Download Mechanics Lecture 6 - Newton`s Laws File

Objectives
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Mechanics 6
Newton’s Laws
IFP Monday 10th November
2015
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Recall and apply
Newton’s three laws
of motion
Explain Force,
Momentum, and
conservation of
momentum
Carry out calculations
based on these
principles
Key Words
Newton’s laws, force,
momentum,
conservation
Sir Isaac
Newton
(1643 – 1727)
• Came up with three laws.
• Write down any you already know on whiteboards
Objectives
Newton’s FIRST law
• In the absence of any resultant
force acting on an object, it
remains either at rest or continues
to move at a constant velocity
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•
•
Recall and apply
Newton’s three laws
of motion
Explain Force,
Momentum, and
conservation of
momentum
Carry out calculations
based on these
principles
Key Words
• So - No force ↔ no change of
motion
• Also called law of inertia
• resultant means net force
Newton’s laws, force,
momentum,
conservation
Objectives
Demo
• Hovercraft
• Air track
•
•
•
Recall and apply
Newton’s three laws
of motion
Explain Force,
Momentum, and
conservation of
momentum
Carry out calculations
based on these
principles
Key Words
Newton’s laws, force,
momentum,
conservation
Objectives
Newton’s SECOND law
Force = mass x acceleration
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Unit of force is one kg ms-2 - known
as one newton
F is the resultant force (if several
forces act on an object)
• Also note that Weight: W = m g
• A free-falling object is unsupported
but not weightless (g is always
there)
Recall and apply
Newton’s three laws
of motion
Explain Force,
Momentum, and
conservation of
momentum
Carry out calculations
based on these
principles
Key Words
Newton’s laws, force,
momentum,
conservation
Think about a lift accelerating
• The direction of net (resultant) force on an object is the
same as the direction of acceleration
• Physics of lifts (lift of mass m, supported by a cable):
- Lift at rest: tension T = weight (m g)
- Lift accelerating as it moves upwards: T > mg (a >0)
- Lift accelerating as it moves downwards: T<mg (a <0 )
- Lift decelerating as it moves upwards: T< mg (a < 0)
- Lift decelerating as it moves downwards: T > mg (a>0)
Newton’s THIRD law
• For every action (force) there is an equal and opposite
reaction.
• Then why don’t they add up to zero?
• Answer: because they are exerted on different objects!
• Example: box on a table
• 2 forces on the box: (1) weight and (2) normal force from
table
• Reaction to (1): force that the box exerts on the earth!
• Reaction to (2): force that the box exerts on the table!
• Therefore you cannot add them.
Try the questions on the
sheet in assigned groups
Objectives
Momentum
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Momentum = mass x velocity
p=mv
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Recall and apply
Newton’s three laws
of motion
Explain Force,
Momentum, and
conservation of
momentum
Carry out calculations
based on these
principles
Vector quantity, units: kg ms-1
F= m a (for an object with fixed mass m)
So F = m (vt – v0)/t
= (m vt – m v0)/t
Which is Rate of change of momentum.
This is a more general statement of Newton’s second
law.
Key Words
Newton’s laws, force,
momentum,
conservation
Conservation of momentum
Principle of conservation of momentum:
When two or more bodies interact, then the
total momentum is conserved if no external
forces act on the bodies.
This can be derived from Newton’s third law (action/
reaction exerted on different bodies)
It is a very important principle!
Example – two balls colliding
By law of conservation of momentum, momentum before collision is
equal to momentum after collision.
So try to solve for v2
Example – two balls colliding
m1 u1 + m2 u2 = m1 v1 + m2 v2
v2 = (m1 u1 + m2 u2 – m1 v1)/m2 (Rearrange, to solve for v2)
v2 = (10 x 20 + 8 x 5 – 10 x 10)/ 8 ms-1
= 17.5 ms-1
Conservation example 2
A car crashes into a wall at 25 ms-1 and is
brought to rest in 0.1s. Calculate
(i) the average force exerted on a 75 kg test
dummy by the seatbelt.
(ii) the dummy’s average acceleration
Assume that the seatbelt does all the stopping of the dummy (e.g. ignore
friction from the seats or normal forces from airbags)
Solution
Dummy’s initial and final momentum:
p0 = m v0 = 75 kg x 25 ms-1 = 1875 kg ms-1
pf = m vf = 75 kg x 0 = 0
Therefore change of momentum :
Δp=pf-p0=-1875 kg ms-1
Impulse = Δp = Force x time
F = Δp/Δt = - 18,750 N (opposite direction to
the initial velocity)
Solution
dummy’s average acceleration:
a = f/m = -18,750/75 ms-2 = -250 ms-2
≈ -25.5 g
Similarly: a = Δv/Δt = (0 – 25) ms-1 /0.1 s
= -250 m/s2
Momentum questions
Solutions