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vx
v0y
30o
vx
vx
v0y
v1=4 m/s v2=0
m1=0.5kg m2
F (N)
t (s)
-12
00.50.3
Momentum
Do your work on a separate sheet of paper or notebook. For each problem, draw clearly labeled diagrams
showing the masses and velocities for each object before and after the collision. Don’t forget about
directions – momentum, velocity and impulse are VECTORS.
1. A 2.38 kg projectile is fired with a velocity of 50 m/s at an angle of 30o above the horizontal. Determine the
momentum of the projectile
a) At its maximum height
At max height, vy=0 and v=vx=v0cos30=50cos30=43.3m/s
p=mv=(2.38)(43.3)= 103 kg m/s horizontally, to right
b) just before it lands
ax=0 so vx does not change. Because it lands at the same height that it was launched, vy is equal and
opposite v0y. Therefore impact speed is same as launch speed, direction is 30o below horizontal.
p=mv = (2.38)(50) = 119 N kg m/s 30o below horizontal
2. A sports car of mass m has the same kinetic energy as an SUV with mass 3m as each is driven along the
same road. Which vehicle, if either, has the largest momentum and what is the difference in their momenta?
Express your result as a percentage.
3. A bug and the windshield of a moving car collide. Indicate which of the following statements are TRUE.
a. The impact force on the bug and the car are the same magnitude. TRUE (Newtons 3rd Law)
b. The impulse on the bug and the car are the same magnitude. TRUE (both the force and the time of contact
are the same )
c. The change in momentum of the bug and the car are the same magnitude. TRUE (same impulse so same
p)
d. The change in velocity of the bug and the car are the same. FALSE (Since the change in momentum is
the same, the bug with smaller mass must have a greater change in velocity)
4. A 0.50 kg car traveling at 4 m/s along a horizontal frictionless
track collides with a car at rest. A plot of the force exerted on
car1 (the 0.5 kg car) vs. time is shown on the right.
SUVcar
SUVcar
SUVcar
vv
vmmv
KK
3
)3( 2
212
21
SUVcar
SUV
SUV
SUV
car
SUV
car
pp
v
v
vm
mv
p
p
577.0
577.03
3
))(3(
a. Determine the average force applied to car 1 during the collision.
The impulse on car 1 (the 0.5 kg) is the area under the F-t curve = -3 N s
b. What is the impulse applied to car 2? Because the system of car1 + car2 has no net external force on it,
there is also no net impulse. The internal collision forces and impulses are equal and opposite according
to Newton’s 3rd Law. Therefore the impulse on car 2 is equal and opposite the impulse on car 1
c. If the velocity of car 2 after the collision is +4m/s, what is the mass of car 2?
d. Determine the velocity of car 1 after the collision.
Impulse-Momentum Theorem OR Conservation of Momentum
e. Is the collision elastic or inelastic? Justify your answer with appropriate calculations
Collision is inelastic because kinetic energy is lost
5. A block of mass m traveling to the right at a velocity of 1.5v0, hits a stationary concrete block of mass
10m resting on a frictionless surface. After the collision, the first block recoils and travels to the left
with a velocity of –v0. Express your answers in terms of the given variables, m and v0.
The system of 2 blocks is isolated; in other words, there are no net external forces on the system.
Therefore, momentum of the system does not change. Momentum of the system is conserved.
NtF
tF
net
net
65.0/3/3
3
1
1
NstFtF netnet 312
kgm
mvvm
ppNs
ptFnet
75.0
)04()'(3
'3
2
2222
22
22
smv
vvvm
ppNs
ptFnet
/2'
)4'(5.0)'(3
'3
1
1111
11
11
smv
m
vmvmv
vmvmvm
pppp
pp syssys
/2'
5.0
)4)(75.0()4(5.0''
''
''
'
1
1
22111
221111
2121
JKKK
JvmvmKKK
JvmKKK
beforeafter
after
before
3
7)4)(75.0()2)(5.0(''''
4)4)(5.0(0
2
212
212
22212
1121
21
2
212
1121
21
1.5v0
v2=0
Before10m
m
v0
v2’
After10m
m
a. What is the total momentum of the system of the bullet and block?
b. How fast and in what direction will the concrete block be moving after the first block hits it?
(right)
c. Determine and compare the kinetic energy before and after the event. Use this comparison to determine the
type of collision (elastic, inelastic, or explosion).
K lost, so inelastic
d. In another scenario, the two blocks (one mass m traveling to the right at a velocity of 1.5v0, towards a
stationary concrete block of M) stick together after the collision. For the perfectly inelastic collision,
determine the speed and direction of the blocks after the collision and the kinetic energy dissipated in the
collision.
2
0
2
0
2
0212
0212
212
021
2
0
2
021
315.0'
81.0)25.0)(10(')10(
125.1)5.1(
mvKKK
mvvmmvvmmvK
mvvmK
Mafter
before
00
0
136.011
5.1'
')10(0)5.1(
'
vm
mvv
vmmvm
pp syssys
2
0
2
021 125.1)5.1( mvvmKbefore
2
0
2
0212
21 102.0)136.0(11')10( mvvmvmmKafter
2
002.1' mvKKK
00
2
200
25.010
5.2'
'10)(0)5.1(
'
vm
mvv
vmvmvm
pp syssys
afterbefore pvmp
05.1
vi=0
Fnet=272
t
vf=62.0vi=0
Fnet=272
t
vf=62.0
vx
v0y
30o
vx
vx
v0y
6. Rocio strikes a 0.058 kg golf ball with a force of 272 N and gives it a velocity of 62.0 m/s.
a) How long was Rocio’s club in contact with the ball?
b) If the golf ball’s initial velocity of 62m/s is at an angle of 30o above the level ground, what is the
momentum of the ball at its maximum height? Once the ball is airborne, the only force acting on it is
gravity and it becomes a projectile. At the max height, the vertical component of the ball’s velocity is zero
and so the peak velocity is horizontal. Since there is no horizontal acceleration the speed at the peak is
vx = v0cos30 = 62cos30 = 54 m/s
c) What is the momentum of the ball right before it hits the ground? The final speed is 62 m/s since it lands
at the same level. Therefore the impact momentum is
pf = mvf = 0.058(62) = 3.6 kg m/s
7. A 0.24 kg volleyball approaches
Tina with a velocity of 3.8 m/s.
Tina bumps the ball, giving it a
speed of 2.4 m/s in the opposite
direction. What average force did
she apply if the interaction time
between her hands and the ball was
0.025 s?
ssN
F
vvm
F
pt
ptF
ballnet
if
ballnet
ball
ballballnet
013.0272
.596.3
272
)062(058.0)(
vi=3.8
m=0.24 F
t=0.025s
vf=2.4
m=0.24
vi=3.8
m=0.24 F
t=0.025s
vf=2.4
m=0.24
Fthrust=35
t
Fthrust=35
t
8. Small rockets are used to make tiny adjustments in the speeds of satellites. One such rocket has a thrust of
35 N. If it is fired to change the velocity of a 72,000 kg spacecraft by 63 cm/s, how long should it be fired?
9. The graph shows the force, in meganewtons (x106N), exerted by a
rocket engine on the rocket as a function of time. If the rocket's
mass is 4000 kg, at what speed is the rocket moving when the
engine stops firing? Assume it goes straight up starting from rest,
and neglect the force of gravity, which is much less than a
meganewton. (You will have to estimate the average force or
impulse over the 12 sec).
Area under the graph is Favt = Impulse of engine on rocket.
If gravity on the rocket is neglected, then the impulse of the
engine is the net impulse which is equal to the change in
momentum of the rocket. The area under the curve is approximately the area under the red triangle which is
26MNs = 26x106 Ns
Area =
Nt
vvm
t
pF
sNvvmptF
ifballballnet
ifballballnet
5.59025.0
488.1
025.0
))8.3(4.2(24.0)(
.488.1))8.3(4.2(24.0)(
min6.211296
35
45360
35
)63.0(72000
35
72000)(
st
v
F
vvm
F
pt
ptF
rocketnet
if
rocketnet
rocket
rocketrocketnet
smv
vvvmx
ppx
ptF
rocketrocketrocket
rocketrocket
rocketnet
/6500'
)0'(4000)'(1026
'1026
1
1
6
6
vB’vM’
r=5m
v1=200m/s
v2=0
v’ v’
h=0
h
10. A 730-N man stands in the middle of a frozen pond of radius 5.0 m. He is unable to get to the other side
because of a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 1.2-kg
physics textbook horizontally toward the north shore, at a speed of 5.0 m/s. How long does it take him to
reach the south shore? The system of man and
book before and immediately after the throw
have no net external forces acing on it (no
friction, normal balances gravity). Therefore,
the momentum of the system does not change
– it is conserved.
Since the ice is frictionless, the man will slide
at a constant speed of 0.081 m/s. To reach the
shore, he must travel 5 m.
11. a) A 0.030-kg bullet is fired vertically at 200 m/s into a 0.15-kg block that is initially at rest. How high does
the combination rise after the collision, assuming the bullet embeds itself in the block?
(Note: there are two parts to these problems: the collision part and what happens after the collision. The 2
parts are linked by the velocity after the collision)
If we neglect air resistance and look at the bullet and block system immediately before the collision to
immediately after, then we can consider bullet and block to be an isolated system (no net force) in which the
equal and opposite collision forces are FAR bigger than the force/impulse of gravity on the bullet. Then we
can use conservation of momentum to find the speed of the block and bullet immediately after the collision.
Once the bullet transfers momentum (and energy) to the block, the block and bullet have kinetic energy that
is transformed into gravitational potential
energy as the block rises.
The mechanical energy of the system of block,
bullet and Earth is conserved since there are
no external forces doing work on the system.
The only force acting within the system is the
conservative force of gravity. (We cannot use
conservation of momentum of the bullet
+block while they rise because they lose
momentum to the Earth due to the
force/impulse of gravity. To conserve system
momentum, we would have to include the
Earth in the system)
smv
v
vmvm
pp
M
M
BBMM
syssys
/081.0'
)5(2.1'49.740
''0
'
svdt
tdv
M
M
7.61081.0/5'/
/'
v1=200m/sv2=0
v’
vf=0
x
v’
Fg
FN
fk
Before
After
Part 1 – The collision- CONSERVATION OF MOMENTUM of bullet+block; (1-D problem in y-
direction)
Part 2 – Right after the collision - The rise- CONSERVATION OF ENERGY of the
bullet+block+Earth system
b) Another 0.030-kg bullet is fired horizontally at 200 m/s into a 0.15-kg block that is initially at rest. Again, the
bullet embeds itself in the block after the collision. How far does the block/bullet slide if the coefficient of
friction between the block and surface is 0.30?
There is friction acting on the block, so
block and bullet is NOT an isolated
system since friction is a NET external
force that changes the momentum of
the system. However, if we look at the
block and bullet system immediately
before the collision and immediately
after (before friction has decreased the
momentum), then we can consider
bullet and block to be an isolated
system (no net force).
Then we can use conservation of
momentum to find the speed of the
block and bullet at the instant that
momentum is transferred to the block
and just begins to slide.
Once the bullet transfers momentum (and energy) to the block, the block and bullet slide along the surface and
the net force of friction slows it down. Right after the collision, we can use kinematics and Newtons Laws OR
Work-Energy to find the distance it slides.
smv
v
vmmvm
pppp
pp syssys
/3.33'
'18.0)200(03.0
')(0
''
'
2111
2121
mg
vh
ghmmvmm
EE TB
6.562
'
)(')(
2
21
2
2121
v’=14.5
v1=15.7
m1=1345
m2=1923
63.50
v2=?
t
x
a
v
smv
x
x
x
0
/3.330
2/94.2 smga
ammg
amF
amf
amF
kx
xk
xNk
xk
xx
mx
x
xavv xxx
189
))(94.2(23.330
2
2
2
0
2
Part 1 – The collision- CONSERVATION OF MOMENTUM of the block and bullet (1-D problem in x-
direction)
Part 2 – Right after the collision – kinematics and Newton’s 2nd Law
Find the acceleration
using Newton’s 2nd Law:
Now knowing three variables, use kinematics to find x
12. (2-D problem) A 1345 kg car moving east at 15.7 m/s is struck by a 1923 kg car moving north. They are
stuck together and move with an initial velocity of 14.5 m/s at = 63.50. Was the north moving car
exceeding the 20.1 m/s speed limit? THe system of 2 cars is isolated immediately before to immediately
after the collision; momentum to the system is conserved from immediately before to immediately after.
y-direction
)lim(/1.22
36.424071923
5.63sin'3268')19231345(19230
')2(
'
2
2
2
12211
itexceedssmv
v
vvv
vmmvmvm
pp
y
yyy
ysysysys
smv
vmmvm
pppp
pp syssys
/3.33'
')(0
''
'
2111
2121
1.7m
vBvC=0
1.7m
vB’
BEFORE AFTER
vC’
13. (2-D problem) A stationary billiard ball, with a mass of 0.17 kg, is struck by an identical ball
moving at 4.0 m/s. After the collision, the second ball moves 600 to the left of its original direction. The
stationary ball moves 300 to the right of the original ball’s original direction. What is the velocity of each
ball after the collision? If the surface that the balls travel on is frictionless, then the system of the 2 balls is
isolated since no net force acts on the system. Then momentum of the system of the 2 balls is conserved.
There are 2 unknowns. The 2 equations are conservation of p in x-dir and conservation of p in y-direction
x-direction
y-direction
14. A 0.11-kg tin can is resting on top of a 1.7-m-high fence post. A 0.0020-kg bullet is fired horizontally at the can. It strikes the can with a speed of 900 m/s, passes through it, and emerges with a speed of 720 m/s. When the can hits the ground, how far is it from the fence post? Disregard friction while the can is in
contact with the post. (Note: there are two parts to this problem: the collision part and what happens after
the collision. The 2 parts are linked by the velocity after the collision).
System is the Bullet and Can since there is a collision between them. The system is isolated (no net external
force) Look at momentum in x-direction: Momentum before collision equals the momentum after. Once
some of the bullet’s momentum is transferred to the can, it has some horizontal momentum and velocity,
'866.0'5.04
30cos'60cos'04
''
21
21
22112211
vv
vv
vmvmvmvm
pp
xxxx
xafterxbefore
'5.0'866.00
30sin'60sin'00
''
21
21
22112211
vv
vv
vmvmvmvm
pp
yyyy
yafterybefore
smv
smv
vv
vv
/2'
/46.3'
'5.0)'(866.0
'5.0'866.00
1
2
225.0866.0
5.04
21
and becomes a projectile once in air. To find out how far the can goes before hitting the ground, we need
its initial velocity from the collision.
Can is launched into air with horizontal velocity of 3.27m/s and becomes a projectile:
x y
smv
v
v
vmvmvm
pppp
pp
C
C
C
CCBBBB
CBCB
syssys
/27.3'
'11.044.18.1
'11.0)720)(0020.0(0)900)(0020.0(
''0
''
'
t
x
smvx
?
/27.3
t
my
sma
v
smv
y
y
y
7.1
2/8.9
/00
mtvx
sa
yt
tatvy
x
y
yy
93.1)589.0(27.3
589.02
2
21
0
