LECTURE 15 Ch11 F17 Linear Momentum Impulsefaculty.uml.edu/.../Teaching/documents/LECTURE15Ch11F17LinearMomentumImpulse.pdfDepartment of Physics and Applied Physics PHYS.1410 Lecture

Department of Physics and Applied PhysicsPHYS.1410 Lecture 15 Danylov

Lecture 15

Chapter 11

Linear MomentumImpulse

Course website:http://faculty.uml.edu/Andriy_Danylov/Teaching/PhysicsI

Physics I

That’s what I do for taking me back to vectors


Today we are going to discuss:

Chapter 11: Newton’s 2nd law (more general form): Section 11.1 Linear Momentum: Section 11.1 Impulse: Section 11.1

IN IN THIS CHAPTER, you will learn to use the concepts of impulse and linear momentum.


Linear MomentumCh.11

Now we know how to use the energy approach to solve problems, so let’s go back to our old good Newton’s 2nd law, F=ma.

F=ma worked great for point objects, but it fails to explain systems such as rockets, where mass changes with motion

(Rockets accelerate by ejecting mass backward).

So, we need to modify (make it more general) our N. 2nd law somehow.


Let’s rewrite N. 2nd law in terms of linear momentum

F ma ddt (m

v) m dv

dt d

pdt

The rate of change of momentum is equal to the net force

So, a force is required to change momentum of an object.

That is what actually Newton wrote.It is more general than our

earlier version F=madtpdF

v

m

Let’s pick an object and start describing it using our “old” N. 2nd law.

We got a new structure mv, so let’s give it a nice name and symbol

Linear momentum is defined as the product of an object’s mass and velocity:

vmp

So, finally, the Newton’s 2nd law will look this way:

Units of momentum: smkg

Momentum is a VECTOR!

ConcepTest Two Boxes/Momentum

A) –20 kg m/s

B) –10 kg m/s

C) 0 kg m/s

D) 30 kg m/s

The cart’s change of linear momentum Δpx is

Δpx = 10 kg m/s – (–20 kg m/s)

Negative initial momentum because motion is to the left and vx < 0.

x

x

initial

final

Final momentum

Initial momentum

∆

10 1 10

= 30 kg m/s

2


ImpulseCh.11

Now we have modified Newton’s 2nd law, let’s apply it to get something, i.e. impulse

How?


Consider “Ball and Wall Collision”

The particle leaves with final velocity in the +x-direction.

(A large force exerted for a small interval of time is called an impulsive force)

(Before) The figure shows a particle with initial velocity in the −x-direction

The particle experiences an impulsive force of short duration Δt.

Now, let’s describe the collision using our new modified N. 2nd law (next slide)

F d

pdt

Instant of maximumcompression


Ball and Wall Collision. Impulse

F d

pdt

Impulse= change in momentum

JdtFf

i

t

t

F dt

ti

t f dppipf if pp

Define Impulse as:

pddtF

From Newton’s 2nd law:

So, Impulse= area under F-vs-t curve

J

pJ

Integrate it:

During the collision, objects are deformed because of the large forces involved . How to relate those forces with a change in momentum?

Force exerted on the ball

Befo

re c

ollis

ion

Afte

r col

lisio

n

fp

ip

Recall: Integral Graphical meaning is an area

p

Work-KE Principle

ImpulsepJ

f

i

t

t

dtFJ

Momentum Principle

Similarity


The force exerted by a tennis racket on the ball (mass 56 g) during a serve ( ) can be approximated by the F vs time plot below.What is the impulse on the ball? What is the speed of the serve?

m/s 71kg 056.0sN 4

fv

Forc

e (k

N)

Time (ms) 10

2

Area under force-time curve is an impulse:

JdtFf

i

t

t

pf pi

AreaJ

ff mvpJ

0

mJv f

0iv

Tennis Ball/Racket CollisionExample

1 2 3 7

)2

22(2 mskN sN 4

5A

B

C

)(2 ABCArea

pJ

Momentum Principle

ConcepTest Kicking BallA) 0.5 m/s left

B) At rest

C) 0.5 m/s right

D) 1.0 m/s right

E) 2.0 m/s right

x

A 2.0 kg object moving to the right with speed 0.50 m/s experiences the force shown. What are the object’s speed and direction after the force ends?

x

vi=0.5m/s

vf -?

Δpx = Jxpf - pi = Jxpf = pi + Jx

∆

∆

. .

/


Impulse/Average force

J

The exact variation of F with time is very often not known (too complicated).

How to get at least something?

J

ptFavg

ptFdtF avgt

t

f

i

Let’s keep the same impulse JAnd the same time interval Δt

But instead of a real forcewe will use an average force

It is easier to find an average force.


Having a certain ∆p, a cat by bending its lags tries to increase ∆t (impact time), so that an impact force would be reduced. (intuitive knowledge of Physics )

ptFavg

How to avoid broken legs for a cat?Example

Initial linear momentum

Final linear momentum0

A

Since the cat falls from a certain height, p(initial)=Δp is given and the cat cannot do anything about that during the collision.

F, that is what can break cat’s bones and the cat “feels” that and tries to reduce F as much as it can.

By bending legs and increasing an impact time, Δt.

Collision with a floorFavg

ConcepTest Two Boxes/Momentum

F F light heavy

We know:

In this case F and t are the same for both boxes!Both boxes will have the same final momentum.

A) the heavier one

B) the lighter one

C) both the same

Two boxes, one heavier than the other,are initially at rest on a horizontalfrictionless surface. The same constantforce F acts on each one for exactly1sec. Which box has more linearmomentum after the force acts ?

tpp if

tFp avf 0iptpFav

(In the previous situation)

Which box has the larger

velocity after the force acts?

A) the heavier one

B) the lighter one

C) both the same

ConcepTest Two Boxes/velocity

lfh mvpMv

hl vvthenmMSince ,


The speed of a fastball is about 40 m/s, and the speed of the ball coming off of player’s bat for a home run is about 54 m/s. The ball (0.145kg) is in contact with the bat for 1ms. What is the average Force exerted by the player?

NFaverage 300,136

tpp

F ifaverage

ssmkgFaverage 001.0

/]4054)[145.0(

tvvm if

))((

in the direction of xorv f

x

Pay attention to directions!!!!!!!!

Faverage

p

t

Average Force on a baseballExample

ptFavg


Thank you

Documents

LECTURE 15 Ch11 F17 Linear Momentum Impulsefaculty.uml.edu/.../Teaching/documents/LECTURE15Ch11F17LinearMomentumImpulse.pdfDepartment of Physics and Applied Physics PHYS.1410 Lecture