Lecture 19 1/23

Physics 111Lecture 19 (Walker: 9.1-3)

Momentum & ImpulseConservation of Momentum

March 13, 2009

Lecture 19 2/23

Example: Landing with a Thud

Block of mass m1 = 2.40 kg is on horizontal table with a coefficient of friction µk = 0.450 and is connected to hanging block of mass m2 = 1.80 kg. When blocks are released, they move a distance d = 0.50 m, and then m2hits the floor.

Find speed of the blocks just before m2 hits.

Lecture 19 3/23

1 2 1 2; 0;i i iU m gh m gd K E m gh m gd= + = = +1 12 2

1 2 1 22 2(0); ;f fU m gh m g K mv m v E= + = +

1 2( 1 2 22 )f iE E E m m v m gd∆ = − = + −

1 2( 1 2 2 12 )nc kE W m m v m gd m gdµ∆ = ⇒ + − = −

( ) [ ]22 1

1 2

2(9.81 m/s )(0.50 m) (1.80 kg) (0.45)(2.40 kg)2(2.40 kg) (1.80 kg)

kgd m mv

m mµ −−

= = =+ +

Ef = m1gh + ½m1v2 + ½m2v2

Wnc = -fkd = -µkm1gd

= 1.30 m/s

Lecture 19 4/23

Lecture 19 5/23

Momentum• From Newton’s laws: force

must be present to change an object’s velocity (speed and/or direction)Wish to consider changes in velocity due to collisions, when force varies in time in a complicated way

• Best method is to use concept of linear momentum

×scalar

vector

Linear momentum = product of mass velocityLinear momentum = product of mass velocity

Lecture 19 6/23

Linear Momentum

Momentum is a vector; its direction is the same as the direction of the velocity. •Can write in component form; for two-dimensional motion:

yyxx mvpandmvp ==

Lecture 19 7/23

Momentum Example

• A 0.25 kg baseball is moving in the x-direction at 10 m/s. What is its momentum?

• p = mv = (0.25 kg)(10 m/s x)= 2.5 kg•m/s x

or px = 2.5 kg•m/s ; py = 0

Lecture 19 8/23

Change in Momentum

Change in momentum: ∆py = pyf - pyi

Teddy Bear: ∆py = 0-(-mv) = mvBouncing Ball: ∆py = mv-(-mv) = 2mv

Lecture 19 9/23

Momentum &Newton’s Second Law*

Newton’s second law, as we wrote it before:

is only valid for objects that have constant mass.

Here is a more general form in terms of momentum, also useful when the mass is changing:

vF ma mt

∆= =

∆∑rr r ( )mv p

t t∆ ∆

= =∆ ∆

r r

Lecture 19 10/23

Impulse• In order to change the

momentum of an object (say, baseball), a force must be applied

• The time rate of change of momentum of an object is equal to the net force acting on it

•

– Gives an alternative statement of Newton’s second law

– (F ∆t) is defined as the impulse

tFportpF netnet ∆=∆

∆∆

= :

Lecture 19 11/23

Impulse

Impulse is a vector, in the same direction as the average force.

Lecture 19 12/23

ImpulseWe can rewrite

as

So we see that

The impulse I exerted on an object is equal to the change in object’s momentum ∆p.

Lecture 19 13/23

Question 1A 10 kg cart collides with a

wall and changes its direction. What is its change in x-momentum ∆px?

a. −30 kg m/sb. −10 kg m/sc. 10 kg m/sd. 20 kg m/se. 30 kg m/s

x

y

What was the impulse exerted on the cart by the wall??

Lecture 19 14/23

Momentum and Impulse

Profile of the force during a collision.

Microscopic view of a “bounce”.

Lecture 19 15/23

Graphical Interpretation of Impulse• Usually force is not constant,

but time-dependent

• If the force is not constant, use the average force applied

• The average force can be thought of as the constant force that would give the same impulse to the object in the time interval as the actual time-varying force gives in the interval

( )i

i it

impulse F t area under F t curve∆

= ∆ =∑

If force is constant: impulse = F ∆t

Lecture 19 16/23

Impulse and Average ForceDefinition of Impulse: avI F t≡ ∆

r r

netI p= ∆r r

ext av ext sysI F t P= ∆ = ∆r r r

area under F vs. t curveavI F t= ∆ =r r

Lecture 19 17/23

Problem Solving StrategyPicture: To estimate average force Fav, first estimate impulse I of the force. The impulse is equal to change in object’s momentum, i.e., mass times the change in velocity. (An estimate of velocity change ∆v can be made from estimates of collision time ∆t and displacement ∆r. )

Solve:1. Draw a sketch showing before and after positions of the object. Add coordinate axes and label velocities.2. Calculate the impulse from the momentum change during the collision. (I=∆p=m∆v)3. Use Fav=I/∆t to calculate the average force.

Check: Average force is a vector, and should be in the same direction as ∆v.

Lecture 19 18/23

ImpulseThe same change in

momentum may be produced by a large force acting for a short time, or by a smaller force acting for a longer time.

Lecture 19 19/23

Problem: Teeing Off

A 50-g golf ball at rest is hit by “Big Bertha” club with 500-g mass. After the collision, ball leaves with velocity of 50 m/s.

a) Find impulse imparted to ballb) Assuming club in contact with

ball for 0.5 ms, find average force acting on golf ball

Lecture 19 20/23

Problem: teeing offGiven:mass: m=50 g

= 0.050 kgvelocity: v=50 m/sFind: impulse=?Faverage=?

1. Use impulse-momentum relation:

2. Having found impulse, find the average force from the definition of impulse:

( )( )smkg

smkg

mvmvpimpulse if

⋅=−=

−=∆=

50.2050050.0

Nssmkg

tpFthustFp

3

3

1000.5105.0

50.2,

×=×

⋅=

∆∆

=∆⋅=∆ −

Note: according to Newton’s 3rd law, there is also a reaction force on club.

Lecture 19 21/23

Example: Hitting a Baseball (1)150 g baseball is thrown at speed of 20 m/s. It is hit straight back to the pitcher at speed of 40 m/s. The interaction force is as shown here.

What is the average force Fav that the bat exerts on the ball? (Neglect all other forces on ball during the brief duration of the collision.)

What is the maximum force Fmax that the bat exerts on the ball?

Lecture 19 22/23

Hitting a Baseball (2)

( )

(0.15 kg)(40 m/s + 20 m/s)9.0 kg m/s

x fx ix fx ixp mv mv m v v∆ = − = −

==

maxTherefore, (9.0 kg m/s) /(.003 s) 3,000 NF = =

av(9.0 kg m/s)F = = 1,500 N

(.006 s)xpt

∆=

∆

∆px = Ix

Also, ∆px =Ix=½Fmax(6ms)

Lecture 19 23/23

Example: A Karate CollisionWith a karate blow, you shatter a concrete block. Your hand has a mass of 0.70 kg, is initially moving downward at 5.0 m/s, and stops 6.0 mm beyond the point of contact.

(a) What impulse does the block exert on your hand?

(b) What is the approximate collision time and average force that the block exerts on your hand?

ˆ ˆ(0.70 kg)(5.0 m/s ) 3.5 N s I p m v y y= ∆ = ∆ = =r r r

ˆ5.0 m/s( )v y∆ = 12avy v t v t∆ = ∆ ≈ ∆ ∆

2 2(0.006 m) 0.00240 s5.0 m/s

ytv∆

∆ = = =∆

avˆ(3.5 N s ) ˆ= 1,500 N

(0.00240 s)I yF yt= =

∆

rr

Lecture 19 24/23

Impulse Applied to Auto Collisions• The most important factor is the collision time

or the time it takes the person to come to rest– Increased collision time reduces injury in car crash

• Ways to increase the time– Crumple zone at front of car– Air bags

The air bag increases the time of the collision and The air bag increases the time of the collision and absorbs some of the energy from the bodyabsorbs some of the energy from the body

Lecture 19 25/23

Example: A Crumpled CarCar with a 80 kg crash dummy drives into concrete wall at 25 m/s (about 56 mi/h).

(a) Estimate displacement of dummy during crash.

(b) Estimate average force that seat belt exerts on dummy.If the front 25% of 4.0 m long car crumples, displacement of car and dummy during crash is about 1.0 m.

av

ˆ ˆ(80 kg)( 25 m/s ) 2000 N s I F t p m v

x x= ∆ = ∆ = ∆

= − = −

r r r r

1 1av 2 2/ / (1.0 m) / (25 m/s) 0.080 st x v x v∆ = ∆ = ∆ ∆ = =

av ˆ ˆ/ ( 2000 N s ) / (0.080 s) 25,000 N F I t x x= ∆ = − = −r

Lecture 19 26/23

Isolated System of Particles• Definition: an isolated system is a set of

particles that has no external forces doing work on it

– A collision may be the result of physical contact between two objects

– “Contact” may also arise from the electrostatic interactions of the electrons in the surface atoms of the bodies

Momentum of an isolated system in which a Momentum of an isolated system in which a collision occurs is conserved (regardless of collision occurs is conserved (regardless of the nature of the forces between the objects)the nature of the forces between the objects)

Lecture 19 27/23

Conservation of Momentum

The principle of conservation of momentum states when no external forces do work on a system consisting of two objects that collide with each other, the total momentum of the system before the collision is equal to the total momentum of the system after the collision

Lecture 19 28/23

Before Monday, read Walker 9.3-7. (Skip material on two-dimensional collisions and on Motion of Center of Mass.)

Homework Assignments #9a should be submitted using WebAssign by 11:00 PM on Monday, Mar. 16.

Quiz on Chaps. 7 and 8; probably next Wed.

End of Lecture 19