Lecture Outlines Chapter 9 Physics, 3

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Lecture Outlines Chapter 9

Physics, 3^rd Edition James S. Walker

Chapter 9

Linear Momentum and

Collisions

Units of Chapter 9

• Linear Momentum

• Momentum and Newton’s Second Law

• Impulse

• Conservation of Linear Momentum

• Inelastic Collisions

• Elastic Collisions

• Center of Mass

• (Systems with Changing Mass: Rocket Propulsion)

Units of Chapter 9

9-1 Linear Momentum

Momentum is a vector; its direction is the same as the direction of the velocity.

9-1 Linear Momentum

Change in momentum:

(a) mv (b) 2mv

Example 9-1

Duck, Duck, Goose: Adding Momenta

9-2 Momentum and 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, also

useful when the mass is changing:

9-3 Impulse

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

9-3 Impulse

We can rewrite

as

So we see that

The impulse is equal to the change in momentum.

Figure 9-2

The average force during a collision

9-3 Impulse

Therefore, the 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.

p₁ = - 0,145 kg • 95 mi/h p₂ = 0,145 kg • 115 mi/h

I = Δp = 0,145kg•210 mi/h

= (93,6 m/s)= 13,6 kg•m/s

9-3 Impulse

Om bollen och slagträt är i kontakt under 1,20 ms blir den genomsnittliga kraften (längs x-axeln) F_av = Δp/Δt = I/Δt=

= 13,6 kg•m/s/(1,2•10^-3s)=

= 11,3 kN

Active example 9-1 (p. 246) Find the Final Speed of the Ball

m = 0,144 kg

v_i = - 43,0 m/s (i negativa x-riktningen, p_i<0) F_av = 6,50 kN

Δt = 1,30 ms

Δp = p_f – p_i = I = F_av • Δt = 8,45 kg • m/s p_f = F_avΔt + p_i (kom ihåg att p_i < 0)

v_f = 8,45/0,144 - v_i = 58,7 + 43,0 = 15,7 (m/s)

Conceptual Checkpoint 9–1 Rain Versus Hail (Number of “drops”

and their speed remain the same.) Kraften att hålla paraplyet är den då; densamma, större eller mindre?

Example 9-2 Jumping for Joy (p.247) (m = 72 kg) a) I=? b) Före hoppet är N = mg. Om personen trycker ner golvet under 0,36 s, vilket yttterliga genomsnittlig kraft känner golvet vid hoppet?

Example 9-2 Jumping for Joy

m = 72 kg v_i = 0 m/s

v_f = 2,1 m/s (i positiva y-riktningen) Δt = 0,36 s

I = Δp = p_f – p_i = 72 kg • 2,1 m/s – 0 = 150 kg•m/s F_av = I/Δt = 420 kg • m/s² = 420 N

9-4 Conservation of Linear Momentum

The net force acting on an object is the rate of change of its momentum:

If the net force is zero, the momentum does not change:

9-4 Conservation of Linear Momentum

Internal Versus External Forces:

Internal forces act between objects within the system.

As with all forces, they occur in action-reaction pairs. As all pairs act between objects in the

system, the internal forces always sum to zero:

Therefore, the net force acting on a system is the sum of the external forces acting on it.

9-4 Conservation of Linear Momentum

Furthermore, internal forces cannot change the momentum of a system.

However, the momenta of components of the system may change.

9-4 Conservation of Linear Momentum

An example of internal forces moving components of a system:

Example 9-3 Tippy Canoe: Compairing Velocity and

Momentum (Vad är p₁ och p₂? Utgångshastigheterna = 0

)

m₁ = 130 kg, m₂ = 250 kg, F = 46 N, t = 1,2 s a₁ = - 46 N/130 kg = - 0, 35 m/s²

ger v₁ = - 0,423 m/s

a₂ = 46 N /250 kg = 0,184 m/s² ger v₂ = 0,221 m/s

p₁ = - 55 kg•m/s och p₂ = 55 kg•m/s

Ingen yttre kraft så rörelsemängden är konstant (men inte hastigheterna!)

[Checkpoint 9-3 p.252] K_i< = >K_f?)

Photo 9-4 Eta Carinae

Active Example 9-2 Find the Velocity of the Bee (m_b = 0,150 g, m_s = 4,75 g och v_s = 0,120 cm/s . Rörelsemängden bevaras (den var ursprungligen = 0) , så att v_b = v_s•m_s/m_b = 3,8 cm/s

9-5 Inelastic Collisions

Collision: two objects striking one another

Time of collision is short enough that external forces may be ignored

Inelastic collision: momentum is conserved but kinetic energy is not

Completely inelastic collision: objects stick together afterwards

9-5 Inelastic Collisions

Solving for the final momentum in terms of the initial momenta and masses:

9-5 Inelastic Collisions

A completely inelastic collision: Vad blir v_f?

Exercise 9-2 En bil blir påkörd bakifrån av en lastbil. Om de rör sig tillsammans efter kollisionen vad får de för

gemensam hastighet v? (Försumma yttre krafter

)

m_c = 1200 kg, m_t = 2600 kg v_c = 2,5 m/s

v_t = 6,2 m/s

Rörelsemängden bevaras så att m_cv₂ + m_tv_t = (m₁ + m₂)v

v = (19120 kg•m/s)/(3800 kg) = 5,0 m/s

Conceptual checkpoint 9-3 Hur mycket rörelseenergi går förlorad? Efter kollisionen, är rörelseenergin a) ½ b) 1/3 c) ¼ av den ursprungliga.

(Kom alltid ihåg att K = mv²/2 (kan skrivas) = p²/2m)

Example 9-4 Goal-Line Stand

Example 9-4 Goal-line stand

En “running back” springer mot målområdet då en

“lineback” gör en “head-on” tackling. Om spelarna rör sig tillsammans efter kollisionen vad får de för gemensam

hastighet v? Vad är K_i och K_f ?(Standard y/x)

m_rb = 95,0 kg, m_lb = 111 kg

v_rb = 3,75 m/s , v_lb = - 4,10 m/s

Rörelsemängden bevaras så att m_rbv_rb + m_lbv_lb = (m₁ + m₂)v

v = - 98,85 kg•m/s/(206 kg) = - 0,480 m/s K_i = m_rbv_rb²/2 + m_lbv_lb²/2 = 668 + 933 (J) K_f = (m₁ + m₂)v²/2 = 23,9 (J)

9-5 Inelastic Collisions

Ballistic pendulum: the height h can be found using conservation of mechanical energy after the object is embedded in the block.

Example 9-5 Ballistic Pendulum Bestäm h(m, M, v₀, g) Helt oelastisk kollision

mv₀ = (m+M)v_f

Kinetisk energi ger därefter lägesenergi så att

(m + M)v_f²/2 = (m + M)gh

h = v_f²/2g = {m/(m+M)}²v₀²/2g Practicle problem (p.257)

m = 7,00 g, M = 0,950 kg, h = 0,220 m

v₀ = (2gh)^1/2(m + M)/m = 2,08 m/s •136,7 = 284 m/s

9-5 Inelastic Collisions

For collisions in two dimensions, conservation of momentum is applied separately along each axis:

Example 9-6 (p.258) Bad Intersection: Analyzing a Traffic Accident m₁ = 950 kg, v₁ = 16 m/s, m₂ = 1300 kg, v₂ = 21 m/s

Låt m₁ + m₂ = M = 2250 kg (Inga yttre krafter) (Standard x/y) Efter kollisionen fastnar bilarna I varandra. Beräkna deras gemensamma hastighet v och dess riktning θ (enligt figur).

Inelastisk kollision dvs rörelsemängden bevaras (men inte K)

x: m₁v₁ = Mv_x y: m₂v₂ = Mv_y

v ={(m₁v₁/M)² + (m₂v₂/M)²}^1/2 = 31246/2250 = 13,9 m/s tan θ = v_y/v_x = m₂v₂/m₁v₁

θ = arctan(1,796) = 61°

9-6 Elastic Collisions

In elastic collisions, both kinetic energy and momentum are conserved.

One-dimensional elastic collision:

9-6 Elastic Collisions

We have two equations (conservation of

momentum and conservation of kinetic energy) and two unknowns (the final speeds). Solving for the final speeds:

Conceptual checkpoint 9-4 Speed after Collision En stillastående “hoverfly” (m₂) kolliderar med en elefant (m₁) i en elastisk “head-on” kollision. Om

elefanten först har hastigheten v₀, vad är då flugans hastighet efter krocken? a) v₀ b) 3/2 v₀ c) 2 v₀

9-6 Elastic Collisions

Two-dimensional collisions can only be solved if some of the final information is known, such as the final velocity of one object:

Elastisk kollisionen i två dimensioner (p.262) Curlingstenarna väger 7,00 kg (Standard x/y) v_1,i = 1,50 m/s

v_1,f = 0,610 m/s

K_i = mv_1,i² /2 + 0 = 7,875 J

K_f = mv_1,f²/2 + mv_2,f²/2 = 1,30 J + mv_2,f²/2 v_2,f = 1,37 m/s

y: m • v_1,f • sin(66,0°) = m • v_2,f •sinθ θ = 24,0°

Example 9-7

Two Fruits in Two Dimensions: Analyzing an Elastic Collision

9-7 Center of Mass

The center of mass of a system is the point where the system can be balanced in a uniform

gravitational field.

9-7 Center of Mass

For two objects:

The center of mass is closer to the more massive object.

Exercise 9-4 Balancing a mobile Δx = 0,500 m m1 = 0,260 kg, m2 = 0,170 kg, M = m1 + m2 = 0,430 kg. Hur lång är (hävarmen) x1?

m1x1 = (0,500 - x1) m2

x1 = m2/M • 0,500 = 0,198 (m)

9-7 Center of Mass

The center of mass need not be within the object:

Example 9-8

Center of Mass of the Arm

Example 9-8 Centre of Mass of the Arm (p.265)

X

= (m

x

+ m

x

+ …) = ∑m

x

/M 9-14 Y

= (m

v

+ m

v

+ …) = ∑ m

y

/M 9-15

X_cm = 1,6 kg • 0,12 m + 0,64 kg • 0,40 m + 2,5 kg • 0 m Y_cm = 1,6 kg • 0 m + 0,64 kg • 0 m + 2,5 kg • 0,18 m M = 4,74 kg

X_cm = 0,095 m Y_cm = 0,095 m

dvs tyngpunkten är “utanför” kroppen (Fosbury flop)

9-7 Center of Mass

Motion of the center of mass:

Example 9-9a

Crash of the Air Carts

Example 9-9b

Crash of the Air Carts

Example 9-9 Crash of the Air Carts (p.267)

Vad är tyngdpunkten för CM-hastigheten före och efter (den fullständigt inelastiska) kollisionen

Före

v_cm = (m₁v₁ + m₂v₂ + …) = ∑m_iv_i/M = v₀/2 Rörelsemängden bevaras vid kollisionen mv₀ = (m+m)v_f

Efter

v_cm = (mv_f + mv_f)/M = v₀/2

Hastigheten för de inblandade vagnarna ändras, men inte hastigheten för tyngpunktsystemet

Active Example 9-3 Find the Velocity of the Centre of Masss

Bee (m_b = 0,150 g, m_s = 4,75 g och v_s = - 0,120 cm/s, v_b = 3,8 cm/s)

v_cm = (m₁v₁ + m₂v₂ + …) = ∑m_iv_i/M = ?

Figure 9-12

Motion of the center of mass

9-7 Center of Mass

The total mass multiplied by the acceleration of the center of mass is equal to the net external force:

The center of mass accelerates just as

though it were a point particle of mass M

acted on by

(9-8 Systems with Changing Mass:

Rocket Propulsion)

If a mass of fuel Δm is ejected from a rocket with speed v, the change in momentum of the rocket is:

The force, or thrust, is

Summary of Chapter 9

• Linear momentum:

• Momentum is a vector

• Newton’s second law:

• Impulse:

• Impulse is a vector

• The impulse is equal to the change in momentum

• If the time is short, the force can be quite large

Summary of Chapter 9

• Momentum is conserved if the net external force is zero

• Internal forces within a system always sum to zero

• In collision, assume external forces can be ignored

• Inelastic collision: kinetic energy is not conserved

• Completely inelastic collision: the objects stick together afterward

Summary of Chapter 9

• A one-dimensional collision takes place along a line

• In two dimensions, conservation of

momentum is applied separately to each

• Elastic collision: kinetic energy is conserved

• Center of mass:

Summary of Chapter 9

• Center of mass:

Summary of Chapter 9

• Motion of center of mass:

• (Rocket propulsion