6. Elastic and nonelastic collisions, momentum  concept and calculations and Newton's 2nd law of motion
Doc Brown's Physics Revision
Notes
Suitable for GCSE/IGCSE Physics/Science courses or
their equivalent
This page will help answer questions such as ...
What is the momentum of a an object?
What is an elastic collision?
What is an inelastic collision?
How do we solve calculations involving
momentum?
What is the momentum of an object and how do we calculate it?
All moving objects have what we call
momentum. (sometimes denoted by a lower case p)
The faster an object moves, the bigger
its momentum.
The bigger the mass of a moving object,
the bigger its moment.
The momentum of an object is directly
proportional to its velocity AND its mass.
Momentum (p) is a vector quantity,
it has both size and direction
Therefore the equation for momentum is
...
momentum (kilogram metre per second) = mass (kilogram) × velocity (metre per second)
momentum p (kg m/s) = mass
m (kg) x velocity
v (m/s)
p = m × v
Example calculations using the formula for
momentum
Q1.1 A 120 kg sprinter is travelling at 9.0 m/s, what is the
momentum of the runner?
momentum p = m × v
p = 120 x 9 =
360 kg m/s
Q1.2 If a 20 g bullet has a momentum of 8 kg m/s, calculate its
velocity.
p = m × v, v = p/m
(don't forget to change the g to kg!)
velocity = 8/(20/1000) =
400 m/s
Q1. 3 What mass must an object have if it has a momentum of 1.5 x 10^{6}
kg m/s and a velocity of 30 m/s?
p = m × v, m = p/v
mass = 1.5 x 10^{6}/30 =
5.0 x 10^{4}
kg
The Law of Conservation of Momentum
Here we will consider collisions between
two objects in a closed system.
Here a closed system here means no other
external forces affect the situation e.g. a collision between two objects 
the event.
If an external force like friction is
involved, total momentum cannot be conserved.
The total momentum of an event in a
closed system is the same before and after the event (e.g. a collision between
two objects).
This is called the 'Law of
Conservation of Momentum' and you can use it do lots of calculations!
e.g. total momentum of two colliding
objects = total moment of objects after collision
e.g. for two colliding objects, where
p = momentum: p_{1} + p_{2} = p_{3} + p_{4}
substituting m and v for the mass
and velocity gives ...
m_{1}v_{1} + m_{2}v_{2}
= m_{1}v_{3} + m_{2}v_{4}
where v_{1} and v_{2}
are the initial velocities and v_{3} and v_{4} the
velocities after the collision,
(you are assuming there is no
change in mass, no bits have flown off!)
and if two objects stick together'
after the collision then: p1 + p2 = p3
substituting m and v for the mass
and velocity gives ...
m_{1}v_{1} + m_{2}v_{2}
= m_{3}v_{3}
where v_{1} and v_{2} are the initial
velocities and v_{3} and m_{3} (m_{3} = m_{1}
+ m_{2}) the final velocity and mass after the collision.
In other words the large object
formed by collision has the momentum equal to the two momentums of
the colliding objects added together.
Momentum is conserved for both elastic and
inelastic collisions.
For a perfect elastic collision,
no kinetic energy is lost  kinetic energy conserved.
In an elastic collision, the
total energy in the kinetic energy stores of the colliding objects
is the same as before and after the collision.
For an inelastic collision,
kinetic energy is not conserved  kinetic energy is lost in some form
e.g. heat or sound.
In an inelastic collision, some
of the energy of the moving objects kinetic energy store is lost and
transferred to other energy stores of the objects themselves e.g.
heat or to the environment e.g. sound.
This is because the atoms are
bashed together increasing their kinetic energy store. They 'relax'
to their normal state by losing the energy as heat or sound.
The above examples involve collision but
there are all sorts of ...
Other examples of momentum effects
e.g.
When a gun is fired the bullet goes in
the forward direction (positive momentum), but the gun recoils in the complete
opposite direction (negative momentum).
From Newton's 3rd law, you get an equal
and opposite force reaction.
The bullet goes one way and the gun the
other way from the force of the explosion.
numerically, ignoring +/ signs: m_{gun}v_{gun}
= m_{bullet}v_{bullet} (see Q2.2
below)
Note that the combined momentum = zero,
which it is at the start and immediately after the gun is fired.
Before a rocket is fired it has zero
velocity and zero momentum.
When the rocket's propellant is fired,
the rocket goes one way (positive momentum) and the exhaust gases go in the
opposite direction (negative momentum).
So, in conserving momentum, we can say ..
numerically, ignoring +/ signs: m_{rocket}v_{rocket}
= m_{exhaust gases}v_{exhaust gases}
(see Q2.2 below)
Because the rocket produces gases with
great force and velocity, the rocket must respond according to Newton's 3rd
law of motion  equal and opposite forces in dramatic action!

Questions on momentum
Q2.1 The diagram below shows the sequence of events when a moving
green ball collides with a stationary purple ball.
The green ball has a mass of 2.0 kg and is moving at 4.0 m/s prior
to colliding with the stationary purple ball.
After the collision both balls are moving in
a forward direction.
If the green ball is still moving forward at 1.5 m/s,
calculate
the velocity of the purple ball.
From the law of conservation of momentum
total momentum before collision = total
momentum after collision
m_{1}v_{1} + m_{2}v_{2}
= m_{1}v_{3} + m_{1}v_{4}
(2 x 4) + (3 x 0) = (2 x 1.5) + (3 x v_{4})
8 + 0 = 3 + 3v_{4}, 3v_{4}
= 8  3 = 5, therefore V_{4} = 5/3 =
1.7 m/s (2sf)
Q2.2(a) Why does a gun recoil (move backwards) when fired?
At the start both the gun and the
bullet have zero moment.
After firing the bullet then has its
own momentum in a forward direction from the shooter.
BUT, this must be balanced by a
momentum in the opposite direction because total momentum must be
conserved.
So, on firing, the gun itself must
get its own equal but opposite momentum (remember that momentum is a
vector quantity!).
Therefore the gun's equal and
opposite momentum causes it to recoil backwards.
(b) A 10 g bullet accelerates to 400 m/s
after being fired from a gun with a mass of 2.0 kg.
(i) What is the momentum of the
bullet? (kg = g/1000)
p = m × v = (10/1000) x
400 =
4.0 kg m/s
(ii) What is the momentum of the gun?
From the law of conservation of
momentum, the gun must have a momentum equal and opposite to that of
the bullet.
Therefore gun momentum =
4.0 kg m/s
(the minus sign is important,
particularly in more complex calculations).
(iii) What is the velocity of the
recoiling gun?
The momentum of the gun and
bullet must be equal
(from the law of conservation of
momentum)
therefore numerically: m_{gun}v_{gun}
= m_{bullet}v_{bullet}
m_{gun}v_{gun} =
4.0 = 2.0 x v_{gun}, v_{gun} = 4.0/2.0 =
2.0 m/s
(again, note the minus sign,
because the gun recoils in the opposite direction to the bullet).
(c) Why is it an advantages to fire with
a heavier gun?
From the equation: momentum = mass x
velocity
The bigger the mass of the gun,
for the same momentum, the lower the recoil velocity and the lesser
impact on the person firing the gun.
Q2.3
The diagram below shows the sequence of events when a moving car crashes into a
stationary car and they combine together and move forward.
The green car (1), of mass 1000 kg, crashes into
the stationary blue car (2) of mass 800 kg.
After the collision, the two cars are locked
together.
Calculate the velocity of the wrecked cars
immediately after the condition.
From the law of conservation of momentum
total momentum before collision = total
momentum after collision
m_{1}v_{1} + m_{2}v_{2}
= m_{3}v_{3}
(1000 x 20) + (800 x 0) = (1000 + 800) x
v_{3}
20 000 + 0 = 1800v_{3}, v_{3}
= 20 000/1800 = 11.1 m/s
(3sf)
Loss of KE calculation?
Q2.4 Imagine a 0.2 kg ball moving at 5 m/s colliding with a 0.3 kg
ball moving at 2 m/s in the same direction.
After the collision the 0.2 kg green ball stops moving and the 0.3 kg
purple ball continues moving in the same direction.
Calculate the final velocity of the of the 0.3 kg
purple ball.
First sort out all the masses and velocities
where v_{1} and v_{2}
are the initial velocities and v_{3} and v_{4} the
velocities after the collision
m_{1} = 0.2 kg, initial velocity
v_{1} = 5 m/s, final velocity v_{3} = 0 m/s
m_{2} = 0.3 kg, initial velocity
v_{2} = 2 m/s, final velocity v_{4} = ? m/s
total momentum of two of objects before
collision = total moment of objects after collision
for two colliding objects, where
p = momentum: p_{1} + p_{2} = p_{3} + p_{4}
substituting m and v for the mass
and velocity of the two objects gives ...
m_{1}v_{1} + m_{2}v_{2}
= m_{1}v_{3} + m_{2}v_{4}
Then substitute everything in the equation,
rearrange and deduce the answer (v_{4}).
(0.2 x 5) + (0.3 x 2) = (0.2 x 0) + (0.3
x v_{4})
1.0 + 0.6 = 0 + 0.3v_{4},
0.3v_{4} = 1.6
therefore the final velocity of the 0.3
kg ball = v_{4} = 1.6/0.3 =
5.3 m/s (2 sf)
Q2.5 Imagine a 0.2 kg ball moving at 5 m/s hitting a 0.3 kg ball
moving at 2 m/s in the same direction.
After the collision both balls keep moving in the same
direction, but with different velocities.
If the 0.2 kg ball continues to move at a velocity of 1.5 m/s,
calculate the final velocity of the 0.3 kg ball.
This is an almost identical problem to Q2.4, except there is no
zero momentum term.
So, sorting out all the masses and velocities
where v_{1} and v_{2}
are the initial velocities and v_{3} and v_{4} the
velocities after the collision
m_{1} = 0.2 kg, initial velocity
v_{1} = 5 m/s, final velocity v_{3} = 1.5 m/s
m_{2} = 0.3 kg, initial velocity
v_{2} = 2 m/s, final velocity v_{4} = ? m/s
total momentum of two objects before
collision = total moment of objects after collision
for two colliding objects, where
p = momentum: p_{1} + p_{2} = p_{3} + p_{4}
substituting m and v for the mass
and velocity of the two objects gives ...
m_{1}v_{1} + m_{2}v_{2}
= m_{1}v_{3} + m_{2}v_{4}
where v_{1} and v_{2}
are the initial velocities and v_{3} and v_{4} the
velocities after the collision
Then substitute everything in the equation,
rearrange and deduce the answer (v_{4}).
(0.2 x 5) + (0.3 x 2) = (0.2 x 1.5) +
(0.3 x v_{4})
1.0 + 0.6 = 0.3 + 0.3v_{4},
0.3v_{4} = 1.6  0.3 = 1.3
therefore the final velocity of the 0.3
kg ball = v_{4} = 1.3/0.3 =
4.3 m/s (2 sf)
Note the final velocity of the 0.3 kg
ball is less than in Q2.4 because in Q2.4 it received all the momentum from
the collision.
Q2.6 Snooker players can hit the white ball 'dead centre' into a
stationary red ball. The stationary red ball then flies off at the same velocity
as the incoming white ball and the white ball is left in a stationary position
where the red ball was.
Ignoring friction and assuming the balls
have the same mass, explain this observation.
The law of conservation states the total
momentum must be the same before and after the collision.
m_{white}v_{intitial} + m_{red}v_{initial}
= m_{white}v_{final} + m_{red}v_{final}
m_{red}v_{initial}
is zero, since the red ball is initially stationary with a momentum of zero.
m_{white}v_{final}
is zero, since the white ball is now stationary, also with a momentum of
zero.
Therefore: m_{white}v_{intitial}
= m_{red}v_{final}
and since the masses are equal, m_{white}
= m_{red}, then the final velocity of the red ball must equal
the initial velocity of the white ball.
It should be noted that skilled snooker
players can play all sorts of tricks with the 'physics' of snooker and often
have a total disregard for the law of conservation of momentum!
Q2.7
Change in Momentum and Newton's Second Law of Motion
When a resultant force acts on an object
for any length of time it will cause a change in momentum by changing the
velocity.
The resultant force causes a change in
speed or direction in the direction of the resultant force.
The resultant force causes a change in
momentum in the direction of the resultant force.
A way of stating Newton's 2nd law is to
say the change in momentum is proportional to the size of the resultant
force and the time interval during which the force is acting on the object.
This statement is justified by
following the three mathematical steps below.
A resultant force on an object will cause it
to accelerate or decelerate.
force = mass x acceleration
F = ma (the
mathematical expression of Newton's 2nd Law of Motion)
NOW acceleration is the change in velocity in
a specific change in time.
a = ∆v / ∆t
(the usual formula for acceleration)
a = acceleration in m/s^{2},
v in m/s, time t in seconds s
If you combine these two equations you get
force = mass x change in velocity / time
F = m
∆v / ∆t
but, for a given
object of mass m, m ∆v
= ∆mv = ∆p = change in momentum
so can write Newton's
2nd law equation as
F =
∆mv / ∆t = ∆p / ∆t
(force is equal to the rate of change of momentum)
and
∆p = ∆mv = F x ∆t
(remember ∆t is the time over which the resultant force acts)
This means that a force applied to an
object over any period of time must change that object's velocity.
From the point of view of solving calculation
problems you need to be pretty familiar with all this maths!
In calculations you can then use Newton’s 2nd
law of motion (F = ma) as follows:
force (newton, N) = change in
momentum (kilogram metre per second, kg m/s) / time (second, s)
F = ∆mv / ∆t = (mv  mu) /
∆t, where = mass of object kg, u = initial velocity
m/s, v = final velocity m/s
this means the rate of change of
momentum is a directly related to the resultant force or force applied,
so you can say the force equals the
rate of change of momentum,
the equation can be written as F = mΔv
/ Δt since m is constant for a given object
which can be expressed as ...
change of momentum (kg m/s) =
resultant force (N) × time for which it acts (s)
∆p = ∆mv = F x
∆t (the
product of the resultant force x time is sometimes called the impulse)
(if the force is
varying, then the average force is used in the calculation, but I don't
think this is needed for GCSE)
Consequences of the force
equalling the rate of change of momentum
For any moving object the faster the
change in momentum (Δmv) occurs, the greater the force (F) involved,
and the shorter the time taken (∆t).
You should be able to see this from
the equation ∆mv = F x
∆t
for a given
mv, increase in F means a decrease in ∆t,
for a given
mv, decrease in F means an increase in ∆t,
This has serious implications for e.g.
car crashes.
The faster the crash happens the
bigger the force on the car and its occupants.
This sudden change momentum change in
such a short time results in a great impact force.
The smaller Δt for a given
momentum change, the greater the force involved and the greater the
chance of injury.
If you can make the rapid
deceleration occur over a longer time (increasing Δt) you reduce
the resultant force and decrease the chance of serious injury.
That is what crumple zones are for in
the design of modern cars.
In terms of F = ma, you are
trying to reduce a and so decrease F.
(For more details on this issue see
Stopping distances,
impact forces  example
calculations
Calculations involving momentum and
Newton's 2nd law
Q3.1
What explosive force is required to accelerate a 20 g bullet in a gun
from 0 m/s to 300 m/s in 0.15 seconds.
20 g = 20/1000 = 0.02 kg, u = 0 m/s, v =
300 m/s, ∆t = 0.15 s
F = ∆mv / ∆t = (mv  mu) /
∆t
F = {(0.02 x 300)  (0.02 x 0)} / 0.15
F = (0.02 x 300) / 0.15 = 6.0 / 0.15 =
40 N
It is important that you can solve
these sorts of problems in different ways, depending on what information
you get and how you are expected to deduce things from it e.g.
Using the same information you
can solve this problem using F = ma directly.
a = 300 / 0.15 = 2000 m/s2, m =
0.02 kg, F = ma, F = 0.02 x 2000 = 40 N
This is the approach I took on my
stopping distances, impact forces  example calculations
Q3.2 What force must a tennis player generate with a tennis racket
exert on a 60 g tennis ball to accelerate it from rest to a speed of 50 m/s in
10 milliseconds (10 ms).
mass of tennis ball =
60 g = 60/1000 = 0.06 kg, time force applied = 10 ms = 10/1000 = 0.01 s
∆p = ∆mv = F x
∆t
mv changes from zero
to 0.06 x 50 = 3.0 kg m/s
therefore: 3.0 = F x
0.01 = 0.01, F = 3.0 / 0.01 =
300 N
Q3.3 A 900 kg car
travelling at 20 m/s
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