MOMENTUM LESSON 5
Momentum Conservation in Explosions
Recall, total system momentum is conserved for collisions between objects
in an isolated system.
For collisions occurring in isolated systems, there are no exceptions to
this law.
WHAT IS AN ISOLATED SYSTEM?
The principle of momentum conservation can be applied to explosions.
In an explosion, an internal impulse acts in order to propel the parts of
a system (often a single object) into a variety of directions.
After the explosion, the individual parts of the system (which is often
a collection of fragments from the original object) have momentum.
If the vector sum of all individual parts of the system could be added
together to determine the total momentum after the explosion, then it
should be the same as the total momentum before the explosion.
Just like in collisions, total system momentum is conserved.
DEMO: Consider two low-friction carts at rest on a track. The system
consists of the two individual carts initially at rest. The total momentum of
the system is zero before the explosion.
The spring is compressed between the two carts. The string is cut, the
plunger is released, and an explosion-like impulse sets both carts in motion
along the track in opposite directions. One cart acquires a rightward
momentum while the other cart acquires a leftward momentum.
If 20 units of forward momentum are acquired by the rightward-moving cart,
then 20 units of backwards momentum is acquired by the leftward-moving
cart. The vector sum of the momentum of the individual carts is 0 units.
Total system momentum is conserved.
Equal and Opposite Momentum Changes
Just like in collisions, the two objects involved encounter the same force for
the same amount of time directed in opposite directions.
This results in impulses which are equal in magnitude and opposite in
direction.
And since an impulse causes and is equal to a change in momentum,
both carts encounter momentum changes which are equal in
magnitude and opposite in direction.
If the exploding system includes two objects or two parts, this
principle can be stated in the form of an equation as:
MOMENTUM (before) = MOMENTUM (after)
0 = m1Δv1 + m2Δv2
If the masses of the two objects are equal, then their post-explosion velocity
will be equal in magnitude (assuming the system is initially at rest).
WHAT HAPPENS IF THE MASSES ARE NOT EQUAL?
If the masses of the two objects are unequal, then they will be set in motion
by the explosion with different speeds. Yet even if the masses of the two
objects are different, the momentum change of the two objects (mass •
velocity change) will be equal in magnitude.
In each of these systems, is total system momentum conserved?
In each of the above situations, the impulse on the carts is the same - a value
of 20 kg•cm/s (or cN•s). Since the same spring is used, the same impulse is
delivered.
Since each cart experiences the same impulse, each cart encounters the same
momentum change in every situation - a value of 20 kg•cm/s.
For the same momentum change, an object with twice the mass will
encounter one-half the velocity change. And an object with four times the
mass will encounter one-fourth the momentum change.
Solving Explosion Momentum Problems
EXAMPLE: A 56.2-gram tennis ball is loaded into a 1.27-kg
homemade cannon. The cannon is at rest when it is ignited.
Immediately after the impulse of the explosion, a photogate timer
measures the cannon to recoil backwards a distance of 6.1 cm in
0.0218 seconds. Determine the post-explosion speed of the cannon
and of the tennis ball.
Like any problem in physics, this one is best approached by listing the
known information.
Given:
Cannon:
Ball:
m = 1.27 kg
d = 6.1 cm
m = 56.2 g = 0.0562 kg
t = 0.0218 s
1. The strategy for solving for the speed of the cannon is to recognize that
the cannon travels 6.1 cm at a constant speed in the 0.0218 seconds.
The speed can be assumed constant since the problem states that it
was measured after the impulse of the explosion when the acceleration
had ceased. Since the cannon was moving at constant speed during
this time, the distance/time ratio will provide a post-explosion speed
value.
vcannon = d / t = (6.1 cm) / (0.0218 s) = 280 cm/s (rounded)
2. The strategy for solving for the post-explosion speed of the tennis ball
involves using momentum conservation principles.
Using a momentum table similar to the one used previously for collision
problems, we can determine the velocity of the tennis ball.
Momentum
Momentum
Before Explosion
After Explosion
(1.27 kg) • (-280 cm/s)
Cannon
0
Tennis Ball
0
(0.0562 kg) • v
Total
0
0
= -355 kg•cm/s
The variable v is used for the post-explosion velocity of the tennis ball.
Using the table, one would state that the sum of the cannon and the tennis
ball's momentum after the explosion must sum to the total system
momentum of 0 as listed in the last row of the table. Thus,
-355 kg•cm/s + (0.0562 kg) • v = 0
Solving for v yields 6323 cm/s or 63.2 m/s - that's 141 miles/hour!
Using the table means that you can use the same problem-solving strategy
for both collisions and explosions.
After all, it is the same momentum conservation principle which governs
both situations.
Whether it is a collision or an explosion, if it occurs in an isolated
system, then each object involved encounters the same impulse to
cause the same momentum change.
The impulse and momentum change on each object are equal in
magnitude and opposite in direction. Thus, the total system
momentum is conserved.
MOMENTUM LESSON 5 HOMEWORK
1. Two pop cans are at rest on a stand. A firecracker is placed between the cans and lit.
The firecracker explodes and exerts equal and opposite forces on the two cans. Assuming
the system of two cans to be isolated, the post-explosion momentum of the system ____.
Explain answer.
a. is dependent upon the mass and velocities of the two cans
b. is dependent upon the velocities of the two cans (but not their mass)
c. is typically a very large value
d. can be either a positive, negative or zero value
e. is definitely zero
2. Students of varying mass are placed on large carts and deliver impulses to each other's
carts, thus changing their momenta. In some cases, the carts are loaded with equal mass;
in other cases they are unequal. In some cases, the students push off each other; in other
cases, only one team does the pushing. For each situation, list the letter of the team which
ends up with the greatest momentum. If they have the same momentum, then do not list a
letter for that situation. Enter the four letters (or three or two or ...) in alphabetical order.
Explain answer.
3. Two ice dancers are at rest on the ice, facing each other with their hands together. They
push off on each other in order to set each other in motion. The subsequent momentum
change (magnitude only) of the two skaters will be ____. Explain answer.
a. greatest for the skater who is pushed upon with the greatest force
b. greatest for the skater who pushes with the greatest force
c. the same for each skater
d. greatest for the skater with the most mass
e. greatest for the skater with the least mass
4. A 62.1-kg male ice skater is facing a 42.8-kg female ice skater. They are at rest on the
ice. They push off each other and move in opposite directions. The female skater moves
backwards with a speed of 3.11 m/s. Determine the post-impulse speed of the male
skater.
5. A 1.5-kg cannon is mounted on top of a 2.0-kg cart and loaded with a 52.7 gram ball.
The cannon, cart, and ball are moving forward with a speed of 1.27 m/s. The cannon is
ignited and launches a 52.7 gram ball forward with a speed of 75 m/s. Determine the
post-explosion velocity of the cannon and cart.
6. If a projectile is launched horizontally with a speed of 12.0 m/s from the top of a 24.6meter high building. Determine the horizontal displacement of the projectile.
7. Two identical freight cars roll without friction towards each other on a level track. One
rolls at 2 m/s and the other rolls at 1 m/s. After the cars collide, they couple (attach
together) and roll together with a speed of _____________.
8. Paul D. Trigger fires a bullet. The speed of the bullet will be the same as the speed of
the recoiling gun ______________.
a.
b.
c.
d.
e.
because momentum is conserved
because velocity is conserved
because both velocity and momentum are conserved
only if the mass of the bullet equals the mass of the gun
none of these
9. A karate expert executes a swift blow and severs a cement block with her or his bare
hand. The
a.
b.
c.
d.
e.
impulse on both the block and the expert's hand have the same magnitude.
force on both the block and the expert's hand have the same magnitude.
time of impact on both the block and the expert's hand is the same.
all of the above.
none of the above.
10. Impulses are smaller when bounces take place.
a. True
b. False
11. A piece of putty moving with 1 unit of momentum strikes and sticks to a heavy
bowling ball that is initially at rest. After the putty sticks to the ball, both are set in
motion with a combined momentum that is ___.
a. less than 1 unit
b. more than 1 unit
c. 1 unit
d. not enough information
12. A large force acting for a long amount of time on a small mass will produce a
______.
a. small velocity change
d. small acceleration
b. large velocity change
e. two of the above
c. small momentum change
13. A 5-N force is applied to a 3-kg object to change its velocity from +9 m/s to +3 m/s.
The momentum change of the object is:
a. -2.5 kg*m/s
b. -10 kg*m/s
c. -18 kg*m/s
d. -45 kg*m/s
e. none of these
14. A 5-N force is applied to a 3-kg object to change its velocity from +9 m/s to +3 m/s.
The impulse experienced by the object is:
a. -2.5 N*s
b. -10 N*s
c. -18 N*s
d. -45 N*s
e. none of these
15. A 5-N force is applied to a 3-kg object to change its velocity from +9 m/s to +3 m/s.
The impulse acts for a time period of
a. 1.8 s
b. 2.5 s
c. 3.6 s
d. 10 s
e. none of these
16. A 0.80-kg ball strikes a wall moving at 5.0 m/s and rebounds in the opposite
direction at 3.5 m/s. If the collision with the wall endures for a total time of 0.0080 s, then
determine the average force acting upon the ball.
MOMENTUM LESSON 5 HOMEWORK KEY
1. e
2. none
3. c
4. 2.14m/s
5. 0.16 m/s
6. 26.8 m
7. 0.5 m/s
8. d
9. d
10. false
11. c
12. b
13. c
14. c
15. c
16. –8.5 x 102 N
MOMENTUM LESSON 5 HOMEWORK
1. Two pop cans are at rest on a stand. A firecracker is placed between the cans and lit.
The firecracker explodes and exerts equal and opposite forces on the two cans. Assuming
the system of two cans to be isolated, the post-explosion momentum of the system ____.
Explain answer.
a. is dependent upon the mass and velocities of the two cans
b. is dependent upon the velocities of the two cans (but not their mass)
c. is typically a very large value
d. can be either a positive, negative or zero value
e. is definitely zero
Answer: E
Before the explosion, the cans were at rest. Thus, the pre-explosion momentum of the
system was 0. If the system can be considered isolated (as stated), then the post-explosion
momentum must also be 0.
2. Students of varying mass are placed on large carts and deliver impulses to each other's
carts, thus changing their momenta. In some cases, the carts are loaded with equal mass;
in other cases they are unequal. In some cases, the students push off each other; in other
cases, only one team does the pushing. For each situation, list the letter of the team which
ends up with the greatest momentum. If they have the same momentum, then do not list a
letter for that situation. Enter the four letters (or three or two or ...) in alphabetical order.
Explain answer.
Answer: -- (don't list any letters)
Don't be fooled!! In each of the four situations, the two carts involved in the impulse
encounter the same impulse causing the same momentum change which results in the
same final momentum. Differences in mass will only effect the resulting velocity change.
And if Team A "pushes on" Team B's cart, then there is an equal and opposite reaction
force exerted by Team B's cart upon the legs of the Team A "pusher". So there is never
any situation in the given four in which one of the teams encounters a greater momentum.
3. Two ice dancers are at rest on the ice, facing each other with their hands together. They
push off on each other in order to set each other in motion. The subsequent momentum
change (magnitude only) of the two skaters will be ____. Explain answer.
a. greatest for the skater who is pushed upon with the greatest force
b. greatest for the skater who pushes with the greatest force
c. the same for each skater
d. greatest for the skater with the most mass
e. greatest for the skater with the least mass
Answer: C
In this situation, the force on the first ice dancer is the same as the force on the second ice
dancer (Newton's third law of motion). And these forces act for the same amount of time
to cause equal impulses on each skater. Since impulse is equal to momentum change,
both skaters must also have equal momentum changes. The mass of the individual skaters
will only effect the subsequent velocity change.
4. A 62.1-kg male ice skater is facing a 42.8-kg female ice skater. They are at rest on the
ice. They push off each other and move in opposite directions. The female skater moves
backwards with a speed of 3.11 m/s. Determine the post-impulse speed of the male
skater.
Answer: 2.14 m/s
Momentum
Momentum
Before Explosion
After Explosion
Male Skater
0
(62.1 kg) • v
Female Skater
0
(42.8 kg) • (-3.11 m/s)
= -133.11 kg•m/s
0
Total
0
(62.1 kg) • v + (-133.11 kg•m/s) = 0
v = (133.11 kg•m/s) / (62.1 kg)
v = 2.14 m/s
5. A 1.5-kg cannon is mounted on top of a 2.0-kg cart and loaded with a 52.7 gram ball.
The cannon, cart, and ball are moving forward with a speed of 1.27 m/s. The cannon is
ignited and launches a 52.7 gram ball forward with a speed of 75 m/s. Determine the
post-explosion velocity of the cannon and cart.
Answer: 0.16 m/s (rounded)
Cart &
Cannon
Tennis Ball
Momentum
Momentum
Before Explosion
After Explosion
(3.5 kg) • (1.27 m/s)
(3.5 kg) • v
= 4.445 kg•m/s
(0.0527 kg) • (1.27
m/s)
(0.0527 kg) • (75
m/s)
Total
= 0.06693 kg•m/s
= 3.9525 kg•m/s
4.5119 kg•m/s
4.5119 kg•m/s
4.445 kg•m/s + 0.06693 kg•m/s = (3.5 kg) • v + 3.9525 kg•m/s
4.445 kg•m/s + 0.06693 kg•m/s - 3.9525 kg•m/s = (3.5 kg) • v
0.5594 kg •m/s = (3.5 kg) • v
6. If a projectile is launched horizontally with a speed of 12.0 m/s from the top of a 24.6meter high building. Determine the horizontal displacement of the projectile.
Answer: It is strongly recommended that you begin by listing known values for each of
the variables in the kinematic equations:
x = ???
dxi = 24.6 m
vix = 12.0 m/s
viy = 0.0 m/s (its launched horizontally)
ax = 0 m/s/s (true for all projectiles)
ay = -9.8 m/s/s (true for all projectiles)
Since three pieces of y-information are now known, a y-equation can be employed to find
the time.
dfy - diy = viy*t + 0.5*ay*t2
Plugging a chugging the above values into this equation yields a time of 2.2406 seconds.
Now the t value can be combined with the vix and ax value and used in an x-equation
dfx - dix = vix*t
to yield the answer 26.8 m.
7. Two identical freight cars roll without friction towards each other on a level track. One
rolls at 2 m/s and the other rolls at 1 m/s. After the cars collide, they couple (attach
together) and roll together with a speed of _____________.
Answer: Use 5000 kg as the mass of the freight cars (or any number you wish) and then
set the expression for initial total momentum equal to the expression for the final total
momentum:
(5000 kg)*(2) + (5000 kg) *(-1) = (5000 kg) *v + (5000 kg) *v
Now solve for v using the proper algebraic steps.
(10000 kg•m/s) - (5000 kg•m/s) = (10000 kg) v
5000 kg•m/s = (10000 kg)v
(5000 kg•m/s) / (10000 kg) = v
0.5 m/s = v
8. Paul D. Trigger fires a bullet. The speed of the bullet will be the same as the speed of
the recoiling gun ______________.
a.
b.
c.
d.
e.
because momentum is conserved
because velocity is conserved
because both velocity and momentum are conserved
only if the mass of the bullet equals the mass of the gun
none of these
Answer: d. Since both the bullet and the gun must encounter the same momentum
change, the velocity change would only be the same if their mass was the same.
Otherwise, the smaller-mass object receives a greater velocity change.
9. A karate expert executes a swift blow and severs a cement block with her or his bare
hand. The
a.
b.
c.
d.
e.
impulse on both the block and the expert's hand have the same magnitude.
force on both the block and the expert's hand have the same magnitude.
time of impact on both the block and the expert's hand is the same.
all of the above.
none of the above.
Answer: d. In any collision, there are always four quantities which are the same for both
objects involved in the collision. Each object experiences the same force (Newton's thrid
law) for the same amount of time, leading to the same impulse, and subsequently the
same momentum change. Only the acceleration and the velocity change can ever differ
for the two objects. The lower mass object always receives the greater velocity change
and acceleration.
10. Impulses are smaller when bounces take place.
a. True
b. False
Answer: FALSE
Collisions involving rebounding are associated with a greater velocity change, a greater
momentum change, a greater impulse and a greater force.
11. A piece of putty moving with 1 unit of momentum strikes and sticks to a heavy
bowling ball that is initially at rest. After the putty sticks to the ball, both are set in
motion with a combined momentum that is ___.
a. less than 1 unit
b. more than 1 unit
c. 1 unit
d. not enough information
Answer: c. Since momentum must be conserved, the total momentum of the ball and
putty must be 1 unit. (Before the collision, the total system momentum is also 1 unit - all
due to the motion of the putty.)
12. A large force acting for a long amount of time on a small mass will produce a
______.
a. small velocity change
d. small acceleration
b. large velocity change
e. two of the above
c. small momentum change
Answer: b. A large force on a small mass will result in a large acceleration (a=F/m) and
subsequently a large velocity change (Δv = a*t). This rules out choices A and D. A large
force and for a long time will result in a large impulse and therefore a large momentum
change. This rules out choice C.
13. A 5-N force is applied to a 3-kg object to change its velocity from +9 m/s to +3 m/s.
The momentum change of the object is:
a. -2.5 kg*m/s
b. -10 kg*m/s
c. -18 kg*m/s
d. -45 kg*m/s
e. none of these
Answer: C
Don't make this harder than it is; the momentum change of an object can be found if the
mass and the velocity change are known. In this equation, m=3 kg and the velocity
change is -6 m/s. When finding the velocity change, always subtract the initial velocity
from the final velocity (vf - vi). The momentum change can also be found if the force and
the time are known. Multiplying force*time yields the impulse and the impulse equals the
momentum change.
14. A 5-N force is applied to a 3-kg object to change its velocity from +9 m/s to +3 m/s.
The impulse experienced by the object is:
a. -2.5 N*s
b. -10 N*s
c. -18 N*s
d. -45 N*s
e. none of these
Answer: C. Impulse is defined as a force acting upon and object for a given amount of
time. Impulse can be computed by multiplying force*time. But in this problem, the time
is not known. Never fear - the impulse equals the momentum change. The momentum
change in this problem is -18 kg*m/s. Thus, the impulse is -18 N*s.
15. A 5-N force is applied to a 3-kg object to change its velocity from +9 m/s to +3 m/s.
The impulse acts for a time period of
a. 1.8 s
b. 2.5 s
c. 3.6 s
d. 10 s
e. none of these
Answer: C
Use the impulse momentum change theorem with F=5 N, m=3 kg and Δ v=-6 m/s.
Solving for time involves the following steps.
t = m*(Δ v)/F = (3 kg)*(-6 m/s)/(5 N) = 3.6 s
16. A 0.80-kg ball strikes a wall moving at 5.0 m/s and rebounds in the opposite
direction at 3.5 m/s. If the collision with the wall endures for a total time of 0.0080 s, then
determine the average force acting upon the ball.
Answer: Answer: -8.5 x102 N
Given: m=0.80 kg; vi = 5.0 m/s; vf = -3.5 m/s; and t=0.0080 s
Find: F
Useful Equation: F*t = m*(Δ v)
Rearranging the equation to solve for F yields
F = m*(Δ v)/t
Substituting yields
F = (0.80 kg)*(-8.5 m/s)/(0.0080 s)
F = -850 N
(NOTE: The velocity change is always found by from vf-vi.