before conservation of momentum problems 1 =9000 m …
TRANSCRIPT
v1=3
m1=9000
v2=0
m2=64000
BEFORE
v’
m1+m2=73000
AFTER
Conservation of Momentum Problems
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. The diagrams are essential (they can be very simple – just draw all objects as boxes )
- Define the system you are analyzing and justify why conservation of momentum of the system can be
used.
Don’t forget about directions – momentum, velocity and impulse are VECTORS.
1. In a railroad yard, a train is being assembled. An empty boxcar, coasting at 3 m/s along a frictionless track,
strikes a loaded car that is stationary, and the cars couple together. Each of the boxcars has a mass of 9000 kg
when empty, and the loaded car contains 55,000 kg of lumber.
The 2 cars together are an isolated system because there are no net external forces acting on the system (no
friction, gravity and normal add to zero, and the internal collision forces sum to zero). Therefore, the
momentum of the system of both cars is conserved (psys=0 because Fsys = 0).
a. Write the momentum conservation equation for this system in terms of variables (m,v).
b. Find the speed of the coupled boxcars.
c. Calculate and compare the kinetic energy of the system before and after the collision. Use this comparison
to determine the type of collision (elastic, inelastic, or explosion).
Kinetic energy is lost and trains stick together after collision so collision is perfectly inelastic.
smv
v
vmmvm
pppp
pp afterbefore
/37.0'
'730000)3)(9000(
')(0
''
2111
2121
J
vmvm
KKKbefore
500,40
03)9000( 22
1
2
22212
121
21
1
J
vmmKafter
4997
)37.0)(73000(
')(
22
1
2
2121
')('' 212111 vmmvmvmvm
2. An 80 kg astronaut is on a spacewalk away from the shuttle when his tether line breaks. Starting at rest
(relative to the shuttle) he throws the 9.5kg oxygen tank away from him with a velocity of -2.0 m/s so that he
recoils in the opposite direction, towards the shuttle.
vT=vA=0
mT=10mA=80
vA’=?
mT=9.5
mA=80vT’=2
BEFORE AFTER
a. Write the momentum conservation equation for this system in terms of variables (m,v).
b. Find the speed with which the astronaut moves towards the shuttle.
c. Determine the impulse exerted on the oxygen tank by the astronaut. The impulse ON the TANK
is the momentum change OF THE TANK
d. Determine the impulse exerted on the astronaut by the oxygen tank.
Since the force the astronaut exerts on the tank is equal and opposite to the force the tank exerts on the
astronaut (Newton’s 3rd
), and since the forces act over the same duration of time, the impulses are also equal
and opposite.
Therefore, the impulse on the astronaut is FAt = +19 N s
Because there are no other forces acting on the astronaut and tank (astronaut and tank are an isolated system),
these equal and opposite forces and impulses cause equal and opposite momentum changes which is why the
momentum of the whole system does not change (no Fnet, no p)
e. Calculate and compare the total kinetic energy before and after the event. Use this comparison to determine
the type of collision (elastic, inelastic, or explosion).
K gained so explosion f. Determine the maximum distance the astronaut can be from the shuttle when the line breaks and still return
within 60.0 s (the amount of time he can hold his breath).
Once the astronaut throws the tank and recoils due to the thrust of the tank, the astronaut does not accelerate
smv
v
vmvm
pp
A
A
AATT
afterbefore
/24.0'
80)2(5.90
''0
'
0beforeK
J
vmvmKKK AATTATafter
5.22
)25.0)(80()2)(10(
''''
22
122
1
22
122
1
Nsvm
pp
ptF
TT
TT
TT
19)2(5.90'
'
''''0 AATTAATT vmvmvmvm
v=2
mC=50
vC’=?
mB=70
BEFORE
AFTER
vB’=0 mC=50
mB=70
mvtx
t
xv
4.14)60(24.0
This is the distance traveled in 60 s at constant velocity of 0.24m/s
3. A 50 kg cart is moving across a frictionless floor at
2.0 m/s. A 70 kg boy, riding in the cart, jumps off
so that he hits the floor with zero velocity.
The boy and the cart together are an isolated
system because there are no net external forces
acting on the system (no friction, gravity and
normal add to zero and the internal interaction
forces sum to zero). Therefore, the momentum of
the system of boy and cart is conserved (psys=0
because Fsys = 0).
a. Write the momentum conservation equation in
terms of variables (m,v).
b. What was the velocity of the cart after the boy
jumped?
c. How large an impulse did the boy give to the cart?
finding the impulse ON the CART
which is the momentum change OF THE CART (if the
impulse is the net impulse.
d. Calculate and compare the total kinetic energy before and after the event. Use this comparison to
determine the type of collision (elastic, inelastic, or explosion).
Kinetic energy gained so explosion
smv
v
vmvmvmm
pp
C
C
CCBBCB
syssys
/8.4'
')50(0)2)(5070(
'')(
'
sN
vmvm
ppptF
CCCC
CCCCnet
.140)28.4(50
'
'
J
vmvm
KKK
BC
BCbefore
240
2)7050( 22
1
22
122
1
J
vmvm
KKK
CCBB
BCafter
576
)8.4)(50(0
''
''
22
1
22
122
1
')( cccb vmvmm
v=0
m2=50
v2’=10
m2=50
BEFORE AFTER
m1=70
m1=70
v1’=?
4. Two children with masses of 50 kg and 70 kg are at rest on frictionless in-line skates. The larger child pushes
the smaller so that the smaller child rolls away at a speed of 10 m/s.
The boy and girl together are an isolated system because there are no net external forces acting on the
system (no friction, gravity and normal add to zero and the internal interaction forces sum to zero). Because
there is no net force on the system, the momentum of the system of boy and girl is conserved. (psys=0
because Fsys = 0).
a. Write the momentum conservation equation in terms of variables (m,v).
b. Calculate the velocity of the larger child.
c. Calculate the impulse that each child imparts to the other.
d. Calculate and compare the total kinetic energy of the system before and after the event. Use this
comparison to determine the type of collision (elastic, inelastic, or explosion).
KE gained so explosion
smv
v
vmvm
pp syssys
/14.7'
)10)(50('700
''0
'
1
1
2211
sNtFtF
sN
vvmvmvm
ppptF
netnet
net
.500
.500)014.7(70
)'('
'
12
1111111
1111
0beforeK
J
vmvm
KKKafter
4280
)10)(50()14.7(70
''
''
22
122
1
2
22212
1121
21
''0 1211 vmvm
v1=45m1=0.050
v2=0
BEFORECollision
AFTERCollision
m2=0.8v1’=45
y=1
.9m
v2’
Projectile Motion of Apple
v2’
x=?
5. A 0.8 kg apple is balanced on a circus performer's head 1.9 m above the ground. An archer shoots a 50 g
arrow at the apple with a speed of 45 m/s. The arrow passes through the apple and emerges with a speed of
12 m/s.
The arrow and apple together can be considered an isolated system if we examine the collision immediately
before the collision takes place and immediately after so that the impulse of gravity on the arrow is
negligible. Then no net external forces act on the system (neglect the friction between apple and head, no air
resistance apple gravity and normal add to zero and the internal interaction forces sum to zero). Therefore,
the momentum of the system of arrow and apple is conserved (psys=0 because Fsys = 0).
a. Write the momentum conservation equation in terms of variables (m,v).
b. Find the speed of the apple as it flies off the performer's head.
c. What is the impulse exerted on the apple by the arrow? Impulse ON the APPLE is the momentum change
OF THE APPLE
d. What is the impulse exerted on the arrow by the apple?
Equal and opposite impulses since the collision forces are equal and opposite by Newton’s 3rd
Law and the
time during which they act is also the same.
smv
v
vmvmvm
pp syssys
/06.2'
'8.0)12)(05.0()45)(05.0(
''0
'
2
2
221111
sNvm
ppptFnet
.65.1)06.2(8.00'
'
22
2222
sNtFtF netnet .65.121
'' 121111 vmvmvm
v1=8m/sv2=0
Before
m2=12kgm1= 6kg
v1’=? v2’=5.33
Afterm2=12kgm1= 6kg
ptotal=+48 kg m/s
e. Calculate and compare the K before and after the event. Use this comparison to determine the type of
collision (elastic, inelastic, or explosion).
K lost so inelastic
f. When the apple hits the ground, how far behind the clown is it? Neglect friction when the apple is atop
the clown’s head. Once the collision is over, the arrow has transferred momentum and energy to the apple
and so the apple has horizontal velocity and it undergoes projectile motion (if we neglect resistance). It is
horizontally launched from a height of 1.9m. (As the apple undergoes projectile motion, its momenti=um
changes in magnitude and direction; momentum is NOT conserved during the projectile motion unless we
include the Earth in the system.)
x y
6. An object, which has a mass of 6.0 kg, is moving to the right with a velocity of 8.0 m/s when it collides with
a second mass of 12.0 kg which is initially at rest. After the collision, the 12.0 kg mass moves off to the right
with a new velocity of 5.33 m/s. Neglect friction. Momentum is conserved in the system of the 2 blocks
because there is no net external force on the system; the internal collision forces do not change momentum
of the system because they sum to zero (psys=0 because Fsys = 0)
a. Write the momentum conservation equation in terms of variables (m,v).
J
vm
KKKbefore
6.5045)05.0(
0
22
1
2
1121
21
J
vmvm
KKKafter
72.5
)06.2)(1(12)05.0(
''
''
22
122
1
2
22212
1121
21
t
x
smvx
/06.2
t
my
sma
v
v
y
y
y
9.1
/8.9
0
2
0
mtvx
sa
yt
tatvy
x
y
yy
28.1)623.0)(06.2(
623.02
2
2
10
'' 121111 vmvmvm
v1=12m/sv2=0
Before
m2=6kgm1= 2kg
v1’=? v2’=?
Afterm2=6kgm1= 2kg
(ptotal=+24 kg m/s)
b. Determine the velocity of the 6.0 kg object after the collision
c. Is the collision elastic? How do you know? Support your answer with evidence.
Kbefore = Kafter. Therefore the collision is elastic
7. An object of mass m1=2.0 kg, moving with a velocity of v1=12.0 m/s collides head-on with a stationary
object whose mass is m2=6.0 kg. The collision is elastic. What are the velocities of the two objects after the
collision? Neglect friction. (Note: there are two unknowns here. Elastic collision means that you have 2
conservation equations to use: conservation of momentum and conservation of mechanical energy. You can
also use conservation of momentum along with another equation involving relative velocities; finish and look
at the geogebra simulation to find this third relationship). Momentum is conserved in the system of the 2
blocks because there is no net external force on the system; the internal collision forces do not change
momentum of the system because they sum to zero (psys=0 because Fsys = 0)
There are two unknown variables here, v1’ and v2’. Because the collision is elastic, we have two equations to
use to solve for the two unknowns
1. Conservation of momentum: the 2 block system has no net force on it so there is no change in
momentum
2. Conservation of energy: because the collision is elastic, kinetic energy of the system is also conserved
smv
v
vmvmvm
pp syssys
/66.2'
)33.5)(12('6)8)(6(
''0
'
1
1
221111
J
vm
KKK sys
1928)6(
0
22
1
2
1121
21
J
vmvm
KKK sys
192
)33.5)(12()66.2)(6(
''
'''
22
122
1
2
22212
1121
21
There are many ways to use these 2 equations to solve for the unknowns. One way is to put expression for v1’
from 1st equation into 2nd
There are 2 solutions to the equation
- v2’ = 0: this makes no sense because it would mean that after being struck, object 2 would not move
- v2’ = 6.0 m/s. This is the one that makes physical sense
Check your answer by using these values to make sure that with them pbefore = pafter and Kbefore = Kafter
2nd
METHOD (easier) There is another equation that can be used in elastic equations and therefore another
set of 2 equations that can be used. For elastic collisions only, the magnitude of the relative velocity does not
change before and after the collision. In fact the relative velocity before an elastic collision is equal in mag
and opposite in direction to the rel velocity after collision:
v1 - v2 = -(v1’ – v2’)
This equation can be used along with the conservation of momentum equation to solve for the two
unknowns.
There are many ways to use these 2 equations to solve for the unknowns. One way is to add the two
expressions together to remove v1’ and solve for v2’. Then use either of the two equations to get v1’.
'312'
'3'12
'6'2)12)(2(
''0
''
'
21
21
21
221111
2121
vv
vv
vv
vmvmvm
pppp
pp syssys
2
2
2
1
2
22212
11212
1121
2121
'3'144
''0
''
vv
vmvmvm
KKKK
KK afterbefore
)'6('40
'4'244848
'3)''816(9144
'3)]'4(3['3)'312(144
'3'144
22
2
22
2
2
2
22
2
2
2
2
2
2
2
2
2
2
2
1
vv
vv
vvv
vvvv
vv
smvv
smv
/0.6)6(312'312'
/0.6'
21
2
'3'12
'6'2)12)(2(
''0
''
'
21
21
221111
2121
vv
vv
vmvmvm
pppp
pp syssys
''12
)''(
12
2121
vv
vvvv
v1=2m/s v2=-4m/s
Before
m2=6kgm1= 4kg
v1’=? v2’=?
After
m2m1
m1m2
ptotal=-16kg m/s
Check your answer by using these values to make sure that with them pbefore = pafter and Kbefore = Kafter
8. A particle of mass 4.0 kg, initially moving with a velocity of 2.0 m/s collides elastically with a particle of
mass 6.0 kg. initially moving with a velocity of -4.0 m/s. What are the velocities of the two particles after the
collision? There is no friction or air resistance. Momentum is conserved in the system of the 2 particles
because there is no net external force on the system; the internal collision forces do not change momentum
of the system because they sum to zero.
There are two unknown variables here, v1’ and v2. Therefore, we need to set up a system of 2 equations.
METHOD 1: In elastic collisions of an isolated system, both momentum and kinetic energy are conserved.
There are many ways to use these 2 equations to solve for the unknowns. One way is to put expression for v1’
from 1st equation into 2nd
Find the roots of the quadratic equation to solve this and you get 2 possible solutions:
- v2’ = 0.80 m/s: this is the only solution that makes physical sense
smv
smv
v
/0.6'
/0.6'
'424
1
2
2
'5.14'
'3'28
'6'416
'6'4)4)(6()2)(4(
''
''
'
21
21
21
21
22112211
2121
vv
vv
vv
vv
vmvmvmvm
pppp
pp syssys
2
2
2
1
2
2
2
1
2
22212
11212
22212
1121
2121
'3'256
'3'2488
''
''
'
vv
vv
vmvmvmvm
KKKK
KK syssys
024'24'5.7
'5.7'2432'3)''1216(2'3)'5.14(256
'3'256
2
2
2
2
22
2
2
2
24
92
2
2
2
2
2
2
2
1
vv
vvvvvvv
vv
v1=140m/sv2=0
v’
vf=0
x
v0=v’
Fg
FN
fk
Before
Collision
(ptotal=3.5)
After
collision
Slides with
constant
acceleration
m2=2kg
m2+m2
m1=0.025kg
- v2’ = -4.0 m/s: this makes no sense because it would mean that after being struck, the velocity of object
2 would not change
Check your answer by using these values to make sure that with them pbefore = pafter and Kbefore = Kafter
2nd
METHOD (easier) In elastic collisions of an isolated system, momentum is conserved and the relative
velocity of the 2 objects before the collision is equal and opposite to the relative velocity after the collision
Check your answer by using these values to make sure that with them pbefore = pafter and Kbefore = Kafter
9. A 25 g bullet traveling horizontally at
140 m/s is fired into 2.0 kg block,
initially at rest on a level surface
with a coefficient of kinetic friction
of 0.28. The bullet gets imbedded in
the block. The momentum
transferred to the block causes it to
slide along the surface. Note: There
are 2 parts to this problem: collision
(before to immediately after)
followed by the sliding of the block.
It is important that you draw 3
diagrams here: before collision,
immediately after collision, and the
sliding with constant acceleration
along surface.
a) Determine the velocity of the block-bullet combination immediately after the bullet collides with the
block (before it starts to slide). The bullet and block together can be considered an isolated system (no
smvv
smv
/2.5'5.14'
/80.0'
21
2
'3'28
'6'416
'6'4)4)(6()2)(4(
''
''
'
21
21
21
22112211
2121
vv
vv
vv
vmvmvmvm
pppp
pp syssys
6''
''6
'')4(2
)''(
21
12
12
2121
vv
vv
vv
vvvv
smvv
smv
v
vvvv
/2.56''
/8.0'
'54
'3)6'(2'3'28'
21
2
2
2221
net force or impulse) if we examine the collision immediately before and immediately after it happens so
that before the collision, the impulse of gravity on the bullet is negligible, and right after the collision,
the impulse of friction on the block is also negligible. Then no net external forces act on the system
(block weight and normal add to zero, and the internal interaction forces sum to zero). Therefore, the
momentum of the block+bullet system is conserved from immediately before to immediately after the
collision (psys=0 because Fsys = 0) .
b) Determine how far the bullet and the block slide along the surface before coming to rest. Once the
collision is over, the bullet has transferred its momentum and some energy to the block and so the block
and bullet together have horizontal velocity, v’, causing the block-bullet to slide along the surface. Since
there is friction, the block/bullet will slow down with constant acceleration.
We can find the distance it slides in 2 ways
METHOD 1: Newton’s Laws and kinematics. Once the collision is over, all the forces acting on the
sliding block are constant and therefore the acceleration is constant (we can use kinematics to find x
once we find acceleration (we know v0 and vf)). If you use Newton’s Laws to find acceleration, you must
follow the steps you have learned: FBD, apply Newton’s 2nd
. Diagram above has the FBD
Now we know three variables for kinematics
METHOD 2: Conservation of Energy. Once the collision is over, the bullet and block have kinetic
energy. However, the system of mechanical energy, which includes the bullet+block+Earth are is NOT
isolated because friction does negative work on the system removing kinetic energy from it and
converting the lost kinetic energy to nonmechanical energy such as heat and sound. Although
mechanical energy is not conserved, total energy is (as always) conserved
smv
v
vmmvm
vmvmvm
pppp
pp syssys
/73.1'
'025.2)140)(025.0(
')(
''0
''
'
2111
2111
2121
2
2121
21
21
/744.2
)()(
)(
)(
smaga
ammgmm
ammf
ammF
xkx
xk
xk
xx
t
x
sma
v
smvv
x
x
x
?
/744.2
0
/73.1'
2
0
ma
vvx
xavv
x
xx
xxx
54.0488.5
99.20
2
2
2
0
2
2
0
2
vS=5.0m/s
mS= 1.2kg
v’=?
mC=0.8 kg
BEFORE
AFTER
mS+mC
(ptotal=+6 kg m/s)
c) How much energy was lost as the bullet was lodged in the block? Because the collision was perfectly
inelastic (the 2 objects stuck together), kinetic energy was lost.
d) The collision of the bullet with the block took place over 0.0022s. Determine the average net force
exerted on the bullet. The impulse ON THE BULLET caused the momentum change OF THE BULLET
10. A 1.20-kg skateboard is coasting along the
pavement at a speed of 5.00 m/s when a 0.800-
kg cat drops from a tree vertically downward
onto the skateboard. What is the speed of the
skateboard-cat combination? (Remember that
momentum is a vector and that conservation of
momentum is a vector equation that is applied
separately in x- and y- directions)
The cat falls vertically so it carries no horizontal
momentum. Therefore the initial momentum of
6 kg m/s carried by the skateboard does not
change. What does change is the increased mass
carrying the 6 kg m/s momentum and so the
velocity goes down once the cat’s mass is added
to the skateboard.
mg
vd
vmmgdmm
dfvmm
EWK
EWE
k
k
k
stoppedfi
fNCi
54.0488.5
99.2
2
'
')()(
0')(
2
2
212
121
2
212
1
J
vm
KKKbefore
245140)025.0(
0
22
1
2
1121
21
J
vmm
KKKafter
03.373.1)025.2(
')(
''
22
1
2
2121
21
JKKK beforeafter 242
Nt
vmvm
t
pp
t
pF
ptF
net
net
15700022.0
)14073.1(025.0'' 1111111
11
Horizontal momentum: The cat and skateboard together can be considered an isolated system (no net force or
impulse) because
- the impulse of gravity on the cat is vertical and does not affect horizontal momentum and
- the skateboard moves with constant velocity which means there is no friction acting on it (no net
horizontal force).
Then no net horizontal external forces act on the system (the internal interaction forces sum to zero).
Therefore, the horizontal momentum of the cat+skateboard system is conserved
(psys x=0 because Fsys x = 0)
Vertical momentum (which we are not concerned about in this problem but it is important to consider the
concepts): The impulse of gravity on the cat is a large net impulse that results in a transfer of momentum to
the Earth (py cat = -py Earth) so in order to conserve momentum vertically, we would have to add the earth to
the system of cat and skateboard. Since the force of gravity on the cat due to the earth is equal and opposite
to the force of gravity on the earth due to the cat, then there is no net external force (impulse) on the
cat+skateboard+Earth system
In the x-direction: Before the collision, the cat has no horizontal momentum. All the px is carried by the skateboard. After the
collision, the horizontal momentum is conserved and since it is now carried by the increased mass of
skateboard and cat, the speed of the skateboard goes down (and the cat gains horizontal speed). The both
travel together at v’ (this is a perfectly inelastic collision)
11. A 65.0-kg person throws a 0.045 0-kg snowball forward with a ground speed of 30.0 m/s. A second person,
with a mass of 60.0 kg, catches the snowball. Both people are on skates. The first person is initially moving
forward with a speed of 2.50 m/s, and the second person is initially at rest. What are the velocities of the two
people after the snowball is exchanged? Disregard the friction between the skates and the ice.
The two skaters + snowball together can be considered an isolated system (no net horizontal force or
impulse) because there is no net external force on the system: no air resistance, no friction between skates
and ice, normal and weight on skaters cancel, and the vertical impulse on the snowball does not affect the
horizontal momentum. Therefore, the horizontal momentum of the two skaters + snowball system is
conserved.
smmm
vmv
vmmvm
pppp
pp
sC
sxSx
xsCsxS
SxCxSxCx
xafterxbefore
/3)2.18.0(
)5)(2.1(
)('
')(0
''
m1+mS
=65+0.045 m2=60kg
v1=2.5m/s
v2=0
m1
vS=30m/s
v2=0mS
v1'
m1
v2'v1'
m2+mS
Before
After
(ptotal=+163 kg m/s)
After
Before
12. A 2.00 kg block is sliding at 15.0 m/s to the right. It collides with and sticks to a 1.00 kg block attached to a
spring connected to a wall as shown. The horizontal surface is frictionless. The spring has a spring constant
of 120 N/m. Note: There are 2 parts to this problem: collision (before to immediately after) followed by
blocks compressing spring.
a) Determine the velocity of the double block combination immediately after the collision.
To analyze the collision part of the problem, we must look at momentum (mechanical energy is not
conserved in this collision and the amount of energy lost during the collision is not known). The two
blocks together can be considered an isolated system (no net force or impulse) if we examine the
collision before and immediately after it happens before the compression of the spring begins and the
spring force acts on the blocks as an external force. Then no net external forces act on the system (block
weights and normal add to zero, there is no friction, and the internal interaction forces sum to zero).
Therefore, the momentum of the two block system is conserved from before to immediately after the
collision.
smv
v
v
vmvmvmm
pppp
pp
SSS
SS
syssys
/48.2'
35.1'656.162
)30)(045.0('655.2)045.65(
')(
''
'
1
1
1
1111
11
smv
v
vmmvm
pppp
pp
SSS
SS
syssys
/0225.0'
'045.6030)045.0(
')(
''
'
2
2
22
22
k=120N/m
k=120N/m
v1=15m/s
Before
collision
(ptotal=+30 kg m/s) m1=2kg
m1 m2
m2=1kg
v2=0
m1 m2
v’
After
collision
k=120N/m
m1 m2Spring
compression
b) What will be the maximum distance the spring is compressed? Once the collision is over, the first
block has transferred some of its momentum to the second block so that the two blocks move together
horizontally causing the spring to compress.
(Note: cannot use conservation of momentum for the second part since the spring is attached to the
wall/earth which exerts a large external net force on the blocks/spring system. In order to have
conservation of momentum, the Earth would have to be included in the system making the problem
intractable.)
To analyze the second part of the problem in which the blocks compress the spring, it is easiest to use
conservation of mechanical energy; the system of the two blocks and the spring is energetically isolated
since there are no external forces doing work on the system (no friction, normal force and gravity do no
work). Therefore, the mechanical energy of the blocks/spring system is conserved. Initial point is when
spring is not compressed and blocks have kinetic energy. Final point is when spring is maximally
compressed (all the K was transformed into US)
smv
v
vmmvm
pppp
pp syssys
/10'
'315)2(
')(
''
'
2111
2121
mk
vmmx
xkvmm
UKUK
EE
SffSii
fi
58.1')(
')(
2
21
2
212
2121
m
vm
M
L38o
Before
collision
After
Collision
EB
h=0
h
Lcos38
v’ET
mRvR’
13. A rifle, which has a mass of 5.50 kg, fires a bullet, which has a mass of m=65.0grams, at a ballistic
pendulum. The ballistic pendulum consists of a block of wood, which has a mass M = 5.00 kg, attached to
two strings of length L=1.25m. When the block is struck by the bullet, the block swings backward until the
angle between the ballistics
pendulum and the vertical reaches a
maximum angle of 38o. Note: There
are 2 parts to this problem: collision
(before to immediately after)
followed by the swinging of
pendulum.
a) Determine the maximum gravitational potential energy of the ballistic pendulum-Earth system when it
reaches its maximum angle. If we define the top of the block as the height of zero gravitational potential
energy, then the max height the block swings to is h = L – Lcos38 = 1.25(1-cos38) = 0.265 m. Potential
energy at that max height is
b) Determine the velocity of the block of wood+bullet immediately after being struck
After being struck, the collision is over. Once the collision is over, the bullet has transferred its
momentum (and kinetic energy) to block+bullet so that it swings up.
(Note: we cannot use conservation of momentum for this second part because the block-bullet is
attached to a string which is attached to the wall/earth; the Earth (via the string) therefore exerts an
external net force on the blocks/bullet system which changes its momentum. In order to have
conservation of momentum, the Earth would have to be included in the system making the problem
intractable.)
To analyze the second part of the problem in which the block/bullet swings as a pendulum, it is easiest
to use conservation of mechanical energy; the system of the block+bullet+Earth is energetically isolated
since there are no external forces doing work on the system (no friction, tension force is centripetal and
does no work). Therefore, the mechanical energy of the block+bullet+Earth system is conserved. Initial
point is bottom of pendulum (UG=0). Final point is top at max height (max UG) (all the K was
transformed into US)
JghmMUg 2.13)265.0)(8.9(065.5)(max
smghv
ghMmvMm
UKUK
EE
gffgii
TB
/28.22'
2.13)(')( 2
21
c) Determine the velocity of the bullet, vB, immediately before colliding with the block of wood.
To analyze the collision part of the problem, we must look at momentum (mechanical energy is not
conserved in this collision and the amount of energy lost during the collision is not known). The
bullet+block together can be considered an isolated system (no net force or impulse) if we examine the
collision before and immediately after it happens before the block swings and the tension force acts on
the block+bullet as an external force. Then no net external forces act on the system (block weights and
tension add to zero, the impulse of gravity on bullet does not affect horizontal momentum, and the
internal interaction forces sum to zero). Therefore, the momentum of the bullet+block system is
conserved from before to immediately after the collision.
d) How much work was done on the bullet as it got stopped by the block of wood (in other words, how
much energy was lost in the collision?)
The collision was perfectly elastic, so kinetic energy was lost
The amount of mechanical energy lost in the collision is due to the work done by the block on the bullet
in slowing it down
e) Determine the recoil velocity of the rifle. The interaction between rifle+bullet is a “collision” or
explosion. The system of rifle + bullet is isolated because there are no net forces acting on the system.
Therefore, momentum of the system is conserved.
f) How much energy was released when the bullet was fired (in other words, how much energy was gained
in explosion?)
smm
vMmv
vMmvm
pppp
pp
m
m
MmMm
syssys
/178065.0
28.2)065.5(')(
')(
''
'
J
mv
KKK
m
Mmbefore
1030178)065.0(
0
22
1
2
21
J
vMm
KKK Mmafter
2.1328.2)065.5(
')(
''
22
1
22
1
JKKK beforeafter 1017
smm
vmv
vmvm
pppp
pp
R
mR
mRR
mRmR
afterbefore
/1.25.5
178)065.0('
'0
''
0beforeK
JmvvmKKK mRRmRafter 1042'''' 22
122
1
JKKK beforeafter 1042