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North Berwick High School
Department of Physics
Higher Physics
Unit 1
Section 3
Our Dynamic Universe
Collisions and Explosions
Section 3 Collisions and Explosions
Note Making
Make a dictionary with the meanings of any new words.
Momentum
1.
2.
3.
State the formula and units for momentum.
State that momentum is a vector.
Explain what is meant by the principle of conservation of
momentum.
Momentum in collisions
1.
Copy the five notes and both worked examples.
Kinetic energy in collisions
1.
2.
Describe both types of collision and whether kinetic energy is
conserved or not.
Copy question 7 on page 16 into your notes and find the solutions.
Explosions
1.
2.
Explain why a gun has a recoil velocity.
Copy the worked example.
Momentum and Newton’s third law
1.
Show that conservation of momentum and Newton’s third law
mean the same thing.
Impulse
1.
2.
3.
4.
Show how impulse can be obtained from Newton’s second law.
State the units of impulse.
State that impulse = area under a force–time graph.
Copy the three worked examples.
Section 3 Collisions and Explosions
Contents
Content Statements ............................................................................ 1
Elastic and inelastic collisions .............................................................. 2
Momentum ......................................................................................... 2
Momentum in collisions ...................................................................... 3
Worked examples ................................................................................ 4
Kinetic energy in collisions .................................................................. 6
A special case ...................................................................................... 7
Explosions ........................................................................................... 7
Worked example ................................................................................. 8
Momentum and Newton’s third law .................................................... 9
Impulse ............................................................................................. 10
Momentum and Newton’s second law ............................................... 11
Worked examples .............................................................................. 13
Problems ........................................................................................... 15
Solutions ........................................................................................... 21
Content Statements
content
notes
context
a)
Elastic and
inelastic
collisions.
Conservation of
momentum in one
dimension and in which
the objects may move in
opposite directions.
Kinetic energy in elastic
and inelastic collisions.
Investigations of
conservation of
momentum and energy.
b)
Explosions and
Newton’s Third
Law.
Conservation of
momentum in
explosions in one
dimension only.
Propulsion systems – jet
engines and rockets
c)
Impulse.
Force-time graphs
during contact of
colliding objects.
Impulse can be found
from the area under a
force-time graph.
Investigating collisions
using force sensors and
dataloggers.
Hammers and pile
drivers.
Car safety, crumple
zones and air bags.
1
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 kgms –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, i.e. 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.
2
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 of a system, not individual
momentum.
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) will be considered to be
positive (+) while other momenta (the opposite direction) will be
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 collision
3
Worked examples
1.
Two cars are travelling towards each other as shown below. They
collide, lock together and move forwards (i.e. 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
1000 kg
A
B
1200 kg
1000 kg
Total momentum before collision = total momentum after collision
Direction
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
=
total momentum before
=
2200v
total momentum after
12 000 – 8000
=
2200v
v
=
4000
2200
v
=
ie
1.8 m s –1 to the right
4
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 ms –1 . Calculate the speed of
vehicle B after the collision.
Before
A
12 m s –1
1200 kg
B
A
9 m s –1
11 m s –1
800 kg
After
1200 kg
B
?
800 kg
Total momentum before collision = total momentum after collision
Take motion
as +
m 1 u 1 = 1200 × 12 = 14 400
m 1 v 1 = 1200 × 11
m 2 u 2 = 800 × 9 = 7 200
m 2 v 2 = 800 × v 2
total momentum before
=
total momentum after
14 400 + 7200
=
13 200 + 800v 2
21 600
=
13 200 + 800v 2
v 2 = 8 400
800
v 2 = 10.5 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.
5
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 (E k ) is conserved
i.e.
total E k before = total E k after
This is called an elastic collision.
(b)
those in which kinetic energy is not conserved
i.e.
E k 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 E k 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:
Ek =
1
2
mv 2
6
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 ms –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 ms –1 and the cue ball will stop.
This effect can be seen in ‘Newton’s cradle’.
(You may want to look at some YouTube clips of giant Newton’s cradles.)
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 direction. 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).
7
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 did 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 ms –1 . Calculate
the recoil velocity of the gun.
Before
After
0 ms –1
1 kg
0.005 kg
?
100 ms –1
1 kg
0.005 kg
Total momentum before explosion = total momentum after explosion
Take motion
as +
momentum = 0
m1v1 + m2v2
(1 × v 1 ) + (0.005 × 100)
(1 × v 1 ) + 0.5
v1
ie v 1
=
=
0
=
0
=
0
=
-0.5
0.5 ms –1 in the opposite direction to the bullet
8
Momentum and Newton’s third law
It can be shown that conservation of momentum and Newton’s third law mean
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 1)
–(change in momentum
of object 2)
In other words, in any particular example involving two objects colliding, if the
momentum of one object increases by, for example, 6 kgms –1 , then the
momentum of the other object must decrease by 6 kgms –1 .
Consider again the previous example. We had concluded that:
–m 2 (v 2 – u 2 )
=
m 1 (v 1 – u 1 )
=
v-u
t
–m 2 a 2 t
=
m1a1t
–F 2 t
=
F1t
–F 2
=
F1
Applying
a
as
at = v-u
And as F = ma
so
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.
9
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 ev eryday
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 the student’s deceleration (negative acceleration) 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.
(a)
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.
10
Momentum and Newton’s second law
Newton’s second law:
but
F = ma
a = v-u
t
F = m v-u
t
so
F = m v-u
t
=
mv - mu
t
which is the rate of change of momentum. This is 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 kgms –1 since these are the units of momentum, but they
are also Ns, and these are the units that should be used for impulse.
11
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?
12
Worked examples
1. In a snooker game, the cue ball, of mass 0.2 kg, is accelerated from the r est
to a velocity of 2 ms –1 by a force from the cue which lasts 50 ms. what size of
force is exerted by the cue?
v = 2 m s –1 ,
u = 0,
Ft
=
F × 0.05 =
F × 0.05 =
F
=
t = 50 ms = 0.05 s,
m = 0.2 kg,
F=?
mv - mu
(0.2 × 2) – (0.2 × 0)
0.4
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 contact varies over this
period, as shown in the graph. Determine the speed of the ball as it l eaves
the racquet.
impulse
= area under graph
= ½ × 20 × 10 –3 × 400 = 4 Ns
u = 0,
Ft
4
4
v
m = 100 g = 0.1 kg,
=
=
=
=
v=?
mv - mu
0.1v – (0.1 × 0)
0.1v
40 ms –1
13
3. A tennis ball of mass 0.1 kg travelling horizontally at 10 ms –1 is struck in the
opposite direction by a tennis racket. The tennis ball rebounds horizontally
at 15 ms –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
0.05F
0.05F
0.05F
F
=
=
=
=
=
mv - mu
(0.1 × (–15)) – (0.1 × 10)
–1.5 – 1
–2.5
–50 N (the negative sign indicates force in opposite
direction to the initial velocity)
14
Problems
Collisions and explosions
1.
What is the momentum of the object in each of the following situations?
(a)
2.
(b)
(c)
A trolley of mass 2·0 kg is travelling with a speed of 1·5 m s 1 . The trolley
collides and sticks to a stationary trolley of mass 2·0 kg.
(a)
Calculate the velocity of the trolleys immediately after the
collision.
Show that the collision is inelastic.
(b)
3.
A target of mass 4·0 kg hangs from a tree by a long string. An arrow of
mass 100 g is fired at the target and embeds itself in the target. The speed
of the arrow is 100 m s 1 just before it strikes the target. What is the speed
of the target immediately after the impact?
4.
A trolley of mass 2·0 kg is moving at a constant speed when it collides and
sticks to a second stationary trolley. The graph shows how the speed of the
2·0 kg trolley varies with time.
0.5
v / ms
-1
0.2
0
time / s
Determine the mass of the second trolley.
15
5.
In a game of bowls a bowl of mass 1·0 kg is travelling at a speed of
2·0 ms 1 when it hits a stationary jack ‘straight on’. The jack has a mass of
300 g. The bowl continues to move straight on with a speed of 1·2 m s 1
after the collision.
(a)
(b)
What is the speed of the jack immediately after the collision?
How much kinetic energy is lost during the collision?
6.
Two space vehicles make a docking manoeuvre (joining together) in space.
One vehicle has a mass of 2000 kg and is travelling at 9·0 m s 1 . The second
vehicle has a mass of 1500 kg and is moving at 8·0 m s 1 in the same
direction as the first. Determine their common velocity after docking.
7.
Two cars are travelling along a race track. The car in front has a mass of
1400 kg and is moving at 20 m s 1 . The car behind has a mass of 1000 kg
and is moving at 30 m s 1 . The cars collide and as a result of the collision
the car in front has a speed of 25 m s 1 .
(a)
(b)
8.
Determine the speed of the rear car after the collision.
Show clearly whether this collision is elastic or inelastic.
One vehicle approaches another from behind as shown.
The vehicle at the rear is moving faster than the one in front and they
collide. This causes the vehicle in front to be ‘nudged’ forward with an
increased speed. Determine the speed of the rear vehicle immediately
after the collision.
9.
A trolley of mass 0·8 kg is travelling at a speed 1·5 m s 1 . It collides head-on
with another vehicle of mass 1·2 kg travelling at 2·0 m s 1 in the opposite
direction. The vehicles lock together on impact. Determine the speed and
direction of the vehicles after the collision.
16
10.
A firework is launched vertically and when it reaches its maximum height it
explodes into two pieces. One piece has a mass of 200 g and moves off
with a speed of 10 m s 1 . The other piece has a mass of 120 g. What is the
velocity of the second piece of the firework?
11.
Two trolleys initially at rest and in contact move apart when a plunger on
one trolley is released. One trolley with a mass of 2 kg moves off with a
speed of 4 m s 1 . The other moves off with a speed of 2 ms 1 , in the
opposite direction. Calculate the mass of this trolley.
12.
A man of mass 80 kg and woman of mass 50 kg are skating on ice. At one
point they stand next to each other and the woman pushes the man. As a
result of the push the man moves off at a speed of 0·5 m s 1 . What is the
velocity of the woman as a result of the push?
13.
Two trolleys initially at rest and in contact fly apart whe n a plunger on one
of them is released. One trolley has a mass of 2·0 kg and moves off at a
speed of 2·0 m s 1 . The second trolley has a mass of 3·0 kg. Calculate the
velocity of this trolley.
14.
A cue exerts an average force of 7·00 N on a stationary snooker ball of
mass 200 g. The impact of the cue on the ball lasts for 45·0 ms. What is the
speed of the ball as it leaves the cue?
15
A football of mass 500 g is stationary. When a girl kicks the ball her foot is
in contact with the ball for a time of 50 ms. As a result of the kick the ball
moves off at a speed of 10 m s 1 . Calculate the average force exerted by
her foot on the ball.
16.
A stationary golf ball of mass 100 g is struck by a club. The ball moves off
at a speed of 30 m s 1 . The average force of the club on the ball is 100 N.
Calculate the time of contact between the club and the ball.
17.
The graph shows how the force exerted by a hockey stick on a stationary
hockey ball varies with time.
F/N
40
0
20
time/ms
ms
17
The mass of the ball is 150 g.
Determine the speed of the ball as it leaves the stick.
18.
A ball of mass 100 g falls from a height of 0·20 m onto concrete. The ball
rebounds to a height of 0·18 m. The duration of the impact is
25 ms. Calculate:
(a)
(b)
(c)
(d)
(e)
19.
A rubber ball of mass 40·0 g is dropped from a height of 0·800 m onto the
pavement. The ball rebounds to a height of 0·450 m. The average force of
contact between the pavement and the ball is 2·80 N.
(a)
(b)
20.
the change in momentum of the ball caused by the ‘bounce’
the impulse on the ball during the bounce
the average unbalanced force exerted on the ball by the concrete
the average unbalanced force of the concrete on the ball.
What is the total average upwards force on the ball during impact?
Calculate the velocity of the ball just before it hits the ground and
the velocity just after hitting the ground.
Calculate the time of contact between the ball and pavement.
A ball of mass 400 g travels falls from rest and hits the ground. The
velocity-time graph represents the motion of the ball for the first 1·2 s
after it starts to fall.
6
Speed /
ms -1
0 A
-4
(a)
(b)
(c)
(d)
B
C
0·6
0·8
E
1·2
time / s
D
Describe the motion of the ball during sections AB, BC, CD and DE
on the graph.
What is the time of contact of the ball with the ground?
Calculate the average unbalanced force of the ground on the ball.
How much energy is lost due to contact with the ground?
18
21.
Water with a speed of 50 m s 1 is ejected horizontally from a fire hose at a
rate of 25 kg s 1 . The water hits a wall horizontally and does not rebound
from the wall. Calculate the average force exerted on the wall by the
water.
22.
A rocket ejects gas at a rate of 50 kg s 1 , ejecting it with a constant speed
of 1800 m s 1 . Calculate magnitude of the force exerted by the ejected gas
on the rocket.
23.
Describe in detail an experiment that you would do to determine the
average force between a football boot and a football as the ball is being
kicked. Draw a diagram of the apparatus and include all the measurements
taken and details of the calculations carried out.
24.
A 2·0 kg trolley travelling at 6·0 m s 1 collides with a stationary 1·0 kg
trolley. The trolleys remain connected after the collision.
(a)
Calculate:
(i)
the velocity of the trolleys just after the collision
(ii)
the momentum gained by the 1·0 kg trolley
(iii)
the momentum lost by the 2·0 kg trolley.
(b)
The collision lasts for 0·50 s. Calculate the magnitude of the
average force acting on each trolley.
25.
In a problem two objects, having known masses and velocities, collide and
stick together. Why does the problem ask for the velocity immediately
after collision to be calculated?
26.
A Newton’s cradle apparatus is used to demonstrate conservation of
momentum.
Four steel spheres, each of mass 0.1 kg, are suspended so that they are in
a straight line.
Sphere 1 is pulled to the side and released, as shown in diagram I.
1
1
4
2
Diagram I
3
4
1
1
2
3
Diagram II
19
4
When sphere 1 strikes sphere 2 (as shown by the dotted lines) , sphere 4
moves off the line and reaches the position shown by the dotted lines.
The student estimates that sphere 1 has a speed of 2 m s 1 when it strikes
sphere 2. She also estimates that sphere 4 leaves the line with an initial
speed of 2 m s 1 . Hence conservation of momentum has been
demonstrated.
A second student suggests that when the demonstration is repeated there
is a possibility that spheres 3 and 4, each with a speed of 0·5 m s 1 , could
move off the line as shown in diagram II.
Use your knowledge of physics to show this is not possible.
20
Solutions
Section 3: Collisions and explosions
1.
(a)
(b)
(c)
20 kg m s 1 to the right
500 kg m s 1 downwards
9 kg m s 1 to the left
2.
(a)
0·75 m s
3.
2·4 m s
4.
3·0 kg
5.
(a)
(b)
6.
8·6 m s
7.
(a)
8.
8·7 m s
1
9.
0·6 m s
1
10.
16·7 m s
11.
4 kg
12.
0·8 m s
1
in the opposite direction to the velocity of the man
13.
1·3 m s
1
in the opposite direction to the velocity of the first trolley
14.
1·58 m s
15.
100 N
16.
3·0 × 10
2
17.
2·67 m s
1
1
in the direction in which the first trolley was moving
1
2·7 m s
0·19 J
1
1
in the original direction of travel
23 m s
1
in the original direction of travel of the 1·2 kg trolley
1
in the opposite direction to the first piece
1
s
21
(b)
(c)
(d)
(e)
+ 0·39 kg m s 1 if you have chosen upwards directions to be
positive;
–0·39 kg m s 1 if you have chosen downwards directions to be
positive
+ 0·39 N s if you have chosen upwards directions to be positive
15·6 N downwards
15·6 N upwards
16·6 N upwards
19.
(a)
(b)
v before = 3·96 m s
9·9 × 10 2 s
20.
(b)
(c)
(d)
0·2 s
20 N upwards (or –20 N for the sign convention used in the graph)
4·0 J
21.
1·25 × 10 3 N towards the wall
22.
9·0 × 10 4 N
24.
(a)
18
(a)
(b)
(i)
(ii)
(iii)
8N
1
downwards; v after = 2·97 m s
1
upwards
4·0 m s 1 in the direction the 2·0 kg trolley was travelling
4·0 kg m s 1 in the direction the 2·0 kg trolley was travelling
4·0 kg m s 1 in the direction the 2·0 kg trolley was travelling
22