Download Topic 4 – Momentum, energy, work and power

Topic 4 – Momentum, energy, work and power
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STOPPING DISTANCES
Stopping safely:
When a driver sees a problem ahead, the car travels some distance before the
driver reacts and steps on the brakes – this is called the ‘thinking distance’
Once the driver has stepped on the brakes, the car will travel some distance before
it comes to a complete stop – this is called the ‘braking distance’
The overall ‘stopping distance’ for a car = thinking distance + braking distance
Factors affecting stopping distances:
Note: make sure you realise that some factors affect just the thinking distance,
some just the braking distance, and some affect both
1. Person’s reaction time (i.e time it takes to respond to danger):
o The slower a person’s reaction time, the longer the thinking distance
o Tiredness, illness, or taking drugs or alcohol can all slow down reaction
times
2. Speed of the vehicle:
o The faster the vehicle is travelling, the longer both the thinking distance
and the braking distance
3. State of the car’s brakes and road conditions:
o For a car to come to a stop there must be friction created between the car’s
tyres and the road…
 the more friction between the tyres and the road, the smaller the
braking distance
o The amount of friction between the tyres and the road depends on:
 the condition of the car’s brakes (the better the brakes, the more
friction they createshorter braking distance)
 condition of the road surface (e.g if the road is wet or has loose
gravel, less friction is createdlonger braking distance)
o Note: due to longer braking distances in wet conditions, drivers should
leave more of a gap between their car and the car in front when it is raining
4. Mass of the vehicle:
o If a vehicle has more mass, more force is needed to make it slow down
o the heavier the vehicle, the longer the braking distance
MOMENTUM
Momentum is a measure of how strongly something is moving…
o momentum (kg m/s) = mass (kg) x velocity (m/s)
o I.e the heavier the vehicle and the faster it is travelling, the greater its
momentum
E.g what is the momentum of a monster truck (mass = 4500kg) when travelling at
12m/s, in an eastward direction?
o Momentum = 45000 x 12 = 54,000 kg m/s east
Momentum has a size and direction so is a vector quantity:
o when giving an answer, you should state not only the amount of
momentum an object has (the ‘size’) but also the direction of that
momentum (which will be in the same direction as the velocity)
Conservation of momentum:
When a moving object collides with another object:
o their masses are added togethertheir combined speed will be slower
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o However, the total momentum of both objects is the same before the
collision as it is after the collision - this is known as ‘conservation of linear
momentum’
E.g:
Before collision:
o Object on the left has a momentum = 15 x 4 = 60 kg m/s to the right
o Object on the right has a momentum = 0 x 2 = 0 kg m/s
After collision:
o Combined objects have a momentum = 10 x 6 = 60 kg m/s to the right
the combined momentum of the objects before and after the collision is the
same
This demonstrates that when objects collide, momentum is conserved (i.e it is
transferred to the other object without any being lost)
MOMENTUM AND SAFETY
When travelling in a car, passengers are going at the same speed as the car
If a car brakes suddenly, there’s a rapid change in momentum, and a strong force
is applied to the passengers - this can result in serious injury (e.g passengers can
smash their heads against the windscreen)
The ‘rate of change of momentum’ (i.e the speed at which the momentum of an
object changes) is equal to the force applied to the object…:
o The rate of change of momentum = change in momentum / time taken for
that change
o force (N) = change in momentum (kg m/s) / time taken for that change
(s)
o E.g what is the force needed to change a cyclist’s momentum by 24 kg m/s
in 4 seconds?
 Force = 24 / 4 = 6 N
So in a car crash…:
o the quicker the rate of change of momentum (i.e the quicker the
passenger’s momentum decreases to 0), the more force applied to the
passengersthe more likely the passengers are to get hurt
o safety measures in cars try to reduce the rate of change of momentum of
passengers (i.e they try to increase the time it takes for the passenger’s
momentum to decrease to 0) so that force applied to the passengers is
smaller
Note: the ‘rate of change of momentum’ is basically the same as saying the ‘rate
of change of velocity’ because the time it takes for velocity to decrease to 0 is the
same time it takes for momentum to decrease to 0 (momentum = velocity x mass)
o Indeed, equation above can be re-written as…:
 force = [(final velocity x mass) – (initial velocity x mass)] / time
 F = (mv – mu) / t
Vehicle safety measures:
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1. Seat belts:
o In an accident, seat belts stretchpassenger’s velocity is slowed down
more gradually
o it takes longer for passenger’s momentum to be slowed to zero (i.e the
rate of change of momentum is reduced)
o force applied to the passengers is less
o reduced chance of injury
However, in high speed accidents, the force is so great that the seat belt by itself
could still cause injury…
2. cars are also fitted with airbags - these work in the same way as seat belts,
reducing the rate of change of momentum
3. Cars also have crumple zones:
o In an accident, the material in these crumple zones squashes and folds in a
specific way
o This crumpling and folding reduces the momentum of the car over a longer
period of time (i.e reduces the rate of change of momentum)
o the impact forces on the passengers are less
o reduced chance of serious injury
WORK AND POWER
‘Work’ is done when energy is transferred from one form to another
o E.g car brakes ‘do work’ by transferring kinetic energy to thermal energy
The amount of ‘work’ done is equal to the amount of energy transferred
Energy is measured in joules (J)work is also measured in joules
Calculating work done:
work done (joules, J) = force (newtons, N) x distance moved in the direction of
the force (metres, m)
o E=Fxd
E.g a sailor lifts a 300 N sail from the deck to a height 4 m up. How much work
does the sailor do?
o Work done = 300 x 4 = 1,200 J
Power:
Power is the rate of doing work – measured in watts (W)
Power (watt, W) = work done (joule, J) / time taken (second, s)
o P=E/t
o 1 watt = 1 joule of work done per second
o I.e the more work done per second, the more power generated
E.g a motorbike engine increases the bike’s kinetic energy over 40m, using a force
of 6000 N. It takes 5 seconds to do this. What power did the engine provide?
o Work done = 6000 x 40 = 240,000 J
o power = 240,000 / 5 = 48,000 W
POTENTIAL AND KINETIC ENERGY
Gravitational potential energy:
Gravitational potential energy is energy that is stored because of an object’s
position in a gravitational field
o I.e: on Earth, if something can fall (e.g a person on a diving board) it has
gravitational potential energy
Gravitational potential energy can be calculated using the equation…:
o Gravitational potential energy (J) = mass (kg) x gravitational field strength
(N/kg) x vertical height from ground (m)
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o GPE = m x g x h
E.g on earth the strength of the gravitational field is 10N/kg. What is the
gravitational potential energy gained by a 500kg Mars Rover when it’s lifted 15m
by a test crane on Earth?
o GPE = 500 x 10 x 15 = 75,000 J
Kinetic energy:
Kinetic energy is another name for movement energy
The kinetic energy can be calculated using the equation…:
o kinetic energy (J) = ½ x mass (kg) x (velocity)2 (m/s2)
o KE = ½ x m x v2
E.g what is the kinetic energy of a 65kg girl running at 6m/s?
o KE = ½ x 65 x (62) = 1,170 J
Conservation of energy:
When energy is transferred from one form to another, energy is conserved (i.e
total amount always remains the same…none is lost)
E.g:
o As an object falls, its gravitational potential energy is converted into
kinetic energy
o When an object with gravitational potential energy falls down, the amount
of kinetic energy it has just before it hits the ground is equal to its initial
gravitational potential energy (as all the GPE has been converted to KE)
E.g when a 500kg item was dropped, 75000 J of gravitational potential energy
were all transferred to kinetic energy by the time it hit the ground. How fast did
the item land?
o KE = 75000 J (because as item falls all 75000 J of GPE is transferred to
KE)
o (velocity)2 = 75000 / (½ x 500) = 300
o velocity = √300 = 17.3 m/s
Braking distances, conservation of energy, and work done:
Brakes ‘do work’ by transferring kinetic energy to thermal energy
When the vehicle stops, it has no more kinetic energy (energy is conservedall
the kinetic energy has been converted to thermal energy)
the work done to bring a vehicle to rest is equal to its initial kinetic energy…
o i.e the greater the initial velocity of the vehicle, the greater its initial
kinetic energy (recall: KE = ½ x m x v2)the more work that needs to be
done to bring the vehicle to rest
o This explains why the faster a vehicle is travelling, the longer its braking
distance