# 01 MECHANICS.indb

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```					                          CONSERVATION OF MOMENTUM IN TWO
DIMENSIONS
Lesson
Collisions
2   Learning Outcomes and Assessment Standards
Learning Outcome 1: Practical Scientific inquiry and problem-solving skills
The learner is able to confidently explore and investigate phenomena relevant to Physical
Science by using enquiry, problem solving, critical thinking and other skills
We know this when the learner is able to:
• Conduct an investigation.
• Interpret data to draw conclusions.
• Solve problems.
• Communicate and present information and scientific arguments.
Learning Outcome 2: Constructing and applying Scientific knowledge
The learner is able to access, interpret, construct and use Physical Science concepts to explain
phenomena relevant to Physical Science.
We know this when the learner is able to:
• Recall and state specified concepts.
• Indicate and explain relationships.
• Apply scientific knowledge.
Learning Outcome 3: The Nature of Science and its Relationship to Technology, Society and
the Environment
The learner is able to demonstrate an understanding of the nature of Science, the influence
of ethics and biases in Physical Science and the inter-relationship of Science, Technology,
indigenous knowledge, the environment and society.
We know this when the learner is able to:
• Evaluate knowledge claims and science’s inability to stand in isolation from other fields.
• Evaluate the impact of science on human development.
• Evaluate science’s impact on the environment and sustainable development.

Overview
Requirements for this section of the curriculum
The momentum of a system is conserved when no external forces act on it
●     Calculate x and y components of momentum.
●     Solve problems involving conservation of momentum in both the x-direction
and the y-direction.
●     Know that an external force causes the momentum to change. The impulse
delivered by the force is F.t = p.
●     Solve problems involving impulse and momentum when the applied force is in
the horizontal or vertical direction.
●     Distinguish between elastic and inelastic collisions.
●     Solve problems involving elastic and inelastic collisions for objects moving
parallel or at right angles to each other.
●     Use one equation for components in the x-direction and one equation for
components in the y-direction.

Links to Grade 11 (impulse and the conservation of momentum [p i= pf])
SEE Grade 11 Physical sciences, Lesson 1a

In this lesson …
6                    The conservation of momentum applied to collisions in two dimensions.

01 MECHANICS.indb 6                                                                                                      2008/07/18 03:01:06 PM
Lesson
Conservation of momentum
DVD
In the Grade 11 course we dealt with collisions in one dimension, and so the
conservation of linear momentum. But momentum is also conserved in two
dimensions. One could call this the conservation of planar momentum. An
excellent example of the conservation of momentum in two dimensions is the
motion of the hard and inflexible snooker balls on a snooker or pool table. But
momentum.

An example
This problem was set in the old curriculum examination in October/November
2006. It is an interesting problem because it can be solved in at least two
ways. First, we can use the principle of conservation of mechanical energy to
solve the problem. But it also brings together in one problem the conservation
of momentum and the work done in the previous lesson on projectiles. (When
you have solved the problem using conservation of momentum, try to solve the
problem by using conservation of mechanical energy, which you studied in Grade
10: [potential energy + kinetic energy = constant].)
Problem:
A 1,5 kg block of wood rests on the edge of a table, 1,2 m above the floor. A
bullet with a mass of 0,075 g and moving at an unknown, horizontal velocity
strikes the block and becomes embedded in it. The block (with the embedded
bullet) leaves the table and strikes the floor with a speed of 8 m·s–1. (Ignore all
frictional effects.)

a)    Calculate the magnitude of the velocity with which the block leaves the
table.
b)    Calculate the magnitude of the velocity with which the bullet strikes the
block.
c)    Calculate the magnitude of the impulse of the bullet during impact.

Solve the problem on your own before looking at the calculations and answer.
ub = ?       mb = 0,075 g     MB = 1,5 kg
uB                MB = 1,5 kg
mb = 0,075 g

Solution:
The velocity of the block as it leaves the table (v(B+b)) can be calculated using the
conservation of linear momentum:
mb.ub + MB.uB = (MB + mb). v(B+b)
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01 MECHANICS.indb 7                                                                               2008/07/18 03:01:06 PM
However we do not know the value of ub so we need to find another method of
tackling the problem. This is where a knowledge of projectiles comes in handy.
At the bottom of its flight, this projectile has a speed of 8 m·s–1. This speed (we
don’t know the angle so we cannot talk of velocity) is made up of a vertical and
a horizontal component of velocity. The horizontal component is constant so it is
equal to the velocity with which the block with the embedded bullet left the table.
The vertical component is determined purely by calculating the fall of the body
through 1,2 m off the table top.
Vv2 = uv2 + 2gh
Vv2 = 0 + 2 (10) (1,2)
Vv2 = 24
∴ Vv = 4,9 m·s–1
Now, Vh can be calculated using Pythagoras’ Theorem:
__
a) Vh = √v2 – vv2
__
Vh = √82 – 4,92                                                     Vh = vB + b
__
Vh = √64 – 24
_
Vh = √40
Vh = 6,32 m·s–1
b)                  mb.ub + MB.uB = (MB + mb). v(B+b)
(0,075). ub + (1,5)(0) = (1,5 + 0,075).(6,32)         Vv              8 m.s–1
u = ___
(1,575 × 6,32) – 0
b            0,075
∴ ub = 132,72 m·s–1
c)     (FΔt)   bullet
= Δp   Block
= (mΔv)Block

Meaning: When the bullet strikes the block it provides the impulse that will
change the momentum of the block. Therefore we can calculate the magnitude of
the impulse by calculating the size of the change in momentum of the body
which feels the effect of the impulse!
(FΔt)   bullet
= (mΔv)Block
= (1,5).(6,32 – 0)
= 9,48 N.s
Alternatively, we can also compute the size of the change in momentum of
the bullet:
(FΔt)   Block
= (mΔv)bullet
= (0,075).(6,32 – 132,72)
= -9,48 N.s
Note: The answer in this calculation is negative. Make sure you understand
why. (To do so, start by looking at what was calculated.)

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01 MECHANICS.indb 8                                                                                 2008/07/18 03:01:07 PM
Elastic and inelastic collisions
The difference between elastic and inelastic collisions is that kinetic energy is
only conserved in elastic collisions. However, momentum is conserved in both
elastic and inelastic collisions.

Elastic collisions             Inelastic collisions
Conservation of      Conserved …                        Not conserved
kinetic energy
as no energy is lost during        e.g. traffic accidents: energy
the collision.                     is used up when metal is
bent (deformed); also when
KE   before
= KE   after
heat and sound energy are
generated in the crash
Conservation of      Conserved …                        Conserved …
momentum
as long as there are no            as long as there are no
external forces e.g. frictional    external forces – i.e.
forces; in other words             momentum is conserved in
momentum is conserved in           closed systems.
closed systems.
Most collisions are not            Most collisions are inelastic.
elastic.
Examples                                                ●   In most collisions a
●    However certain collisions        degree of deformation
between fast moving               of one or both of the
atoms are almost elastic.         colliding bodies takes
●    When a satellite bounces          place.
off the atmosphere of a       ●   An inelastic collision
planet that collision is          often produces heat or
almost perfectly elastic.         sound.
●    Very hard, inflexible steel
balls, as in the desktop
ornament called a
almost elastically.

Example
A collision which is REALLY in 2-D!
A 7 500 kg truck is travelling at a constant velocity of 5 m·s–1 due east when it
collides, almost head on, with a 1 500 kg car moving at 20 m·s–1 in a direction
30° South of West. After the collision the two vehicles remain tangled together.
At what speed and in what direction will the wreckage move immediately after
the collision?
Understanding the problem
To solve this problem, choose two axes to represent the two dimensions. It is
normal to choose a Cartesian plane which is described using x-y axes. In this
problem we will call the axes “W-E” and “N-S” to correspond with the directions
used in the problem.
East ≡ “positive x direction” (i.e. West is the “negative x”) and North is the
“positive y direction”. We use components of the momentum (p) along these
directions and then combine them using Pythagoras to determine the magnitude
and direction of the final momentum. Remember (1) that linear momentum is                  9

01 MECHANICS.indb 9                                                                                2008/07/18 03:01:08 PM
conserved in both the x (W-E) and y (N-S) directions. Remember (2) that we are
dealing with vector quantities.
N

Ptruck
W                                                                           E

Pcar

PRESULTANT

Components:                       S

N

PW-E = –20 cos 30°                       Ptruck is PW-E only
W                                                                                E
θ = 30°
P N-S = –20 sin 30°
Pcar

Linear momentum in the W-E direction BEFORE:
ptruck = mtruck.vtruck
ptruck = 7 500.(+5)
ptruck = +37 500 kg.m·s–1                    (due East)
pcar = mcar.vcar
pcar = 1 500.(–20 cos 30°)
pcar = –30 000 × 0,866
pcar = –25 980 kg.m·s–1                      (due West)
ptot = ptruck + pcar
ptot = +37 500 kg.m·s–1 + (–25 980 kg.m·s–1)
P   tot
= +11 520 kg.m·s–1                 (due East)
Linear momentum in the N-S direction BEFORE:
ptruck = mtruck.vtruck
ptruck = 7 500 . (0)                         (no N-S component for the truck’s velocity)
ptruck = 0 kg.m·s        –1

pcar = mcar.vcar
10         pcar = 1 500.(–20 sin 30°)
pcar = –30 000 × 0,500

01 MECHANICS.indb 10                                                                                         2008/07/18 03:01:08 PM
pcar = –15 000 kg.m·s–1                 (due South)
ptot = ptruck + pcar
ptot = 0 kg.m·s–1 + (–15 000 kg.m·s–1)
ptot = –15 000 kg.m·s–1                 (due South)
Because we know that linear momentum is conserved, the total linear
momentum of the wreck (the truck and car combined) will have an W-E
component of +11 520 kg.m·s–1 and a N-S component of –15 000 kg.m·s–1 .
When these are combined using Pythagoras’ theorem we get:
___
pfinal = √pN – S2 + pW – E2
____
∴ pfinal = √(–15 000)2 + (+11 520)2
___
∴ pfinal = √357 710 400
∴ pfinal = 18 913 kg.m·s–1
Since the two masses are now combined we can find the speed (the magnitude of
the velocity) of the wreckage as it moves away from the point of impact:
pfinal
v = _ = __
18 913 kg.m·s–1
mtotal         9 000 kg                                   px = 11 520 kg.m.s–1
v = 2,1 m·s–1                                                        θ
py
θ = tan–1 _      px
∴ θ = tan __
–15 000
–1
+ 11 520
∴ θ = 52,4°
py = 15 000 kg.m.s–1
V

Since tan < 0 the angle is in the second quadrant, i.e. 52,4° South of East.

Observation
When we do the vector diagrams for the initial momentum, the resultant
momentum for BEFORE the crash also gives us the resultant momentum for
AFTER the crash – because momentum is conserved. Because momentum is a
vector quantity, the total initial momentum that we calculate gives us the size and
direction of the final momentum.

Activity 2.1
An explosion blows a rock into three parts. Two pieces fly off at right angles to one           PAIRS
another: a 1,0 kg piece at 12 m·s–1 and a 2,0 kg piece at 8 m·s–1. The third piece
flies off at 40 m·s–1.
BASELINE
a)     Draw a diagram to show the direction of travel of the third piece of rock.              ASSESSMENT

b)     Determine the mass of the third piece of rock.

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01 MECHANICS.indb 11                                                                                         2008/07/18 03:01:09 PM
Activity 2.2
A neutron (mass = 1,7 × 10–27 kg) travelling at 6,26 × 105 m·s–1 collides with
a helium nucleus which is at rest. The mass of the helium nucleus is 4 times the
mass of a neutron. The helium nucleus rebounds at an angle of 45° and moves
at 1,80 × 105 m·s–1. After the collision the neutron’s speed is 5,12 × 105 m·s–1.
What angle (θ) will the neutron rebound at?

5,1 × 105 m.s–1

θ

6,2 × 105 m.s–1   45°

1,8 × 105 m.s–1

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01 MECHANICS.indb 12                                                                              2008/07/18 03:01:09 PM

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