July 3, Week 5 Today: Finish Chapter 9, Begin...

July 3, Week 5

Momentum 3rd July 2014

Today: Finish Chapter 9, Begin Chapter 10,Work

No Office Hours on Friday.

Homework Assignment #5 - Due Monday, July 7. (Homeworkassignment #6 will be due Friday, July 11)

Test #5 on Tuesday, July 8

There will be a reading quiz due Monday.

Using Conservation of Momentum II


Before

A

−→vAi

B

−→vBi

A

−→vAf

B

−→vBf

After

mA−→vAi +mB

−→vBi = mA

−→vAf +mB

−→vBf



Before

A

−→vAi

B

−→vBi

A

−→vAf

B

−→vBf

After

mA−→vAi +mB

−→vBi = mA

−→vAf +mB

−→vBf

Component Form:



Before

A

−→vAi

B

−→vBi

A

−→vAf

B

−→vBf

After

mA−→vAi +mB

−→vBi = mA

−→vAf +mB

−→vBf

Component Form:

mA (vAx)i +mB (vBx)i = mA (vAx)f +mB (vBx)f



Before

A

−→vAi

B

−→vBi

A

−→vAf

B

−→vBf

After

mA−→vAi +mB

−→vBi = mA

−→vAf +mB

−→vBf

Component Form:

mA (vAx)i +mB (vBx)i = mA (vAx)f +mB (vBx)fmA (vAy)i +mB (vBy)i = mA (vAy)f +mB (vBy)f

Completely-Inelastic Collisions


When the colliding objects stick together, the collision is calledcompletely inelastic.




Before

A

−→vAi

B

−→vBi




Before

A

−→vAi

B

−→vBi

mA−→vAi +mB

−→vBi




Before

A

−→vAi

B

−→vBi

After

mA−→vAi +mB

−→vBi




Before

A

−→vAi

B

−→vBi

B

A

After

mA−→vAi +mB

−→vBi




Before

A

−→vAi

B

−→vBi

B

A−→vf

After

mA−→vAi +mB

−→vBi




Before

A

−→vAi

B

−→vBi

B

A−→vf

After

mA−→vAi +mB

−→vBi (mA +mB)

−→vf




Before

A

−→vAi

B

−→vBi

B

A−→vf

After

mA−→vAi +mB

−→vBi = (mA +mB)

−→vf




Before

A

−→vAi

B

−→vBi

B

A−→vf

After

mA−→vAi +mB

−→vBi = (mA +mB)

−→vf

Component Form: mA (vAx)i +mB (vBx)i = (mA +mB) (vx)f

mA (vAy)i +mB (vBy)i = (mA +mB) (vy)f

Conservation Exercise IV


A 1-kg mass sliding to the right with speed 1m/s on a frictionlessfloor collides with a 2-kg mass going to the left at 2m/s. If themasses stick to each other, how fast is the combo going after?

BEFORE




BEFORE AFTER

?




BEFORE AFTER

?

(a) (5/3) m/s




BEFORE AFTER

?

(a) (5/3) m/s (b) − (5/3) m/s




BEFORE AFTER

?

(a) (5/3) m/s (b) − (5/3) m/s (c) −1m/s




BEFORE AFTER

?

(a) (5/3) m/s (b) − (5/3) m/s (c) −1m/s

(d) 3m/s




BEFORE AFTER

?

(a) (5/3) m/s (b) − (5/3) m/s (c) −1m/s

(d) 3m/s (e) −3m/s




BEFORE AFTER

?

(c) −1m/s

Conservation: (1 kg)(1m/s) + (2 kg)(−2m/s) = (1 kg + 2 kg)(vx)f(1 kg ·m/s)− (4 kg ·m/s) = (3 kg)(vx)f ⇒ −(3 kg ·m/s) = (3 kg)(vx)f

2D-Conservation Exercise


A 6 kg box-shaped firecracker explodes into two unequal pieces. Ifthe first piece of mass 2 kg has velocity 20m/s at 45◦, what speedand direction must the other piece have?

BEFORE

6 kg




BEFORE

6 kg

AFTER

2 kg

45◦

20m/s




BEFORE

6 kg

AFTER

2 kg

45◦

20m/s

(a) 10m/s at 225◦




BEFORE

6 kg

AFTER

2 kg

45◦

20m/s

(a) 10m/s at 225◦ (b) 20m/s at 225◦




BEFORE

6 kg

AFTER

2 kg

45◦

20m/s

(a) 10m/s at 225◦ (b) 20m/s at 225◦ (c) 40m/s at 225◦




BEFORE

6 kg

AFTER

2 kg

45◦

20m/s


(d) 10m/s at 135◦




BEFORE

6 kg

AFTER

2 kg

45◦

20m/s


(d) 10m/s at 135◦ (e) 40m/s at 135◦




BEFORE

6 kg

AFTER

2 kg

45◦

20m/s

(a) 10m/s at 225◦

0 = mA−→vAf +mB

−→vBf ⇒

−→vBf = −

(

mA

mB

)

−→vAf = −

(

2

4

)

−→vAf

2D Exercise II


A block with MA = 1 kg and velocity 3m/s to the right has aperfectly inelastic collision with MB = 2 kg that has velocity 3m/sup. How fast must the masses be going the instant after theircollision?

1 kg3m/s

2 kg

3m/s

2D Exercise II



1 kg3m/s

2 kg

3m/s

(a) 9m/s

2D Exercise II



1 kg3m/s

2 kg

3m/s

(a) 9m/s

(b)√45m/s = 6.7m/s

2D Exercise II



1 kg3m/s

2 kg

3m/s

(a) 9m/s

(b)√45m/s = 6.7m/s

(c) 3m/s

2D Exercise II



1 kg3m/s

2 kg

3m/s

(a) 9m/s

(b)√45m/s = 6.7m/s

(c) 3m/s

(d)√5m/s = 2.236m/s

2D Exercise II



1 kg3m/s

2 kg

3m/s

(a) 9m/s

(b)√45m/s = 6.7m/s

(c) 3m/s

(d)√5m/s = 2.236m/s

(e) 1m/s

2D Exercise II



1 kg3m/s

2 kg

3m/s

(d)√5m/s = 2.236m/s

2D Exercise II



1 kg3m/s

2 kg

3m/s

(d)√5m/s = 2.236m/s

(vAx)i = 3m/s (vAy)i = 0

2D Exercise II



1 kg3m/s

2 kg

3m/s

(d)√5m/s = 2.236m/s


(vBx)i = 0 (vBy)i = 3m/s

2D Exercise II



1 kg3m/s

2 kg

3m/s

(d)√5m/s = 2.236m/s



x-Component: (1 kg)(3m/s) + 0 = (3 kg) (vx)f ⇒ (vx)f = 1m/s

2D Exercise II



1 kg3m/s

2 kg

3m/s

(d)√5m/s = 2.236m/s




y-Component: 0 + (2 kg)(3m/s) = (3 kg) (vy)f ⇒ (vy)f = 2m/s

2D Exercise II



1 kg3m/s

2 kg

3m/s

(d)√5m/s = 2.236m/s




y-Component: 0 + (2 kg)(3m/s) = (3 kg) (vy)f ⇒ (vy)f = 2m/s

vf =√

(vx)2

f + (vy)2

f=

√

1m2/s2 + 4m2/s2

Work and Energy


Energy - The ability to do work or results from work being done tosomething.

Work and Energy



Work - a measure of much effort goes into causing motion.

Work and Energy




−→F

Work and Energy




−→F

−→d

−→d = displacement= distance anddirection traveled.

Work and Energy




−→F

−→d


Work done by the force: W = Fd

Work and Energy




−→F

−→d


Work done by the force: W = Fd

Unit: N ·m = kg ·m2/s2 = J

Joule

Restrictions


−→F

−→d

Work, W = Fd

This equation is correct only in the situation that:

Restrictions


−→F

−→d

Work, W = Fd

This equation is correct only in the situation that:−→F is constant

Restrictions


−→F

−→d

Work, W = Fd

This equation is correct only in the situation that:−→F is constant−→d is a straight line

Restrictions


−→F

−→d

Work, W = Fd

This equation is correct only in the situation that:−→F is constant−→d is a straight line−→F and

−→d are in the same direction.

Work Exercise I


A 10N block is pulled 0.5m upwards with constant speed by amassless rope. How much work is done by the tension force?

10N

0.5m

Work Exercise I



10N

0.5m

(a) 0 J

Work Exercise I



10N

0.5m

(a) 0 J

(b) 0.5 J

Work Exercise I



10N

0.5m

(a) 0 J

(b) 0.5 J

(c) 5 J

Work Exercise I



10N

0.5m

(a) 0 J

(b) 0.5 J

(c) 5 J

(d) 10 J

Work Exercise I



10N

0.5m

(a) 0 J

(b) 0.5 J

(c) 5 J

(d) 10 J

(e) Not enough information to determine

Work Exercise I



10N

0.5m

−→T

−→w = 10N

(c) 5 J

∑

Fy = may ⇒

T − 10N = 0 ⇒ T = 10N

W = Td = (10N)(0.5m)

Work Exercise I



10N

0.5m

−→T

−→w = 10N

(c) 5 J

∑

Fy = may ⇒

T − 10N = 0 ⇒ T = 10N

W = Td = (10N)(0.5m)

Note: W = work and w = weight

Perpendicular Force


A force perpendicular to the displacement does no work.

Perpendicular Force



−→F

Perpendicular Force



−→F

−→d

Perpendicular Force



−→F

−→d

This force cannot be responsiblefor this motion ⇒ W = 0

Arbitrary Direction


Only the component of the force parallel to the displacement doeswork.

Arbitrary Direction



−→F

Arbitrary Direction



−→F

−→d

Arbitrary Direction



−→F

−→dφ

φ = angle between−→F and

−→d

Arbitrary Direction



−→F

−→dφ


−→d

F‖

Arbitrary Direction



−→F

−→dφ


−→d

F‖

W = F‖d = Fd cosφ

Arbitrary Direction



−→F

−→dφ


−→d

F‖

W = F‖d = Fd cosφF⊥

Arbitrary Direction



−→F

−→dφ


−→d

F‖

W = F‖d = Fd cosφF⊥ −→

F⊥ does no work

Arbitrary Direction



−→F

−→dφ


−→d

F‖


F⊥ does no work

W = Fd cosφ

Arbitrary Direction



−→F

−→dφ


−→d

F‖


F⊥ does no work

W = Fd cosφ

Only correct for Constant force & Straight-line displacement

Work Exercise II


A 10N block is pulled 0.5m upwards with constant speed by amassless rope. How much work is done by gravity?

10N

0.5m

Work Exercise II



10N

0.5m

(a) 0 J

Work Exercise II



10N

0.5m

(a) 0 J

(b) 5 J

Work Exercise II



10N

0.5m

(a) 0 J

(b) 5 J

(c) 5 J cos 180◦ = −5 J

Work Exercise II



10N

0.5m

(a) 0 J

(b) 5 J

(c) 5 J cos 180◦ = −5 J

(d) 10 J cos 180◦ = −10 J

Work Exercise II



10N

0.5m

(a) 0 J

(b) 5 J

(c) 5 J cos 180◦ = −5 J

(d) 10 J cos 180◦ = −10 J

(e) Not enough information to determine

Work Exercise II



10N

0.5m

10N

180◦

(c) 5 J cos 180◦ = −5 J

W = Fd cosφ

Work Exercise II



10N

0.5m

10N

180◦

(c) 5 J cos 180◦ = −5 J

W = Fd cosφ

⇒ W = (10N)(0.5m) cos 180◦

Work Exercise II



10N

0.5m

10N

180◦

(c) 5 J cos 180◦ = −5 J

W = Fd cosφ

⇒ W = (10N)(0.5m) cos 180◦

Negative work slows an object down

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