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Collisions
*
See Momentum and impulse on
page 480.
A collision occurs when one object strikes another object at speed. Collisions
are of great interest to physicists because they represent one of the main ways
that force and energy are transferred from one object to another.
A collision is an interaction that involves a change in the momentum of the
objects involved. Momentum is the product of an object’s mass and velocity.
The fundamental principle underlying all analyses of collisions is the law of
conservation of momentum.
The law of conservation of momentum states that the sum of the initial
momenta of all the objects involved in a collision will equal the sum of the final
momenta of the same objects. That is, if we find the momentum of each of the
objects before the collision and add these together, we will get the same result
as we do when we add up the momentum of each of the objects after the
collision. Mathematically this is often expressed as:
Σp i = Σpf
initially
m1
u1
object 1
m2
u2
collision
object 2
m1
v1
finally
object 1
m2
v2
object 2
figure 9.1 In this example,
two objects collide head-on.
where Σp i is the sum of all the initial momenta, and Σpf is the sum of all the final
momenta.
Let us consider a simple system in which two objects are involved in a
straight-line collision (see figure 9.1). Object 1 has an initial velocity of u1 and a
mass of m1. Object 2 has an initial velocity of u2 and a mass of m2. In this case,
the law of conservation of momentum could be expressed as in equation 9.1.
This simple law allows us to make some powerful predictions about the results
of certain collisions.
equation 9.1 m1u1 + m2u2 = m1v1 + m2v2
where m1 and m2 are the masses (kg), u1 and u2 are the initial velocities
(m s–1), and v1 and v2 are the final velocities (m s–1) of objects 1 and 2,
respectively.
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Worked example 9.1
One of the skaters in figure 9.2, with mass 80 kg, is skating in a straight
line with a velocity of 6.0 m s–1 while the other, of mass 70 kg, skating in
the opposite direction, also with a speed of 6.0 m s–1.
a The two skaters collide and the heavier skater comes to rest.
Assuming that friction can be ignored, what will happen to the lighter
skater after the collision?
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6.0 m s–1
figure 9.2 When
two skaters collide
on a nearfrictionless skating
rink they exert
equal and opposite
forces on each
other.
Practical activity
Conservation of
60
momentum in collisions
b What would happen if the two skaters had hung on to each other and
stayed together after the collision?
Solution
a As both velocity and momentum are vector quantities, a positive
direction should be established. Adopting the direction of the
heavier skater’s motion as the positive direction we have:
u1 = 6.0 m s–1, u2 = –6.0 m s–1, v1 = 0 m s–1, m1 = 80 kg and
m2 = 70 kg. Substituting into equation 9.1:
m1u1 + m2u2 = m1v1 + m2v2
m1u1 + m2u2 – m1v1
⇒ v2 = m2
80 × 6.0 + 70 × (–6.0) – 80 × 0 60
= = = 0.857
70
70
Hence, the 70 kg skater bounces back in the opposite direction with a
speed of 0.86 m s–1.
b Treating the two skaters as one mass after the collision gives a final
mass of 150 kg and v1 = v2 = v. Substituting into equation 9.1 gives:
m1u1 + m2u2 80 × 60 + 70 × (–6.0)
60
= = = 0.4
⇒v=
mtotal
1
50
150
In this case, the skaters move off together in the direction of the
heavier skater at a speed of 0.4 m s–1.
A different outcome after the collision results in a different velocity for each
skater in worked example 9.1. There is no unique answer when applying the
idea of conservation of momentum. The final velocity of any object depends on
what happens to all the objects involved in the collision.
Conservation of momentum also applies to collisions involving more than
two objects and to collisions that occur in two and three dimensions. In each
case, the total initial momentum is the vector sum of the momenta of all the
objects, and the vector sum of the final momenta of the objects is exactly equal
to the total initial momentum.
Practical activity
Conservation of
54
momentum in explosions
An explosion of a
stationary object can be
thought of as a special
type of collision where the
total initial momentum is
equal to zero. This means
that momentum produced
in any particular direction
must be balanced out
exactly by momentum in
the opposite direction.
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m1u1 + m2u2 = mtotalv
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Elastic
..... and inelastic collisions
*
See Energy and work on
page 379.
The law of conservation of momentum is another expression of Newton’s laws
of motion. Therefore, it must apply to every collision in nature. Another
quantity that must be conserved in collisions is energy.
A collision involves moving objects. Moving objects have kinetic energy.
According to the law of conservation of energy, energy cannot be created or
destroyed, but it can be changed from one form to another. It is often useful to
study what happens to the kinetic energy that objects take into a collision.
Elastic collisions
In an elastic collision, the
sum of the initial kinetic
energies of all the objects
involved in a collision will
equal the sum of the final
kinetic energies of the same
objects.
*
See Kinetic energy on
page 419.
The kinetic energy taken into a collision may be conserved as kinetic energy. If
this occurs, then the collision is called an elastic collision.
In the microscopic world of interactions between atoms and molecules, all
collisions are elastic. This means that conservation of kinetic energy is a very
useful tool for studying these interactions. In the macroscopic world of
everyday life, however, perfectly elastic collisions are rare. Most collisions
involve some conversion of kinetic energy into heat, sound and other forms of
‘waste’ energy. Fortunately, there are some macroscopic interactions (for
example, collisions between billiard balls) that are so close to being elastic that
the law of conservation of kinetic energy can still be useful.
The kinetic energy of each object can be calculated using E k = 12 mv 2 where
m is the mass of the object (kg) and v is its velocity (m s–1). When dealing
mathematically with the law of conservation of kinetic energy, it is important to
remember that energy is a scalar and not a vector quantity.
Let us consider what would happen if the skaters in worked example 9.1
were involved in an elastic collision. Remember we had: m1 = 80 kg,
u1 = 6.0 m s–1, m2 = 70 kg, and u2 = –6.0 m s–1. The initial kinetic energy is
given by:
1
2
m1u12 + 12m2u22 = 0.5 × 80 × 6.02 + 0.5 × 70 × (–6.0)2
= 1440 + 1260 = 2700
Hence, the initial kinetic energy is 2700 J, and, if the collision is elastic, the
total final kinetic energy must also equal 2700 J. Using this fact, we can show
that neither of the outcomes explored in worked example 9.1 are elastic
collisions.
When the heavier skater stopped on colliding with the lighter skater, the
final velocities were v1 = 0 m s–1 and v2 = 0.86 m s–1. In this case:
1
2
m1v12 + 12m2v22 = 0.5 × 80 × 02 + 0.5 × 70 × 0.862
= 0 + 26 = 26
The total final kinetic energy is 26 J. Almost all of the kinetic energy would
have been changed into other forms of energy, like heat or sound, during the
collision.
If the two skaters stuck together after colliding, then v1 = v2 = 0.4 m s–1. This
would result in an even smaller final kinetic energy:
m1v12 + 12m2v22 = 0.5 × 80 × 0.42 + 0.5 × 70 × 0.42
= 6.4 + 5.6 = 12
Clearly, neither of these two collisions would be elastic.
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2
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We can determine the speeds of the skaters if they did have an elastic
collision because, in an elastic collision, both the momentum and the kinetic
energy are conserved. For conservation of momentum:
Σpi = Σpf
m1u1 + m2u2 = m1v1 + m2v2
80 × 60 + 70 × –60 = 80 × v1 + 70 × v2
60 = 80 × v1 + 70 × v2
6 = 8 × v1 + 7 × v2
Since there are two unknowns in this equation, we cannot proceed any
further without some extra piece of information. We can formulate another
equation that describes this situation by considering conservation of kinetic
energy:
0.5 × 80 ×
m1u12 + 12m2u22 = 12m1v12 + 12m2v22
+ 0.5 × 70 × (–6.0)2 = 0.5 × 80 × v12 + 0.5 × 70 × v22
2700 = 40 × v12 + 35 × v22
540 = 8 × v12 + 7 × v22
1
2
6.02
We now have two equations involving the same two unknowns, v1 and v2.
Solving these simultaneously:
6 = 8 × v1 + 7 × v2
(6 – 7v2)
⇒ v1 = 8
Substituting into 540 = 8 × v12 + 7 × v22:
(6 – 7v2)2
+ 7v22
⇒ 540 = 8
(36 – 84v2 + 49v22)
⇒ 540 = + 7v22
8
⇒ 4284 + 84v 2 – 105v 22 = 0
⇒ 105v 22 – 84v 2 – 4284 = 0
Applying the general formula for the solution of quadratics produces two
sets of solutions:
v1 = 6.0 m s–1, v22 = –6.0 m s–1 and v1 = –5.2 m s–1, v2 = 6.8 m s–1
The first pair of solutions describes the set of initial velocities. The second
set of solutions represents the final velocities for an elastic collision. This means
that the 80 kg skater will rebound with a velocity of 5.2 m s–1 and the 70 kg skater
will bounce back at 6.8 m s–1 after an elastic collision.
To check this solution, we can calculate the total final kinetic energy:
1
2
⇒ 4320 = 36 – 84v2 + 49v22 + 56v22
The general formula for the
solution of a quadratic
equation, ax2 + bx + c = 0, is:
x=
*
–b ± b 2 – 4ac
2a
See Mathematical skills on
the ePhysics CD.
m1v12 + 12m2v22 = 0.5 × 80 × (–5.2)2 + 0.5 × 70 × 6.82
= 1082 + 1618 = 2700 J
This result is the same as the initial kinetic energy, therefore this collision
would be elastic. This is an unlikely solution for two skaters colliding, but nearly
elastic collisions can be observed when hard objects like billiard balls collide.
Collisions
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Inelastic collisions
By far the majority of collisions that we observe in everyday life are inelastic.
This means that kinetic energy is not conserved in these collisions. Since energy
cannot be created or destroyed, the kinetic energy that is lost in an inelastic
collision is actually transformed into different forms of energy such as heat or
sound. Even when dealing with inelastic collisions, it can be useful to consider
conservation of energy. An inelastic collision will usually result in a net loss of
kinetic energy. In other words, the total final kinetic energy will be less than the
total initial kinetic energy:
ΣEk,initial > ΣEk,final
This fact can help eliminate the possibility of certain outcomes from a
collision. For example, if we once again consider the example of the two skaters
in worked example 9.1 colliding, it can be proven that it is impossible for either
skater to come out of the collision with a final speed of more than 9 m s–1. If the
heavier skater had a final speed of 9 m s–1 then her final kinetic energy would be:
Ek = 12mv 2 = 0.5 × 80 × 92 = 3240 J
This is more than the total initial kinetic energy for the collision—2700 J—
and is, therefore, impossible. Similarly, if the lighter skater had a final speed of
9 m s–1, her final kinetic energy would be 2835 J, which is also more than the
total initial kinetic energy for the situation.
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Questions
356
a Is this consistent with the law of conservation
of momentum? Explain.
b Using reasonable estimates for the initial
speed, and mass of the truck and Superman,
describe what will happen.
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mass 2.0 kg. What is their combined speed in
m s–1 after the collision?
6 Which, if any, of the situations in questions 1–5
involve elastic collisions? Use calculations to
justify your answer.
7 Superman stops a truck simply by blocking it with
his outreached arm.
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Use g = 9.8 m s–2 when required.
1 A white billiard ball of mass 100 g travelling at
2.0 m s–1 across a low-friction billiard table has a
head-on collision with a black ball of the same
mass initially at rest. The white ball stops while
the black ball moves off. What is the velocity of the
black ball?
2 A girl with mass 50 kg, running at 5 m s–1, jumps
onto a 4 kg skateboard travelling in the same
direction at 1.0 m s–1. What is their new common
velocity?
3 A man of mass 70 kg steps forward out of a boat
and onto the nearby riverbank with a velocity,
when he leaves the boat, of 2.5 m s–1 relative to
the ground. The boat has a mass of 400 kg and
was initially at rest. With what velocity, relative to
the ground, does the boat begin to move?
4 A railway car of mass 2 tonnes, moving along a
horizontal track at 2 m s–1, runs into a stationary
train and is coupled to it. After the collision the
train and car move off at a slow 0.3 m s–1. What is
the mass of the train alone?
5 A trolley of mass 4.0 kg and moving at 4.5 m s–1
collides with, and sticks to, a stationary trolley of
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8 Newton’s cradle (pictured) involves nearly perfect
elastic collisions between hanging ball bearings.
Predict what will happen when the ball is released.
Use calculations to back up your prediction.
Assume that the mass of each ball bearing is 100 g
and the ball in the person’s hand will strike the
other balls with a speed of 2 m s–1 when released.
9 A 150 g ice puck collides head on with a smaller
100 g ice puck, which was initially stationary, on a
smooth, frictionless surface. The initial speed of
the 150 g puck is 3 m s–1. Calculate the final
velocities of the two pucks assuming that the
collision is elastic.
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