AP Physics
Chapter 9 Review
Momentum
et. al.
1.
m
A 2000-kg truck traveling at a speed of 3.0sec
makes a 90¯
turn in a time of 4.0 seconds and emerges from this turn
m
with a speed of 4.0 sec
. What is the magnitude of the
average resultant force on the truck during this turn?
€
Δv
€ = v f − vi
Δv = 0iˆ + 4 ˆj
− 3iˆ + 0 ˆj
(
) (
Δv = (−3iˆ + 4 ˆj )
m
sec
)
m
sec
m
sec
m
Δv = iˆ 2 + ˆj 2 = 5 sec
€
2
#1.2
FΔt = mΔv
mΔv
F=
Δt
m
2000kg
5
(
)
(
sec )
F=
4 sec
F = 2500N
€
3
m
2. A 1.5-kg object moving with a speed of 4.0 sec
collides
perpendicularly with a wall and emerges with a speed of
m
6.0 sec
in the opposite direction. If the object is in contact
with the wall for 5.0 msec, what is the magnitude of the
average force on the object by the wall? €
€
FΔt = mΔv
mΔv
F=
Δt
m
m
1.5kg
6
−
−4
(
)( sec ( sec ))
F=
0.005sec
F = 3000N
4
m
3. A 1.5-kg playground ball is moving with a velocity of 3.0 sec
directed 30¯ below the horizontal just before it strikes a
horizontal surface. The ball leaves this surface 0.50 s later
m
with a velocity of 4.0 sec
directed 60¯ above the horizontal.
What is the magnitude of the average resultant €
force on the
ball?
€ ˆ
v i = 2.6iˆ −1.5
j
(
)
m
sec
v f = 2iˆ + 3.46 ˆj
(
)
m
sec
Δv = v f − v i
Δv = 2iˆ + 3.46 ˆj
− 2.6iˆ −1.5 ˆj
(
) (
Δv = (−0.6iˆ + 4.96 ˆj )
m
sec
)
m
sec
m
sec
2
2
m
ˆ
ˆ
Δv = i + j = 5 sec
5
#3.2
FΔt = mΔv
mΔv
F=
Δt
m
1.5kg
5
(
)
(
sec )
F=
0.5sec
F = 15N
€
6
16
F (N)
8
€
€
€
−8
€
€
€
0
1 2
3 4
5 t (sec)
€
€ € €
4. The only force acting on a 2.0-kg object moving
€ the x axis is shown. If the velocity vx is +2.0 secm
along
at t = 0, what is the velocity at t = 4.0 sec?
€
7
€
16
F (N)
8
€
€
€
1 2
−8
€
[
0
€
€ € €
3 4
t
(sec)
5
€
]
1
1
€
−8
+
1sec
−8N
+
( ) [ 2 ( )( )] [ 2 (2sec)(16N )] + 16 N ⋅ sec
€
Δp = 20N ⋅ sec
9
16
Δp = m(v f − v i )
8
Δp = mv f − mv i
F (N)
€
€
€
0
Δp + mv i = mv f
1 2
−8
€
€
3 4
€
€ € €
€
€
5 t (sec) Δp + mv
i
= vf
m
m
m
20 Kg⋅
+
2kg
2
(
)
(
sec
sec )
vf =
2kg
m
v f = 12 sec
#4.4
€
10
5.
m
A 3.0-kg ball with an initial velocity of (4i + 3j) sec
collides
m
with a wall and rebounds with a velocity of (-4i + 3j) sec
.
What is the impulse exerted on the ball by the wall?
€
Δp = p f − pi
Δp = 3Kg(−4i + 3j)
€
m
sec
− 3Kg( 4i + 3 j)
m
sec
Δp = −12i + 9 j −12i − 9 j
Δp = −24i N ⋅ sec
11
€
6.
m
A 0.15-kg baseball is thrown with a speed of 40 sec
. It is hit
m
straight back at the pitcher with a speed of 80 sec
. What is
the magnitude of the impulse exerted on the ball by the
bat.
€
€
Δp = p f − pi
m
m
Δp = 0.15Kg(80) sec
− (0.15Kg)(−40 sec
)
Δp = 18 N ⋅ sec
€
12
7. A 12-g bullet is fired into a 3.0-kg ballistic pendulum
initially at rest and becomes embedded in it. The pendulum
subsequently rises a vertical distance of 12 cm. What was
the initial speed of the bullet?
E Bottom = E Top
K B = UTop
1
2
2
B
mv = mgh
v B = 2gh
vo
v B = 2g(0.12m)
m
v ′B = 1.53 sec
€
13
€
#7.2
pi = p f
mv o = ( m + M )v ′
m+ M
vo =
v′
m
3.012kg
m
vo =
1.53 sec
(
)
0.012kg
m
v o = 384 sec
€
14
8. A 10-g bullet moving 1000
m
sec
strikes and passes through a
2.0-kg block initially at rest, as shown. The bullet emerges
m
from the block with a speed of 400 sec
. To what maximum
height will the block rise above its initial position?
€
€
pi = p f
v
pb = p′b + p′B
€
15
pi = p f
#8.2
pb = p′b + p′B
mb v o = mb v ′b + Mv ′B
m
0.01kg
600
mb (v o − v ′b )
(
)
(
sec )
= v ′B =
M
2kg
m
v ′B = 3 sec
This is the velocity of the Block,
by itself, after the collision!
€
16
#8.3
From conservation of Energy!
v ′B )
(
h=
2
2g
m 2
sec
3 )
(
h=
2g
h = 0.46m
€
17
9. A 12-g bullet moving horizontally strikes and remains in a
3.0-kg block initially at rest on the edge of a table. The
block, which is initially 78.4 cm above the floor, strikes the
floor a horizontal distance of 100 cm from its initial position.
What was the initial speed of the bullet?
2y
t=
= 0.4 sec
g
vo
78.4cm
€
€
€
€
x 1m
v ′x = =
= 2.5 secm
t 0.4sec
18
vo
€
t
h
d
u
v ′ = 2.5 secm
€
pi = p€f €
€
€
mb vo = (m b + M )v ′B
m
′
3.012kg
2.5
m
+
M
v
(
)( sec )
(
)
(
b
B)
vo =
=
mb
0.012kg
m
v o = 627.5 sec
€
19
10. A 6.0-kg object moving 5.0
collides with and sticks to a
2.0-kg object. After the collision the composite object is
m
moving 2.0 sec
in a direction opposite to the initial direction
of motion of the 6.0-kg
€ object. Determine the speed of the
2.0-kg object before the collision.
€
6kg
€
€
m
5.0 sec
€
vo
2kg
€
€
€
€€
m
sec
t
€
h
u
m
−2 sec
6kg 2kg
d
€ €
20
6kg
€
€
m
5.0 sec
vo
2kg
p€1i + p2i =€p′
€
t
€
h
d
u
m
−2 sec
6kg 2kg
€ €
€ 1 + m 2 )v ′
m1v1i + m2v 2i€ = (€m
m2v 2i = ( m1 + m2 )v ′ − m1v1i
v 2i
m1 + m2 )v ′ − m1v1i
(
=
m2
m
m
8kg
−2
−
6kg
5
(
(
(
)
(
)
sec )
sec )
v 2i =
2kg
m
v 2i = −23 sec
21
6kg
m
5.0 sec
p = 30 Kg⋅m
sec
€
€
€
vo
€
€
t
Kg⋅m
€ sec €
d
u
m
−2 sec
€
€€
6kg 2kg
p′ = −16 Kg⋅m
sec
p = −46
Kg⋅m
′
p = −16 sec€
€
2kg
h
€ €
€
Kg⋅m
sec
p −46
m
vo = =
= −23 sec
m
2kg
€
22
11. A 2.0-kg object moving 5.0
m
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.
m€
5.0 sec
2kg
p =10
kg⋅m
sec
€
€ €
Ki =
€
1
2
m 2
€
sec
(2kg)(5 )
8kg
€€
€
= 25J
t
h
d
u
€
Kf =
2kg
€
p′ =10 kg⋅m
sec
1€10kg
2
(
8kg
m
1 sec
m 2
sec
)(1 )
= 5J
€
K Lost = K i − K f = 20J
€
€
€
23
12. A 1.6-kg block is attached to the end of a 2.0-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
10-g bullet moving horizontally in the opposite direction.
The bullet remains 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 just before hitting the
block?
1.6kg
v Bi = 2gh
2m
m
v Bi = 6.26 sec
€
v bi
€
1.6 k
g
t
€
€
€
€
€€
h
d
u
v′ = 0
24
#12.2
pi = p f
pBi + pbi = 0
pBi = − pbi
mB v Bi = −m b v bi
mB v Bi
(1.6Kg)(6.26 secm )
= v bi = −
−m b
(0.010kg)
m
v bi = −1002 sec
25
€
€
13. A 3.0-kg mass sliding on a frictionless surface has a
m
velocity of 5.0 sec
east when it undergoes a one-dimensional
inelastic collision with a 2.0-kg mass that has an initial
m
velocity of 2.0 sec
west. After the collision the 3.0-kg mass
m
has a velocity of 2.2 sec
east. How much kinetic energy does
€
the two-mass system lose during the collision?
m
−2 sec
m
5.0
sec
€
3kg €
1
€2
2
1 1
1
€2
2
1 2€
Ki = m v + m v
€
K i = ( 3kg)(5
1
2
m 2
sec
)
1€
2
2kg
t
€
€€
+ (2kg)(2
m €2
sec
)
h
d
u
3kg
2kg
m
2.2 sec
2
1
K
=
mv
€ f
€ 2
1
m 2
K f = 2 (5Kg)(2.2 sec )
K f = 12.1J
K i = 37.5J + 4J = 41.5J
K Lost = K i − K f = 29.4J
€
26
14. A 4.0-kg mass is released from rest at point A of a circular
frictionless track of radius 0.40m as shown in the figure.
The mass slides down the track and collides with a 2-kg
mass that is initially at rest on a horizontal frictionless
surface. If the masses stick together, what is their speed
after the collision?
v B = 2gh
v B = 2g(0.40m)
m
v B = 2.8 sec
A
v
40 cm
€
€
2kg 4kg
€
€
€
€
€
€
€€
h
t
d
m
2.8 sec
u
€
€
2kg
4kg
27
€
2kg 4kg
p
€
v=
€
€
m € €€
m
11.2 kg⋅
sec
v=
6kg
m
v = 1.86 sec
€
h
t
m
2.8 sec
u
2kg
d
€
€
4kg
p = mv
€
p =11.2
kg⋅m
sec
€
28
15. A 70kg man who is ice skating north collides with a 30kg
boy who is ice skating west. Immediately after the
collision, the man and boy are observed to be moving
together with a velocity of 2.0 secm , in a direction 37¯ north
of west. What was the magnitude of the boy's velocity
before the collision?
(
p′ = 200
37°
€
t
€
h
d
v boy
u
€€
€
30kg
v boy
€
€
) (
30kg
70kg
€
, at 143° = −160iˆ + 120 ˆj
kg⋅m €
sec
€
70kg
)
kg⋅ m
sec
p
=
m
Kg⋅m
160 sec
=
30kg
m
v boy = 5.3 sec
29
€
m
16. A 4.0-kg mass has a velocity of 4.0 sec
, east when it explodes
into two 2.0-kg masses. After the explosion one of the
m
masses has a velocity of 3.0 sec
at an angle of 60¯ north of
east. What is the magnitude of the velocity of the other
mass after the explosion? €
€
m
3 sec
60°
4 kg
m
4 sec
o
b
€ o
m
€
€
€
€
€
€
2kg
2kg
€
€
€
30
€
m
3 sec
m
6 Kg⋅
sec
60°
4
4 kg
H
€
V
€
€ €
€
m
sec
16iˆ Kg⋅secm €€
€
b
€
€
0
€
2
2
ˆ
ˆ
p= i + j
p = 13 + 5.2
m
p = 14 Kg⋅
sec
2
o
m€
€
€
2€
o
2kg
2kg
−21.8°
60°
m
3 Kg⋅
sec
€
3iˆ Kg⋅m
sec
13iˆ Kg⋅secm
5.2 ˆj Kg⋅secm
−5.2 ˆj Kg⋅m
sec
€
€
€
p
v=
m
€
m
14 Kg⋅
sec
v=
2kg
m
v = 7 sec
€
5.2 Kg⋅m
sec
€
€
 j
θ = tan  
i
−1 −5.2 j 
θ = tan 

 13i 
θ = −21.8°
−1
31
17.
A 4.2-kg object, initially at rest, "explodes" into three
objects of equal mass. Two of these are determined to
m
have velocities of equal magnitudes (5.0 sec
) with directions
that differ by 90¯. How much kinetic energy was released
in the explosion?
m
5 sec
o
b o
€ m
4.2kg
90°
1.4kg
1.4kg
€ 1.4kg
€
€€
€ €
H 0
€
V
€ €
€ €
−7 ˆi kg⋅secm
−7 ˆj kg⋅secm
0
€
€
m
5 sec
p = 7 2 + 72
p = 9.9
Kg⋅m
sec
p
m
9.9 Kgsec⋅ m
v=
1.4kg
v=
m
v = 7.07 sec
€
€
€
p€= iˆ 2 + ˆj 2
€
7iˆ kg⋅secm
€
€
7 ˆj kg⋅m
sec
32
€
1
2
2
1
2
2
K = mv = 17.5J
K = mv =17.5J
€
1
2
2
K = mv = 35J
€
K Net = ∑ K
€
K Net = 70J
K Net = 17.5J + 17.5J + 35J
€
33
18. A
3.0-kg mass, initially at rest on a frictionless surface,
explodes into three 1.0-kg masses. After the explosion the
m
velocities of two of the 1.0-kg masses are: (1) 5.0 sec
, north
m
and (2) 4.0 sec
, 30¯ south of east. What is the magnitude of
the velocity of the third 1.0-kg mass after the explosion?
5
€
3 kg
o
m
sec
€
90°
1 kg
b o
€ m
1 kg
−30°
€ 1 kg
m
4 sec
€
€
Kg⋅m
€ iˆ Kg⋅ m€
H €0 o −3.46
ˆ
3.46
i
sec
€ € €o
sec
€ bm
€
Kg⋅ m
ˆ
−3 j sec
−2 ˆj Kg⋅secm
5 ˆj Kg⋅secm
V 0
€ €
€€ €
€
€ €
€
€
31
€
€
€
−3.46iˆ Kg⋅secm
−3 ˆj Kg⋅secm
€
€
p = iˆ 2 + ˆj 2
2
p = 3.46 + 3
p = 4.58
2
Kg ⋅ m
sec
p
v=
m
m
4.58 Kg⋅
sec
v=
1.0kg
m
v = 4.58 sec
, at 221°
€
€
32
19.
Three particles are placed in the xy plane. A 40-g
particle is located at (3, 4)m, and a 50-g particle is
positioned at (-2, -6)m. Where must a 20-g particle be
placed so that the center of mass of this three-particle
system is located at the origin?
20g
40g
x CofM
mx
∑
=
∑m
€
40(3) + 50(−2) + 20( x )
0=
110g
100 −120
x=
= −1m
20
€
50g
yCofM
€
€
20g ⇒ (−1, 7)m
my
∑
=
∑m
40( 4 ) + 50(−6) + 20( y )
0=
110g
300 −160
y=
= 7m
20
36
20.
€
m
At the instant a 2.0-kg particle has a velocity of 4.0 sec
in
the positive x direction, a 3.0-kg particle has a velocity of
m
5.0 sec
in the positive y direction. What is the speed of the
center of mass of the two-particle system? €
v xCofM
mv
∑
=
∑m
x
m
2kg( 4 sec
)
=
5kg
m
v x = 1.6iˆ sec
v yCofM
€
mv
∑
=
∑m
x
m
3kg(5 sec
)
=
5kg
m
v y = 3 ˆj sec
37
€
2
2
ˆ
ˆ
v= i + j
2
2
v = 1.6 + 3
vCofG = 3.4
m
sec
€
38