UNIT G484

Module 1
4.1.2
Collisions
Candidates should be able to :

State the principle of conservation of momentum.

Apply the principle of conservation of momentum to solve
problems when bodies interact in one dimension.

Define a perfectly elastic collision and an inelastic collision.

Explain that whilst the momentum of a system is always
conserved in the interaction between bodies, some change
in kinetic energy usually occurs.


CONSERVATION OF MOMENTUM





The principle of conservation of momentum is deduced from 1
NEWTONS second and third laws.













If two bodies collide, then by NEWTON III each body
exerts an equal and opposite force (F) on the other.

Since force (F) acts on each body for the same time (t),
the two bodies will experience an equal and opposite
impulse (Ft).

Each body therefore experiences an equal and opposite
change of momentum, which means that the total change
of momentum of the two bodies is zero (i.e. TOTAL
MOMENTUM REMAINS CONSTANT).



The principle of conservation of momentum is universally true and
it applies as much to the motion of galaxies as to the interaction
of sub-atomic particles.

FXA 2008



THE PRINCIPLE OF CONSERVATION OF MOMENTUM
When bodies in a system interact, the TOTAL MOMENTUM
REMAINS CONSTANT provided no external force acts on
the system.

FOR A COLLISION




FOR AN EXPLOSION

momentum before collision = momentum after collision
momentum after explosion = momentum before explosion
CAR A CAR B
F
BA
F
AB

F
AB
= F
BA
= F
UNIT G484
Module 1
4.1.2
Collisions

2
FXA 2008



















The apparatus shown above can be used to test the validity of
the principle of conservation of momentum.


Cart 1 is pushed towards stationary cart 2 and as it passes
through light-gate 1 the computer automatically calculates its
speed (v1).

v1 = length of card passing through the light-gate = L
time taken for card to pass through light-gate t


When cart 1 collides with cart 2, the velcro fasteners cause
the carts to stick together.


The two carts then pass through light-gate 2 and their common
speed (v2) is also calculated by the computer.


The mass of each cart is measured using an electronic balance.
EXPERIMENTAL VERIFICATION OF THE PRINCIPLE
OF CONSERVATION OF MOMENTUM

computer
interface unit
light-gate 1 light-gate 2
smooth track
cart 2 cart 1
velcro fasteners
card of
length (L)


Mass of cart 1, m1 =

Mass of cart 2, m2 =

Speed of cart 1 before collision, v1 =

Speed of carts (1 + 2) after collision, v2 =





Momentum before collision = m1 v1 =

Momentum after collision = (m1 + m2) v2 =
RESULTS
kg
kg
m s
-1

m s
-1

CALCULATIONS
x = kg m s
-1

x = kg m s
-1


Is the total momentum after the collision equal to
the total momentum before the collision ?

Has the principle of conservation of momentum been
verified ?

Did any external force act on the system of colliding
carts ? If so, name the external force.
SOME QUESTIONS
UNIT G484
Module 1
4.1.2 Collisions

PRACTICE QUESTIONS (1) 3

1 A fullyladen Range Rover of total mass 2200 kg travelling at
20 m s
-1
collides with a Mini Cooper of mass 850 kg.

(a) Assuming that the two vehicles separate after the impact,
calculate the Range Rovers velocity after impact if :

(i) The Mini is initially stationary and its velocity after the
collision is 10 m s
-1
.

(ii) The Minis initial velocity is 8 m s
-1
in the same direction
as the Range Rover and its velocity after the collision is
18 m s
-1
.

(b) If the Minis initial velocity is 12 m s
-1
in the opposite direction
to that of the Range Rover and the two vehicles become stuck
together as a result of the impact, calculate their combined
velocity after the collision.



2 An astronaut who is about to
attach a replacement part to
the Hubble Space Telescope
suddenly realises that the tether
which keeps him linked to his
space vehicle has broken loose.
In an attempt to get back to his
vehicle, he throws the replacement
part into space.


The astronauts total mass is 175 kg and that of the part is 5.5 kg.
Calculate the astronauts velocity if he throws the part with a velocity
of 1.5 m s
-1
.
FXA 2008




A railway truck (A) of mass 2 x 10
4
kg travelling at 2.0 m s
-1
collides
with a second truck (B) of mass 1.0 x 10
4
kg moving in the opposite
direction at 1.5 m s
-1
. If the trucks couple automatically on impact,
calculate the common velocity with which they move after the collision.




A bullet of mass 50 g is fired from a rifle of mass 2.0 kg with a muzzle
Velocity of 250 m s
-1
. What is the initial recoil velocity of the rifle ?
MOMENTUM CONSERVATION - WORKED EXAMPLES
COLLISION








(2.0 x 10
4
x 2.0) + (1.0 x 10
4
x 1.5) = (3.0 x 10
4
x v)

v = 2.5 =
3.0
2.0 x 10
4
kg
A A B B
3.0 x 10
4
kg 1.0 x 10
4
kg
2.0 m s
-1
1.5 m s
-1
v m s
-1

BEFORE COLLISION AFTER COLLISION
0.83 m s
-1

EXPLOSION








0 = (0.05 x 250) + (2.0 x v)

v = 12.5 =
2.0
2.0 kg 0.05kg
250 m s
-1

-v m s
-1

BEFORE FIRING
AFTER FIRING
6.25 m s
-1

UNIT G484
Module 1
4.1.2 Collisions
TYPES OF COLLISION

In any collision between two objects :

TOTAL MOMENTUM IS ALWAYS CONSERVED
(so long as no external force acts).




TOTAL ENERGY IS ALWAYS CONSERVED
(The principle of conservation of energy states that energy
cannot be created or destroyed, only converted from one
form into another).




KINETIC ENERGY MAY OR MAY NOT BE CONSERVED
This depends on whether the collision is ELASTIC or
INELASTIC.
4
FXA 2008
i.e. TOTAL MOMENTUM AFTER = TOTAL MOMENTUM BEFORE
COLLISION COLLISION
i.e. TOTAL ENERGY AFTER = TOTAL ENERGY BEFORE
COLLISION COLLISION
A PERFECTLY ELASTIC COLLISION is one in which
kinetic energy is conserved.
i.e. TOTAL KINETIC ENERGY = TOTAL KINETC ENERGY
AFTER COLLISION BEFORE COLLISION

An INELASTIC COLLISION is one in which kinetic energy is
not conserved.
i.e. TOTAL KINETIC ENERGY < TOTAL KINETIC ENERGY
AFTER COLLISION BEFORE COLLISION



The collisions between gas molecules in an ideal gas and those
between electrons and the molecules of a superconductor are
assumed to be PERFECTLY ELASTIC.



The collision between two
steel ball bearings in the
executive toy called Newtons
Cradle is a close approximation
to a PERFECTLY ELASTIC
collision, but some kinetic
energy is transferred to
sound energy, heat energy
and work done in plastic
deformation. Once the toy is set in motion, the balls continue to move
for some time, but eventually the initial kinetic energy is transferred
to other forms and the balls come to rest.


Another close approximation
To a PERFECTLY ELASTIC
collision is that between two
snooker balls.


POINTS TO NOTE
In a PERFECTLY ELASTIC COLLISION :
m m m m
v v -v -v
BEFORE AFTER
VELOCITY OF APPROACH OF = VELOCITY OF SEPARATION OF
COLLIDING OBJECTS COLLIDING OBJECTS
UNIT G484
Module 1
4.1.2 Collisions

PRACTICE QUESTIONS (2) 5

1 In a school experiment, a cart of mass 2.5 kg moving with a velocity
of 0.5 m s
-1
collides with a second, stationary cart of mass 3.0 kg.
The velcro fastenings attached to the cart ends causes them to stick
Together on impact and they move off with a common velocity of
0.227 m s
-1
.

(a) Show that momentum is conserved in this collision.

(b) Calculate the kinetic energy before and after the collision.

(c) Explain whether the collision is elastic or inelastic.


2 A tennis ball of mass 0.06 kg is
dropped from a height of 1.5 m
above a hard surface and it rebounds
to a height of 1.0 m above the
surface.

(a) Calculate the kinetic energy
lost on impact with the
surface.


(b) Explain whether this collision
is elastic or inelastic.


FXA 2008
In practice all collisions are, to some extent, INELASTIC
(i.e. some of the initial kinetic energy is transferred to other
energy forms).



Car CRUMPLE ZONES are
specifically designed to make
a collision more INELASTIC
and so absorb the kinetic
energy in a crash.



The same thinking is used in
the design of MOTORWAY
CRASH BARRIERS.






A close approximation to a PERFECTLY INELASTIC collision
is that which would occur between two pieces of soft dough of
equal mass moving towards each other with equal speed. On
collision, all the kinetic energy of each piece is transferred to
sound energy, heat energy and work done in plastic deformation.
So the pieces splat into each other, deform into one mass of
dough, heat up and come to a standstill.


A PERFECTLY INELASTIC collision is one in which ALL
the initial kinetic energy is transferred to other energy
forms.
KINETIC ENERGY AFTER COLLISION = 0

1.5 m
1.0 m
UNIT G484
Module 1
4.1.2 Collisions
3 A stationary radioactive
nucleus of mass
3.98 x 10
-25
kg undergoes
radioactive decay in which
it emits an alpha particle
of mass 6.68 x 10
-27
kg.


(a) If the speed of the
alpha particle is
2.5 x 10
7
m s
-1
, what is the speed of the recoiling nucleus ?

(HINT : The mass of the recoiling nucleus = mass of the parent
nucleus - mass of the alpha particle).

(b) If all the energy released in this decay is the sum of the kinetic
energies of the alpha particle and the recoil nucleus, calculate
the alpha particles kinetic energy as a fraction of the total
energy released.

4







A 3 kg sphere moving with a velocity of 5.0 m s
-1
collides head-on
with a stationary sphere of mass 2 kg. If the 2 kg sphere moves
off with a velocity of 4.2 m s
-1
, calculate :

(a) The velocity (v) of the 3 kg sphere after the collision.

(b) The kinetic energy lost as a result of the collision.
5 6



















A bullet of mass 25 g, travelling horizontally at a speed of 250 m s
-1

strikes and embeds itself in a stationary wooden block which is
suspended so that it can swing freely.

(a) Calculate : (i) The common speed of the block and embedded
bullet immediately after impact.

(ii) The maximum vertical height through which the
block rises as a result of the impact.

(b) How much of the bullets initial kinetic energy is lost as a result
of the impact and into what forms is this lost kinetic energy
transferred ?

(c) Explain whether this collision is elastic or inelastic.
FXA 2008
recoiling nucleus
Emitted
Alpha particle
Parent nucleus
v = 2.5 x 10
7
m s
-1

vN

3 kg
2 kg
3 kg
2 kg
5.0 m s
-1
4.2 m s
-1
v m s
-1

BEFORE COLLISION AFTER COLLISION

h

Bullet of
mass = 25 g
Wood block of
Mass = 975 g
250 m s
-1

v
BEFORE IMPACT AFTER IMPACT
UNIT G484
Module 1
4.1.2 Collisions
HOMEWORK QUESTIONS

1 (a) (i) Define the momentum of a body.

(ii) A body, initially at rest, explodes into two unequal fragments
of mass m
1
and m
2
. Mass m
1
has a velocity v
1
and mass m
2

has a velocity v
2
. Using the principle of conservation of
momentum, derive an expression for v
1
/v
2
.

(b) An isolated nucleus of mass 4.0 x 10
-25
kg is initially at rest.
It decays, emitting an alpha particle of mass 6.7 x 10
-27
kg
with a kinetic energy of 1.2 x 10
-14
J.

(i) Show that the speed of the alpha particle is about
2 x 10
6
m s
-1
.

(ii) Calculate the momentum of the alpha particle.

(iii) Hence find the speed of the recoiling nucleus.

(c) Before the decay described in (b) The nucleus is at point P as
shown below.






(i) Place a small cross at a possible position, to full scale, of the
alpha particle 8.0 x 10
-9
s after emission.
(ii) Indicate with an arrow, starting at P, the direction of
movement of the recoiling nucleus.
(iii) Estimate how far the recoiling nucleus has moved in
8.0 x 10
-9
s.
(OCR A2 Physics - Module 2824 - June 2004)
2 This question is about the interactions between three identical, 7
perfectly elastic, solid cubes.

(a) Fig 1 shows three numbered cubes, each of mass m, placed in
contact on a frictionless, horizontal surface. A steady horizontal
force P is applied to the end surface of cube 1.







(i) Show that the acceleration of cube 3 is P/3m.


(ii) Write down an expression for the resultant force F
3

on cube 3.


(iii) Write down expressions in terms of P for :

1. The acceleration, a
2
of cube 2.

2. The resultant force, F
2
on cube 2.


(iv) Hence write down an expression for the magnitude of the
force applied by :

1. Cube 3 on cube 2, F
32
.

2. Cube 1 on cube 2, F
12
.
FXA 2008
P

P
1 2 3
Fig. 1
UNIT G484
Module 1
4.1.2
Collisions
(b) Cube 3 is removed. Cube 1 is moved to the left and then
projected towards the stationary cube 2 with speed u. Suppose
that after collision cube 1 moves with speed v
1
and cube 2 moves
with speed v
2
. See Fig. 2.









(i) Write down equations for the conservation of momentum
and of energy in terms of m, u, v
1
, and v
2
.

(ii) Put the values v
1
= 0 and v
2
= u into your equations in (i)
and show that they are solutions.

(c) Cube 3 is replaced in its original position close to cube 2. Cube 1
is moved to the left and projected towards cube 2 with speed u.
See Fig. 3.






(i) After the collision, cubes 1 and 2 remain at rest and cube 3
moves with speed u. Explain these observations.

(ii) Cubes 2 and 3 are now glued together. Describe without
calculation or explanation what happens in this situation
when cube 1 is projected towards cube 2 as shown in Fig. 3.
3 In a motorway collision, a lorry of mass 3500 kg moving at a 8
speed of 24 m s
-1
rammed the rear end of a car of mass 1000 kg
which was travelling in the same direction as the lorry with a speed
of 14 m s
-1
. After the impact, the car shot forward at a speed of
20 m s
-1
.

(a) Use the principle of conservation of momentum to calculate the
speed of the lorry after the collision.
(b) (i) Calculate the total kinetic energy before and after the
collision.

(ii) Explain whether the collision is elastic or inelastic.

4 A man of mass 95 kg jumps off a landing stage into a rowing boat of
mass 250 kg which is untied and lying stationary in the water.
If the mans velocity as he lands in the boat is 4.0 m s
-1
at an angle
of 30 to the vertical, calculate the speed with which the man and
boat move off together.

5 (a) Define the momentum of a particle. State the principles of
conservation of linear momentum and of the conservation of
energy as applied to head-on collisions between particles. Explain
the conditions under which linear momentum and kinetic energy
are conserved.
(b) Use one or both of the principles in (a) to explain why in elastic
collisions a small particle bounces back from a massive particle
(Fig. 1), whereas a large massive particle incident on a small
particle does not bounce back (Fig. 2).
FXA 2008

1 2
u
1 2
v1 v2
Fig. 2

1 2 3
u
Fig. 3

BEFORE AFTER
Fig. 1
BEFORE AFTER
Fig. 2