IMPULSE: F = ∆(mv)/∆t ⇒ F ∆t = ∆ (mv)

IMPULSE:
Newton’s 2nd law:
F = ∆(mv)/∆t ⇒ F ∆t = ∆ (mv)
This product is called “Impulse” of
the force
It tells us how much a force acting over a time ∆t
changes the momentum of an object.
Example:
“Egg toss”: You are trying to catch an egg. How
would you do it to keep it from breaking ? Why ?
You would try to move with the egg to increase
the time over which it is accelerated. Since the
change in momentum is fixed, this would reduce
the force on the egg.
F = ∆(mv)/∆t If ∆t is longer, F will be smaller.
Examples
ƒ A rocket + its fuel one system a jet is expelled
from a nozzle carries a momentum – the same
momentum is added to the rocket.
ƒ A cart with a child on it – one system again – if a
child jumps off the cart, the cart moves in the
opposite direction to conserve the total
momentum.
ƒ Seatbelts: if a car collides with an obstacle, say
a tree, the car looses its momentum in an
inelastic collision which causes damage to the car
and the tree. What happens to a driver? A driver
has certain momentum before a collision; if the
belt is not fastened, the driver continues motion
until he hits the window in front of him (which
has already (almost) stopped with respect to the
ground. If the belt is fastened, the momentum is
damped with a car – maybe the car is damaged a
little more, but not a driver!
The net constant force F starts acting on the point of
mass m, initially at rest. It passes a displacement d.
Analyze this situation.
F
a =
constant
m
v
1 2
d = at , v = at at the end
2
F = ma
1
1
1
2 2
2
F ⋅ d = ma t = m ( at ) = mv 2
2
2
2
t
As a result of action of the force F in the direction of d
through displacement d, an object acquires quantity called
kinetic energy: The work done by the external force F
changes the kinetic energy of the system: F d = ∆KE
Work done by force is defined as:
W=Fd
Where d is the displacement in the direction of the
force over which the force F acts.
Example: Lift a 1-kg box by 1 m versus 2 m with a
constant speed:
F=mg =1 kg×9.8m/s2=9.8 N - lifting force is just equal to
the force of gravity, acceleration is zero
W = 9.8 N × 1 m = 9.8 N m = 9.8 J
W = 9.8 N × 2 m = 19.6 N m = 19.6 J
N × m = Joule
unit of work
What forces do, and what forces do not do work?
ƒ Reaction of the surface never does work on the
object if it moves along the surface because it is
always perpendicular to the surface – there is
always a zero component along the surface.
ƒ Friction force usually does negative work because
its directed opposite to the displacement.
ƒ Gravity does positive work when the object is
moving down and negative when it is moving up
Mg
N
WMg=0
WN=0
In W = F d, the displacement, d, has to be in the
direction of the force.
Example:
How much work is done to accelerate a 1000 kg car
from 0 to 20 m/s in 5 seconds ?
W=Fd
We don’t know F and we don’t know d
Find F: Acceleration a = ∆v/∆t = (20 m/s)/(5s) = 4 m/s2
F = m a = 1000 kg x 4 m/s2 = 4000 N
Find d: d = ½ a ∆t2= ½ (4 m/s2) (5s)2 = 50 m
W = F d = (4000 N) x (50 m) = 200,000 J=200 kJ
In the last example,
Let us compare the initial and final states of the object (a
car).
Initial state: v=0, p=mv=0, Kinetic Energy =½mv2=0
Final state: v=20m/s, p=mv=1000kg×20m/s=20,000 kg m/s =
4000N × 5s = F∆t (2nd N. law)
KE = ½mv2= =½ 1000kg × 400(m/s)2=200,000J
= 4,000 N × 50 m = F d = WF
As a result of work done by the force, the kinetic energy
of the particle changed.
W = F d = ∆(½ m v2)
Work-Energy theorem: the change of kinetic energy is
equal to the work done by the net force
Kinetic energy
Work-energy theorem tells us that the work done on the
object is equal to the change of its kinetic energy.
W = F d = ∆(½ m v2)
By Newton’s third law the force acting on the object is
equal and opposite to the force exerted by the object on
the external agent! Therefore, the work done by that force
– by the object is equal to the work done on it with a
negative sign, since the displacement of the object is still
the same
Then if the object has some kinetic energy and it reduces,
the work done on the object is negative, but the work done
by the object is positive.
Kinetic energy is a measure of a system’s capacity to do
work.
ƒ When we accelerate a car (doing work on it) , we are actually
transferring energy to the car. The moving car now has this
energy, which could be used to do further work (e.g. it can
crash to a wall, and damage it!)
ƒ The energy of a moving object is called the
Kinetic Energy (KE) defined as:
KE = ½ m v2
ƒ It is dependent on the frame of reference as the velocity is
(if you are sitting in a car your kinetic energy with respect to
the car is zero, but it is not with respect to the ground)
ƒ Kinetic energy is measured in Joules (J) as well as work.
ƒ Kinetic energy is a scalar
ƒ Kinetic energy is always positive
∆KE = ∆(½ m v2) = F ∆d
Doesn’t it look like ∆(mv) = F ∆t ?
Consider an example: I take an object of mass m = 1 kg
and raise it to the height of 2 m above the ground. As I
do it, the velocity of the object is approximately
constant. What work did I do?
The net force was zero… There were two forces, however:
F upwards, and mg downwards.
1. The work done by the net force is zero, ∆KE=0
2. The work done by mg is WG= (-mg) d = - mgd = -19.6 J
3. The work done by my hand is (F=-mg) WF = F d = mgd
=19.6 J
I worked against gravity and now the object has something
it did not have before…
What is it?
I now let it go… I do not do any work now.
y
It falls down. The only force is mg, the distance is –d,
The work done by gravity is WG = (-mg)(-d) = mgd = 19.6 J
Where did it go?
The object was accelerating (before reaching the floor),
Let’s calculate its final velocity:
v = gt
But we do not know t...
t=
2d
g
1 2
gt Since it passed d over t...
2
2d
⇒ v=g
= 2 gd
g
d=
Now let’s find KE = ½mv2 = mgd = WG !!!
The work done by the force gravity went to KE
One more time:
We did the positive work against gravity,
then the gravity did the same positive work, that
eventually went to the kinetic energy of the object.
In the beginning, the gravity was not “capable” of doing
that positive work!
The result of our positive work against the gravity was
changing a position of the object, so it became capable…
We call this capability a POTENTIAL ENERGY.
Still more…
We do the work against the gravity, the object’s
potential energy increases by the amount of this work.
Then the potential energy is converted into kinetic as a
result of work done by gravity.
1. Wext = ∆PE, 2. ∆PE + ∆KE = 0
In general,
Wnet, external = ∆KE + ∆PE
Conservation of mechanical energy:
In a closed system without losses (no friction, resistance,
etc.)
KE + PE = constant,
(KE + PE)
before
= (KE + PE) after
Simplified previous example:
You drop an object which then falls freely from height
d. What is its speed just before it hits the ground?
∆PE = PEafter − PEbefore = 0 − mgd = − mgd
∆KE = KEafter − KEbefore
1 2
1 2
= mv − 0 = mv
2
2
1 2
∆KE = −∆PE ⇒
mv = −(−mgd ) = mgd
2
v 2 = 2 gd
⇒ v = 2 gd
Notice, it’s the same result we got using Newton’s 2nd
law and kinematics!
Another example:
A sled slides from a hill of a height d. What is its
speed at the bottom of the hill?
∆PE = PEafter − PEbefore = 0 − mgd = − mgd
∆KE = KEafter − KEbefore
1 2
1 2
= mv − 0 = mv
2
2
1 2
∆KE = −∆PE ⇒
mv = −(−mgd ) = mgd
2
v 2 = 2 gd
⇒ v = 2 gd
Same result !
Back to collisions:
Elastic collisions are such collisions in which the KE is
conserved. If it is not conserved, the collision is called
inelastic.
An example of elastic collision: A point particle of
mass 1 kg and velocity of 2 m/s, hits a similar point
particle initially at rest. If the collision is elastic,
find the velocities of both particles after collision.
mv = mv1 + mv2
v = v1 + v 2
1 2 1
1
2
2
mv = mv1 + mv2
2
2
2
v 2 = v1 + v 2 + 2 v1v 2
They EXCHANGE velocities…
2
2
v 2 = v1 + v 2
2
v1 = 0
v1 = 0 ,
or
2
v2 = 0
v2 = v
If any similar objects collide elastically head on, they
exchange their velocities.
Another example: two carts of equal mass have equal
opposite velocities, they elastically collide head on. Find
their velocities after collision.
Initial momentum = (mv)1 + (mv) 2 = 0
After collision : 0 = mv1 + mv2 ⇒ v1 = −v2
⎛1
⎞
KE before = 2⎜ mv 2 ⎟ = mv 2
⎝2
⎠
1
1
1
1
2
2
2
2
KE after = mv1 + mv2 = mv1 + mv1
2
2
2
2
2
= mv1
Therefore : v1 = −v,
v2 = v
Last example: A particle of mass 2 kg and velocity of 1
m/s hits another particle of 2 kg initially at rest, head on.
If the first particle’s velocity after collision is 0.2 m/s,
Find the velocity of the second particle.
m1v0 = m1v1 + m2 v2
2 ⋅1 = 2 ⋅ 0.2 + 2 ⋅ v2
v2 = 1 − 0.2 = 0.8 m/s
What happened to the kinetic energy?
1
KE before = 2 ⋅12 = 1 J
2
1
1
2
KE after = 2 ⋅ 0.2 + 2 ⋅ 0.82 = 0.04 + 0.64 = 0.68 J
2
2
∆KE = 0.68 − 1 = −0.32 J
Where did the energy go?
Other forms of energy:
• Internal energy – we will discuss it very soon
• Chemical energy (energy stored in a compound, for
example wood)
• Electrical energy
• Elastic potential energy
Nuclear energy
• Nukilar
A spring gun:
Potential elastic energy of the spring
Kinetic energy of the projectile
Potential energy of the projectile
Kinetic energy of the projectile
Sound waves and “heat” (internal energy)
A climber on the rope…
The work done by his muscles
Potential energy in the g field
Kinetic energy and heat
Sound, heat, arms damage
CONSERVATION OF ENERGY:
“Energy is never created or destroyed, it is just
converted from one form into another.”
“ The total energy in a isolated system is constant.”
The only problem with this is that sometimes it is
difficult to follow the energy conversions similar to
follow the forces. If a car hits a tree, for example, it
is not easy to tell how much energy went to damaging a
car, and how much into damaging a tree, and so on…
QUESTION:
Think of these examples. What type of energy is
converted into what different type of energy ?
• Car Engine
TYPES Of ENERGY:
• Elevator
1. Chemical energy
• Battery
2. Gravitational potential
energy
• Generator
• Hydroelectric power
plant
3. Electrical energy
• You on your Bicycle
5. Elastic potential energy
4. Kinetic energy
6. Nuclear Energy