Energy Revision
Kinetic
Ek = ½ mv2
Potential
Ep = mgh
Heat
EH = cmT
(changing temperature)
Heat
EH = ml
(changing state)
Electrical
EE = Pt = IVt
light
Work done
Ew = Fd
energy
sound
nuclear
E = Pt
Energy is conserved, this means the total energy remains the same
Problems using conservation of energy.
At bottom of
slope
Ek = ½mv2
At top of
slope
Ek = 0
h
Ep = 0
Ep = mgh
Assume no friction
Total energy is conserved
Ep at top = Ek at bottom
mgh = ½mv2
gh = ½ v2
v = 2gh
ie. mass is not important
With friction
Total energy is conserved
Ep at top = Ek at bottom + Ew
mgh = ½mv2 + Fd
Tutorial questions page 23/24 Qu 1 to 6
Purple book Ex 2.9
Small mass, fast
Or
Large mass, slow
Which would do the most damage?
Momentum = mass x velocity
momentum = mv
Momentum is a vector quantity measured in kgm/s or kgms-1
Tutorial questions page 24 Ou1
Collisions – trolleys stick together afterwards
At start trolley 2 is stationary.
m1
kg
u1 Momentum before m2
ms-1
m1u1 kgm/s
kg
v
(m1 + m2) v
m/s Total momentum after
kgm/s
0.5
0.5 x
=
0.5
(0.5 + 0.5) x
0.5
0.5 x
=
1
(0.5 + 1) x
0.5
0.5 x
=
1.5
(0.5 + 1.5) x
1
1
1x
1x
=
=
0.5
1
(1 + 0.5) x
(1 + 1) x
before
after
=
=
=
=
=
The Principle of Conservation of momentum
Total momentum remains the same provided there are no
outside forces.
Total momentum before = total momentum after
Elastic collisions – where the total kinetic energy is
conserved.
Inelastic collisions – where the kinetic energy is not
conserved eg. Some of the kinetic energy is changed into
heat and sound energy
A
B
1. Toy car A with a mass of 2kg and a velocity of 1m/s,
collides with stationary car B, mass 1kg. Velcro causes
them to stick together.
(a) What is their speed after the collision?
(b) Is it an elastic or inelastic collision?
Tutorial Questions Page 24/25 Qu 2 to 4
2. Two rubber balls collide head on as shown.
The red ball rebounds at 1m/s.
A
B
mass 4kg
mass 2kg
velocity 2m/s
velocity -3m/s
(a)What is the velocity of the blue ball?
(b) Is it an elastic or inelastic collision?
Tutorial Questions Page 25/26 Qu 5 to 9
Explosion – velocity before explosion is zero.
Momentum m1 v1
before
kg m/s
m1v1 m2
kgm/s kg
0
0.5
0.5
0
0.5
1
0
0.5
1.5
0
1
0.5
0
1
1
v2
m 2 v2
Total
m/s kgm/s momentum
after
3. A firework explodes and breaks into two pieces, 500g
and 250g.
(a) If the 500g part travels at 15 ms-1, what will the velocity
of the other part be immediately after the explosions?
(b) Why does it not stay at this velocity?
(c) Can kinetic energy ever be conserved in explosions?
Tutorial Questions Page 27 Qu 10 to 13
4. A field gun of mass 1000kg fires a shell of mass
5kg with a velocity of 100m/s.
Calculate the recoil velocity of the gun.
When you jump, what’s the least
painful way to land?
If you fall off something, what
sort of surface would you prefer
to land on? Explain!
In each case you have the same momentum
to loose to come to a stop.
Increasing the time it takes you to stop
decreases the force.
So the relationship F × t is an important
one.
Impulse = Ft
Impulse – on one object
1. Impulse = average force x time
= Ft
Impulse is a vector quantity measured in newton seconds (Ns)
2. Impulse = change in momentum of the one object
= mv - mu
Impulse is also measured in kgm/s or kgms-1
3. Impulse = area under the force time graph
F (N)
0
This means that
Impluse = Ft = mv – mu = area under force time graph
Things to beware of
F is average force, not the maximum.
Direction and sign of velocities
ie rebound
Impact time is often very short and can be given in
milliseconds (ms)
Mass is often given in grams
1. A car with mass 600 kg and velocity of 40 m/s skids and
crashes into a wall.
The car comes to rest 50 ms after hitting the wall. Calculate
the average force on the car during the collision.
2. During a game of hockey a stationary ball of mass 150 g
is struck by a player. The graph shows how the force on the
ball varies with time.
(a) Calculate the impulse on the ball.
F/N
(b) Calculate the speed which it
leaves the stick.
1200
3
6
t/ms
(c) A softer ball is hit and leaves the
stick with the same velocity. Sketch
its force time graph.
Tutorial Questions Page 28/29 Qu 14 to 20
Aim: To find the average force exerted by a cue on a snooker ball.
Measurements
d, diameter of ball =
m
t1, contact time =
s
t2, time to go through light gate =
m, mass of ball =
kg
s
Calculations
Velocity of ball after collision, v = d/t2 =
Average force, F = (mv – mu)/t1 =
Tutorial Questions Page 28/29 Qu 14 to 20
Open Ended Question
Which ball would you prefer to be
dropped on your foot and why?
Use the words force, time, change in
momentum and impulse in your
explanation!
Wearing no seatbelt
The person continues to move forward at a
constant speed. Newton’s first law
Until they collide with the dashboard etc,
stopping them suddenly.
F = (mv – mu)/t so short time means large
average force
Wearing a seat belt
The person is brought to a stop at the same time as the
car is stopping. The stopping time is increased as the car
crumples and the seatbelt has some give. The force is also
on the parts of the body where it will do the least harm
F = (mv-mu)/t the same change in momentum but a
longer time means a smaller average force.
Points to note
1. The rate the fuel is ejected at is 30 kg/s.
This means in a time of 1 second there is a mass of 30 kg
2. change in momentum of the rocket =
- change in momentum of fuel
Newton’s Third Law
Tutorial Questions Page 30/31 Qu 21 to 26
Open Ended Question
324 m
A careless school pupil
drops a 1p coin from his
pocket at the top of the
Eiffel Tower.
Find the average force
exerted by the coin on the
ground.
You will need to estimate
some values.
Calculate the velocity of
the coin on impact on the
ground.
(Estimate the mass of the
coin and use equations of
motion)
Open Ended Question
Calculate the velocity of the
coin on impact on the
ground. (Estimate the mass
of the coin and use
equations of motion)
324 m
Calculate change in
momentum. What value
would you give to the
rebound velocity?
Calculate average force.
What value would you
give to the contact time?