Linear Momentum p is mass times velocity vector p kg ms a 1 kg object moving at 1000 ms has the same momentum as a 1000 kg object moving at 1 ms p 1000 kg ms ID: 647576
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Slide1
LINEAR
MOMENTUM
&
IMPULSESlide2
(Linear) Momentum, p
●
is mass times velocity vector! ● (p) = kg m/s► a 1 kg object moving at 1000 m/s has the same momentum as a 1000 kg object moving at 1 m/s (p = 1000 kg m/s)► a roller skate rolling has more momentum than stationary truck. Slide3
Momentum is a commonly used term in sports.
When a sports announcer says that a team has the momentum they mean that the team is
really on the move and is going to be hard to stop. The term momentum is a physics concept. Any object with momentum is going to be hard to stop or change. To change the motion of such an object, it is necessary to apply a force against its motion for a given period of time. The more momentum that an object has, the harder that it is to stop. Thus, it would require a greater amount of force or a longer amount of time or both to bring such an object to a halt. As the force acts upon the object for a given amount of time, the object's velocity is changed; and hence, the object's momentum is changed.Slide4
Actually, Newton formulated his second law as:
Force = time rate of change of momentum
How can the momentum of an object be changed?
or, more usually, by exerting a force causing an acceleration that changes its velocity.
Δ
p
is the change in momentum produced by the force
F
in time
Δ
t
If the mass doesn’t change, then
By changing its mass,
Let’s go back to Newton’s second law: F = ma. Slide5
is in fact the form in which you should remember the second law of motion since the law in the form
F
= ma is actually, as we have seen mathematically, a special case – it can not be applied to situations in which mass can change.
Watch me,
please
If the mass
does
change, then
Slide6
→ we can get a very useful form of Newton’s 2. law:
F∆t = ∆p Dp = mv - mu
F∆t
is called the
impulse
of the force.
impulse (action of a force F over time
D
t
) will produce change in momentum Dpunits: (
F∆t) = Ns Ns = kg m/s
REMEMBER: Although we write F for simplicity, we actually mean Fnet , because only Fnet and not individual forces can change momentum (by producing an acceleration)Slide7
● Achieving the same change in momentum over a long time requires smaller force and over a short time greater force.
Let’s think about the time it takes to slow the truck to zero.
You could stop it with your own force – just if you exert it over a long, long period of time.
or, you could exert a huge force over very short period of time.
∆p =
F∆t
EXAMPLES
10 kg m/s =
(10N)(1s) =
(5N)(2s) =
(1N)(10s) =
(0.1N)(100s) =
And the story goes on….Slide8
For better understanding we’ll do another example:
h = 2 m
m
= 30 g = 0.03 kg
both eggs fall the same distance, so the velocity of both eggs just before impact is:
h
h
Impact: before impact u = 6 m/s; just after impact is v = 0
∆p = mv – mu = – 0.18 kg m/s
If
you look at the impulse-momentum relation
F∆ t = ∆p, you see that for the same change in momentum (– 0.18 kg m/s in this case), if the time is smaller the ground must have exerted greater force on the egg. And vice versa. The pillow will exert smaller force over greater period of time.
EXAMPLES
In both cases momentum is reduced to zero during impact/interaction with the floor. But the time of interaction is different. In the case of concrete, time is small while in the case of pillow, the stopping time is greatly increased.Slide9
● Often you want to reduce the momentum of an object to zero but with minimal impact force (or injury). How to do it? Try to maximize the time of interaction; this way
stopping force is decreased.Getting smart and smarter by knowing physics:
► Car crash on a highway, where there’s either a concrete wall or a barbed-wire fence to crash into. Which to choose? Naturally, the wire fence – your momentum will be decreased by the
same amount, so the impulse to stop you is the same, but with the wire fence, you extend the time of impact, so decrease the force.
►
Bend your knees when you jump down from high! Try keeping your knees stiff
while landing – it hurts! (only try for a small jump, otherwise you could get
injured…) Bending the knees extends the time for momentum to go to zero,
by about 10-20 times, so forces are 10-20 times less.
EXAMPLESSlide10
►
Safety net used by acrobats, increases impact time, decreases the forces.
► Catching a ball –let your hand move backward with the ball after contact… ► Bungee jumping ► Riding with the punch, when boxing, rather than moving into
By moving away, the time of contact is extended, so force is less than if he hadn’t moved.
By moving into the glove, he is lessening the time of contact, leading to a greater force, a bigger ouch!
►
Wearing the gloves when boxing versus boxing with bare fists.
EXAMPLESSlide11
● Sometimes you want to increase the force over a short time
This is how in karate (tae kwon do), an expert can break a stack of bricks with a blow of a hand:
Bring in arm with tremendous speed (large momentum), that is quickly reduced on impact with the bricks. The shorter the time, the larger the force on the bricks.EXAMPLESSlide12
Formulas we had are for the constant force
.
What if the force changes over time ∆t ? The graph shows the variation with time of the force on the football of mass 0.5 kg.
ball was given an impulse of approximately 100x0.01 = 1Ns during this 0.01s.
area under graph is the total impulse given to the ball
≈ 2x(100x0.05)/2 = 5 Ns
∆p =
m∆v
→ ∆v
= 10 m/s
F
∆t
= ∆p → ∆p = 5 kg m/s
v = u +
∆v
Determine the change in impulse due to a time varying force
Change in momentum,
Δ
p, in time
Δ
t is the area under the graph force vs. time.Slide13
In actuality one is much more likely to use the
measurement of the speed of the football to estimate the
average force that is exerted by the foot on the football.The time that the foot is in contact with the ball can be measured electronically. Favg = m ∆v/ ∆t Slide14
Till now we were concentrated on ONE object.
Now we move to the system of (usually) two objects
exerting strong forces over a short time intervals on each otherlike: collisions, explosions, ejectionsSlide15
●
collisions can be very complicated
● two objects bang into each other and exert strong forces over short time intervals which are very hard to measure● fortunately, we can predict the future without going into pesky details of force. ● What will help us is the law of conservation of linear momentum:Collisions - conservation of momentumSlide16
Law of Conservation of Momentum
● consider system:
particle 1 and particle 2 collide with one another.
m
1
u
1
m
2
u
2
velocities
just before
interaction (collision)
velocities
just after
interaction (collision)
v
1
v
2
The total linear momentum of a system of interacting particles is conserved - remains constant, provided there is no resultant external force.
Such a system is called an “isolated system”.
p
after
=
p
before
(p
1
+ p
2
= p)
m
1
v
1
+ m
2
v
2
= m
1
u
1
+ m
2
u
2Slide17
Certain situations (collisions, explosions, ejections) do not allow detailed knowledge of forces (strength, direction, duration) or acceleration. Of course that these situations must follow Newton’s laws. The only problem is that it is difficult to see exactly how to apply them. One cannot easily measure neither forces involved in the collision nor acceleration (velocity appears to be instantaneously acquired).
The law of conservation of momentum gives us an easy and elegant way to predict the outcome without knowing forces involved in process. It is much easier to measure velocities and masses before and after interaction.
WE CAN APPLY THE LAW OF CONSERVATION OF MOMENTUM TO COLLISIONS AND EXPLOSIONS (EJECTIONS) IF DURING INTERACTION THE NET EXTERNAL FORCE IS ZERO OR IT CAN BE NEGLECTED.
Example: baseball is struck with a bat – duration of the collision is about 0.01 s, and the average force the bat exerts on the ball is several thousand
Newtons
what is much greater than the force of gravity, so you can ignore it. And as we consider velocities just before and just after interaction, there is no much change due to gravity. The system can be considered isolated and momentum is conserved. Slide18
beauty of the law of conservation of momentum
●
if we know what the objects were doing before they collided, we can figure out what can happen after they collide.● We can work backward sometimes to figure out from the collision scene what was going on before the collision. Slide19
Momentum is conserved in every isolated system. Internal forces can never change momentum of the system. Slide20
Example how to use law of conservation of momentum in the case of ejections or explosions.Slide21
A 60.0-kg astronaut is on a space walk when her tether line breaks. She throws her 10.0-kg oxygen tank away from the shuttle with a speed of 12.0 m/s to propel herself back to the shuttle. What is her velocity?
p
before = pafter70
before
u = 0
after
10
12.0 m/s
60
v
1
= ?
0
=
m
1
v
1
+ m
2
v
2
0 = 60.0 v
1
+ 10.0 (12.0)
v
1
= − 2.0 m/s
moving in the negative direction means toward shuttle
Slide22
Very similar case is spaceship propulsion which is actually example of conservation of momentum. Since no outside forces act on the system (spaceship + its fuel) or it is very small compared to the explosion, the momentum gained by fuel ejected in the backward direction must be balanced by forward momentum gained by the spaceship.
hot gas ejected at
very high speed
p
before
=
p
after
0 = m
1
v
1
+ m
2
v
2m1 v1 = - m2
v2
● the same as untied balloon. Slide23
Similar examples are: recoil of the firing gun, recoil of the firing cannon, ice-skater’s recoil, throwing of the package from the boat etc.Slide24
●
Two stationary ice skaters push off
● both skaters exert equal forces on each otherp
before = p
after0 = m1 v
1
+ m
2
v
2
m
1
v
1
= - m2
v2 ● however, the smaller skater acquires a larger speed (due to larger acc.) than the larger skater.● momentum is conserved!Slide25
If you consider momentum: before = 0, so after must be zero too, therefore the speeds gained (while the force of interaction acted) are pretty different.Slide26
Example how to use law of conservation of momentum in the case of collisions.Slide27
There are two fish in the sea. A 6 kg fish and a 2 kg fish. The big fish swallows the small one. What is its velocity immediately after lunch?
the big fish swims at 1 m/s toward and
swallows the small fish that is at rest.
Net external force is zero.
Momentum is conserved.
p
before lunch
= p
after lunch
momentum is vector, direction matters; choose positive direction in the direction of big fish.
+
Mu
1
+ mu
2
= (M + m)v
1 m/s
before lunch
after lunch
(6 kg)(1 m/s) + (2 kg)(0 m/s) = (6kg + 2 kg) v
6 kg m/s = (8 kg) v
v
= 0.75 m/s
in the direction of the large fish before lunch
v = ?Slide28
p
before lunch
= p after lunch +
Mu
1
+ mu
2
= (M+m)v
1 m/s
before lunch
after lunch
v
= 0.25 m/s
in the direction of the large fish before lunch
v = ?
b. Suppose the small fish is not at rest but is
swimming toward the large fish at 2 m/s.
-
2 m/s
(6 ) (1 ) + (2 ) (—2 ) = (6 + 2
) v
6 — 4
= 8
v
The negative momentum of the small fish is very effective in slowing the large fish.
Slide29
p
before lunch
= p after lunch +
Mu
1
+ mu
2
= (M+m)v
1 m/s
before lunch
after lunch
v
= 0 m/s
v = ?
c. Small fish swims toward the large fish at 3 m/s.
-
3 m/s
(6 ) (1 ) + (2 ) (—3 ) = (6 + 2
) v
6 — 6
= (8 )
v
fish have equal and opposite momenta. Zero momentum before lunch is equal to zero momentum after lunch, and both fish come to a halt.Slide30
p
before lunch
= p after lunch +
Mu
1
+ mu
2
= (M+m)v
1 m/s
before lunch
after lunch
v
=
—
0.25 m/s
v = ?
d. Small fish swims toward the large fish at 4 m/s.
-
4 m/s
(6 ) (1 ) + (2 ) (—4) = (6 + 2
) v
6 — 8
= 8
v
The minus sign tells us that after lunch the two-fish system moves in a direction opposite to the large fish’s direction before lunch.Slide31
A red ball traveling with a speed of 2 m/s along the x-axis hits the eight ball. After the collision, the red ball travels with a speed of 1.6 m/s in a direction 37
o
below the positive x-axis. The two balls have equal mass. At what angle will the eight ball fall in the side pocket? What is the speed of the blue (8th) ball after collision. before collision: after collision:
8
8
u
1
u
2
= 0
θ
2
37
0
the point of collision
v
2
v
1
p
before
= p
after
in x – direction
m u
1
+ 0 = m v
1
cos 37
0
+ m v
2
cos
q
2
v
2
cos
q
2
= u
1
- v
1
cos 37
0
= 0.72 m/s (1)
in y
–
direction
0 = - m v
1
sin 37
0
+ m v
2
sin
θ
2
v
2
sin
θ
2
= v
1
sin 37
0
= 0.96 m/s
(2)
direction of v
2
; (2)/(1)
tan
θ
2
= 1.33
θ
2
= 53
0
(2) → v
2
= 0.96 / sin 53
0
v
2
= 1.2 m/s Slide32
Derivation of the Law of Conservation of Momentum
● consider system: particle 1 and particle 2 collide with one
another with no net external force acting on neither of them.
m
1
u
1
m
2
u
2
velocities
just before
interaction (collision)
F
1
F
2
velocities
just after
interaction (collision)
forces
during
collision
v
1
v
2
● During the time interval the collision takes place, ∆t, impulse F
1
∆t
given to particle 1 will cause its momentum change ∆p
1
. During
the same time interval impulse F
2
∆t will change particle’s 2
momentum by ∆p
2
.
Slide33
∆
p
1 + ∆p2 = 0 → ∆(p1 + p2 ) = 0 → ∆p = 0 → pafter = pbefore Total momentum of a system before and after collision is the same.
particle 1
: F
1
∆t = ∆p
1
particle 2
: F
2
∆t = ∆p
2
F
1 = – F2 (N3.L) → ∆p2 = – ∆p1
What one object loses in the collision the other one gains.
(p1 + p2 = p)
Conservation of Momentum: if no external force act on a system, the total momentum of the system is conserved – it will not change. Such a system is called an “isolated system”.
This argument can be extended up to any number of interacting particles so long as the system of particles is still isolated.