Chapter 7 – Momentum

We’ve talked the last few weeks about forces and how they relate to the

change in an object’s motion. We’ve also defined the motion of an object

through its momentum, or the product of its mass and velocity. Symbolically we
! !
write this as p = mv in units of kg⋅m
s
. Then the net force is time rate of change of

the objects momentum, and we call these Newton’s laws of motion.

! !
dp !
Fnet = = manet (1)
dt

Consider an object of mass m . If net external force acting on the object is zero,

then for any non-zero time interval dt ≠ 0 , equation 1 becomes

! !
dp ! ! !
Fnet = 0 = → dp = 0 → p f = pi (2)
dt

and the objects momentum is a constant and the object travels in a straight line at a

constant speed. Equation (2) is, of course, Newton’s 1st law of motion.

We call equation (1) an interaction and the object has to interact with something

in its environment in order for a force to be imparted on the object. Thus the object

interacts with something in its environment and the interaction of the object with the

environment exerts a force on the object. And by equation (1), this changes the

momentum of the object. We call the time interval in equation (1) the interaction

time. Thus we could write equation (1) as

! ! ! ! ! ! !
dp = p f − pi = Fnet dt → p f = pi + Fnet dt (3).

Equation (3) is called the impulse-momentum theorem, where we define the impulse
! ! !
I given to an object as I = Fnet dt . Equation (3) implies that the change in the
momentum of the object is due to an interaction of the object with its environment
!
for a time dt . By assumption, the interaction time dt ≠ 0 and if Fnet = 0 then

equation (3) is merely a restatement of Newton’s 1st law of motion, equation (2). If
!
Fnet ≠ 0 then equation (1) is a measure of the interaction and we define the

interaction to be equation (1), which we have called Newton’s 2nd law of motion.

Equation (1) gives us a way to quantify the interaction, or the change in motion and
!
this change in motion of the mass m we have called the acceleration anet of the

system.

Suppose instead of a one mass system, that we have two objects of masses m1

and m2 interacting. That is, m1 and m2 are exerting forces on one another. At some

! !
time, let the momenta p1 and p2 of masses m1 and m2 respectively be given as shown

in figure 1 below. Masses m1 and m2 are interacting and the forces involved in their
! ! !
interactions are given as F1,2 and F2,1 . F1,2 is the force exerted on mass m1 due to its

!
interaction with mass m2 , while similarly F2,1 is the force exerted on m2 due to its

interaction with mass m1 . The object with mass m1 experiences a change in its

! !
dp1
momentum given by equation (1) as F1,2 = due to its interaction with mass m2 .
dt
!
p1
!
p2

!
m1 F1,2 ! !
F2,1 = − F1,2
m2

Figure 1: Two objects interacting, exerting equal and opposite

forces on each other.
 Analogously, the object with mass m2 experiences a change in its momentum

! !
dp2
given by equation (1) as F2,1 = due to its interaction with mass m1 . By Newton’s
dt
! !
3rd law of motion we have F1,2 = − F2,1 . Or, equivalently we can write

! ! ! !
dp1 dp2
F1,2 = − F2,1 → =− . This implies that the change in the momentum of the
dt dt

system of masses m1 and m2 does not change for any non-zero interaction time dt ,

which is of course the same for both objects interacting. Thus we can write
! ! !
dp1 dp2 d ! ! dpsystem
+ = ( p + p2 ) = = 0 . Therefore, using equation (1) we have that
dt dt dt 1 dt
! ! ! ! !
dpsystem = p f ,system − pi,system = 0 → p f ,system = pi,system (4) .

Equation (4) states that the total momentum of the system is constant and does

not change in time and thus the net force, by equation (1), that acts on the system

must be zero. We call equation (4) a statement of conservation of momentum. Total

momentum of the system is a conserved quantity and the total momentum of the

system does not change in during the interaction. The individual momenta of each of

the masses may change due the interaction, but the entire momentum of the system is

constant. We have to be careful of how we define our system.

Example 1: Suppose that a ball of mass m = 200g is thrown at a wall at an angle of

θ = 600 , measured with respect to the vertical, as shown in Figure 2. What
are the components of the changes in the ball’s momentum, Δpx and Δp y
!
respectively? What is the change in the ball’s momentum, Δp , if the ball
has an initial speed of vi = 5 ms and a final speed approximately equal to that
of the initial speed. If the ball is in contact with the wall for a time of
Δt = 0.2s , what force does the wall exert on the ball? What force does the
ball exert on the wall?
 m vi
θ

θ
m vi

Figure 2: A ball is thrown off at a wall and

bounces off at approximately the same speed as
it was incident.

Solution: The change in the components of the momentum for the ball, assuming that
the wall doesn’t move is given as

Δpx = p fx − pix = −mvix − mvix = −2mvi sin θ = −2 × 0.2kg × 5 ms × sin60 = −1.73 kg⋅m
s

Δp y = p fy − piy = mv fy − mviy = mvi cosθ − mvi cosθ = 0 kg⋅m

s

Thus the change in momentum of the ball is

! ⎛ Δp ⎞
Δp = Δpx2 + Δp 2y @φ = tan −1 ⎜ y ⎟
⎝ Δpx ⎠
⎛ 0 kg⋅m ⎞
!
(1.73 ) + (0 )
2 2
Δp = kg⋅m kg⋅m
@φ = tan −1 ⎜ s
kg⋅m ⎟
⎝ −1.73 s ⎠
s s

!
Δp = 1.73 kg⋅m
s
@φ = 1800

The ball experiences a change in momentum horizontally but continues to

move vertically upward.

The wall exerts a force on the ball given by

! !
Δp 1.73 kg⋅m
Fball ,wall = = s
@φ = 1800 = 8.65N @φ = 1800 .
Δt 0.2s
By Newton’s third law, the ball exerts a force of equal magnitude but
!
opposite direction on the wall. Thus Fwall ,ball = 8.65N @φ = 00 .

Example 2: Consider the following system, shown in Figure 3, in which two blocks of
masses m and 3m respectively are placed on a horizontal, frictionless
surface. A light (i.e. massless) spring is attached to one block and the
blocks are squeezed together and tied by a light cord. If the cord is cut and
the block of mass 3m moves to the right with a speed of a v3m = 2 ms , what is
the speed of the block of mass m ?
 vm v3m

m 3m m 3m

Figure 3: Two masses on a horizontal surface for Example 2.

Solution: Since the momentum of the system is conserved we have, from equation (4),
Δpsystem = p f − pi = 0 , with pi = 0 . Thus we can write our statement of conservation of
momentum as:

Δpsystem = p f − pi = 0 → p f = mvm + ( 3m) v3m = 0

∴vm = −
(3m) v = −3v3m = −3× 2 ms = −6 ms
3m
m

Therefore the block of mass m moves to the left (as expected if the block of
mass 3m moves right) at a speed of 6 ms .

Example 3: In Example 2, how much energy was initially stored in the compressed
spring, if the mass of the smaller block is m = 0.35kg ?

Solution: If we take the system as the two masses, the spring, and the world, then
change in energy of the system is zero. We have
ΔEsystem = 0 = ΔK m + ΔK 3m + ΔU g + ΔU s = ( 1
2
mvm2 − 0 + ) ( (3m) v
1
2
2
3m ) (
− 0 + U S , f − U S ,i )
∴U S ,i = 12 mvm2 + 12 ( 3m) v3m (0.35kg ) ⎡⎣⎢( −6 ) + 3( 2 ms ) ⎤⎥ = 8.4J
2 2
2
= 1 m
2 s
⎦

Collisions

Returning to Figure (1) we have two objects coming together, interacting and then perhaps

moving apart again. Whether the objects make physical contact or not, this is an example of a

collision. To begin our study collisions between two objects let’s take two objects and make

them interact (collide) in one dimension. Consider Figure 4 below, which shows an object of

mass m1 moving with a velocity v1i . In order to make the problem less algebraically intensive,

let us take the mass m2 to be initially at rest. Of course, mass m2 could be moving in the same

direction as mass m1 or mass m2 could be moving directly at mass m1 . We’ll worry about that
in a bit. Our goal here is to determine the velocities of the both masses after the collision.

This of course depends on the type of collision.

vi1
v2i = 0
m1 m2

Figure 4: Mass m1 moves to the right at an initial
velocity v1i and collides with mass m 2 initially at rest.

There are two types of collisions that we will investigate and they are called inelastic and

elastic. No matter which type of collision we investigate, if the collision time is small and we

assume that there are no external forces acting on the masses during the collision, then

momentum is conserved, as is given by equation (4). What separates the collision types is

whether the energy due to the motion of the objects, that is the kinetic energy, is conserved or

not. In general, the total kinetic energy of the system of objects is not a conserved quantity

since there are energy losses to sound (you can hear the objects collide) and deforming the

objects (think cars crumpling when they collide) but we can approximate situations in which

the kinetic energy is conserved.

Inelastic Collisions

Inelastic collisions are those in which the momentum is conserved but the kinetic energy is

not. Total energy is always conserved. Consider the following situations in which we have

two objects colliding. Returning to Figure 4 above suppose that the two objects stick together

after the collision, as shown in Figure 5 below. We would like to determine the velocity of the

system after the collision? To determine the velocity we apply conservation of momentum, and

assuming that to the right is the positive x-direction, we have

Δpx = p fx − pix = 0
∴ pix = p fx → m1vi1 = m1v1 f + m2 v2 f = ( m1 + m2 )V
 vi1
v2i = 0 V

m1 m2 m1 m2

Figure 5: Mass m1 moves to the right at an initial velocity v1i and collides with
mass m2 initially at rest. After the collision, masses m1 and m2 move off together
with a common velocity V.

⎛ m1 ⎞
Solving for the velocity of the system after the collision we get V = ⎜ ⎟ v1i . If m1 ≫ m2
⎝ m1 + m2 ⎠

then the velocity of the system after the collision V is approximately equal to the velocity of

m1 before the collision, v1i . If m1 ≪ m2 then the velocity of the system after the collision V is

very small compared to the velocity of m1 before the collision, v1i , but it is not zero. Next, let

us calculate the kinetic energy before and after the collision and then the change in kinetic

energy. If kinetic energy is conserved, then ΔK = 0 ; otherwise it is not. The kinetic energy

before the collision is K i = 12 m1v1i2 . The kinetic energy after the collision is

⎛ m12 ⎞ ⎛ m12 ⎞ 2
K f = ( m1 + m2 )V = ( m1 + m2 ) ⎜
1 2 1
⎟v =
2 1
v . Taking the difference,
2 2
⎜⎝ ( m + m )2 ⎟⎠ i1 2 ⎜⎝ m1 + m2 ⎟⎠ i1
1 2

which I’m not going to physically write out, we get that the change in kinetic energy is not

equal to zero and thus kinetic energy is not conserved. The difference between K f and K i is

the energy lost to the collision, as heat, deforming the objects, and sound.

Before we move on to elastic collisions let’s do one more example, but with some numbers.

Again, as in Figure 4, let the cars collide but this time let them not stick together after the

collision, but rather move separately but in the same direction, as seen in Figure 6. Let the cars

have masses m1 = 1200kg and m2 = 9000kg with velocities v1i = 25 ms and v2i = 20 ms , initially,

respectively. After the collision let the velocity of car 1 be v2 f = 18 ms while the velocity of car
 2 is unknown, call it v2 f . Let’s determine the velocity of car 2 by applying conservation of

momentum.

Δpx = p fx − pix = 0
∴ pix = p fx → m1vi1 + m2 v21 = m1v1 f + m2 v2 f

v2 f =
m1vi1 + m2 v21 − m1v1 f
=
(1200kg × 25 ) + (9000kg × 20 ) − (1200kg × 18 ) = 20.9
m
s
m
s
m
s m
s
m2 9000kg
vi1 v2i v1f v2f

m1 m2 m1 m2

Figure 6: Mass m1 moves to the right at an initial velocity v1i and collides with
mass m2 moving to the right with velocity v2i. After the collision, masses m1 and m2
move off separately with final velocities v1f and v2f.

If this is an inelastic collision then the change in kinetic energy should be zero. To see

whether the change in kinetic energy is zero or not, calculate the initial and final kinetics

energies before and after the collision. The initial kinetic energy is

K i = 12 m1v1i2 + 12 m2 v2i2 = 12 × 1200kg × ( 25 ms ) + 12 × 9000kg × ( 20 ms ) = 2.18 × 106 J , while the final

2 2

kinetic energy,

K f = 12 m1v12f + 12 m2 v22 f = 12 × 1200kg × (18 ms ) + 12 × 9000kg × ( 20.9 ms ) = 2.16 × 105 J . The

2 2

change in kinetic energy is ΔK = K f − K i = ( 2.18 − 2.16 ) × 106 J = 20000J is lost to the

collision. This is an inelastic collision, but not by that much.

Elastic Collisions

Elastic collisions conserve both momentum and kinetic energy. For macroscopically sized

objects, completely elastic collisions are an approximation. Consider Figure 7 below in which

mass m1 is moving to the right at an initial speed v1i , while mass m2 , located to the right of
 mass m1 , is initially at rest. Mass m1 collides elastically with mass m2 and we would like to

calculate the final velocities of each of the masses after the collision, v1 f and v2 f . Assuming

that no external forces act during the collision, we apply conservation of momentum and

kinetic energy to determine the two unknown velocities.

vi1 v2f
v2i = 0 v1f

m1 m2 m1 m2

Figure 7: Mass m1 moves to the right at an initial velocity v1i and collides with
mass m2 at rest. After the collision, masses m1 and m2 move off separately with
final velocities v1f and v2f.

Here we assume that both masses are moving to the right after the collision. In

particular, mass m2 will most likely move to the right after the collision while mass

m1 may move to the right or it could move to the left. We’ll determine the actual

directions by solving the equations for momentum and kinetic energy. Applying

conservation of momentum and kinetic energy we get

Δpsystem = 0 → pi,system = p f ,system → m1v1i = m1v1 f + m2 v2 f

ΔK = 0 → K i,system = K f ,system → 12 m1v1i2 = 12 m1v12f + 12 m2 v22 f

We have two equations and two unknowns. From the equation for momentum, we

solve for the velocity of mass m2 after the collision v2 f and substitute this into the equation

⎛ m2 ⎞
m1
( ) ( ).
2
for kinetic energy. Thus v2 f = v − v and 2 m1v1i = 2 m1v1 f + 2 m2 ⎜ 12 ⎟ v1i − v1 f
1 2 1 2 1
m2 1i 1 f ⎜⎝ m ⎟⎠
2
 After some algebra we can determine expressions for the final velocities of mass m1 and m2

after the collision. We find

⎛ m − m2 ⎞
v1 f = ⎜ 1 ⎟ v1i (5a)
⎝ m1 + m2 ⎠

⎛ 2m1 ⎞
v2 f =⎜ ⎟ v1i (5b)
⎝ m1 + m2 ⎠

Let’s check some limiting cases of the masses to see if equations 5a and 5b seem

reasonable. Suppose that m1 ≪ m2 . Equations (5a) and (5b) become

⎛ m − m2 ⎞ ⎛ 2m1 ⎞
v1 f = ⎜ 1 v
⎟ 1i ~ −v and v = ⎜ m + m ⎟ v1i ~ 0 respectively. Here the lighter
⎝ m1 + m2 ⎠
1i 2f
⎝ 1 2⎠

mass m1 bounces off of the heavier mass m2 in the opposite direction with very little

loss in speed and mass m2 remains stationary. If m1 ≫ m2 , equations (5a) and (5b)

⎛ m − m2 ⎞ ⎛ 2m1 ⎞
become v1 f = ⎜ 1 ⎟ v1i ~ v1i and v2 f = ⎜ ⎟ v1i ~ 2v1i . Here the heavier
⎝ m1 + m2 ⎠ ⎝ m1 + m2 ⎠

mass m1 keeps going in the same direction with little loss of speed while the lighter

mass m2 gets a large kick in speed. If the two masses are approximately equal

⎛ m − m2 ⎞
m1 ~ m2 , equations (5a) and (5b) become v1 f = ⎜ 1 ⎟ v1i ~ 0 and
⎝ m1 + m2 ⎠

⎛ 2m1 ⎞
v2 f = ⎜ ⎟ v1i ~ v1i . Here the incident mass m1 comes to rest and mass m2
⎝ m1 + m2 ⎠

leaves with the speed of the incident mass. Equations (5a) and (5b) therefore seem

reasonable.

Example 4: Consider the frictionless track ABC as shown below in Figure 8. A block
of mass m1 = 5kg is released from point A. It makes a head-on collision at
point B with a block of mass m2 = 10kg , initially at rest. What is the
 maximum height to which mass m1 rises back up the track after the
collision? In addition, suppose that just past point C there is a rough
region in which the coefficient of friction is µ = 0.9 . After how much
distance would block come to rest?

A m1

m2
B C
Figure 8: Two masses on a frictionless track.

Solution: Assuming mass m1 and the world are the system, energy is conserved.
Applying conservation of energy between points A and B (just before m1
collides with m2 ) we have
ΔE = ΔK1 + ΔU g1 + ΔU s = 0

→ ( m v − 0) + (0 − m gy ) + (0 − 0) = 0
1
2 1
2
1f 1 i1

v1 f = 2gy1i = 2 × 9.8 sm2 × 5m = 9.9 ms

During the collision between blocks m1 and m2 at point B, total momentum
of the system is conserved and if we model this as a completely elastic
collision, kinetic energy is also conserved. Applying conservation of
momentum and kinetic energy during the collision we have

Δpsystem = 0 → pi,system = p f ,system → m1v1i = m1v1 f + m2 v2 f

ΔK = 0 → K i,system = K f ,system → 12 m1v1i2 = 12 m1v12f + 12 m2 v22 f

The solutions are given by equations 5a and 5b. We have

⎛ m − m2 ⎞ ⎛ 5kg − 10kg ⎞
v1 f = ⎜ 1 ⎟ v1i = ⎜ ⎟⎠ × 9.9 s = −3.3 s
m m

⎝ 1
m + m2⎠
⎝ 5kg + 10kg
⎛ 2m1 ⎞ ⎛ 2 × 5kg ⎞
v2 f = ⎜ ⎟ v1i = ⎜ ⎟⎠ × 9.9 s = 6.5 s
m m

⎝ 1
m + m2⎠
⎝ 5kg + 10kg
 To determine the height mass m1 rises back up the track after the collision
we apply conservation of energy and we have
ΔE = ΔK1 + ΔU g1 + ΔU s = 0

( ) ( )
→ 0 − 12 m1v1i2 + m1gy f 1 − 0 + ( 0 − 0 ) = 0

v1i2 ( −3.3 ms )
2

yf1 = = = 0.56m
2g 2 × 9.8 sm2

Mass m2 moves to the right after the collision and when it encounters the
rough surface, friction does work on mass bringing it to rest. The distance
traveled by mass m2 is given by

W fr = ΔK → F fr Δx cosθ = µ FN Δx cosθ = 12 m2 v22 f − 12 m2 v2i2

( 6.5 ms )
2
v2i2
Δx = = = 2.4m
2 µ g 2 × 0.9 × 9.8 sm2

Collisions in two-dimensions

The development of equations (1) – (4) apply whether the motion is in one or

more than one dimension. Consider the arrangement of masses shown in Figure 9.

Let mass m1 be incident along the x-axis with velocity v1i and let mass m1 make a

glancing collision with mass m2 initially at rest. Take the origin of the coordinate

system be at mass m2 . After the collision mass m1 scatters at an angle θ measured

with respect to the x-axis while mass m2 scatters at angle φ , also measured with

respect to the x-axis.

v1f
y
m1

v1i

m1 θ
m2 x
ϕ

m2
v2f

Figure 9: Two masses undergo a glancing collision in two-dimensions.

 We can see where mass m1 goes after the collision, so we assume that we

know of can measure θ . We would like to calculate the velocities of each of the

masses after the collision ( v1 f and v2 f ) and the scattering angle of mass m2 , φ .

Assume the collision is elastic and we conserve momentum in the x- and y-directions

as well as kinetic energy. Conservation of momentum horizontally and vertically

gives us

x : m1v1i = m1v1 f cosθ + m2 v2 f cosφ (6a)

y : 0 = m1v1 f sin θ − m2 v2 f sin φ (6b)

while conservation of kinetic energy gives

1
2
m1v1i2 = 12 m1v12f + 12 m2 v22 f (7) .

In theory the above three equations (6a), (6b), and (7) are solvable for the

three unknowns that we have. However, the algebra can be challenging. Let’s do

this by using some numbers. First, let m1 = m2 = m and v1i = 3 ms . What are the final

velocities of each mass after the collision and at what angle φ does mass m2 scatter if

mass m1 scatters at θ = 350 ? Conservation of momentum horizontally and vertically

becomes

x : v1i = v1 f cosθ + v2 f cosφ

y : 0 = v1 f sin θ − v2 f sin φ

and conservation of kinetic energy is given as

v1i2 = v12f + v22 f .

To solve we first square the x- and y-momentum equations, add the results together

and then use conservation of kinetic energy.

x : v1i2 = v12f cos 2 θ + v22 f cos 2 φ + 2v1 f v2 f cosθ cosφ
y : 0 = v12f sin 2 θ + v22 f sin 2 φ − 2v1 f v2 f sin θ sin φ

( ) ( )
→ v1i2 = v12f cos 2 θ + sin 2 θ + v22 f cos 2 φ + sin 2 φ + 2v1 f v2 f ( cosθ cosφ − sin θ sin φ )

Using conservation of energy we get, noting that sin 2 α + cos 2 α = 1

v1i2 = v12f + v22 f = v12f + v22 f + 2v1 f v2 f ( cosθ cosφ − sin θ sin φ ) .

The expression in parentheses is a trigonometric identity,

cosθ cosφ − sin θ sin φ = cos (θ + φ ) .

Now we can determine the scattering angle φ of m2 . We have, assuming that

v1 f ≠ 0 and v2 f ≠ 0

0 = 2v1 f v2 f cos (θ + φ ) → cos (θ + φ ) = 0 → θ + φ = 350 + φ = 900

∴φ = 550

Here we note that IF and ONLY IF m1 = m2 the masses scatter at right angles

to each other. If the masses are not equal then we have to go back equations (6a),

(6b), and (7) and solve the problem with the unequal masses. Now that we have the

scattering angle we can use equations (6a) and (6b) to solve for the final speeds of

each mass after the collision. From equation (6b) we solve, say, for v2 f , and get

⎛ sin θ ⎞ ⎛ sin35 ⎞
v2 f = ⎜ ⎟ v1 f = ⎜ v = 0.7v1 f . Inserting this into equation (6a) we can
⎝ sin φ ⎠ ⎝ sin55 ⎟⎠ 1 f

determine v1 f . We have

( )
v1i = v1 f cos35 + 0.7v1 f cos55 = 1.22v1 f → v1 f =
v1i 3m
= s = 2.46 ms . And lastly
1.22 1.22

v2 f = 0.7v1 f = 0.7 × 2.46 ms = 1.72 ms .

 Of course we’ve also looked at inelastic collisions in which momentum is

conserved, but not kinetic energy. How about we look at an inelastic collision in two

dimensions. Consider the traffic accident shown in Figure 10 below in which two

vehicles, a car (in blue) and a van (in green) approach an intersection. Both fail to

stop and the cars collide, stick together and move off with a common final velocity.

What is the final velocity of the cars after the collision?

V
y mc
mv

vic θ
x
mc

viv mv

Figure 10: Inelastic collision of two vehicles at an intersection. The

vehicles move off together after the collision.

Let’s take up the page as the positive y-direction and to the right as the

positive x-direction. Applying conservation of momentum horizontally and

vertically we have

x: mc vic = ( mc + mv )V cosθ
.
y: mv viv = ( mc + mv )V sin θ

Dividing these two expressions will allow us to determine the angle θ that the

velocity vector makes with the horizontal. Once we know the angle we can use

either momentum equation to determine the final speed V of the car and van after the

collision. The angle is determined as

mv viv ( m + mv )V sinθ = tanθ
= c
mc vic ( mc + mv )V cosθ

mv viv 2500kg × 20 ms
tan θ = = = 1.33 → θ = 53.10
mc vic 1500kg × 25 s
m

The final speed of the car and van after the collision is

mc vic 1500kg × 25 ms
mc vic = ( mc + mv )V cosθ → V = = = 15.6 ms
( mc + mv ) cosθ (1500kg + 2500kg ) cos53.1
Summary

We’ve looked at momentum and its conservation in one and two dimension.

We’ve also looked at two types of collisions between objects, namely inelastic and

elastic. Elastic collisions conserve momentum and kinetic energy. Inelastic

collisions conserve momentum only, with some of the energy of motion being lost to

sources like heat, light, sound and deforming the objects. Both collisions conserve

total energy. Unless explicitly stated, in order to truly tell if a collision is elastic or

not, one needs to calculate the change in kinetic energy. If the change is zero, the

collision is elastic. If the change is not zero then the collision is inelastic.

Let’s try a few more examples for practice.

Example 5: Rutherford Backscattering Spectroscopy

Alpha particles are routinely accelerated using a particle accelerator and are
directed with the use of magnets into targets composed of various elements.
A famous experiment called Rutherford’s experiment has a beam of alpha
particles incident on a target of gold. An alpha particle (a helium nucleus) is
accelerated to a certain speed and makes an elastic head-on collision with a
stationary gold nucleus. What percentage of its original kinetic energy is
transferred to the gold nucleus?
Solution:
The percent kinetic energy lost is given by
⎡ KE after ⎤ ⎡ 12 mα vα2 ,after collision + 12 mv Au,
2
after collision
⎤
% lost = ⎢1 − ⎥ × 100% = ⎢ 1 − 2 ⎥ × 100%
⎣ KE initial ⎦ 2 mvi ,α
1
⎣⎢ ⎦⎥

We need to determine both the final speed of the alpha particle and the gold
nucleus after the collision. To do this we apply conservation of momentum
and kinetic energy. From conservation of momentum we have
mα vi ,α = mα v f ,α + m Au v f , Au and from conservation of kinetic energy
1
2 mα vi2,α = 12 mα v 2f ,α + 12 mAu v 2f , Au .

Here we have two equations and two unknowns. From momentum we solve
m
for the final velocity of the alpha particle and obtain v f ,α = vi ,α − Au v f , Au .
mα
We square this result and substitute into the equation for kinetic energy we
obtain a quadratic equation in the final velocity of the gold nucleus. The
⎛ m2 ⎞
quadratic equation is 0 = ⎜⎜ Au + mAu ⎟⎟v 2f , Au − (2mAu vi ,α )v f , Au . Using the
⎝ mα ⎠
⎧⎛ 2mα ⎞ ⎫
⎪⎜ ⎟v ⎪
quadratic formula we find the solutions v f , Au = ⎨⎜ mα + m Au ⎟ i ,α ⎬ and reject
⎝ ⎠
⎪ 0 ⎪
⎩ ⎭
the zero speed solution. Substituting this result into our equation for the
final speed of the alpha particle and we calculate the final speed to be
m ⎛ m − mAu ⎞
v f ,α = vi ,α − Au v f , Au = ⎜⎜ α ⎟⎟vi ,α . So the percent of the initial kinetic
mα m
⎝ α + m Au ⎠
energy lost is

⎡ KEafter ⎤ ⎡ 12 mα vα2 ,after collision + 12 mvAu,

2
after collision
⎤
% lost = ⎢1 − ⎥ ×100% = ⎢1 − 2 ⎥ ×100%
⎣ KEinitial ⎦ 2 mvi ,α
1
⎢⎣ ⎥⎦
⎡ 2
⎛ mα − m Au ⎞ 2 1 ⎛ 2mα ⎞ 2 ⎤
2

⎢ 2 mα ⎜⎜
1 ⎟ vi ,α + 2 m Au ⎜⎜ ⎟ vi ,α ⎥
⎢ ⎝ mα + m Au ⎟⎠ ⎝ mα + m Au ⎟⎠ ⎥
% lost = ⎢1 − 2 ⎥ ×100%
1
mv
⎢ 2 i ,α
⎥
⎢⎣ ⎥⎦
⎡ ⎛ m − m ⎞2 ⎛ 2 ⎞ ⎤
2

% lost = ⎢1 − ⎜⎜ α Au
⎟⎟ + m Au ⎜⎜ ⎟⎟ ⎥ ×100%
⎢⎣ ⎝ mα + m Au ⎠ ⎝ mα + m Au ⎠ ⎥⎦

Using the mass of an alpha particle of 4u and of gold 197u, we have

 ⎡ ⎛m −m ⎞
2
⎛ 2 ⎞
2
⎤
% lost = ⎢1 − ⎜⎜ α Au
⎟⎟ + m Au ⎜⎜ ⎟⎟ ⎥ × 100%
⎢⎣ ⎝ mα + m Au ⎠ ⎝ mα + m Au ⎠ ⎥⎦
⎡ ⎛ 4 − 197 ⎞ 2 ⎛ 2 ⎞ ⎤
2

= ⎢1 − ⎜ ⎟ + 197⎜ ⎟ ⎥ × 100% = 9.8%

⎣⎢ ⎝ 4 + 197 ⎠ ⎝ 4 + 197 ⎠ ⎦⎥

Example 6: Ice Hockey

A hockey puck traveling at 1.2 ms collides with a second stationary equal mass puck
and, after the collision, moves with a speed of 0.8 ms deflected by an angle of 300 .
What is the velocity (magnitude and direction) of the other puck after the collision?
In addition, what is the fraction of the initial energy lost in the collision?

Solution: Using a standard Cartesian coordinate system we apply conservation of

momentum in the horizontal and vertical directions and we have
pix = p fx → mv1ix = mv1 cosθ + mv 2 cos φ → v1ix = v1 cosθ + v 2 cos φ and
piy = p fy → 0 = mv1 sin θ − mv 2 sin φ → 0 = v1 sin θ − v 2 sin φ .

Here we have two equations and two unknowns, v2 and φ . Inserting the
numbers from the problem we have for the horizontal and vertical directions
v1ix = v1 cosθ + v2 cosφ → v2 cosφ = 0.51 and v2 sin φ = 0.4 .

Dividing these two expressions we solve for the unknown angle and find
0.4
tan φ = = 0.7843 → φ = tan −1 (0.7843) = 38.1o . Therefore the unknown
0.51
velocity is v2 sin φ = v2 sin 38.1 = 0.4 → v2 = 0.65 ms .

The fraction of the initial energy lost is

⎡ KEafter ⎤ ⎡ 12 mv1,2 after collision + 12 mv2,2 after collision ⎤ ⎡ v1,2 after collision + v2,2 after collision ⎤
%lost = ⎢1 − ⎥ × 100% = ⎢ 1 − ⎥ × 100% = ⎢1 − ⎥ × 100%
⎣ KEinitial ⎦ ⎣⎢
1
2
mv12ix ⎦⎥ ⎣⎢ v12ix ⎦⎥
⎡ m ⎤ ⎡ (0.8 m )2 + (0.65 m )2 ⎤
%lost = ⎢1 − ⎥ × 100% = ⎢1 − s s
⎥ × 100% = 26.2%
⎣ m+M⎦ ⎢⎣ (1.2 ms )2 ⎥⎦

Example 7: A bullet in a block on a horizontal surface

A 10g projectile is fired at 500 ms into a 1kg block sitting on a

frictionless horizontal surface. The projectile lodges in the center of
the block, and both move off together.

a. What is the final velocity of the block after the collision?

b. The block slides along the frictionless surface some distance and
then encounters a ramp, which slopes up at an angle of 600 . What
distance does the block travel along the surface of the ramp before
coming to a stop?

c. If the coefficient of friction between the block and the ramp is µ = 0.2 ,
how far does the block slide up the ramp before stopping?

Solution:
a. Assuming that the positive x-direction is to the right we apply conservation
of momentum. We find for the velocity after the collision
mb vi 0.01kg × 500 ms
pix = p fx → mb vi = (mb + mbl )V → V = = = 4.95 ms
(mb + mbl ) 1.01kg
to the right.

b. Define d as the distance the block slides along the ramp and h as the
height the block rises above the horizontal, we have from the geometry
h
sin θ = → h = d sin θ . Applying conservation of energy between the
d
bottom of the ramp and where the block comes to rest we have
(
ΔE = 0 = ΔU g + ΔKE = mgy f − mgyi + 12 mv 2f − 12 mvi2 ) ( )
0 = mgh − 12 mvi2 = mgd sin θ − 12 mvi2

( 4.95 ms ) = 1.44m
2
vi2
∴d = =
2g sin θ 2 × 9.8 sm2 sin60
c. In the presence of friction, energy is lost to heat between the surfaces. To
calculate the new distance we use
ΔU g + ΔKE = −ΔE friction
(mgy f − mgyi ) + ( 1
2 )
mv 2f − 12 mvi2 = − F friction × d new
mgh − 12 mvi2 = mgd new sin θ − 12 mvi2 = − µ k mg cosθd new

∴ d new =
vi2
=
(4.95 ms ) 2

= 1.29 m
2 g (sin θ + µ k cosθ ) 2 × 9.8 sm2 (sin 60 + 0.2 cos 60 )
Example 8: Fireworks

A rocket used for fireworks explodes just when it reaches its highest point in a
vertical trajectory. It initially bursts into three fragments with masses of m , 3m ,
and 4m , each of these to explode slightly later. If the 4m fragment falls
vertically downward with an initial velocity of 8 ms , and the 3m fragment is
ejected with a velocity of 10 ms at an angle of 300 above the horizontal, what is
the velocity of the third fragment?

Solution: At the highest point of the rocket’s motion, its velocity is zero. Therefore the
initial x- and initial y-momenta are both zero when the rocket explodes. After
the explosion we apply conservation of momentum in the vertical and horizontal
directions. Assuming that the piece of mass m has a momentum in the same
quadrant as the 3m piece.

In the vertical direction 0 = −4m(8 ms ) + 3m(10 ms )sin 30 + mv sin φ → 17 = v sin φ .

In the horizontal direction we have

0 = 3m(10 ms )cos 30 + mv cosφ → −25.98 = v sin φ .

Here we have that the x- component of the velocity is negative, while the y-
component is positive, so the momentum vector lies in the 2nd quadrant, our
guess was incorrect, but that’s ok. Now we know. Taking the ratio of these two
equations produces an angle of 33.20 above the negative x-axis. Then using any
one of the above equations we find for the magnitude of the velocity to be 31 ms .

Example 9: Proton scattering

A proton moving with an initial velocity vix in the x-direction, as shown in Figure
11, collides elastically with another proton that is initially at rest. If the two protons
have equal speeds after the collision, what is the speed of each proton after the
collision in terms of vix , and what are the directions of the velocity vector after the
collision?

Figure 11: An incident proton scattering off a stationary proton.

Solution: We break up the momentum into x and y-components and use conservation of
momentum in each direction. We have in the x-direction
pix = p fx → mvix = mv cosθ + mv cos φ , while in the y-direction
piy = p fy → 0 = mv sin θ − mv sin φ → sin θ = sin φ → θ = φ .

Using the results from the vertical motion we rewrite the x-momentum as
vix = v cosθ + v cosφ = 2v cosφ .

Next we use the kinetic energy to obtain an expression for vix in terms of v so that
we can determine the unknown angle φ . Conservation of energy gives
KEi = KE f → 12 mvix2 = 12 mv 2 + 12 mv 2 → vix2 = 2v 2 → vix = 2v .

vix
Therefore we have the magnitude of the velocity after the collision as v = . And
2
from the x-momentum we calculate the angle to be
2
vix = 2v = 2v cos φ → cos φ = → φ = 45o . The angle of the velocity vector after
2
the collision is φ = θ = 45 .
0