1.3a Elastic and inelastic collisions
Momentum
The phrase ‘gathering momentum’ is used in everyday life to indicate that
something or someone has got going and is likely to prove difficult to stop.
There are echoes of this in sport, where the phrase ‘the momentum has turned in
the other team’s favour’ can often be heard. In Physics momentum is an
indication of how difficult it would be to stop something. The faster or more
massive an object is the more momentum it will have. The precise definition of
momentum is:
The momentum (p) of any object is the product of its mass ( m) and its velocity
(v):
p  mv
Since mass is measured in kilograms and velocity in metres per second, the units
of momentum are kg m s –1 .
Momentum is a vector quantity, and the direction of the momentum is the same
as the direction of the velocity. It is useful because it is a conserved quantity, ie
the total momentum is the same before and after a collision, in the absence of
external forces. This is called the principle of conservation of momentum.
Momentum in collisions
Notes
1.
You should learn the statement of the principle of conservation of
momentum: the total quantity of momentum before a collision is the same
as the total quantity of momentum after the collision in the absence of an
external force.
2.
This is a fundamental law of physics and applies to all collisions: road
accidents, collisions between meteors and planets, collisions between
atoms.
3.
The law applies to total momentum, not individual momentum.
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4.
Since momentum is a vector quantity we cannot add momenta (plural of
momentum) like ordinary numbers; we must take account of direction. For
the problems that we will consider this means that some momenta (usually
in the original direction) may be positive (+) while other momenta (the
opposite direction) are negative (–).
5.
In problems it is essential to demonstrate that you know the conservation
law. This should be stated as part of your working as:
Total momentum before collision = total momentum after collis ion
Worked examples
1.
Two cars are travelling towards each other as shown below. They collide,
lock together and move forwards (ie to the right) after the collision.
Find the speed of the cars immediately after the collision.
Before
After
10 m s–1
8 m s–1
1200 kg
A
1000 kg
B
?
1200 kg
1000 kg
Take motion  as +
m 1 u 1 = 1200 × 10 = 12,000
(m 1 + m 2 )v = (1200 + 1000)v
m 2 u 2 = 1000 × –8 = - 8000
= 2200v
total momentum before = total momentum after
12,000 – 8000 = 2200v
v = 4000 = 1.8
2200
ie = 1.8 m s –1 to the right
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2.
One vehicle (vehicle A) approaches another (vehicle B) from behind as
shown below. The vehicles are moving with the speeds shown. After the
collision the front vehicle is travelling at 11 m s –1 . Calculate the speed of
vehicle B after the collision.
Before A
B
12 m s–1
A
9 m s–1
1200 kg
After
B
11 m s–1
800 kg
1200 kg
?
800 kg
Take motion  as +
m 1 u 1 = 1200 × 12 = 14,400
m 1 v 1 = 1200 × v 1 11
m 2 u 2 = 800 × 9 = 7200
m 2 v 2 = 800 × v 2
total momentum before = total momentum after
14,400 + 7200 = 1320 + 800v 2
21,600 = 1320 + 800v 2
v2 
20,280
 25.35
800
ie v 2 = 25.4 ms –1 to the right
Note: In momentum problems it can help to lay out your working under the
headings ‘before’ and ‘after’. Always draw a diagram of the situation before and
after, including all relevant details such as masses, velocities and directions of
motion. Include the statement of conservation of momentum.
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Kinetic energy in collisions
Momentum is always conserved in collisions and explosions. By the law of
conservation of energy, the total energy is also conserved in collisions and
explosions, but kinetic energy is not necessarily conserved.
There are two kinds of collision:
(a)
those in which kinetic energy (KE) is conserved
ie total KE before = total KE after
This is called an elastic collision.
(b)
those in which kinetic energy is not conserved
ie KE is lost during the collision to other forms of energy, such as heat
energy
This is called an inelastic collision.
If after a collision the objects stick together, this is always an inelastic collision.
If the objects bounce apart the collision may be elastic; the only sure way of
finding out is to calculate the total KE before and after the collision. Usually this
will involve using conservation of momentum first to calculate all the relevant
velocities. Remember, momentum is always conserved in the absence of external
forces.
Reminder: E K  1 mv 2
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A special case
If two objects of the same mass collide elastically, they exchange velocities after
the collision. For example, if a cue ball travelling at 2 m s –1 collides with a
second snooker ball of the same mass (a head-on collision, with no spin
involved), the second ball will move off at 2 m s –1 and the cue ball will stop.
This effect can be seen in ‘Newton’s cradle’.
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1.3b Explosions and Newton’s third law
Explosions
Explosions are treated in the same way as collisions, in that total momentum is
conserved. For example, in the case of a bullet being fired from a gun, the total
momentum before firing is zero, since nothing is moving.
After firing, the bullet has momentum in the forward dir ection. The gun must
therefore have the same magnitude of momentum in the opposite direction so the
two momenta cancel each other out, leaving the total momentum still equal to
zero. For this reason the gun must have a recoil velocity after the explosion ( i.e.
the gun ‘jumps’ backwards).
It should be obvious that in an explosion kinetic energy is not conserved.
Think about this. If a bomb explodes leading to an overall gain in kinetic energy.
What kind of energy does the bomb have before the explosion?
Worked example
A gun of mass 1 kg fires a bullet of mass 5 g at a speed of 100 m s –1 . Calculate
the recoil velocity of the gun.
Before
After
0 m s–1
1 kg
0.005 kg
?
100 m s–1
0.005 kg
1 kg
Take motion  as +
momentum = 0
m1v1 = 1 × v1
m 2 v 2 = 0.005 × –100
= –0.5
total momentum before = total momentum after
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0 = (1 × v 1 ) – 0.5
v 1 = 0.5
ie v 1 = 0.5 m s –1 in the opposite direction to the
bullet
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Momentum and Newton’s third law
It can be shown that conservation of momentum and Newton’s third law mea n
the same thing.
Starting from conservation of momentum:
total momentum before
Rearrange:
or:
ie
= total momentum after
m1u1 + m2u2
= m1v1 + m2v2
m2u2 – m2v2
= m1v1 – m1u1
m 2 (u 2 – v 2 )
= m 1 (v 1 – u 1 )
–m 2 (v 2 – u 2 )
= m 1 (v 1 – u 1 )
–(change in momentum of
object 2)
= (change in momentum of
object 1)
In other words, in any particular example involving two objects colliding, if the
momentum of one object increases by, for example, 6, then the momentum of the
other object must decrease by 6.
Consider again the example given above. We had concluded that:
–m 2 (v 2 – u 2 ) = m 1 (v 1 – u 1 )
Applying the impulse relationship:
–F 2 t = F 1 t
so
–F 2 = F 1
Newton’s third law states that if one body exerts a force on a second body, the
second body exerts a force on the first body that is equal in size and opposite in
direction.
Note: These forces operate on different bodies so they do not cancel each other
out.
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1.3c Impulse
Obvious question: you’re stranded by a fire on an upper floor and have to jump.
Would you rather jump onto grass or concrete? Why? Does what you land on
affect your speed of fall? What difference does a different material make when
you land on it? Try to use precise physics terminology rather than everyday
language.
What is the physics behind this situation? Let’s look at some numbers.
A student jumps from a window ledge 2 m high. Find their velocity when they
reach the ground. This will be the same whatever surface they land on.
When the student lands, the different surfaces have a different amount of ‘give’.
This means the time for their deceleration will be different. For this example we
will take the time of landing on the concrete to be 0 .01 s and on the grass to be
0.3 s, as the grass gives way underneath.
Estimate the average force exerted on the student by the concrete and grass.
(b)
Find the deceleration and therefore the average force:
concrete
grass
What do you think about these results?
If you had no choice but to jump onto concrete, what could you do to help
minimise the chance of injury?
In each of these examples, the change in momentum is the same since the speed
at which the student hits the ground is the same. The difference is the time in
which the change in momentum takes place. Rearranging Newton’s second law
helps to make this more explicit.
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Momentum and Newton’s second law
Newton’s second law: F  ma
but
so
a
vu
t
 v  u  mv  mu
F  m

t
 t 
which is the rate of change of momentum. This how Newton himself defined his
second law.
The law can be rearranged:
Ft  mv  mu
The product Ft is called the impulse. Impulse is a vector quantity.
The units of Ft could be kg m s –1 since these are the units of momentum, but they
are also Ns, and these are the units that should be used for impulse.
The force calculated from the impulse relationship is the average force. The
force involved is rarely constant, eg consider a tennis ball hit by a racquet:
The force exerted by the racquet on the ball will change with time. The impulse
is calculated from the area under the graph, so:
impulse = area under a force–time graph
This is the case regardless of the shape of the graph. Leading on from this, what
else is represented by the area under a force–time graph?
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Worked examples
1.
In a snooker game, the cue ball, of mass 0.2 kg, is accelerated from the rest
to a velocity of 2 m s –1 by a force from the cue which lasts 50 ms. what
size of force is exerted by the cue?
u = 0, v = 2 m s –1 , t = 50 ms = 0.05 s, m = 0.2 kg, F = ?
Ft  mv  mu
F × 0.05 = (0.2 × 2) – (0.2 × 0)
F × 0.05 = 0.4
F=8N
2.
A tennis ball of mass 100 g, initially at rest, is hit by a racquet. The racquet
is in contact with the ball for 20 ms and the force of cont act varies over
this period, as shown in the graph. Determine the speed of the ball as it
leaves the racquet.
impulse
= area under graph
= ½ × 20 × 10 –3 × 400 = 4 N s
u=0
m = 100 g = 0.1 kg
v=?
Ft  mv  mu
4 = 0.1v – (0.1 × 0)
4 = 0.1v
v = 40 m s –1
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3.
A tennis ball of mass 0.1 kg travelling horizontally at 10 m s –1 is struck in
the opposite direction by a tennis racket. The tennis ball rebounds
horizontally at 15 m s –1 and is in contact with the racket for 50 ms.
Calculate the force exerted on the ball by the racket.
m = 0.1 kg
u = 10 m s –1
v = -15 m s –1 (opposite direction to u)
t = 50 ms = 0.05 s
Ft  mv  mu
= (0.1 × (–15)) – (0.1 × 10)
= –1.5 – 1
= –2.5
F = –50 N (the negative sign indicates force in opposite direction to
the initial velocity)
0.05F
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