Students are responsible for ALL of the material in this presentation . Use this as an initial guide through the material. Please consult your text for clarification and ask questions in class.



Linear Momentum

 a vector quantity describing the product of an object’s mass and velocity.

  Momentum is represented by a lowercase “p” Mathematically, momentum is represented: p = mv Momentum (kgm/s) = (Ns) Mass (kg) Don’t believe me? Check for yourself… Ns = kgm/s recall that N = kgm/s/s so if you multiply N x s you will get (kgm/s/s) x s = kgm/s  Velocity at a given moment (m/s)

Question:



Consider applying a net force (F) to an object as shown. If this force is applied for a time frame of Δt, then the box will accelerate to some resulting (final) velocity.

F



If the time frame is doubled, what happens to the resulting velocity?

Obviously, the resulting velocity will increase (because it experiences an acceleration for a longer period of time.)



Impulse is a

vector quantity describing the product of the (average) force (applied to an object) and the time over which that force is applied.

  Impulse is represented by a capital “J” Mathematically, impulse is represented: J = F Δt Impulse (Ns) Force applied (N) Time during which the force is applied (s)

 Objects respond to an impulse (J) by changing their momentum (p).

* A large impulse will produce a large response (i.e. a baseball will move faster if subjected to a larger impulse) to a given mass….BUT…the same impulse will have a different effect on different masses (i.e. That same impulse applied to a bowling ball will not result in the same speed as the baseball had).

J J

 We know that F = ma and F = m ( So

v f



v o

)

t a



v f



v o t

Ft = m(v f – v o )

We already stated that the quantity Ft is impulse (J). We can also rewrite the right side of the equation giving us:

J = mv f - mv o

The quantity mass x velocity is known as momentum. This gives us:

J = p f – p o = Δp



Impulse-Momentum Theory

This is a HUGE DEAL! Think about it….

Consider that you get a job as a stuntman (or stunt woman as it might be) in the next Marvel Comic movie. You will be filming a scene in which you crash a speeding car into a wall. The director asks you, “would you rather perform the stunt by actually crashing into a brick wall or would you rather crash into a Styrofoam replica placed in front of a bunch of hay bales?”

…obviously you would choose the Styrofoam replica.

You intuitively know that the Styrofoam/hay combination will provide a “softer” stop than the actual brick wall. But, did you ever stop to think why ?

The “ why ” is because of the impulse-momentum theorem.

Let’s assign some values. Let us say that the 1100-kg car is initially traveling east at 30 m/s. No matter which object the car hits, it will stop in both cases. So the change in momentum is the same for both cases….

v v f o =30 m/s = 0 m/s m = 1100-kg Δp = p f – p o Δp = mv f – mv o = -33000 kgm/s FOR BOTH THE BRICK WALL AND THE STYROFOAM WALL.

The change in momentum will be -33000 kgm/s for either case. But HOW they stop will indicate whether a large or small force was responsible.

The brick wall would stop the car VERY QUICKLY…let’s say it brings the car to rest in 2.1-seconds.

The Styrofoam replica would stop the car SLOWLY….let’s say it brings the car to rest in 5.8-seconds.

Because the TIME of IMPACT is different, the FORCE applied will be different as well.

Hitting the Brick Wall….

Δp = -33000 kgm/s t = 2.1 s F = ?

Δp = Ft -33000 = F(2.1) F = -15714.3 N Hitting the Styrofoam….

Δp = -33000 kgm/s t = 5.8 s F = ?

Δp = Ft -33000 = F(5.8) F = -5689.7 N

NOTE – the negative signs on the forces indicate the direction of the force.

The Shorter the time frame, the greater the force .

The Longer the time frame, the weaker the force .

THIS IS A BIG DEAL!

Work on the following practice problems BEFORE you continue to the next topic. Page 199: 1  3 Page 201: 1  4 Page: 203: 1  3 You can find solutions to these problems on the class webpage.

You should (as always) use the solutions to check your work…not as a substitute for trying it on your own.

Before we get into the law, let’s look at a few definitions: 

Internal Forces

  forces that act within the system objects within the system exert forces on each other 

External Forces

 forces exerted on the objects (of the system) by external agents.

System

 the collection of objects being studied.

 an isolated system is one for which all external forces are neglected.

 Consider an isolated system of two masses as shown.

 When they collide, the objects will exert equal and opposite forces on each other (F 12 = -F 21 ) for a period of time, t. Thus, there is an impulse on each object which will change their momentum.

Since F12 = F21 J = Δp J 1 + J 2 = Δp 1 F 12 t + F 21 t = (p 1f – p + Δp 1o 2 ) + (p 2f – p 2o ) 0 = p 1f

p 1o

– p 1o

+ p 2o

+ p 2f

= p 1f

– p 2o

+ p 2f

This is the conservation of momentum.

For an isolated system, the total initial momentum of the system MUST BE CONSERVED. That is the total initial momentum (p1o + p2o ) must equal the total final momentum (p1f + p2f).

…these forces do NOT cancel each other out on in individual object b/c they are acting on different objects. HOWEVER, when dealing with a system we are looking at the combination of the objects (as if they were one object) so F 12 + F 21 = F 12 -F 21 = 0 The net force on the SYSTEM is zero even though individual forces exist!



A collision is an interaction between masses in which

there is a transfer of momentum and energy.

 For ALL collisions, (total) momentum is ALWAYS conserved!

 There are two MAIN categories of collisions (but three types if you consider a sub-group).

 Elastic Collisions  Inelastic Collisions  Perfectly inelastic collisions

Collisions are classified according to changes in the kinetic energy.

1) elastic collisions  total kinetic energy IS CONSERVED!

This is the defining characteristic.

 the objects will move independently after they collide (this, however, does not make it elastic) 2) inelastic collisions  total kinetic energy is NOT conserved!

This is the defining characteristic.

 the objects may move independently after they collide (this, however, does not make it inelastic) 3) Completely (or perfectly) inelastic collisions  Because it is an inelastic collision, KE is not conserved. What makes it special is that the objects will stick together and move as one object after colliding. (We will be able to represent this mathematically.)

Hang on…what is kinetic energy?

Energy is the capacity to do work.

 Energy is the ability for an object to produce a change in itself or in the environment.

 Mechanical Energy (ME) – the forms of energy due to an object’s position, orientation and or motion.

Kinetic Energy (KE) – the energy associated with the motion of a mass

KE = ½ mv 2

Kinetic energy is a scalar quantity (no direction) and has units of kgm 2 /s 2 = Joule This is what we are covering in class right now. 

A cue ball (mass m A = 0.400kg) moving with speed v A 1.80m/s strikes ball B, initially at rest, of mass m B = = 0.500kg. As a result of the collision, the cue ball travels (in the same direction) with a speed of 0.20 m/s. A) What is the velocity of ball B?

B) Is this collision elastic or inelastic?

Cue Ball (A) m A V Ao V Af = 0.40-kg = 1.80 m/s = 0.20 m/s Eight Ball (B) m B V Bo = 0.50-kg = 0 m/s V Bf = ?

Momentum is conserved…

0

m

A

v

Ao

+ m

B

v

Bo

= m

A

v

Af

+ m

B

v

B f

m

A

v

Ao

- m

A

v

Af

= m

B

v

Bf

(m

A

v

Ao

– m

A

v

Af

) / m

B

= v

Bf

1.28 m/s= v

Bf

Cue Ball (A) m A V Ao V Af = 0.40-kg = 1.80 m/s = 0.20 m/s Eight Ball (B) m B V Bo = 0.50-kg = 0 m/s V Bf = 1.28 m/s If the collision is elastic then KE o = KE f ….check it….

0

½ m

A

v

Ao 2

+ ½ m

B

v

Bo 2

= ½ m

A

v

Af 2

+ ½ m

B

v

Bf 2

½ m

A

v

Ao 2

= ½ m

A

v

Af 2

+ ½ m

B

v

Bf 2

0.648 J = 0.418 J

These are NOT equal….

kinetic energy was lost . Therefore, this collision is inelastic .

A cue ball (mass m A = 0.400kg) moving with speed v A 1.80m/s strikes ball B, initially at rest, of mass m B = = 0.500kg. As a result of the collision, ball B is deflected off at an angle of 30.0o with a speed v What is the velocity of ball A?

B = 1.10m/s. 30  ???

The motion occurs in TWO DIMENSIONS….thus we must (mathematically) analyze BOTH DIMENSIONS.

Cue Ball (A) m A = 0.40-kg V Aox = 1.80 m/s V Aoy = 0 m/s V Afx = ?

V Afy = ?

Eight Ball (B) m B = 0.50-kg V V Box V Bfx V Bfy Boy = 0 m/s = 0 m/s = 1.10cos30

 = 1.10sin30



These are the components of the velocities only in the x direction (East-West)

These are the x and y

These are the components of the velocities only in the y direction (North-South)

Cue Ball (A) V m A Aox = 0.40-kg = 1.80 m/s V Aoy V = 0 m/s V Afx Afy = ?

= ?

Eight Ball (B) m B = 0.50-kg V Box = 0 m/s V V Bfx V Bfy Boy = 0 m/s = 1.10cos30

 = 1.10sin30

 Analyze the motion in the x-direction (East-West) m A v Aox + m B v Box = m A v Afx + m B v Bfx m A v Aox – m B v Bfx = m A v Afx (m A v Aox –m B v Bfx ) / m A = v Afx 0.609 m/s= v Afx

Cue Ball (A) V m A Aox = 0.40-kg = 1.80 m/s V Aoy V = 0 m/s V Afx Afy = ?

= ?

Eight Ball (B) m B = 0.50-kg V Box = 0 m/s V V Bfx V Bfy Boy = 0 m/s = 1.10cos30

 = 1.10sin30

 Analyze the motion in the y-direction (North-South) m A v Aoy + m B v Boy = m A v Afy + m B v Bfy – m B v Bfy = m A v Afy (–m B v By ) / m A = v Afy -0.688 m/s= v Afy

We now know that the resulting motion has two dimensions. Now we just need to combine these motions.

0.608 m/s a 2 + b 2 = c 2 (0.608) 2 + (0.688) 2 = c 2 c = 0.92m/s = resulting speed 0.688 m/s -1 (0.688/0.608)  = 48.5

 So, the velocity is 0.92 m/s 48.5

 SE

A Ford F-150 (4690 – Lbs) is stopped at a red light when it is rear-ended by a Honda Civic (2600-Lbs) traveling north at 20.0 m/s. The Honda becomes wedged beneath the rear bumper of the Ford, causing the two vehicles to travel together after the crash. A) What is the resulting velocity of the wreckage?

B) Verify (mathematically) that the collision is inelastic.

Honda Civic (A) m A = 1181.82-kg V Ao = 20 m/s Ford F150 (B) m B = 2131.82-kg V Bo = 0 m/s V f = ?

m

A

v

Ao

+ m

B

v

Bo

= m

A

v

Af

+ m

B

v

B f The vehicles get STUCK TOGETHER, so they will have the SAME FINAL VELOCITY.

m

A

v

Ao

= (m

A

+ m

B

) v

Bf

7.13 m/s= v

f

Work on the following practice problems BEFORE you continue to the mixed review. All of these problems deal with the conservation of momentum and collisions.

Page 214: 1, 3, 5 Page 216: 1, 3 Page 219: 1,3 You can find solutions to these problems on the class webpage.

You should (as always) use the solutions to check your work…not as a substitute for trying it on your own.

Work on the following practice problems BEFORE you continue to the next topic. This is a mixed review of all

of the problems covered in this section.

Page 223 – Theory: 2, 3, 4, 6, 8, 24, 25, Page 223 – Problems: 11, 12, 14, 22, 23, 28, 30, 40 Worksheet You can find solutions to these problems on the class webpage.

You should (as always) use the solutions to check your work…not as a substitute for trying it on your own.

 http://www.youtube.com/watch?v=T9lehHxv-C8 The clip of the “Rube Goldberg Device” featuring the Honda Accord is from an actual commercial. To view the commercial in its entirety, click here .

 Don’t think it’s real??? Read this  Do you like Rube Goldberg Machines??? Me too! Check this one out!