11/11/2017 1 v p m = JG G v v v p m p m p m = = = .... P p p p p = + + ...

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Physics 141: Chapter 6, Pg 1

Chapter 6: Linear Momentum and Collisions

6.1 Linear Momentum

6.2 Impulse

6.3 Conservation of Linear Momentum

6.4 Elastic and Inelastic Collisions

6.5 Center of Mass

6.6 Jet Propulsion and Rocket


• Momentum p: For a single object, the momentum is defined as:

• Momentum is a vector—it has both a magnitude and a direction.

• SI unit of momentum: kg • m/s.

6.1: Linear Momentum

vp m=vv

v

x x

y y

z z

p mp m

p m

==

=

• For a system of objects, the total momentum is the vector sum of each.

1 2 3 .... iP p p p p= + + + =∑

Momentum

Capital P



Force and Momentum• The change in momentum is the difference between the momentum

vectors.

• If an object’s momentum changes, a force must have acted on it.Physics 141: Chapter 6, Pg 4


• If an object’s momentum changes, a force must have acted on it.

Force and Momentum

( )0 0 0v v v v

net

m p pm m pF mat t t t

− −− Δ= = = = =

Δ Δ Δ Δ

• Newton’s second law of motion in terms of momentum:

(Eq.6.3)

It states that the net force is equal to the rate of change of the momentum

0net

p ppFt t

−Δ= =Δ Δ p

: Before action (Initial state) 0p

: After action (Final state)

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A 0.150-kg baseball traveling with a horizontal speed of 4.50 m/s is hit by a bat and then moves with a speed of 34.7 m/s in the opposite direction. What is the change in the ball’s momentum?

Exercises 6.5 (p215)


6.2: Impulse

Impulse

• Impulse I is defined as the change in momentum:

00 v vaveI F t p p p m m= Δ = Δ = − = −

• Actual contact times may be very short.

netpFt

Δ=Δ

aveF t pΔ = ΔFave: Average force

• SI unit of Impulse: [N • s] or [kg • m/s]

• Impulse-Momentum theorem: The impulse acting on an object is equal to the change in the object’s momentum


6.2: Impulse

Multiple choose question:

You drop an egg onto 1) the floor 2) a thick piece of foam rubber. In both cases, the egg does not bounce.In which case is the impulse greater?A) Floor; B) Foam; C) the same

Impulse = ΔPSame change in momentum

Two identical balls are dropped from the same height onto the floor. In case 1 ball 1 bounces back up, and in case 2 ball 2 sticks to the floor without bouncing. In which case is the impulse given to the ball by the floor the biggest? 1. Case 1 2. Case 2 3. The same

Bouncing BallImpulse = Δp

= mvfinal – mvinitial

= m( vfinal – vinitial)= 2m( |vfinal|+|vinitial|)

Sticky BallImpulse = Δp= mvfinal – mvinitial

= m( 0 – vinitial)= m |vinitial|


Example 6.5 (p188): A 70.0-kg worker jumps stiff-legged from a height of 1.00 m onto a concrete floor.

(a) What is the magnitude of the impulse he feels on landing, assuming a sudden stop in 8.00 ms?

(b) What is the average force?

6.2: Impulse

Conceptual questions:• When jumping from a height to the ground, it is advised to land

with the legs bent rather than stiff-legged, why is this?

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Exercises 6.18 (p216) : A ball of mass 200-g is released from rest at a height of 2.00 m above the floor and it rebounds straight up to a height of 0.900 m.

(a) Determine the ball’s change in momentum due to its contact with floor.(b) If the contact time with the floor was 0.095 s, what was the average force

the floor exerted on the ball, and in what direction?


6.3: Conservation of Linear Momentum

Law of conservation of momentum: If there is no net force acting on a system, its total momentum cannot change (i.e, momentum is conserved).

Conservation of Linear Momentum

netpFt

Δ=Δ

If 0pΔ =0netF = Then:

0P P=Initial state Final state

Total Initial momentum of a system = Total Final momentum of a system

1 2 3 ... kP p p p p= + + + =∑Total momentum

i fP P=or


Momentum is Conserved

• Momentum is “Conserved” meaning it can not be created nor destroyed – Can be transferred

• Total Momentum does not change with time.• This is a BIG deal!


• Draw “before”, “after” • Define system so that Fext = 0• Set up axes• Compute Ptotal “before”• Compute Ptotal “after”• Set them equal to each other

Procedure for solving problems:



Two masses, m1 = 1.0 kg and m2 = 2.0 kg, are held on either side of a light compressed spring by a light string joining them, as shown in Fig. 6.9. The string is burned (negligible external force), and the masses move apart on the frictionless surface, with m1 having a velocity of 1.8 m/s to the left. What is the velocity of m2?

Example 6.6: Momentum Conservation

• Initial total momentum Pi = 0• Final total momentum Pf

= - m1 v1 + m2 v2

• Set x-axis (+ve, to right direction)

x

• Momentum conservation: Pi = Pf

- m1 v1 + m2 v2 = 01

2 12

v v 0.9m/smm

= =

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Question: Reverse collisions IA man and a block rest on a frictionless frozen lake. The man has a mass of 50 kg. The man pushes off from the block and slides away at a speed of 30 m/s. The speed of the box is:

A. 10 m/sB. 15 m/sC. 30 m/sD. 90 m/s

150 kg shore



Question: Reverse collisions IIThe man pushes off from the block and slides away at a speed of 30 m/s. After the push, the sum of the momenta of the man and the block is:A. 0 kg m/sB. 1500 kg m/sC. 3000 kg m/sD. 6000 kg m/s



Question:A bucket full of water is slid across a sheet of ice in the rain. As the bucket fills, what happens to its velocity and its momentum? (Assume no friction)

A. both decreaseB. both remain constantC. velocity remain constant; momentum increasesD. velocity decreases; momentum remains constant



Exercises 6.40:An object initially at rest explodes and splits into three fragments. The first fragment flies off to the west, and the second fragment flies off to the south. (a) The third fragment will fly off toward a general direction of (1) southwest, (2) north of east, (3) either due north or due east. Why? (b) If the object has a mass of 3.0 kg, the first fragment has a mass of 0.5 kgand a speed of 2.8 m/s, and the second fragment has a mass of 1.3 kg and a speed of 1.5 m/s, what are the speed and direction of the third fragment?

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Solution to Exercises 6.40:


6.4: Elastic and Inelastic Collisions

Elastic vs. Inelastic Collisions

• A collision is said to be elastic when kinetic energy is conserved before and after the collision.

– Carts colliding with a spring in between, billiard balls, etc.vi

• A collision is said to be inelastic when kinetic energy is not conserved

– Car crashes, collisions where objects stick together, etc.

i fK K=

i fK K>

i: initialf: final



Elastic vs. Inelastic Collisions

• Total kinetic energy of a system is conserved in an elastic collision; Total kinetic energy of a system is not conserved in an inelastic collision.

• Total momentum of a system is conserved in both elastic and inelastic collisions

i fP P=

Elastic Inelastic

i fK K= i fK K>

i fP P=

Total Kinetic energy

Total Momentum

21 v2 j jK m=∑

vj jjP p m= =∑ ∑

Total Kinetic energy:

i: initialf: final

Total Momentum:



Question: Two masses in space move toward each other and collide. Assume no rotational motion occurs before or after the collision. If the two masses stick together after colliding, what is the final horizontal velocity of the masses?

A. 9 m/s to the leftB. 7 m/s to the leftC. 5 m/s to the leftD. 0 m/s

10 m/s

10 m/s

600

5kg

20kg

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Two-body Inelastic collision• A completely inelastic collision is one where the objects stick together afterwards.

01 1

1 2

vv =

mm m+

Final speed :

Final Kinetic Energy :

0

21 12 1

f 1 2 1 2 i1 2 1 2

v1 1= ( )v ( )2 2

m mK m m m m Km m m m

⎛ ⎞ ⎛ ⎞+ = + =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

Initially at rest

Initial Kinetic Energy :

0

2i 1 1

1= v2

K m

i fK K>Obviously :Physics 141: Chapter 6, Pg 22


Two-body Elastic Collisions

Before After collision

• Final velocities in terms of initial velocities can be found by solving 2 Eqs above:

2 2 2 21 1 2 2 1 1 2 2

1 1 1 1v v v v2 2 2 2i i f fm m m m+ = +Conservation of Kinetic energy:

1 1 2 2 1 1 2 2v v v vi i f fm m m m+ = +Total Momentum:Assume: both objects in the positive x-direction

1 2 21 1 2

1 2 1 2

2v = v vf i im m mm m m m

⎛ ⎞ ⎛ ⎞−+⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

1 1 22 1 2

1 2 1 2

2v = v vf i im m m

m m m m⎛ ⎞ ⎛ ⎞−

−⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

(Signs are used to indicate the directions of velocities)



• A block of mass M is initially at rest on a frictionless horizontal surface. A bullet of mass m is fired at the block with a muzzle velocity (speed) v. The bullet lodges in the block, and the block ends up with a speed V. In terms of m, M, and V :– What is the initial speed of the bullet v?– What is the initial energy of the system?– What is the final energy of the system?– Is kinetic energy conserved?

vV

before after

x

Example1 Stuck together: completely inelastic collision!



Solution:

• Consider the bullet & block as a system. After the bullet is shot, there are no external forces acting on the system in the x-direction. Momentum is conserved in the x direction!

– Px (initial) = Px (final)– mv = (M+m)V

vV

initial final

v M mm

V=+⎛

⎝⎜⎞⎠⎟

x

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• Now consider the kinetic energy of the system before and after:

• Before:

• After:

• So

( )E mv m M mm

V M mm

M m VB = =+⎛

⎝⎜⎞⎠⎟

=+⎛

⎝⎜⎞⎠⎟

+12

12

12

22

2 2

( )E M m VA = +12

2

Kinetic energy is NOT conserved! (friction stopped the bullet)However, momentum was conserved, and this was useful.

E mM m

EA B=+

⎛⎝⎜

⎞⎠⎟

v M mm

V=+⎛

⎝⎜⎞⎠⎟


Solution:


Two balls of equal mass are thrown horizontally with the same initial velocity. They hit identical stationary boxes resting on a frictionless horizontal surface. The ball hitting box 1 bounces back, while the ball hitting box 2 gets stuck.– Which box ends up moving faster?

(a) Box 1 (b) Box 2 (c) same

1 2

Example2:6.4: Elastic and Inelastic Collisions


• Since the total external force in the x-direction is 0, momentum is conserved along the x-axis.

• In both cases the initial momentum is the same (mv of ball).• In case 1 the ball has negative momentum after the collision, hence the box

must have more positive momentum if the total is to be conserved.• The speed of the box in case 1 is biggest!

x

1 2V1 V2

Solution:

mv0 = MV1 - mv

V1 = (mv0 + mv ) / M

Box 1mv0 = (M+m)V2

V2 = mv0 / (M+m)

Box 2

V1 numerator is bigger and its denominator is smaller than that of V2.V1 > V2



What physical quantities are conserved in the above collision? 1. Only momentum is conserved 2. Only total mechanical energy is conserved 3. Both are conserved 4. Neither are conserved

Is it possible for a system of two objects to have zero total momentum and zero total kinetic energy after colliding, if both objects were moving before the collision?

1. YES

2. NO if both objects are moving in opposite directions with the same mass and velocity they would have a resulting velocity of zero.

Continue…..


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The precollision conditions for two elastic collisions are shown in the figure below. What are the final velocities in each case?

Example 6.11 (p201):

1 2 1 2=4.0, =1.0, v = 10 m/s, v = 50 m/si im m + +

1 1 22 1 2

1 2 1 2

2v = v v 13 (m/s)f i im m m

m m m m⎛ ⎞ ⎛ ⎞−

− =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

Given:

1 2 21 1 2

1 2 1 2

2v = v v 8.0 (m/s)f i im m mm m m m

⎛ ⎞ ⎛ ⎞−+ =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

(a)

By using:

right right

DirectionsBefore after

right right




Continue…

Given: (b)1 2 1 2=2.0, =4.0, v = 6 m/s, v = 6 m/si im m + −

1 1 22 1 2

1 2 1 2

2v = v v 2.0 (m/s)f i im m m

m m m m⎛ ⎞ ⎛ ⎞−

− =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

1 2 21 1 2

1 2 1 2

2v = v v 10 (m/s)f i im m mm m m m

⎛ ⎞ ⎛ ⎞−+ = −⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠

DirectionsBefore after

right left

left right




Question: Bob m1 is swung from a height h and allowed to collide with bob m2, which is at rest. The bobs are the same size but may have different masses. The velocity of m1 just before striking m2 is v. If the bobs collide elastically, and m1 is heavier than m2, which of the following is true?A. m1 and m2 will both move to the right at the same speedB. m1 and m2 will both move to the right, but m1 will move more slowlyC. m2 will move to the right, and m1 will move to the leftD. m2 will move to the right, and m1 will stop

m1 m2h



Exercise 6.63 (p219): A 1.0-kg object moving at 10 m/s collides with a stationary 2.0-kg object as shown in the figure below. If the collision is perfectly inelastic, how far along the inclined plane will the combined system travel? (Neglect friction.)

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Exercises 6.69: A freight car with a mass of 25 000 kg rolls down an inclined track through avertical distance of 1.5 m. At the bottom of the incline, on a level track, the carcollides and couples with an identical freight car that was at rest. What percentageof the initial kinetic energy is lost in the collision?



6.5: Center of Mass

System of Particles

• Until now, we have considered the behavior of very simple systems (one or two masses).

• But real life is usually much more interesting!• How can we analyze the overall motion of a system of particles?

• We use an equivalent single particle to represent a system of particles, through the concept of center mass (CM)

A system of 3 particles:


6.5: Center of Mass

Center of Mass of the System

1 2 31 2 3 1

1 2 3

1

...X =...

i n

iinn i

CM i nn

ii

m xm x m x m x m x

m m m m m

=

==

=

+ + + +=

+ + + +

∑

∑• Using signs to indicate the directions of x, the center of mass can be given by:

X i i i iCM

i

m x m xm M

= =∑ ∑∑

• How do we describe the “position” of a system made up of many parts?• Define the Center of Mass:

The center of mass is the point at which all of the mass of an object or system may be considered to be concentrated, for the purposes of linear or translational motion only.

ii

M m=∑ (total mass of the system)


6.5: Center of Mass

Example 6.15 (p205):

A dumbbell has a connecting bar of negligible mass. Find the location of the center of mass

(a) if m1 and m2 are each 5.0 kg and (b) If m1 is 5.0 kg and m2 is 10.0 kg.

The center of mass is where the system is balanced!

X i i i iCM

i

m x m xm M

= =∑ ∑∑

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How to find center of mass?

+ + +

6.5: Center of Mass

• For a solid body the center of mass is also the balance point. The body could be suspended from its center of mass and it would not rotate, i.e. not be out of balance.

• For homogenous solid bodies that have a symmetrical shape, the center of mass is at the center of body's symmetry, its geometrical center.

• The center of mass of a solid body does not have to lie within the body. The center of mass of a hula-hoop is at its center where there is no hoop, just hula.

The center of mass of a flat object can be found by suspension.


6.6: Jet Propulsion and Rockets

• Jet propulsion: application of jets (a stream of liquid or gas) to produce motion.

• Examples: Air balloons, Rifles, Rockets, etc • Jet propulsion is explained by Newton’s 3rd Law and in absence of

external forces, the conservation of momentum applies.


Summary of Chapter 6

• Momentum of a point particle is defined as its mass multiplied by its velocity.

• The momentum of a system of particles is the vector sum of the momenta of its components.

• Newton’s second law in terms of momentum:

vp m=

1 2 3 ... kP p p p p= + + + =∑

• Impulse-Momentum theorem: The impulse acting on an object is equal to the change in the object’s momentum

ave f iI F t p p p= Δ = Δ = −

netpFt

Δ=Δ


Summary of Chapter 6

0pΔ =

• Law of conservation of momentum: If there is no net force acting on a system, its total momentum cannot change (i.e, momentum is conserved).

0netF = or i fP P=• In an elastic collision, the total kinetic energy is conserved.

• In an inelastic collision, the total kinetic energy is not conserved

• Momentum is conserved for both elastic and inelastic collision

i fK K=

i fK K>

• The center of mass of an object is the point where all the mass may be considered to be concentrated.

• Coordinates of the center of mass:

:M total mass of the system

X i i i iCM

i

m x m xm M

= =∑ ∑∑

11/11/2017 1 v p m = JG G v v v p m p m p m = = = .... P p p p p = + + ...

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Transcript of 11/11/2017 1 v p m = JG G v v v p m p m p m = = = .... P p p p p = + + ...