d

CW

500gg30cm  300gg10cm  600ggd

530cm  310cm  6d 
150cm  30cm  6d
120cm
 d  20cm  the 70cm mark!
6
31
27. The three blocks are released from rest. Determine the
acceleration of the system if no friction is present.
T2
2M
T1
T1
T2
M
F
Net  ma
Mg  3Mg  Mg  6Ma
3g  g  6a

g
a
3

3M
3Mg
32
28. A person weighing 600N rides min an elevator that has a
downward acceleration of 0.75 sec . What is the magnitude
of the force of the elevator floor on the person?
2
FNet  ma
mg  N  ma
mg  ma  N
N

mg  a  N
mg



60kg 9.25 kgN  N
555N  N
33
29. A 0.20-kg object attached to the end of a string swings in a vertical
circle of radius = 80 cm. At the top of the circle the speed of the
object is 4.5 . What is the magnitude of the tension in the string
at this position?

v
mg  T  FC
T mg
T  FC  mg
v 2

T  m  g
 R

 4.5 m 2



sec
m 
T  0.2kg
10
sec 2 
 0.8m


T  3.06N
34
30. A roller-coaster car has a mass of 500 kg when fully loaded with
passengers. At the bottom of a circular dip of radius
40 m (as
m
shown in the figure) the car has a speed of 25 sec . What is the
magnitude of the force of the track on the car at the bottom of the
dip?
N  mg  FC
40m
m
25 sec

N  FC  mg
v 2

N  m  g
 R

 25 m 2



sec
m 
N  500kg

10
sec 2 
 40m


N  12812.5N
35
31. Daniel’s apparent weight at the top of an inverted loop of
radius 30 m while riding a roller coaster is 3W. What is
Daniel’s apparent weight at the bottom of the loop?
At the top!
FC W  3W  4W
W 3W
At the bottom!
 

N
W

FC  N  W
FC  W  N
4W  W  N
5W  N
36
32. As shown in the top view above, a disc of mass m is moving
horizontally to the right with speed v on a table with negligible
friction when it collides with a second disc of mass 2m. The
second disc is moving horizontally to the right with speed v/2 at
the moment of impact. The two discs stick together upon impact.
The speed of the composite body immediately after the collision is
pi  p f

mv  2m v2  p f  2mv
pf
2mv 2
v 

 3v
m
3m
37
33. A 0.30-kg mass attached to the end of a string swings in a vertical
circle (R = 1.4 m), as shown. At an instant when  = 30°, the
speed of the mass is 6.0. What is the magnitude of the resultant
force
v on the mass at this instant?
m
R

g
mv
FC 
R



FC
2
0.3kg6 


m 2
sec
1.4m
FC  7.71N
38
m
34. A race car traveling at 100 sec enters an unbanked turn of 400 m
radius. The coefficient of friction between the tires and the track is
1.1. The track has both an inner and an outer wall. Which

statement is correct?
2
mv
FC 
 mg
R
v  Rg 

m
1.1
400m
10
 
 sec2

m
v  66.3 sec
The driver and car crash into the outer wall!

39

35. A force acting on an object moving along the x axis is given
by Fx  14 x  3x 2 N where x is in meters. If the force is
applied at an angle of 37° with respect to the horizontal,
how much work is done by this force as the object moves
from x = -1 m to x = +2 m?


W  F  x  Fx cos 
 F xdx
W   14 x  3x dx
W  7x  x  cos
W  28  8  7  1 45
W 
2
2
1
2
3 2
1

W  12 45  9.6J
40

36. A particle of mass m moves along a straight path with a
2
speed v defined by the function v  bt  c , where b and c
are constants and t is time. What is the magnitude F of the
net force on the particle at time t = t1 ?
dv
a
 2bt
dt
F  ma
F  2bmt1

41
37. A 0.50-kg particle moves under the influence of a single
conservative force. At point A where the particle has a speed
m
of 10 sec
, the potential energy associated with the
conservative force is +40 J. As the particle moves from A to
B, the force does +25 Joules of work on the particle. What
 is the value of the potential energy at point B?
2
1
K i  2 mvi
ET  K U  65J
1
m 2
K i  2 0.5kg 10 sec
If the KE increases by +25J,
then the U must decrease by
K i  25J
the same amount.



U  40J  25J 15J
42
38. To stretch a certain nonlinear spring by an amount x
requires a force F given by F  40x  6x 2 , where F is in
newtons and x is in meters. What is the change in potential
energy when the spring is stretched 2 meters from its
equilibrium position?

U    F x dx
U    0 40x  6x dx
2
2
U  20x  2x
2

3 2
0
U  80 16
U  64J
43
39. A skier weighing 550N comes down a frictionless ski run
that is circular
(R = 30 m) at the bottom, as shown. If her
m
speed is 12 sec at point A, what is her speed at the bottom of
the hill (point B)?
EA  EB
KA  UA  KB

1
2
mvA2  mgh  12 mvB2
v  2gh  v
2
A
2
B
v A2  2gh  v B
h  R  Rcos   7.02m
12

m 2
sec


 19.6 secm 2 7.02m  v B
m
v B  16.8 sec
44
40. A block (mass = 4.0 kg) sliding on a horizontal frictionless
N
surface is attached to one end of a horizontal spring ( k =100 m )
which has its other end fixed. If the maximum distance the block
slides from the equilibrium position is equal to 20 cm, what is the

speed of the block at an instant when it is a distance of 16 cm
from the equilibrium position?
U S  U16cm  K16cm
1
2
kA  kx  mv
2
2
1
2
1
2
kA  kx  mv
2
2
v
2
k
m
A
2
x
v
2
k A  x  mv
2
2
2




100 N
m 

4kg
0.2m  0.16m 
2
2
m
v  0.6 sec
2
45
41. A 1.2-kg mass is projected down a rough circular track
(radius =m 2.0 m) as shown. The speed
m of the mass at point A
sec
sec
is 3.2 , and at point B, it is 6.0 . How much work is
done on the mass between A and B by the force of friction?


K A  UA  W f  K B
K A  UA  K B  W f
A
R
B


1
2
mv  mgR  mv  W f
2
A
1
2
2
B
6.144J  23.52J  21.6J  W f
8.06J  W f
46
42. A 25-kg block on
a horizontal surface is attached to a light
kN
spring ( k = 8.0 m ). The block is pulled 10 cm to the right from
its equilibrium position and released from rest. When the block
has moved 5.0 cm toward its equilibrium position, its kinetic
energy 
is 12 J. How much work is done by the frictional force
on the block as it moves the 5.0 cm?
U10cm W f  U 8cm  K
U10cm U 8cm  K  W f
1
2
kA2  12 kx 2  K  W f
1
2
k A 2  x 2  K  W f


8000 Nm
2
2
0.1m  0.05m 12J  W f
2
30J 12J  W f 18J
47
43. As an object moves from point A to point B only two
forces act on it: one force is nonconservative and does -30
Joules of work, the other force is conservative and does
+50 Joules of work. Between A and B,
48
m
sec
44. A 2-kg object moving with a speed of 6.0 collides
perpendicularly with a wall and emerges with a speed of
6.0 secm in the opposite direction. If the object is in contact

with the wall for 2.0 msec, what is the magnitude of the
average force on the object by the wall?

p
F
t
mv f  v i 
F
t
2kg6 secm   6 secm 
F
0.002sec
F  12000N
49
m
sec
45. A 1.0-kg playground ball is moving with a velocity of 6.0
directed 30° below the horizontal just before it strikes a
horizontal surface. Them ball leaves this surface 0.50 s later
sec

with a velocity of 4.0 directed 60° above the horizontal.
What is the magnitude of the average resultant force on the

ball?
m
m

vi  5.2iˆ  3 ˆj

sec

v f  2iˆ  3.46 ˆj

sec
v  v f  v i


 
v  3.2iˆ  6.46 ˆj 
v  2iˆ  3.46 ˆj
m
sec
 5.2iˆ  3 ˆj

m
sec
m
sec
m
v  iˆ 2  ˆj 2  7.2 sec
50
Ft  mv
mv
F
t
m
1.0kg7.2 sec


F
0.5sec
F  14.4N

51
46. The only force acting on a 2.0-kg object moving along the
m
x axis is shown.
If the velocity vx = -2.0 sec at t = 0,
what is the velocity at t = 4.0 seconds?
Fx (N)
p 
4
1
kg m
Fdt

4

2

8
   sec

kg m
p

2
t sec
sec
2 3 4
p  p f  pi
8
    
p  pi  p f


2 kgsecm  4 kgsecm  p f  6 kgsecm
p f 6 kgm
sec
m
vf  
 3 sec
m
2kg

52
m
sec
m
sec
47. The speed of a 2.0-kg object changes from 30 to 40 during a
5.0-second time interval. During this same time interval, the
velocity of the object changes its direction by 90¯. What is the
magnitude of the average total forceacting 
on the object during
this time interval?
v  v f  v i

 
v  30iˆ  40 ˆj 
v  0iˆ  40 ˆj
m
sec
 30iˆ  0 ˆj

m
sec
2
2
m
ˆ
ˆ
v  i  j  50 sec

m
sec
Ft  mv
mv
F
t
m
2kg50 sec


F
5sec
F  20N
53
m
48. A 2.0-kg object moving 5.0 sec collides with and sticks to an 8.0-kg
object initially at rest. Determine the kinetic energy lost by the
system as a result of this collision.

pi  p f
mv  m  M v 
m
v  v 
m M
2kg
m
5 sec
 v 


10kg
m
1 sec
 v 
K  K i  K f
K  12 mvi2  12 m  M v 2f
K 
1
2
2kg5 
m 2
sec
 10kg1
1
2

m 2
sec
K  25J  5J
K  20J

54
49. A 2-kg block is attached to the end of a 3.0-m string to form a
pendulum. The pendulum is released from rest when the string is
horizontal. At the lowest point of its swing when it is moving
horizontally, the block is hit by a 25-gram bullet moving
horizontally in the opposite direction. The bullet remains
embedded in the block and causes the block to come to rest at the
low point of its swing. What was the magnitude of the bullet's
velocity 3m
just before hitting the block?
2kg
v Bi  2gh
m
v Bi  7.75 sec
2k g

v bi
t


h
d
u
v 0
55
pi  p f
pBi  pbi  0
pBi   pbi
mB v Bi  mb v bi
m
2Kg7.75 sec
mB v Bi


 v bi  
mb
0.025kg
m
v bi  620 sec

56
50. A 3.0-kg
mass sliding on a frictionless surface has a velocity of
m
5.0 sec east when it undergoes a one-dimensional inelastic m
sec
collision with a 2.0-kg mass that has an initial velocity of 2.0 m
sec
west. After the collision the 3.0-kg mass has a velocity of 1.0
east. How much kinetic energy does the two-mass system lose


during the collision?

m
m
m
1 sec
2 sec
h
5.0 sec

3kg
15

2kg
kgm
sec
4




t
kgm
sec



d
u


3kg
2kg
4 kgm
sec
m
7 kg
sec


kgm
p 7 sec
v 
m 2kg
m
v  3.5 sec
57
K i  12 m1v12  12 m2v 22
Ki 
1
2
3kg5 
m 2
sec

1
2
2kg2 
m 2
sec
K i  37.5J  4J  41.5J
2
K i  12 m1v12  12 m2v 
2
Kf 
1
2
m
5Kg1 sec

2
1
2
m
2Kg3.5 sec

2
K f  2.5J  12.25J
K f  14.75J
K Lost  K i  K f  26.75J
51. A 80-kg man who is ice skating south collides with a 40-kg boy
who is ice skating east. Immediately after the collision, the man
and mboy are observed to be moving together with a velocity of 2.0
sec
, in a direction 37¯ south of east. What was the magnitude of the
boy's velocity before the collision?
kg m
, at  37

p 192iˆ 144 ˆj 
p 240
80kg

sec
kg m
sec
t
40kg

h
d
40kg
80kg 37



u




v boy
v boy
v boy
p

m
192 Kgsec m

40kg
m
 4.8 sec
59
52. A 6-kg object, initially at rest, "explodes" into three objects of
equal mass. Two of these
are determined to have velocities of
m
equal magnitudes (4.0 sec ) with directions that differ by 90¯. How
much kinetic energy was released in the explosion?
 4
6kg

 H 0

V 0

o
b o
 m

90
2kg
m
4 sec
2kg
 2kg


m
sec
p  iˆ 2  ˆj 2

kgm
ˆ
8iˆ kgm
8
i
sec
sec
8 ˆj

kgm
sec
 kgm
8 ˆj
sec
p  82  82
p  11.3 Kgm
sec
p
v
m
11.3 Kgsec m
v
2.0kg
m
v  5.66 sec
60



K  mv 16J
2
K  mv 16J
2
K  mv  32J
1
2
1
2
1
2
2
K Net  K
K Net 16J 16J  32J
K Net  64J
61
53. A particle moves in the xy plane with a constant
acceleration given by a  4.0 j secm . At t = 0 its position
m
and velocity are 10i m and 2i  8 j ,sec
respectively.
What is the distance from the origin to the particle at
t = 2.0 s?
2

y
 v yi t  ay t
y  8
1
2
m
sec
2
2sec 
1
2
4 2sec
m
sec2
2
y  8 ˆj m
rˆ  6i  8 j m

ˆr  iˆ 2  ˆj 2  10.0m
62
54. A ball is thrown horizontally from the top of a building 80 meters
high. The ball strikes the ground at a point 100 meters
horizontally away from and below the point of release. What is
the speed of the ball just before it strikes the ground?
t
t
2y
g
x 100m
m
vx  
 25 sec
t 4 sec
280m 
m
vy  gt  40 sec
10 
m
sec 2
t  4.0sec
v  v  v  47.2 secm
2
x
2
y

63
A 5-kg mass starts from rest under the influence of variable
force F as a function of distance x as shown on the graph below.
30
F(N)
20
10
0
2
4
6
8
10
12
x (m)
-10
55. During which time interval(s) will the velocity of the mass
be constant?
When the acceleration = 0 and the Force = 0.
Between x = 6 m and x = 8 m.
64
56.
During which time interval will the most Work be
done?
30
F(N)
20
10
0
2
4
6
8
10
12
x (m)
-10
W
 F xdx
The areas between x = 0 and x 4 m, between x = 4 m
and x = 6 m, and between x = 8m and x = 12 m are all
the same. So the same work is done in all three areas.
65
57.
What is the velocity of the object at x = 12 m?
30
F(N)
20
10
0
2
4
6
8
10
12
x (m)
-10
W  K   F x dx
K  40  40  40J

K  40J
K  12 mv 2f  12 mvi2
K  12 mv 2f
240J 
2K
m
vf 

 4 sec
m
5kg
66
An object of mass m, initially at rest, slides down from a height of
1.25 meters on a frictionless ramp, collides and sticks to an identicle
particle 2 of mass m at rest as shown. Then particle 1 and 2 together
collide elastically with particle 3 of mass 2m which is also initially at
rest.
58. The speed of particle
m
1
1 after the collision
with particle 2 would
1.25m
2
3
be?
m
2m
  

v o  2gh

vo  5

m

sec
Double the mass = half the velocity!
m
2.5 sec

67
m
1
1.25m

2
3
m
2m




59. The speed of particle 1 and
2 together after the elastic
collision with particle 3
would be?
m1  m2
v1 f 
v1i
m1  m2
2m  2m
m
vf 
2.5 sec
2m  2m
vf  0
68
60. The same force F is applied horizontally to bodies 1, 2, 3 and 4,
of masses m, 2m, 3m and 4m, initially at rest and on a frictionless
surface, until each body has traveled distance 2d. The correct
listing of the magnitudes of the velocities of the bodies, v1, v2, v3,
and v4 is
v 2f  v i2  2a2d
v f  4ad
a1 
F
m
a2 
F
2m
a3 
F
3m
a4 
F
4m

69




v1 f  2 mF d
F d
v1 f  2 ad  2 m
 
 
F 2d  2 F d
v 2 f  2 2m
2m
 
2F d  v
2m 
F d
2v 2 f  2 2 
2m
2v 2 f  2
 
1f
 
 2v 2 f
F 2d  2 F d
v 3 f  2 3m
3m
 
F 2d 
v 4 f  2 4m
 d
F
m
70

v1 f  2 mF d
3mF d

F d
 2 3 3m

3F d
 2 3m

F d v
 2 m

v3 f  2
3v 3 f
3v 3 f
3v 3 f
1f
 d
F d
 2 m

v4 f 
2v 4 f
F
m
2v 4 f  v1 f
71
v1f  2v 2 f  3v 3 f  2v 4 f

72
61. A constant force F is applied to a body of mass m that initially is
headed east at velocity vo until its velocity becomes 2vo. The
total time of travel is 3t. The totalto distance the body travels in
that time is
v0
v  vo
m
m
3t

v  2vo
 a   F
m

0v

a


o
t
F
v
 o
m
t
F
vo  t
m

vo
t

m
1t
2t
v  vo 
x

 vot 
1
2
at
2


74
x  v o t  12 at 2
x  v o 3t 
x
 3 mF
t
2
 
 12 a 3t
2
9
F
2mt
x   32 mF t 2

2
x
3 F
2m
t2 
F t2
x  72 m
F
m
t2 
F
m
t2
F
m
0  vo
v
a
 o
t
t
F
vo
2
2
 
v

v
m
t
f
i  2ax
F
2
vo  t
F
Ft
2
t

m
m
m
a

  
2a
x

3 F
2m
2
t2 
F t2
x  72 m
  3Ft
3 mF t
 x  F
2m
F
m
t2 
F
m
2
t2
2m
2
F
m
0  vo
v
a
 o
t
t
F
v
  o
m
t
F
vo  t
m
a

x  v o 3t  a3t 
2
1
2
F t2 
x  62 m
9
2
 
F t2
m
2
3
F
x 2 mt
2
F
x total  2x  m t
75