1 19 Dynamics B Momentum II

Conservation of Momentum

Conservation of Momentum

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A PowerPoint Presentation stolen by

A PowerPoint Presentation stolen by

Bro. Nigel from a professor at

Bro. Nigel from a professor at

Southern Polytechnic State University

Southern Polytechnic State University

Momentum is conserved in this

rocket launch. The velocity of the rocket and

its payload is determined by

the mass and velocity of the expelled gases.

Photo: NASA

Objectives: After completing this

Objectives: After completing this

module, you should be able to:

module, you should be able to:

• State the law of conservation of momentum State the law of conservation of momentum and apply it to the solution of problems.

and apply it to the solution of problems. • Distinguish by definition and example Distinguish by definition and example

between elastic and inelastic collisions.

between elastic and inelastic collisions.

• Predict the velocities of two colliding bodies Predict the velocities of two colliding bodies when given the coefficients of restitution,

when given the coefficients of restitution,

masses, and initial velocities.

A Collision of Two Masses

A Collision of Two Masses

When two masses m₁ and m₂ collide, we will use the symbol

u

to describe velocities before collision.

The symbol

v

will describe velocities

after

collision.

Before

Before

_m

1

u

₁

m

₂

u

₂

m

₁

v

₁

m

₂

v

2 After

A Collision of Two Blocks

A Collision of Two Blocks

m

₁

m

_2B

“u”

=

Before “

v

” = After

m

₁

u

1

u

2

m

₂

Before

Before

m

₂

v

2

m

₁

v

₁ After

After

Conservation of Energy

Conservation of Energy

m

₁

u

1

u

2

m

₂

The kinetic energy beforebefore colliding is

equal to the kinetic energy afterafter colliding plus the energy lostlost in the collision.

2 2 2 2

1 1 1 1

1 1 2 2 1 1 2 2

2

m u



2

m u



2

m v



2

m v



Loss

2 2 2 2

1 1 1 1

1 1 2 2 1 1 2 2

Example 1.

Example 1. A A 2-kg2-kg mass moving at mass moving at 4 m/s4 m/s collides with a

collides with a 1-kg_1-kg mass initially at rest. After _{mass initially at rest. After}

the collision, the 2-kg mass moves at

the collision, the 2-kg mass moves at 1 m/s1 m/s and the 1-kg mass moves at

and the 1-kg mass moves at 3 m/s3 m/s. What . What energy was lost in the collision?

energy was lost in the collision?

It’s important to draw and label a sketch with

It’s important to draw and label a sketch with

appropriate symbols and given information.

appropriate symbols and given information.

m₂

u₂ = 0

m₁

u₁ = 4 m/s

m₁ = 2 kg m₁ = 1 kg

BEFORE BEFORE

m₂

v₂ = 2 m/s

m₁

v₁ = 1 m/s

m₁ = 2 kg m₁ = 1 kg

Example 1 (Continued).

Example 1 (Continued).

What energy

_{What energy}

was lost in the collision? Energy is

was lost in the collision? Energy is

conserved.

conserved.

m₂

u

u₂₂ = 0= 0

m₁

u

u_{1 1} = = 4 m/s4 m/s

m

m_{1 1} = = 2 kg2 kg mm_{1 1} = = 1 kg1 kg

m₂

v

v₂₂ = = 2 m/s2 m/s

m₁

v

v_{1 1} = = 1 m/s1 m/s

m

m_{1 1} = = 2 kg2 kg m_{m1 1} = = 1 kg1 kg

BEFORE:

BEFORE: 12 m u1 12  12 m u2 22  12 (2 kg)(4 m/s)2  0 16 J

2 2 2 2

1 1 1 1

1 1 2 2

2 m v  2 m v  2 (2 kg)(1 m/s)  2 (1 kg)(2 m/s)  3 J

AFTER AFTER

Energy Conservation: K(Before) = K(After) + Loss

Impulse and Momentum

Impulse and Momentum

A uA uB B

A B

v_{A v}

B

B

--

F

F

_AA



_

t

t

F

F

_{B B}



_

t

t

Opposite but Equal

F



t

F



t = mv

_f

– mv

_o

F

_B



t = -F

_A



t

Impulse = p

m

_B

v

_B

- m

_B

u

_B

= -(m

_A

v

_A

- m

_A

u

_A

)

m

_A

v

_A

+ m

_B

v

_B

= m

_A

u

_A

+ m

_B

u

_B

m

_A

v

_A

+ m

_B

v

_B

= m

_A

u

_A

+ m

_B

u

_B

Conservation of Momentum

Conservation of Momentum

A uA uB B

A B

v_{A v}

B

B

--

F

F

_AA



_

t

t

F

F

_{B B}



_

t

t

The total momentum AFTER a collision is equal to the total momentum BEFORE.

Recall that the total energy is also conserved:

K

K_A0A0 + K + K_B0B0 = K = K_AfAf + K + K_BfBf + Loss + Loss

K

K_A0A0 + K + K_B0B0 = K = K_AfAf + K + K_BfBf + Loss + Loss

Kinetic Energy: K = ½mv

Kinetic Energy: K = ½mv

22

m

_A

v

_A

+ m

_B

v

_B

= m

_A

u

_A

+ m

_B

u

_B

Example 2:

Example 2: A A 2-kg2-kg block block A A and a and a 1-kg1-kg block block B

B are pushed together against a spring and _{are pushed together against a spring and}

tied with a cord. When the cord breaks,

tied with a cord. When the cord breaks,

the

the 1-kg1-kg block moves to the right at block moves to the right at 8 m/s8 m/s. . What is the velocity of the

What is the velocity of the 2 kg2 kg block? block?

A _B

The initial velocities are

The initial velocities are

zero

zero, so that the total _{, so that the total}

momentum

momentum before_before release is _{release is}

zero.

zero.

m

m

_AA

v

v

_AA

+ m

+ m

_BB

v

v

_BB

= m

= m

_AA

u

u

_AA

+ m

+ m

0 _BB

u

u

_BB 0

m

_A

v

_A

= - m

_B

v

_B

_v

_A

= -

m

B

v

B

Example 2 (Continued)

Example 2 (Continued)

m

m

_AA

v

v

_AA

+ m

+ m

_BB

v

v

_BB

= m

= m

_AA

u

u

_AA

+ m

+ m

0 _BB

u

u

_BB0

m

_A

v

_A

= - m

_B

v

_B

v

_A

= -

m

_B

v

_B

m

_A

A _B

2 kg

1 kg _A

B

8 m/s

v

_A2

v

_A

= -

(1 kg)(8 m/s)

Example 2 (Cont.):

Example 2 (Cont.):

Ignoring friction, how

Ignoring friction, how

much energy was released by the spring?

much energy was released by the spring?

A _B

2 kg

1 kg _A

B

8 m/s 4 m/s

Cons. of E:

½

_½

kx

kx

22

=

₌

½

_½

m

_m

A

A

v

v

AA22

+ ½

+ ½

m

m

BB

v

v

BB22

½

½

kx

kx

2 2

= ½

= ½

(2 kg)(4 m/s)_{(2 kg)(4 m/s)}22 + ½(1 kg)(8 m/s) + ½(1 kg)(8 m/s)22

½

Elastic or Inelastic?

Elastic or Inelastic?

An elastic collision loses no energy. The deform-ation on collision is fully restored.

Completely Inelastic Collisions

Completely Inelastic Collisions

Collisions where two objects stick together and have a common velocity after impact.

Collisions where two objects stick together and have a common velocity after impact.

Example 3:

Example 3:

A

A

60-kg

60-kg

football player stands

football player stands

on a frictionless lake of ice. He catches a

on a frictionless lake of ice. He catches a

2-kg

2-kg

football and then moves at

football and then moves at

40 cm/s

40 cm/s

.

.

What was the initial velocity of the football?

What was the initial velocity of the football?

Given:

u

_B

= 0

; m

_A

=

2 kg

; m

_B

=

60 kg

;

v

_A

= v

_B

= v

_C

v

_C

=

0.4 m/s

A

A

B

B

m

m

_AA

v

v

_AA

+ m

+ m

_BB

v

v

_BB

= m

= m

_AA

u

u

_AA

+ m

+ m

_BB

u

u

_BB

Momentum: 0

(m

(m

_AA

+ m

+ m

_BB

)v

)v

_CC

= m

= m

_AA

u

u

_AA

(2 kg + 60 kg)(0.4 m/s) = (2 kg)

u

_A

Inelastic collision:

u

u

_AA= 12.4 m/s = 12.4 m/s

u

Example 3 (Cont.):

Example 3 (Cont.):

How much energy

How much energy

was lost in catching the football?

was lost in catching the football?

0

½(2 kg)(12.4 m/s)2 = ½(62 kg)(0.4 m/s)2 + Loss

154 J = 4.96 J + Loss Loss = 149 J_{Loss = 149 JLoss = 149 JLoss = 149 J}

97% of the energy is lost in the collision!!

2 2 2

1 1 1

General: Completely Inelastic

General: Completely Inelastic

Collisions where two objects stick together and have a common velocity v_C after impact.

Conservation of Momentum:

Conservation of Momentum:

Conservation of Energy:

Conservation of Energy:

(

m

_A



m v

_B

)

_c



m u

_{A A}



m u

_{B B}

(

m

_A



m v

_B

)

_c



m u

_{A A}



m u

_{B B}

2 2 2

1 1 1

2

m u

A A



2

m u

B B



2

(

m

A



m v

B

)

c



Loss

2 2 2

1 1 1

Example 4.

Example 4. An An 87-kg87-kg skater skater BB collides with a collides with a 22-kg22-kg skater

skater AA initially at rest on ice. They move together initially at rest on ice. They move together after the collision at

after the collision at 2.4 m/s2.4 m/s. Find the velocity of the . Find the velocity of the skater

skater BB before the collision. before the collision.

A

A BB

u

u_BB = ?= ?

u

u_AA = 0= 0

Common speed after

Common speed after

colliding:

colliding: 2.4 m/s.2.4 m/s.

22 kg

22 kg

87 kg

87 kg

(

)

A A B B A B C

m u

A A



m u

B B



(

m

A



m v

B

)

C

m u



m u



m



m v

v

v

_BB

= v

= v

_{A A}

= v

= v

_CC = = 2.4 m/s2.4 m/s

(87

(87 kg)kg)_uuBB = (87 kg + 22 kg)(2.4 m/s) = (87 kg + 22 kg)(2.4 m/s)

(87 kg)

(87 kg)uu_BB =262 kg m/s =262 kg m/s

Example 5:

Example 5:

A

A

50 g

50 g

bullet strikes a

bullet strikes a

1-kg

1-kg

block, passes all the way through, then

block, passes all the way through, then

lodges into the

lodges into the

2 kg

2 kg

block. Afterward,

block. Afterward,

the 1 kg block moves at

the 1 kg block moves at

1 m/s

1 m/s

and the

and the

2

2

kg

kg

block moves at

block moves at

2 m/s

2 m/s

. What was the

. What was the

entrance velocity of the bullet?

entrance velocity of the bullet?

2 kg 1 kg

1 m/s _{2 m/s}

1 kg 2 kg

2 kg 1 kg

1 m/s _{2 m/s}

1 kg 2 kg Find entrance velocity of

bullet:

m

_A= 0.05 kg;

u

_A= ?

(0.05 kg)

u

_u

_AA =(1 kg)(1 m/s)+(2.05 kg2.05 kg)(2 m/s)

m

_A

u

_A

+ m

_B

u

_B

+ m

_C

u

_C

= m

_B

v

_B

+ (m

_A

+m

_C

) v

_AC

Momentum After = Momentum Before

50 gA

C B

0 0

(0.05 kg)

u

_u

_AA =(5.1 kg m/s)

u

_A= 102 m/s

Completely Elastic Collisions

Completely Elastic Collisions

Collisions where two objects collide in such a way that zero energy is lost in the process.

Velocity in Elastic Collisions

Velocity in Elastic Collisions

A B

A B

u

_B

u

_A

v

_A

v

_B

1. Zero energy lost.

2. Masses do not change. 3. Momentum conserved.

(Relative



v

After) = - (Relative



v

Before) Equal but opposite impulses (

F



t

) means that:

Example 6:

Example 6:

A

A

2-kg

2-kg

ball moving to the

ball moving to the

right at

right at

1 m/s

_{1 m/s}

strikes a

_{strikes a}

4-kg

_4-kg

ball moving

_{ball moving}

left at

left at

3 m/s

_{3 m/s}

. What are the velocities

_{. What are the velocities}

after impact, assuming complete

after impact, assuming complete

elasticity?

elasticity?

A B

A B

3 m/s 3 m/s 1 m/s

1 m/s

v

v_AA 1 kg1 kg 2 kg2 kg vv_BB

v

v

_AA

- v

_{- v}

_BB

= - (u

_{= - (u}

_AA

- u

_{- u}

_BB

)

₎

v

v

_{A A}

- v

_{- v}

_BB

= u

_{= u}

_BB

- u

_{- u}

_AA

v

v

_AA

- v

_{- v}

_BB

=

(-3 m/s)

-

(1 m/s)

From conservation of energy (relative v):

Example 6 (Continued)

Example 6 (Continued)

A B

A B

3 m/s 3 m/s 1 m/s

1 m/s

v

v_AA 1 kg1 kg 2 kg2 kg vv_BB

m

m

_AA

v

v

_AA

+ m

+ m

_BB

v

v

_BB

= m

= m

_AA

u

u

_AA

+ m

+ m

_BB

u

u

_BB

Energy:

Energy:

v

_v

_AA

- v

- v

_BB

=

=

- 4 m/s- 4 m/s

(1 kg)

v

_v

_AA+(2 kg)

v

_v

_BB=(1 kg)(1 m/s)+(2 kg)(-3 m/s)

v

v

_AA

+ 2v

+ 2v

_BB = -5 m/s

Momentum also conserved:

v

v

_AA

- v

- v

_BB

=

=

- 4 m/s- 4 m/s

Example 6 (Continued)

Example 6 (Continued)

A B

A B

3 m/s 3 m/s 1 m/s

1 m/s

v

v

_AA 1 kg1 kg 2 kg2 kg vv_BB

v

_A + 2

v

_B = -5 m/s

v

v

_AA

- v

- v

_BB

=

=

- 4 m/s- 4 m/s

Subtract: 0 + 3

v

v

_B2B2

= -

= -

1 m/s1 m/s

v

_B = - 0.333 m/s

v

_B = - 0.333 m/s

Substitution:

v

v

_AA

- v

- v

_BB

=

=

- 4 m/s- 4 m/s

v

v

_A2A2

-

-

(-0.333 m/s)(-0.333 m/s)

=

=

- 4 m/s- 4 m/s

v

_A= -3.67 m/s

Example 7.

Example 7. A A 0.150 kg0.150 kg bullet is fired at bullet is fired at 715 m/s715 m/s into a into a 2- 2-kg

kg wooden block at rest. The velocity of block afterward is wooden block at rest. The velocity of block afterward is 40 m/s

40 m/s. The bullet passes through the block and emerges . The bullet passes through the block and emerges with what velocity?

with what velocity?

A A B B A A B B

m v

A A



m v

B B



m u

A A



m u

B B

m v



m v



m u



m u

B

B

A

A

u

u

_{B B} = 0= 0

(0.150 kg)

(0.150 kg)

v

_v

_AA+ + (2 kg)(40 m/s) =_{(2 kg)(40 m/s) =} (0.150 kg)(715 m/s)(0.150 kg)(715 m/s)

0.150

0.150

v

_v

_AA+ + (80 m/s) =_{(80 m/s) =} (107 m/s)(107 m/s)

0.150

0.150

v

_v

_AA = = 27.2 m/s)27.2 m/s) vA  27.2 m/s_0.150

Example 8a:

Example 8a:

Inelastic collision: Find v

_{Inelastic collision: Find v}

_CC

.

_.

A

A BB

5 kg

5 kg 7.5 kg7.5 kg

u

u

_BB=0=0 2 m/s

2 m/s

A

A BB

Common

Common

v

v_CC after _after

v

v_CC

( )

A A B B A B C

m uA A  m uB B ( mA  m vB ) C

m u  m u  m  m v

After hit:

After hit:

v

_v

_BB

= v

= v

_AA

= v

= v

_CC

(5

(5 kg)(2 m/s) = (5 kg + 7.5 kg)kg)(2 m/s) = (5 kg + 7.5 kg)

v

v

_CC

12.5

12.5

v

_v

_CC =10 m/s =10 m/s

v

_C = 0.800 m/s

v

_C = 0.800 m/s

In an completely inelastic collision, the two balls stick together and move as one after colliding.

Example 8.

Example 8. (b) Elastic collision: Find (b) Elastic collision: Find vv_A2A2 and v and v_B2B2

A

A BB

5 kg

5 kg 7.5 kg7.5 kg v

v_B1B1=0=0 2 m/s

2 m/s

A A A A B B

m vA A  m vA A  m vB B

m v  m v  m v

Conservation of Momentum:

Conservation of Momentum:

(5

(5 kg)(2 m/s) = (5 kg)kg)(2 m/s) = (5 kg)vv_A2A2 + (7.5 kg) + (7.5 kg)

v

v

_BB

A

A BB

v

v_AA vvBB

5

v

_A + 7.5

v

_B = 10 m/s

( )

A B A B v  v   u  u

For Elastic Collisions:

For Elastic Collisions:

2 m/s

A B

v

A



v

B

 

2 m/s

v



v

 

Continued . . .

Example 8b (Cont).

Example 8b (Cont). Elastic collision: Find Elastic collision: Find

v

_v

_AA & &

v

_v

_BB

A

A BB 5 kg

5 kg 7.5 kg7.5 kg

v

v_BB=0=0 2 m/s

2 m/s

A

A BB

v

v_AA vvBB

Solve simultaneously:

Solve simultaneously:

5 v_A + 7.5 v _B = 10 m/s

2 m/s

A B

v

A



v

B

 

2 m/s

v



v

 

5

5 vv_AA + 7.5 + 7.5 vv_BB = 10 m/s= 10 m/s -5

-5 vv_AA + 5 + 5 vv_BB = +10 m/s= +10 m/s x (-5)

x (-5)

12.5

12.5 vv_BB = 20 m/s = 20 m/s

20 m/s

1.60 m/s 12.5

B

v  

v

v_AA - 1.60 m/s = -2 m/s- 1.60 m/s = -2 m/s

v

_A = -0.400 m/s

v

_A = -0.400 m/s

General: Completely Elastic

General: Completely Elastic

Collisions where zero energy is lost during a collision (an ideal case).

Conservation of Momentum:

Conservation of Momentum:

Conservation of Energy:

Conservation of Energy:

2 2 2 2

1 1 1 1

2 2 2 2

A A B B A A B B

A B B A

m u

m u

m v

m v

Loss

v

v

u

u















2 2 2 2

1 1 1 1

2 2 2 2

A A B B A A B B

A B B A

m u

m u

m v

m v

Loss

v

v

u

u















A A B B A A B B

m v

A A



m v

B B



m u

A A



m u

B B

Example 9:

Example 9: A A 50 g50 g bullet lodges into a bullet lodges into a 2-kg2-kg block of clay hung by a string. The bullet and

block of clay hung by a string. The bullet and

clay rise together to a height of

clay rise together to a height of 12 cm_{12 cm}. What _{. What}

was the velocity of the

was the velocity of the 50-g50-g mass just before mass just before entering?

entering?

u

u

_AA

B A

B

A 12 cm

Example (Continued):

Example (Continued):

B

A 12 cm 50 g

u_A

2.05 kg

2 kg Collision and Momentum:

m

_A

u

_A+0= (

m

_A

+m

_B)

v

_C

(0.05 kg)

u

_A = (2.05 kg)

v

_C

To find

v

_A we need

v

_C .

After collision, energyenergy is conserved for masses.

v

_C = 2

gh

v

_C = 2

gh

2 1

Example (Continued):

Example (Continued):

B

A 12 cm 50 g

u_A

2.05 kg

2 kg

m

_A

u

_A

+0= (m

_A

+m

_B

)v

_C

(0.05 kg)

u

_A = (2.05 kg)(1.53 m/s)

v

_C = 2gh = 2(9.8)(0.12)

After Collision:

v

_C = 1.53 m/s

u

_A = 62.9 m/s

u

_A = 62.9 m/s

Summary of Formulas:

Summary of Formulas:

Conservation of Momentum:

Conservation of Momentum:

Conservation of Energy:

Conservation of Energy:

2 2 2 2

1 1 1 1

2

m u

A A



2

m u

B B



2

m v

A A



2

m v

B B



Loss

2 2 2 2

1 1 1 1

2

m u

A A



2

m u

B B



2

m v

A A



2

m v

B B



Loss

A A B B A A B B

m v

A A



m v

B B



m u

A A



m u

B B

m v



m v



m u



m u

For elastic only:

For elastic only:

A B B A

v



v



u



u