Chapter 8: MOMENTUM Hints-Updated 5/4/12
Momentum (p=mv) is an important concept because it’s another quantity that is conserved in the
universe. We can relate the change in momentum of a body to what is called the impulse (I=∫Fnet dt) as
follows. As a result of Newton’s 3rd law, whenever two bodies interact each one applies an equal and
opposite force and impulse to the other (FAB =-FBA). Newton’s 2nd law (Fnet=ma) can then be rewritten as
∫Fnet dt =p (“impulse equals change in momentum”); also the area under a Fnet vs. time graph.
Therefore, during an interaction momentum is transferred from one body to the other in equal amounts
and the total amount of momentum of a “closed system” doesn’t change.
The Conservation of Momentum Principle is most useful in analyzing collisions and explosions. In
general, when the interactive forces between two (or more) bodies are much larger than any other force
acting on the bodies, those other forces are negligible and the sum total of the momentum of those bodies
is conserved during the interaction. The total momentum just before an interaction can be compared to the
total momentum just after the interaction in order to solve problems. Finally, since momentum is a vector
quantity, the principles apply independently to the x, y, and z-directions and you have one equation for
every dimension.
1. Consider two gliders (A=0.2 kg and B=0.3 kg) that collide on an air track. During the collision the
gliders apply an average force of 6 N for 0.05 sec to each other. We will look at the situation just before
the collision takes place, during the collision, and just after the collision. Information about the initial
velocities is given in the chart below. Use dynamics and kinematics principles to fill in the rest of the
Collision
chart.
4 m/s
5 m/s
A
forces=6N
B
Glider
Velocity just
before collision
Average acceleration
during collision (F/m)
Change in velocity
during collision
Velocity just after
collision
A
5 m/s
-6/0.2 = -30 m/s2
v=at= -1.5 m/s
v=at + vo= 3.5 m/s
B
4 m/s
6/0.3 = 20 m/s2
v=at= 1.0 m/s
v=at + vo= 5.0 m/s
A+B
9 m/s
-10 m/s2
-0.5 m/s
The point here is to show that adding v’s or a’s is not fundamentally
8.5 m/s
meaningful.
Now fill in the following chart about the momentum and impulses in this collision:
Glider
Momentum just before
collision
Average impulse
during collision
Change in momentum
during collision
Momentum just
after collision
A
1.0 kg m/s
I=Ft =-0.3 Ns
p =-0.3 Ns
0.7 kg m/s
B
1.2 kg m/s
I=Ft =0.3 Ns
p = 0.3 Ns
1.5 kg m/s
0
fundamentally
0
meaningful since ptotal
A+B
2.2 kg m/s
Here adding p’s is
2.2 kg m/s
is “conserved”
By comparing these two charts you may begin to appreciate the usefulness of the momentum-impulse
concept in analyzing collisions.
a) Are there any quantities in the first chart that are “conserved”? No!
b) Are there any quantities in the second chart that are “conserved”? Yes! ∑p.
2. Consider a child with a mass of 40 kg who is playing on a trampoline.
During one bounce the child hits the trampoline with a vertical speed of
10 m/s. A graph of the net, impulsive force is shown. Assume that the
trampoline applied only a vertical force on the child.
Fnet
640 N
a) Using the graph estimate the change in the momentum of the child.
b) What was the average force that the trampoline applied to the child?
1.0 s 2.0 s 3.0 s
Don’t forget that the force applied by the trampoline is not the only
force on the child; you must also consider the weight of the child to get the net force.
c) Determine the velocity of the child after the bounce. Is this a positive or negative velocity?
d) If the child misses the trampoline and instead falls on the ground, what is the change in his
momentum? Why is this more dangerous than bouncing off the trampoline?
e) On the graph above sketch a curve representing the interaction with the ground. What would the
two curves have in common?
f) Momentum is supposedly a conserved quantity, yet here the momentum of the child has changed,
what other object is also changing momentum? How much does the momentum of that other object
change? Why do we ignore this part of the problem?
a) The area under the graph is the impulse and also the momentum change during that time interval. It
can be estimated as a triangle and has a value of ~640 N·s.
b) The average net force is Imp/∆t =640/2=320 N upwards. This includes the weight of the child of 400 N
downward. So the actual force applied by the trampoline is 720 N upwards (do the force diagram if you
are not sure: Fnet =N – mg).
c) Using ∆p=p-po the speed after the bounce is found to be 6 m/s upwards. The sign of the velocity
depends on how you defined the downward motion in the first place. But you need to be consistent. The
initial momentum is down, the impulse is up, and the final momentum is also up.
d) The child doesn’t bounce so the final momentum is zero and p =400 N·s. This is less
of a change in momentum than when he bounced (p =640 N·s) but it is more dangerous Fnet
because this happens in less time and the impulsive force is therefore greater.
e) The curve should be higher and narrower. The area under the graph will measure the
momentum change just as in the bounce, that is 400 N·s.
f) The other object changing momentum is, of course, the earth. The change in the
momentum of the earth is equal and opposite but because the earth is so much more
massive, it hardly moves at all.
3. A 200-g puck sliding on ice strikes a barrier at an angle of 53o and it bounces off at an angle of 45o. The
speed of the puck before the bounce was 15 m/s and after the bounce it is 12 m/s. The time of contact
during the bounce is 0.05 sec.
a) Write these velocities in rectangular (x&y) components relative to the barrier.
b) Determine the x and y components of the impulsive force. Which one can be
53o
45o
identified as a normal force and which can be identified as a friction force?
c) What was the direction of the impulsive force with respect to the barrier?
d) How much energy was “lost” (actually turned into internal energy) as heat in the bounce?
e) If the impulsive force had the same normal component but the friction component were zero, what
would be the final velocity of the puck?
a) vo =9î -12j; v =8.5î +8.5j —±’·°‡flfi›‹€⁄`Œ„´‰ˇÁ¨ˆØ∏”’»¿˘¯¯¯
b) px =-0.1 N·sFx =-2 N (friction); py =4.1 N·sFy =82 N (normal)
c) Reconstruct the net force vector to get angle with the horizontal: arctan (Fy/Fx)=89o.
d) ∆K= 8.1 J of kinetic energy is due to heat.
e) The horizontal component of the velocity (vx) would not change: v =9î +8.5j. The final v value would
be a bit larger, [(92) +(8.52)] 1/2 =12.4 m/s, and the angle [tanø=8.5/9] would be a bit smaller (43.4o).
4. On a horizontal plane, mass m moves uniformly in a circle of radius R in a time T. Determine:
p
a) The average impulse on the mass over: (i) one quarter a revolution,
(ii) one half a revolution,
(iii) a full revolution.
b) Draw a vector diagrams illustrating the relationship between momenta and impulse in (a).
c) The average impulsive force on the mass during each interval in (a). Don’t forget to give directions.
d) We learned that centripetal forces can do no work. Can they apply impulses? Explain.
Magnitude of the momentum is always p=m(2R/T) but the direction of the momentum changes. Starting
from 12 o’clock position in a circle: (a&b) Recall that Impulse =p = p – po
(i)
(ii)
(iii) 0
p=p√2
p
p=2
p
-po
p
-po
c) (i) Fnet =Impulse/∆t =[p(2)1/2]/(T/4) with the same direction as the impulse (45o WoS or 225º).
(ii) Fnet =Impulse/∆t =[2p]/(T/2 with the same direction as the impulse (W or 180º).
(iii) Fnet = 0
d) Since I=Fnet ∆t and W=F·∆s, centripetal forces apply impulses all the time but do no work, as seen
above. Impulse can average to zero over a complete rotation due to its vector nature as illustrated here.
5. Consider the following one-dimensional, inelastic collision between two cars. One car, with a mass of 2
kg and a speed of 5 m/s, collides with another car with a mass of 3 kg moving with a speed of 2 m/s in the
same direction. After the collision the 2-kg car has a speed of the -1 m/s. Even though energy is not
conserved during this collision, momentum is conserved if we restrict the analysis to a comparison of the
momenta of the cars just before and just after the collision.
a) What conditions are required for the momentum conservation principle to be valid? Why
do we restrict our analysis to the moments just before and just after the collision?
The conditions are that the collision (action-reaction) forces be large enough over the time of the
interaction that all other forces are negligible. If you wait too long after a collision, forces like friction
with the ground will transfer momentum to the earth and momentum conservation would not apply
between the two carts only.
b) Determine the velocity of the 3-kg car just after the collision. See chart below.
c) If the collision had been perfectly inelastic, what would be the final speed of the carts?
The cart would “stick” together and move with speed v =[total p/ total m]=16/5=3.2 m/s
d) It is instructive to look at a collision from different “frames of reference”. The information in this
problem was given with respect to the ground frame of reference. Now redo the same problem from two
other different frames of reference:
(1) a frame of reference moving with the initial velocity of the 2-kg car (5 m/s), and
(2) the center of mass frame of reference (here the center of mass speed is 3.2 m/s).
In each case determine the initial and final speeds of the cars relative to each frame. To do this, imagine
yourself looking at the colliding cars from a third car that is moving parallel to the track at the speed of
the frame. Remember that the relative velocity formula for velocities is: vA/B + vB/C = vA/C. In this equation
“A” represents the car being looked at, “B” represents the moving frame, and “C” represents the ground,
so that the relationship becomes: vcart/frame + vframe/ground = vcart/ground vcart/frame = vcart/ground - vframe/ground
The following chart may help you summarize your results. Answers in chart.
Frame of
Reference
Velocity just before
collision (m/s)
Momentum just before Velocity just after
collision (kg·m/s)
collision (m/s)
Momentum just after
collision (kg-m/s)
Ground
2-kg
3-kg
2-kg
3-kg
2-kg
3-kg
2-kg
3-kg
5 m/s
2 m/s
10
6
-1 m/s
6
-2
18
-1-5=-6
6-5=1
-12
3
Cart moving
0
2-5=-3
0
-9
at v=5 m/s
Center of Mass 1.8 m/s 2-3.2=-1.2 3.6
-3.6
*(vcm= 3.2m/s)
*To be done after center-of-mass discussion.
-1-3.2=-4.2 6-3.2=2.8 -8.4
8.4
e) Does the Momentum Conservation Principle apply in any inertial frame of reference?
Note that the total momentum before is equal to the total momentum after, even though the total amount
may differ in different frames of reference. In the center of mass frame you may notice that the total
momentum is zero before and after (that’s because in the center of mass frame the velocity of the entire
system is, of course, zero!).
6. A child is playing on a swing. He swings down from an angle of 53o. The mass of the child is 40 kg and
the seat of the swing has a mass of 5 kg. The length of the swing is 5 m. At the bottom of the swing the
boy quickly picks up his 5-kg dog from rest and continues to swing up with it on his lap.
53o
a) How far will the child and dog swing up before coming to rest?
b) Instead of picking up the dog, the child slips off the swing at the bottom
with zero speed relative to the swing. How far will the swing rise all by itself?
c) In a third version of this problem, the child slips of the swing and the swing rises
to an angle of 37o all by itself. What was the speed of the child relative to the ground when he slipped
off? …relative to the swing?
d) In which part of this problem is energy conserved? …is momentum conserved?
In this problem we see the interplay between energy and momentum conservation. You can use energy
conservation to determine the speed of the child and swing at the bottom [vbottom=(2gh)1/2, where h=LLcosø]. At the bottom a quick interaction (collision or explosion-like) occurs that conserves momentum.
Then the swing moves upwards and you can determine the maximum height using energy conservation.
a) Just before picking up dog the speed of child and swing is: vbottom=6.32 m/s. After inelastic collision
with dog the speed of system is 5.69 m/s. Swing rises 1.62 m and makes angle of 47.5o with vertical.
b) This a tricky question. If the child slips off the swing with zero relative velocity, it means that both the
child and the swing retain their speed of 6.32 m/s as separate bodies. Since the swing has the same
speed with or without the child it will simply rise to its original height at angle of 53o.
c) Do the problem in reverse. Use the final height to determine the speed of the swing relative to the
ground after the child slipped off (vswing= 4.47 m/s). Use momentum conservation at the bottom to find
speed, relative to ground, with which the child slipped off (vchild=6.55 m/s). The relative velocity of
child and swing as separate objects at the bottom is (6.55-4.47= 2.08 m/s).
d) Energy is conserved when the swing is coming down or going up. Momentum is conserved at the
bottom when the child quickly interacts with the dog or the swing seat.
7. In this problem we compare the momentum and kinetic energy of bodies in motion.
a) Show that in general K=p2/2m.
b) If a 0.04 kg marble and a 0.16 kg ball have the same kinetic energy, which has more momentum?
Determine the momentum ratio of the two objects.
c) If 700 N man and 500 N woman had the same momentum, which has more kinetic energy? Determine
the kinetic energy ratio of the two objects.
a) K=mv2/2= m2v2/2m = p2/2m
b) p is proportional to the square root of the mass for equal K: p1/p2=(m1/m2)1/2= (0.04/0.16)1/2 =1/2
c) K is inversely proportional to mass for equal momentum: K1/K2= m2/m1= 50/70=5/7
8. Two cars are headed for a collision with equal speeds v. Cart A, however, has a larger mass M and cart B a
smaller mass m. The two cars stick together after the collision so this is a perfectly inelastic collision.
v
v
a) If M=2m, show that the combined speed of the cars is v/3.
A
B
b) Determine the change in the velocity of each car, the change in the kinetic
energy of each car, and the change in momentum of each car.
c) Which car exerted the larger force on the other car? Which exerted a larger impulse? Which did more
work? Which car is likely to sustain more damage? Justify your answers.
d) Determine the speed of the cars after the collision in the extreme case of M being much larger than m
(m/M0). This would be like a car colliding with a wall.
e) Determine the speed of the center of mass of the two cars before the collision and compare to the speed
in (a). Does the comparison make sense? You may have to wait until we discuss this topic in class.
a) 2mv – mv=3mv´ v´=v /3
b) Answers are given for all masses in terms of “m” the mass of B: vA = -2v/3 ;vB =+4v/3 ; pA = -4mv/3
;pB =+4mv/3 ;KA = -8mv2/9 ;KB = -4mv2/9.
c) You should know that the carts exert the same force and impulse (but in opposite direction) on each other
because of Newton’s 3rd law which is the basis for the momentum conservation principle!! This doesn’t
mean that the carts will suffer the same amount of damage. A total amount of “lost K”= 4mv2/3 is turned
into “internal energy”, and the car that ends up with a greater share of that internal energy will most
likely surfer greater damage, but that depends on many factors, such as materials, car design, etc….. Here
the smaller car (B) accelerates more (vB >vA ), but it’s not obvious that one car is more damaged than
the other.
d) If M>>>m, then v’= v. The large mass will not change at all while the small mass changes direction 180o
and undergoes the greatest amount of change.
e) Using velocity of the center of mass formula: vcom= vo/3, the same as after the collision. This makes sense
because the velocity of the center of mass doesn’t change during a collision.
9. Determine the center of mass of the following objects or system of objects.
System
a)
M2
M1
L
c)
d)
X center of mass
Ycenter of mass
Two unequal point masses
Origin added
to illustration
M2 D/(M1 +M2 )
0
Uniformly dense bent rod
Origin added
to illustration
L/4
L/4
Uniformly dense surface:
b=2a
Origin added
to illustration
2.64a
0.786a
Uniformly circular surface Origin added
with hole
to illustration
5/6R
0
Triangular uniform surface Origin added
to illustration
2a/3; Requires
calculus.
b/3
Done in class.
Non-uniform linear
density: o=constant
0
(10/18)oL;
Requires calculus.
L
0
a
a
Chosen origin
D
0
b)
Needed info
a
b
b
0
0 R
R
Hole
e)
b
0
a
f)
0
L
= o(1+x/L)
l
Origin added
to illustration
9d) This requires a “trick”. Think of the object as a circular solid of “+mass” (πR2) and a circular hole
of “-mass” (-πR2) [ is the area mass density dm/dA]. Then Xcom=[(πR2)R +(-πR2RπR2
+(-πR2 Simplifying the expression gives the answer in the chart.
9e) Check the class notes.
9f) The integral is: Xcom=∫xdm =∫xdx =o∫L xo(1+x/L)dx…just do the integral.
10. A toy cannon mounted on top of a cart fires a ball at an angle of 37o to the horizontal. The initial
speed of the ball is 20 m/s. The mass of the ball is 0.5 kg and the mass of the cannon and cart
combined is 2 kg.
a) Determine the recoil velocity of the cart due to the cannon firing.
b) How do you account for the change in the y-component of the momentum of the ball?
c) The cart is attached to a spring of constant k=400 N/m that is initially relaxed. What will be the
maximum stretch of the spring?
Do this problem in two parts: (1) during the short timed “explosion” that the cannon fires, momentum
is conserved between the cart and the ball in the x-direction only, giving the cart some motion; then
(2) the cart moves back more slowly stretching the spring and storing potential energy.
a) (0.5)20cos37º=-2 vcart  vcart =-4 m/s.
b) The y-component of the momentum comes from the vertical interaction with the earth.
c) initial Kcart = final Uspring =ks2/2 smax =28 cm.
11. Two blocks of unequal masses (m and 5m) are compressing a spring of constant k between them.
When the masses are released, the spring pushes the masses apart over a frictionless surface as it
returns to equilibrium.
m1
m2
a) After the spring is decompressed determine the following ratios:
Ratio
Velocity
v1/v2
-5/1
Momentum
p1/p2
-1/1
Kin. Energy
K1/K2
5/1
Impulse
I1/I2
-1/1
Work
W1/W2
5/1
Power
P1/P2
5/1
b) What fraction of the energy stored in the spring went to the smaller mass?
c) When a rifle is fired the change in the momentum of the bullet is the same as the change in the
momentum of the rifle. Explain why it is the bullet that is the more deadly part.
b) K1/Uo=K1/(K1+K2)=5/6=83%. In explosions the smaller mass gets the larger share of the energy
released.
c) Even though the bullet has the same momentum as the recoiling gun, it carries most of the energy of the
explosion in the gun firing, hence the bullet can do more damage.
12. A block of mass M moving with speed v hits a surface at an angle . The block lands on a flat
side and doesn’t bounce but it slides a distance D on the horizontal surface slowed by friction during a
time interval “t” until it stops. The coefficient of kinetic friction here is k.
v
a) Determine the value of the speed v in terms of the other given terms using

µk
momentum and impulse principles.

b) What fraction of the original kinetic energy was changed to internal energy
D, t
(or “dissipated”) due to heat by the sliding?
c) What happened to the rest of the original K that was not changed to internal energy by friction?
All answers should be in terms of the “givens”. Here the x-component of the momentum is changed by
the impulse applied by the force of friction.
a) px=ft  Mvcos=kMgt  v =kgt/cos.
b) Wfriction/Ko= 2Dcos2/kgt2; Wfriction equals the amount of internal energy generated by the slide.
c) It was “internalized” and also produced sound during the impact before the mass started to slide.
13. A projectile is fired with speed vo at an angle ø with respect to the horizontal. Assume
no air resistance. The projectile returns to the same level and it has a range R and altitude H.
vx
vo
H
a) In going from the ground to the maximum height, determine the change in
ø
momentum in the x-direction and in the y-direction.
R
b) In going from ground back to the ground to the, determine the change in
momentum in the x-direction and in the y-direction.
c) What is the impulsive force in the case of a projectile? Draw a graph of F vs. t for the projectile.
2vx
d) Now suppose that, at the maximum height H, the projectile breaks into two
equal pieces. One piece ends up with no velocity and moves directly downward
vo
H
taking a time “t” to hit the ground. How much time will it take the other piece
ø
to hit the ground? How far apart will the two pieces be on the ground?
∆x=R
Fnet
(a) px =0 ; py =-mv sinø(b) px =0 ; py =-2mv sinø
t
(c) The impulsive force here is, of course, the weight of the mass
-mg
itself and it can only change the vertical momentum component.
d) Both pieces have no initial vy so they take the same time “t” to hit the ground. After the explosion the
second piece ends up with twice the vx speed of the combined mass before the explosion, and therefore
goes twice as far horizontally so the pieces will be ∆x =R apart when they hit the ground.
14. In lab we will experiment with a ballistic pendulum. This is a massive bob hanging from a light,
pivoting rod that is set to catch a rapidly moving projectile. After catching the projectile the rod rises to a
maximum height “h=L-Lcos”. This can be used to determine the initial speed of the projectile. Assume
that the mass of the bob is M, that the mass of the projectile is m, and that the length of the rod is L, but
has negligible mass. The initial speed of the projectile is vo.

a) In what part of this problem is momentum conserved?
In what part is energy conserved?
m
M
h
b) Determine the maximum angular swing of the pendulum
in terms of the other given quantities.
vo
c) If the bob fails to catch the projectile and the projectile bounces
back, does the pendulum swing higher or lower? Justify your answer.
Here are some variations on this problem:
d) Now assume that, when it catches the projectile, the bob gets enough speed to swing all
the way to the top ( =180º and h=2L). What was the speed of the projectile before it was
imbedded in the bob? Assume the bob reaches the top with zero speed, which is possible
when connected to a rigid rod.
e) Now assume that the rod in the apparatus is replaced with a string. In this case the bob
needs a minimum speed at the top to keep the string extended. With what speed must the
projectile strike the bob to allow it to reach the top if it’s connected to a string?
h
m
M
vo
a) Review the theory of the Ballistic Pendulum lab. Momentum is conserved during the collision between
the bob and the projectile. Energy is (mostly) conserved during the rise of the pendulum.
b) cos=1- [mvo/(M+m)]2/2gL
c) The bob will rise higher because a “bounce” imparts a larger impulse than a “stick” and the pendulum
bob would end up with a larger speed after the interaction.
d) We want the bob to get enough kinetic energy from the projectile to rise 2L and reach the top. Since the
rod is rigid the bob doesn’t need a minimum speed at the top as would be the case with a string, so vtop=0.
Using energy conservation it’s found that the pendulum bob needs a speed at the bottom v’=(4gL)1/2.
Using momentum conservation it’s found that the projectile needs to have a speed vo=(1+M/m) (4gL)1/2.
e) If the rod were replaced with a string, there is a minimum speed [(Lg)1/2] required at the top of the
swing. So the bob needs more speed at the bottom [v’=(5gL)1/2] and the projectile speed required will be
vo= (1+M/m) (5gL)1/2.
15. In a perfectly elastic collision both momentum and energy are conserved. Consider two carts of mass
mA and mB with initial velocities vA and vB and final velocities v´A and v´B, all with respect to the ground.
a) Write the momentum conservation principle for this collision and the kinetic energy conservation
principle.
b) Derive the very useful relative velocity formula for this collision: (vA - vB) = -(v´A– v´B).
c) Explain the significance of the relationship. Why is the difference (vA - vB) called the relative velocity
between the two carts?
d) Describe the collision as seen by an observer in one of the carts.
e) Describe the collision as seen by an observer moving with the center of mass.
(a&b)This was done in class so refer to your notes. The basic idea is that you can divide the energy
conservation equation into the momentum conservation equation to cancel out the masses.
c) The relationship says that (if the collision is perfectly elastic) the colliding objects approach each other
with the same relative speed with which they will separate after the collision. The relative speed is the
speed of one object from the frame of reference of the other object.
d) To an observer in either cart the collision will look like the other cart approaches before the collision
with the same speed with which it retreats after the collision.
e) The carts will move toward the center of mass with equal and opposite momentum and then they will
separate with the same magnitude of momentum in the opposite direction.
16. Consider an elastic collision between two carts. The collision is elastic because an ideal spring of
constant 425 N/m is attached to one of the carts. Cart A has a mass of 3 kg and cart B has a mass of 1 kg.
Initially cart A is moving with a velocity of 1 m/s toward cart B that is at rest. The graph below plots the
positions of both carts vs. time from the beginning to the end of the collision:
0.20
Position (m)
Car B
Collision ends
0.10
Collision
starts
Car A
0.0
0.04
0.08
0.12
0.16
0.20
time (sec)
Use the information in the graph to fill in the relevant values in the chart below: (Note: This graph may
not print out properly so your answers may not totally agree with mine!)
Before Collision
Maximum Compression at ~0.08 s After Collision
cart
A
B
A
B
A
B
velocity
1 m/s
0
3/4 m/s
3/4 m/s
1/2 m/s
3/2 m/s
momentum
3 kg·m/s
0
9/4 kg·m/s
3/4 kg·m/s
3/2 kg·m/s
3/2 kg·m/s
Kinetic energy
3/2 J
0
27/32 J
9/32 J
3/8 J
9/8 J
a) Verify that the relative velocity before the collision is the same (but negative) as the relative velocity
after the collision. Also explain why the carts have equal velocities when spring is most compressed.
b) Verify that momentum is the same before, during, and after the collision.
c) Verify that kinetic energy is the same before and after, but not during the collision. What happens to the
“missing kinetic energy” during the collision?
d) Determine the maximum compression of the spring.
e) Sketch a velocity vs. time graph for both carts throughout the interaction.
a) vAB before =(0-1) and vAB after =(3/2 -1/2)=1.
b&c) Easily shown from info in chart.
d) Uspring=Kbefore – Kmax comp=0.375=ks2/2 smax=4.2 cm
e) something
like this:
v (m/s)
1
A
B
t
17. For the following 2-D collisions or explosions draw the appropriate vector diagram relating the
momentum vectors before and after the event. Also write the momentum conservation relationships and
give an expression for the velocity of the center of mass.
Example
Exploding object:
Inelastic Collision
Elastic Collision
Before
Before
Before
After
Illustration
v
After
3m
vA
2m
v√3
After
m
v
v0
v=0
m
2m
m
5m

2m
m
v
v’
2v
2m
vB
Vector diagram
pA=4mv



p1=2mv

 p2=mv

p3=mv√3
pAB=5mv’
p´A =mvA
p´B =2mvB

total po=mv0


 pB=3mv

Momentum
Conservation
Expressions
Center of Mass
Velocity
2
2
2
p1 = (p2 ) + (p3) ;
p1cos1 =p3;
p1sin1 =p2;
tan 1 =p2 /p3=1/√3 ;
1 =30o
0
(pAB) = (pA) +(pB) ;
(5mv´)2 = (4mv)2 +(3mv)2;
tan AB =pB /pA=3/4 ;
AB=37o
p0 = (p´A)2 +(p´B)2
2p´A p´B cos(180);
p0 =p´A cosp´B cos
p´A sinp´B sin
Vcm =v´= v
Vcm =mvo/(m +2m) =vo/3
2
2
2
2
18. Atomic particles collide all the time. Consider a helium nucleus which is also called an alpha particle
(4He) striking a gold nucleus (197Au) head on. In nuclear notation, the superscript denotes the atomic mass
of the particle in what are called atomic mass units (1 “amu” is about the mass of a proton). Assume the
collision is perfectly elastic and use “amu” as the mass unit.
He
vo
a) Assuming a one-dimensional collision, determine expressions for the final
velocities of the two nuclei in terms of the initial speed vo of the alpha particle.
b) What fraction of the alpha particle’s energy is transferred to the gold nucleus?
c) Now assume the collision is not 1-dim so that the particles move on in different directions.
If the alpha particle is deflected at right angles to its initial direction, as shown, determine
the speeds of the two atoms after the collision and the direction of motion of the Au nucleus.
(You can simplify this problem a lot if you assume that the speed of the smaller particle
changes only negligibly. The answers you get are very close to the correct ones).
Au
a) Since this is a basic elastic collision problem, use momentum conservation (4vo=4vHe +197vAu) and the
relative velocity formula (4-0)=(vAu - vHe). I got vHe =-0.96vo ; vAu =0.04vo. Note that since the gold
nucleus is so much more massive it gains very little speed while the speed of the helium atom is almost the
same as before the collision but in the opposite direction.
b) KAu / Ko = 0.796 =7.9 %
4vHe
197vAu
c) Using the mom-conservation vector diagram on the right: (4vo)2 +(4vHe)2 =(197vAu)2 
ø
4vo
Also K is conserved, so: 4(vo)2 =4(vHe)2 +197(vAu)2. Combining these you get the
answers: vAu =0.028vo ; ø =43.5o. If you simplify the problem by assuming vHe ~ vo, you get very similar
answers: vAu = 0.029vo ; ø =45o
For the more general case of a mass striking another mass at rest and then moving off in different
directions, the following momentum vector diagram applies:
P2´
P1´
The primed (´) quantities denote the “after collision” vectors.


P1
Using the Law of Cosines :
2
2
2
2
2

p1 =(p´1 ) +(p´2) 2p´1 p´2 cos(180)=(p´1 ) +(p´2) 2p´1 p´2 cos()
Since energy is also conserved: m1v12/2 = m2v´22/2 + m1v´12/2
These equations are combined to solve for any two unknowns.
18A. An interesting special case of a perfectly elastic collision occurs when the colliding balls have the
same mass. It can be proven that the angle between their paths after the collision will always be 90o
regardless of the actual initial and final velocities of the masses. See if you can prove this combining the
energy and momentum conservation principles. This fact helps pool players predict the path of a ball after
collision.
For the case of equal masses the general cases equations above simplify to equations involving only the
velocities (masses cancel out). When you combine the simplified equations the result is that: cos(180 
)=0, which is only true if the two angles add up to 90o.
19. Two stars start from rest very far apart (think infinity) and
are attracted together by gravity. Assume that the stars masses
are M1 and M2 and that no other mass will interfere with their
interaction. Consider the starts when they are a distance D apart.
a)
b)
c)
d)
e)
D
Write the energy conservation formula that applies here and explain why it is valid here.
Write the momentum conservation formula that applies here and explain why it is valid here.
Determine the velocities of each star when they are a distance D apart.
Determine the center of mass of the system and its velocity.
Couldn’t you simply use the acceleration of each mass to determine the speeds? Explain.
a) The only force acting is the force of gravity, which is conservative, so energy is conserved. Here we
compare infinity to the position D apart: 0 = -GM1M2/D + M1v12/2 + M2v22/2
b) Since the force of gravity is equal and opposite on the two stars, the net force is zero so momentum is
conserved. Again we compare infinity to the position D apart: 0 = M1v1 + M2v2
c) Combining the two conservation equations give: v1 = M2 [2G/D(M1+M2)] 1/2; v2=M1[2G/D(M1+M2)] 1/2.
The velocities have opposite directions of course.
d) Measured from M1: Xcm =DM2/ (M1 +M2); Vcm =0
e) You could, but since the accelerations are not constant you would have to perform two integrals to
determine the velocities. The energy and momentum method is easier.
20. A bullet of mass m moving with velocity vo crashes into a block of mass M at rest on a frictionless
horizontal surface. The block is connected to a spring of constant k. Assume that the time of the
bullet/block interaction is very short compared to the time it takes the spring to respond. In each of the
following cases determine maximum compression of the spring. To get some numerical values make m=
20 g, M=0.5 kg, vo = 300 m/s, and k= 400 N/m.
a) In the first case assume that the bullet sticks to the block.
b) In the second case assume that the bullet bounces back
directly with half its original speed.
m
M
c) In the third case the bullet crashes through the block emerging on the other side with half its
original speed but still moving in the same direction.
d) Which has the bigger impact on the large mass, the sticky or the bouncy interaction? Explain.
In all these cases momentum is conserved during the collision between the bullet and the block, which
happens so quickly that the spring is not involved. After the collision, energy is conserved between the
block and the spring as spring is compressed.
a) pblock= 6 kg·m/s ; sspring= 43 cm
b) pblock= 9 kg·m/s ; sspring= 64 cm
c) pblock= 3 kg·m/s ; sspring= 21 cm
d) It should be obvious from the results that when the bullet bounces more momentum (and energy) is
transferred to the block.
21. In another version of the problem above, assume that the block is not connected to the spring but it is
sitting at the edge of a table that is H high. When the bullet hits the block it flies off the table and it lands a
distance X from the bottom of the table. In each case below, determine the initial velocity of the bullet in
the terms of the given quantities.
m
M
a) In the first case assume that the bullet sticks to the block.
b) In the second case assume that the bullet bounces back
directly with half its original speed.
c) In the third case the bullet crashes through the block emerging on the other side with half its
original speed but still moving in the same direction.
d) Which makes the block move farther, the sticky bullet or the bouncy bullet? Explain.
This is similar to the problem above except that after the collision the block becomes a projectile with
horizontal velocity given to it by the collision. You can use kinematics to relate the height, range and
velocity so that vx =X(g/2H)1/2.
a) vo =(1 + M/m)vx.;
b) vo =(2M/3m)vx.;
c) vo =(2M/m)vx.
d) Again, we see that the “bounce” is stronger than the “stick”. This always seems counterintuitive to
students so you need to realize that when something bounces the momentum change is greater because the
object goes from positive to negative momentum (or vice-versa).
22. A child stands on a cart initially at rest. The cart and child together have mass of 60 kg. Also on the
cart are three large blocks 20 kg each. The child throws each block horizontally away from the cart with a
speed of 2 m/s relative to the cart always.
Vblock/cart =2 m/s
a) Determine the speed of the cart and child, relative to the
ground, after each block is tossed out by the child.
b) If the child could manage to throw all the blocks at once,
what would be the recoil velocity of the cart and child? Explain why this answer is not the same as
the final velocity in (a).
c) Assume that, instead of individual blocks, there is a tank carrying 60-kg of water. The child uses a
hose to eject the water horizontally away from the tank at a rate of 1 kg/sec. The water leaves the
hose with a speed of 2 m/s relative to the child. Determine the final speed of the cart and child after
all the water is ejected.
d) Does the rate at which the water is ejected in (c) affect the final speed of the cart? Explain.
This problem aims to prepare you for the rocket problem. The important idea here is to properly relate the
velocities with respect to the cart and to the ground. A momentum conservation formula must contain
velocities measured with respect to the same frame of reference in order to be valid.
a) v1=-0.4 m/s ; v1,2 = -0.9 m/s ; v1,2,3 = -1.57 m/s. Solution will be done in class.
b) v1,2,3 = -2 m/s. The answer is larger than before because all the tossed blocks have the same 2 m/s
velocity relative to the ground. In the previous problem each successive toss was with respect to the cart
and blocks 2 & 3 were therefore slower relative to the ground.
c) Here the hose is equivalent to fuel being ejected by a rocket. So the rocket formulas apply: vfinal
=ve=ln(Minitial/Mfinal)=2 ln2 = 1.39 m/s in the negative direction.
d) The rate doesn’t affect the final speed, but it determines how long the process takes. Here for example it
took 60 sec for the water to be ejected.
23. Consider a rocket in outer space away from any gravitational attraction. It is ejecting fuel at a rate of 5
kg/s with a speed of 200 m/s relative to the rocket. Assume that 40% of the original 40-kgrocket mass is
fuel.
Vexhasut/rockt
a)
b)
c)
d)
e)
Determine the thrust of the rocket.
Determine the initial acceleration and the final acceleration of the rocket.
Determine the final speed of the rocket and the time that it took to attain that speed.
Show that you can derive the basic rocket equations from the momentum conservation principle.
Redo parts (b) and (c) above assuming that the rocket is actually lifting off from the surface of the
earth so that the weight of the rocket must be taken into consideration. Assume that “g” is constant
during the burning of the fuel.
f) If you run out of fuel to eject in outer space is there another way to move your rocket? Explain.
g) How would you determine the distance traveled by the rocket? (You don’t need to get the final
answer)
a) Fthruest =ve dm/dt =1,000 N
b) ainitial =F/Mo =25 m/s2 ; afinal =42 m/s2
c) vfinal = ve ln(Mo/M) =102 m/s
d) This was done in class so review your notes. The textbook also has a similar derivation.
e) Fnet =Fthrust - Mg ; anet =athrust - g
f) There is not other way to move the rocket except to eject something from it. You must have something to
push against that becomes a separate object. Newton’s 3rd law doesn’t allow for exceptions. You would
have to wait for something to rescue you.
g) You could integrate the velocity formula with respect to time. Here the mass is the quantity that varies
with time.
24. A raft of length L and mass M is a distance d from the dock. At the end closer to the dock stands a
person of mass m. For the purpose of getting some numerical values assume that M=200 kg, m= 50 kg,
L=10 m, d= 4 m.
Dock
d
a) As the person walks the center of mass of the system doesn’t change. Explain why.
b) The person walks to the other end of the raft. How far did the raft move relative to the dock? How
far did the person move relative to the dock?
c) If the person accelerated until it reaches the end of the raft with a speed 6 m/s relative to the raft,
what would be the speed of the raft relative to the water? What would be the speed of the person
relative to the water?
a) The net force on the system is zero so the center of mass of the system of the person and the raft cannot
move. The person and the raft can move relative to each other though.
b) The center of mass doesn’t move so the expressions for Xcom before and after the person moves must be
equal Xcom before =Xcom after [50(4)+200(9)]/250=[50(4+10+∆xraft)+200(9+∆xraft)]/250. Solving
gives xraft =-2 m andxperson =8 m. Note that when the person is on the left end the center of mass is 4 m
from that end, when he moves to the other end the center of mass is 4 m from the other end. Altogether the
raft moved (L – 8 =-2)...
c) This problem is basically an explosion but you need to be careful with the speed of the person, 6 m/s is
relative to the raft, the speed of the person relative to the water is (6 + vraft/water). So: 0 = m(vperson/raft +
vraft/water) +M(vraft/water). After substituting numerical values: (vraft/water) =-1.2 m/s
Note: You could also use this velocity to find the distance traveled by the raft without resorting to the
center of mass. You would need to find the time involved in the process using kinematics: xperson
=vaveraget t =10/3. The same time applies to the raft: xraft =vaveraget = (-1.2/2)(10/3) = -2 m.
25. A canoe with a mass of 200 kg is moving along with a speed of 12 m/s. When it passes by a second
canoe with a mass of 300 kg, the second canoe transfer 50 kg of its mass to the first canoe directly across
(perpendicular to) the motion (lowering its mass to 250 kg). Assume that the canoes cannot move
sideways, so all changes in velocity occur parallel to their motions. Determine the final velocities of both
canoes in the following cases.
a) Explain why the motion of the second canoe is not affected by the
12 m/s
200 kg
mass transfer, but the motion of the first canoe is affected by it.
b) In the first case the second canoe is at rest when it transfers the
50 kg
50 kg mass.
250 kg
c) In the second case the second canoe is moving with the same
velocity as the first canoe when it transfers the 50 kg mass.
d) In the third case the second canoe is moving with a speed of
10 m/s in the same direction as the first canoe when it transfers the 50 kg mass.
e) In the fourth case the second canoe is moving with the same speed as the first canoe but in the
opposite direction when it transfers the 50 kg mass.
a) Since the canoes cannot move sideways, the second canoe cannot change its motion by discarding mass
in a sideways direction. The discarded mass simply carries away the same momentum it had as part of
the canoe. The first canoe, however, ends up colliding with a mass that has a different velocity and that
affects its motion. The general momentum conservation equation that applies here is: M1v1+mtransferv2
=(M1+mtransfer)v’; here M1 is the mass of the first canoe, m is the mass transferred by the second canoe,
v1 and v2 are the velocities of the two canoes before the transfer, and v’ is the final velocity of the first
canoe after transfer is completed. So the speed of the second canoe is not altered by the transfer, the
final speed of the first canoe depend on the motion of the transferred mass:
b) v’=9.6 m/s; c) v’=12 m/s
d) v’=11.6 m/s; e) v’=7.6 m/s
More practice problems:
26. A tennis ball moving at 18 m/s strikes the 45o hatchback of a car moving away at 12 m/s. Both speed
are given with respect to the ground. The ball rebound elastically from the car. Assume that the car is
much more massive than the ball, which is reasonable.
Car frame of
reference
a) Determine the speed of the ball relative to the car before the bounce.
b) If the ball bounces elastically, how fast and in what direction
will it be moving with respect to the car after the bounce?
c) What is the velocity of the ball with respect to the ground after the bounce? Where did the lost
kinetic energy of the ball go?
d) Explain why in this problem it was safe to assume that the speed of the car is not noticeably
affected by the ball striking it.
The key to solving this problem is to do it from the frame of reference of the cart. Also make the
reasonable assumption that since the ball is so much smaller than the car, the motion of the car is not
changed significantly by the impact of the ball.
a) 6 m/s i;
b) 6 m/s j
c) 12 m/s i +6 m/s j. In polar coordinates: 13.4 m/s r + 270 
dThe lost kinetic energy was transferred to the car but it was negligible compared to the original energy
of the car so the speed of the car wasn’t affected very much.
27. A frame with a mass of 0.15 kg stretches a spring a distance of 0.50 m
when it is hung from it. A lump of putty with a mass of 0.20 kg is dropped
from rest onto the frame from a height of 30 cm.
a) Find the spring constant.
b) In which part of this problem is momentum conserved?… in energy conserved?
c) Determine the maximum stretch of the spring.
d) Determine the stretch of the spring when the system comes to a final state of equilibrium.
30 cm
a) 3 N/m
b) Momentum is conserved when the lump putty collides with the frame. Energy is conserved when the
putty is in free-fall and then the spring stretches to stop the motion of the frame.
c) 2.0 m; d) 1.17 m
28. Two identical billiard balls are initially at rest when they are struck symmetrically by a third identical
ball moving with velocity vo in the x-direction. The collision is perfectly elastic.
a) Explain why the struck balls must end up with the same speed
and also be deflected by 300 to the horizontal.
b) Draw an appropriate vector diagram relating the initial
and final momenta of the masses.
c) Find the velocities of all three balls after the collision in terms of vo.
d) Determine the impulses on each ball during the collision.
vo
a) Since the impulsive forces between the balls are symmetric and the balls are
identical in size and mass, the line of action of the forces form an equilateral
triangle as shown on the right and the forces must be equal and 600 apart.
b) Note that since the original ball bounces back,the base of this triangle represents the sum of the initial
and final momentum of the original ball.
c) v1’= -0.2 vo ; v2’= 0.69 vo ; v3’= 0.69 vo . These are
P2’
P3’


directed 30º above and below the direction of the first ball.


d) Impulses are equal to momentum changes of each ball.
p1
p1’

29. A small mass m is moving with velocity vo on a horizontal frictionless surface. The small makes
contact with a curved shaped wedge of mass M initially at rest which it free to move on the frictionless
surface. As the two masses interact, the small mass rises on the curved side of the wedge to a height h and
the wedge accelerates to the right. Finally the mass slides down and returns the way it came but with onethird its original speed and the wedge moves on with a constant speed.
v
vo/3
vo
h
Before
Middle
After
a) In which part of this problem is momentum conserved?...energy conserved?
b) When the mass reaches the height h what is the relative velocity between the mass and the wedge?
c) Determine the speed of the mass and wedge together when the mass reached the height h.
d) Determine the height h in terms of the other quantities.
e) Determine the final velocity of the wedge after its interaction with the mass.
f) Determine the ratio m/M.
a) Momentum and energy are both conserved at every stage of this problem.
b) The relative velocity is zero, which means that both the mass and the wedge have the same velocity
c) v =mvo/(m +M)
d) h =[vo2 +(1 + M/m) v2]/2g, where v is the speed in (c).
e) vwedge=2vo/3;
f) m/M=1/2