General Physics I

Spring 2011

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Momentum

Impulse• Consider a constant force that

acts on an object over a time interval ∆t. The object moves in a straight line. The impulse delivered by the force is defined as

• Note that the impulse is equal to the area under the force-versus-

Fx

Force

.x xJ F t= ∆

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• Note that the impulse is equal to the area under the force-versus-time graph. If the force is negative (pointing in the negative x direction), then the impulse will be negative. The area will be below the time axis, which indicates a negative impulse.

time∆t

Impulse• A collision is an interaction that

lasts for a short time. During a

collision between two objects, a

force acts on each object over the

time period of the collision. The

forces are an action/reaction pair.

The magnitude of each force

varies in a complex way during

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varies in a complex way during

the collision. The figure shows the

force-versus-time graph for one

object during the collision. Note

that the impulse is equal to the

area under the force-versus-time

curve even when the force varies

with time.

Impulse and Average Force• Since the force usually varies

during a collision (or other kind

of interaction), it is useful define

an average force such that this

constant force gives the same

impulse (area under graph) over

the same time interval. Thus,

, .x x avgJ F t= ∆

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• Note the that impulse must be a

vector because force is a vector.

Thus, in two or three

dimensions, we have

,

,

.

In component form:

.

.

avg

x x avg

y y avg

J F t

J F t

J F t

= ∆

= ∆

= ∆

� �

Momentum• Consider an object on which a constant force acts. We assume

one-dimensional motion for simplicity. The impulse is given by

.

From Newton's second law, . Thus,

. Now, . Hen ,

.

ce

x

x x xf

x

x x

x x x x xi

i

f

J F t

F ma

F t ma t v v a t

F t m v m v

= ∆

=

∆ = ∆ − =

−

∆

∆ =

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• We define the x-component of the momentum of the object by

• Since velocity is a vector, the momentum is also a vector. Thus,

• This vector equation embodies all the components:

.x xp mv=

.p mv=� �

; .x x y yp mv p mv= =

Impulse-Momentum Theorem• From the previous slide, we see that there is a relationship

between impulse and change in momentum:

( ) ( ) .

( ) (

It holds for the -component as well:

) .

A single vector equation describes both relations:

.

x x x x if

y y y y if

J p m v m v

J p m v m v

J p mv m

y

v

=∆ = −

=∆ = −

=∆ = −�� � �

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These equations tell us that an impulse delivered to an object

causes its momentum to change. The equations above are a

statement of the impulse-momentum theorem.

Note that though we used a constant force to obtain the

relation, it holds for all forces, whether they are varying or not.

.if

J p mv mv=∆ = −�� � �

Impulse

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Airbags and Seatbelts• Imagine that you are in your car traveling at 40 mph and you

are involved in a head-on collision. Whether you are using a seatbelt or not, your final velocity is zero. Thus, your momentum change is the same whether or not you are using your seatbelt. According to the impulse-momentum theorem, the impulse is also the same, with or without the seatbelt. Now,

Since ∆px is fixed, you can decrease the average force your experience during the collision if you increase the time interval over which you are brought to rest. This is what

, .x avg xF t p∆ =∆

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time interval over which you are brought to rest. This is what seatbelts and airbags do. They increase the time over which a person is brought to rest (∆t), so that that Fx,avg must decrease.

For a person or object subjected to an impulse, the smaller the magnitude of the average force, less likely injury or damage will occur.

There is a story of a skydiver who failed to open his parachute and survived because he fell into deep snow.

Consider two carts, of masses m and 2m, at rest on an air

track. If you first push one cart for 3 s and then the other

for the same length of time, exerting equal forces on each,

the momentum of the lighter cart is

1. four times

2. twice

3. equal to

4. one-half

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4. one-half

5. one-quarter

the momentum of the heavier cart.

Total Momentum• If there is a collection of objects that may or may not be

interacting, the total momentum of the system is just the vector

sum of the individual momenta:

• In terms of components, we have

1 2 3...

iP p p p p= = + + +∑�

� � � �

.....P p p p= + + +

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1 2 3

1 2 3

.....

.....x x x x

y y y y

P p p p

P p p p

= + + +

= + + +

Problem Solving: Impulse and Momentum• Make one sketch depicting the situation immediately before the

interaction and one sketch depicting the situation immediately

after the interaction.

• Set up a coordinate system. It is important to identify the correct

sign of the momentum, so a coordinate system is essential.

Remember: momentum is a vector!

• List known quantities (e.g., (vx)i for object 1, (vx)i for object 2,

etc.)

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etc.)

• Solve for the desired unknown quantity using the impulse-

momentum theorem.

Workbook: Chapter 9, Questions 6, 7, 8, 13

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Conservation of Momentum• Consider a system of interacting

particles. The particles

experience forces from outside

the system as well as exert

forces on each other, which are

action/reaction pairs. These

interaction forces are called

internal forces because they

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internal forces because they

originate inside the system. The

forces that originate outside the

system are called external forces.

• The net internal force on the

entire system is zero because

each action/reaction pair cancels

out when accounting for forces

on the whole system.

Conservation of Momentum• Thus, the net impulse delivered

to the entire system due to all

internal forces is zero.

• We can see this more clearly by

considering a collision between

two balls.

• Note that impulse that ball 1

delivers to ball 2 is equal in

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