AP Physics - cf.edliostatic.comcf.edliostatic.com/tC5dEpzDtCo0vxcsQRI60AvEetZ1sEht.pdf3. A 1.5-kg...

AP Physics

Chapter 9 Review

Momentumet. al.

1. A 2000-kg truck traveling at a speed of 3.0 makes a 90¯ turn in a time of 4.0 seconds and emerges from this turn with a speed of 4.0 . What is the magnitude of the average resultant force on the truck during this turn?

Δv = v f − vi

Δv = 0ˆ i + 4 ˆ j ( ) msec − 3ˆ i + 0 ˆ j ( ) m

Δv = −3ˆ i + 4 ˆ j ( ) msec

Δv = ˆ i 2 + ˆ j 2 = 5 msec

FΔt = mΔv

F =mΔvΔt

F =2000kg( ) 5 m

sec( )4 sec

F = 2500N

2. A 1.5-kg object moving with a speed of 4.0 collides perpendicularly with a wall and emerges with a speed of 6.0 in the opposite direction. If the object is in contact with the wall for 5.0 msec, what is the magnitude of the average force on the object by the wall?

FΔt = mΔv

F =mΔvΔt

F =1.5kg( ) 6 m

sec − −4 msec( )( )

0.005secF = 3000N

3. A 1.5-kg playground ball is moving with a velocity of 3.0 directed 30¯ below the horizontal just before it strikes a horizontal surface. The ball leaves this surface 0.50 s later with a velocity of 4.0 directed 60¯ above the horizontal. What is the magnitude of the average resultant force on the ball?

vi = 2.6ˆ i −1.5 ˆ j ( ) msec v f = 2ˆ i + 3.46 ˆ j ( ) m

Δv = v f − vi

Δv = 2ˆ i + 3.46 ˆ j ( ) msec − 2.6ˆ i −1.5 ˆ j ( ) m

Δv = −0.6ˆ i + 4.96 ˆ j ( ) msec

Δv = ˆ i 2 + ˆ j 2 = 5 msec

FΔt = mΔv

F =mΔvΔt

F =1.5kg( ) 5 m

sec( )0.5sec

F =15N

4. The only force acting on a 2.0-kg object moving along the x axis is shown. If the velocity vx is +2.0 at t = 0, what is the velocity at t = 4.0 sec?

t sec( )

−8( ) + 12 1sec( ) −8N( )[ ] + 1

2 2sec( ) 16N( )[ ] +16[ ]N ⋅ sec

Δp = 20N ⋅ sec

t sec( )

#4.410

Δp = m v f − vi( )Δp = mv f −mvi

Δp + mvi = mv f

Δp + mvi

m= v f

v f =20 Kg⋅m

sec + 2kg( ) 2 msec( )

2kg v f = 12 m

t sec( )

Δp = pf − pi

Δp = 3Kg −4i + 3 j( ) msec − 3Kg 4i + 3 j( ) m

Δp = −12i + 9 j −12i − 9 jΔp = −24i N ⋅ sec

A 3.0-kg ball with an initial velocity of (4i + 3j) collides with a wall and rebounds with a velocity of (-4i + 3j) . What is the impulse exerted on the ball by the wall?

Δp = pf − pi

Δp = 0.15Kg 80( ) msec − 0.15Kg( ) −40 m

sec( )Δp = 18 N ⋅ sec

A 0.15-kg baseball is thrown with a speed of 40 . It is hit straight back at the pitcher with a speed of 80 . What is the magnitude of the impulse exerted on the ball by the bat.

7. A 12-g bullet is fired into a 3.0-kg ballistic pendulum initially at rest and becomes embedded in it. The pendulum subsequently rises a vertical distance of 12 cm. What was the initial speed of the bullet?

E Bottom = ETop

KB = UTop

12 mvB

2 = mgh

vB = 2gh

vB = 2g 0.12m( )′ v B =1.53 m

pi = pf

mvo = m + M( ) ′ v

vo =m + M

m′ v

vo =3.012kg0.012kg

1.53 msec( )

vo = 384 msec

8. A 10-g bullet moving 1000 strikes and passes through a 2.0-kg block initially at rest, as shown. The bullet emerges from the block with a speed of 400 . To what maximum height will the block rise above its initial position?

pi = pf

pb = ′ p b + ′ p B

pi = pf

pb = ′ p b + ′ p Bmbvo = mb ′ v b + M ′ v Bmb vo − ′ v b( )

M= ′ v B =

0.01kg( ) 600 msec( )

2kg′ v B = 3 m

This is the velocity of the Block, by itself, after the collision!

#8.3 From conservation of Energy!

h =′ v B( )2

h =3 m

sec( )2

2gh = 0.46m

9. A 12-g bullet moving horizontally strikes and remains in a 3.0-kg block initially at rest on the edge of a table. The block, which is initially 78.4 cm above the floor, strikes the floor a horizontal distance of 100 cm from its initial position. What was the initial speed of the bullet?

78.4cm

t =2yg

= 0.4 sec

′ v x =xt

0.4sec= 2.5 m

′ v = 2.5 msec

pi = pf

mbvo = mb + M( ) ′ v B

vo =mb + M( ) ′ v B( )

=3.012kg( ) 2.5 m

sec( )0.012kg

vo = 627.5 msec

10. A 6.0-kg object moving 5.0 collides with and sticks to a 2.0-kg object. After the collision the composite object is moving 2.0 in a direction opposite to the initial direction of motion of the 6.0-kg object. Determine the speed of the 2.0-kg object before the collision.

5.0 msec

−2 msec

p1i + p2i = ′ p m1v1i + m2v2 i = m1 + m2( ) ′ v

m2v2i = m1 + m2( ) ′ v −m1v1i

v2i =m1 + m2( ) ′ v − m1v1i

v2i =8kg( ) −2 m

sec( ) − 6kg( ) 5 msec( )

2kgv2i = −23 m

5.0 msec

−2 msec

5.0 msec

−2 msec

′ p = −16 Kg ⋅msec

p = 30 Kg⋅msec

p = −46 Kg⋅msec

vo =pm

=−46 Kg ⋅m

2kg= −23 m

11. A 2.0-kg object moving 5.0 collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy lost by the system as a result of this collision.

5.0 msec

1 msec

p =10 kg⋅msec

′ p =10 kg ⋅msec

Ki = 12 2kg( ) 5 m

sec( )2= 25J

K f = 12 10kg( ) 1 m

sec( )2= 5J

KLost = Ki −K f = 20J

12. A 1.6-kg block is attached to the end of a 2.0-m string to form a pendulum. The pendulum is released from rest when the string is horizontal. At the lowest point of its swing when it is moving horizontally, the block is hit by a 10-g bullet moving horizontally in the opposite direction. The bullet remains in the block and causes the block to come to rest at the low point of its swing. What was the magnitude of the bullet's velocity just before hitting the block?

′ v = 0

1.6kg€

vBi = 2ghvBi = 6.26 m

pi = pf

pBi + pbi = 0 pBi = −pbi

mBvBi = −mbvbi

= vbi = −1.6Kg( ) 6.26 m

sec( )0.010kg( )

vbi = −1002 msec

13. A 3.0-kg mass sliding on a frictionless surface has a velocity of 5.0 east when it undergoes a one-dimensional inelastic collision with a 2.0-kg mass that has an initial velocity of 2.0 west. After the collision the 3.0-kg mass has a velocity of 2.2 east. How much kinetic energy does the two-mass system lose during the collision?€

5.0 msec

−2 msec€

2.2 msec

Ki = 12 m1v1

2 + 12 m1v2

Ki = 12 3kg( ) 5 m

sec( )2+ 1

2 2kg( ) 2 msec( )2

Ki = 37.5J + 4J = 41.5J

K f = 12 mv 2

K f = 12 5Kg( ) 2.2 m

sec( )2

K f = 12.1J

KLost = Ki −K f = 29.4J

40 cm€

14. A 4.0-kg mass is released from rest at point A of a circular frictionless track of radius 0.40m as shown in the figure. The mass slides down the track and collides with a 2-kg mass that is initially at rest on a horizontal frictionless surface. If the masses stick together, what is their speed after the collision?

2kg€

vB = 2gh

vB = 2g 0.40m( )vB = 2.8 m

2.8 msec

p = mv

p =11.2 kg⋅msec

v =11.2 kg ⋅m

6kgv = 1.86 m

15. A 70kg man who is ice skating north collides with a 30kg boy who is ice skating west. Immediately after the collision, the man and boy are observed to be moving together with a velocity of 2.0 , in a direction 37¯ north of west. What was the magnitude of the boy's velocity before the collision?

70kg€

30kg€

′ p = 200 kg ⋅msec , at 143°( ) = −160ˆ i + 120ˆ j ( ) kg ⋅m

vboy =pm

vboy =160 Kg⋅m

30kgvboy = 5.3 m

16. A 4.0-kg mass has a velocity of 4.0 , east when it explodes into two 2.0-kg masses. After the explosion one of the masses has a velocity of 3.0 at an angle of 60¯ north of east. What is the magnitude of the velocity of the other mass after the explosion?

4 msec

3 msec

4 msec

3 msec

16ˆ i Kg⋅msec

3ˆ i Kg⋅msec

13ˆ i Kg⋅msec

5.2 ˆ j Kg ⋅msec

−5.2ˆ j Kg ⋅msec

v =14 Kg⋅m

2kgv = 7 m

−21.8°

p = ˆ i 2 + ˆ j 2

p = 132 + 5.22

p =14 Kg⋅msec

6 Kg ⋅msec

3 Kg⋅msec

5.2 Kg ⋅msec

θ = tan−1 ji

θ = tan−1 −5.2 j13i

θ = −21.8°

17. A 4.2-kg object, initially at rest, "explodes" into three objects of equal mass. Two of these are determined to have velocities of equal magnitudes (5.0 ) with directions that differ by 90¯. How much kinetic energy was released in the explosion?

5 msec

7 ˆ j kg⋅msec

7ˆ i kg⋅msec

−7ˆ i kg⋅msec

−7ˆ j kg⋅msec

p = ˆ i 2 + ˆ j 2

p = 72 + 72

p = 9.9 Kg⋅msec

v =9.9 Kg ⋅m

1.4kgv = 7.07 m

K = 12 mv 2 =17.5J

K = 12 mv 2 = 35J

KNet = K∑KNet =17.5J +17.5J + 35JKNet = 70J

18. A 3.0-kg mass, initially at rest on a frictionless surface, explodes into three 1.0-kg masses. After the explosion the velocities of two of the 1.0-kg masses are: (1) 5.0 , north and (2) 4.0 , 30¯ south of east. What is the magnitude of the velocity of the third 1.0-kg mass after the explosion?

5 msec

4 msec

−30°

5 ˆ j Kg⋅msec

−2ˆ j Kg⋅msec

3.46ˆ i Kg ⋅msec

−3.46ˆ i Kg⋅msec

−3ˆ j Kg⋅msec

−3.46ˆ i Kg⋅msec

−3ˆ j Kg⋅msec

p = ˆ i 2 + ˆ j 2

p = 3.462 + 32

p = 4.58 Kg ⋅msec

v =4.58 Kg⋅m

1.0kgv = 4.58 m

sec , at 221°

19. Three particles are placed in the xy plane. A 40-g particle is located at (3, 4)m, and a 50-g particle is positioned at (-2, -6)m. Where must a 20-g particle be placed so that the center of mass of this three-particle system is located at the origin?

xCofM =mx∑m∑

0 =40 3( ) + 50 −2( ) + 20 x( )

x =100−120

20= −1m

yCofM =my∑m∑

0 =40 4( ) + 50 −6( ) + 20 y( )

y =300 −160

20= 7m

20g ⇒ −1, 7( )m

20. At the instant a 2.0-kg particle has a velocity of 4.0 in the positive x direction, a 3.0-kg particle has a velocity of 5.0 in the positive y direction. What is the speed of the center of mass of the two-particle system? €

vxCofM =mvx∑m∑

=2kg 4 m

sec( )5kg

vx =1.6ˆ i msec

vyCofM=

mvx∑m∑

=3kg 5 m

sec( )5kg

vy = 3ˆ j msec

v = ˆ i 2 + ˆ j 2

v = 1.62 + 32

vCofG = 3.4 msec

AP Physics - cf.edliostatic.comcf.edliostatic.com/tC5dEpzDtCo0vxcsQRI60AvEetZ1sEht.pdf3. A 1.5-kg...

Documents

Transcript of AP Physics - cf.edliostatic.comcf.edliostatic.com/tC5dEpzDtCo0vxcsQRI60AvEetZ1sEht.pdf3. A 1.5-kg...

Playground Fun

Discovery Playground

Bucks playground

Kids´ Playground

Kids Playground

'Playground' evaluation

Baseband Playground

Primary Playground. You may also like · 2020. 2. 11. · Primary Playground. cprimary playground The Solar System Earth Jupiter Mars Mercury. cprimary playground Neptune Uranus Saturn

Playground( individual)

Primary Playground. You may also like · 2020. 5. 29. · Primary Playground. cprimary playground. cprimary playground. cprimary playground. SPACE COLOR BY CODE WC WORDS coor CVC

Things Every Playground Owner Should Know About … the Accessible Playground: 7 Things Every Playground Owner Should Know About the Accessibility of Their Playground Surfaces From

cf.edliostatic.comcf.edliostatic.com/m6576r52x2Z2OCzGBJBIjbW8zMyVF2sK.docx · Web viewcf.edliostatic.com

Playground Player

Elias playground!!!!!

Eda Playground

cf.edliostatic.comcf.edliostatic.com/gTQZGrQPdnJAQ7PHoCeuiPAQTpTNiveB.docx · Web viewThis course will include intensive study of representative works from various genres and periods,

MES- - flashcards: playground · PDF fileMES- Flash Cards the playground . MES- Flash Cards the playground . MES-English

Playground powerpoint

Civic Playground

Making Butter - Primary Playground - Primary Playground