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?

€

msec

€

msec

€

Δv = v f − vi

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

sec

Δv = −3ˆ i + 4 ˆ j ( ) msec

Δv = ˆ i 2 + ˆ j 2 = 5 msec

2

#1.2

€

FΔt = mΔv

F =mΔvΔt

F =2000kg( ) 5 m

sec( )4 sec

F = 2500N

3

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?

€

msec

€

msec

€

FΔt = mΔv

F =mΔvΔt

F =1.5kg( ) 6 m

sec − −4 msec( )( )

0.005secF = 3000N

4

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?

€

msec

€

msec

€

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

sec

€

Δv = v f − vi

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

sec

Δv = −0.6ˆ i + 4.96 ˆ j ( ) msec

Δv = ˆ i 2 + ˆ j 2 = 5 msec

5

€

FΔt = mΔv

F =mΔvΔt

F =1.5kg( ) 5 m

sec( )0.5sec

F =15N

#3.2

6

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( )

€

0€

8€

16

€

−8

€

2

€

3

€

4€

F(N)

€

1

€

msec

7

€

5

€

−8( ) + 12 1sec( ) −8N( )[ ] + 1

2 2sec( ) 16N( )[ ] +16[ ]N ⋅ sec

Δp = 20N ⋅ sec

9

€

t sec( )

€

0€

8€

16

€

−8

€

2

€

3

€

4€

F(N)

€

1

€

5

#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

sec

€

t sec( )

€

0€

8€

16

€

−8

€

2

€

3

€

4€

F(N)

€

1

€

5

5.

€

Δp = pf − pi

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

sec

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?

€

msec

€

msec

11

6.

€

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.

€

msec

€

msec

12

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?

13

€

vo

€

E Bottom = ETop

KB = UTop

12 mvB

2 = mgh

vB = 2gh

vB = 2g 0.12m( )′ v B =1.53 m

sec

#7.2

14€

pi = pf

mvo = m + M( ) ′ v

vo =m + M

m′ v

vo =3.012kg0.012kg

1.53 msec( )

vo = 384 msec

15

€

v

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?

€

msec

€

msec

€

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

sec

16

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

#8.2

17

#8.3 From conservation of Energy!

€

h =′ v B( )2

2g

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?

€

vo

€

78.4cm

€

t =2yg

= 0.4 sec

€

′ v x =xt

=1m

0.4sec= 2.5 m

sec

18

€

vo

€

t

€

h

€

u

€

d

€

′ v = 2.5 msec

€

pi = pf

mbvo = mb + M( ) ′ v B

vo =mb + M( ) ′ v B( )

mb

=3.012kg( ) 2.5 m

sec( )0.012kg

vo = 627.5 msec

19

20

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.

€

msec

€

msec

€

t

€

h

€

u

€

d

€

6kg

€

6kg

€

2kg

€

2kg

€

5.0 msec

€

−2 msec

€

vo

€

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

m2v2i = m1 + m2( ) ′ v −m1v1i

v2i =m1 + m2( ) ′ v − m1v1i

m2

v2i =8kg( ) −2 m

sec( ) − 6kg( ) 5 msec( )

2kgv2i = −23 m

sec

€

t

€

h

€

u

€

d

€

6kg

€

6kg

€

2kg

€

2kg

€

5.0 msec

€

−2 msec

€

vo

21

€

t

€

h

€

u

€

d

€

6kg

€

6kg

€

2kg

€

2kg

€

5.0 msec

€

−2 msec

€

vo

€

′ p = −16 Kg ⋅msec

€

′ p = −16 Kg ⋅msec

€

p = 30 Kg⋅msec

€

p = −46 Kg⋅msec

€

vo =pm

=−46 Kg ⋅m

sec

2kg= −23 m

sec

22

23

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.

€

msec

€

t

€

h

€

u

€

d

€

8kg

€

8kg

€

2kg

€

2kg

€

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

24

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?

€

t

€

h

€

u

€

d

€

′ v = 0

€

2m

€

1.6kg€

1.6kg€

vBi = 2ghvBi = 6.26 m

sec

€

vbi

25

#12.2

€

pi = pf

pBi + pbi = 0 pBi = −pbi

mBvBi = −mbvbi

mBvBi

−mb

= vbi = −1.6Kg( ) 6.26 m

sec( )0.010kg( )

vbi = −1002 msec

26

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?€

msec

€

msec

€

t

€

h

€

u

€

d

€

3kg

€

2kg

€

5.0 msec

€

2kg

€

3kg

€

−2 msec€

msec

€

2.2 msec

€

Ki = 12 m1v1

2 + 12 m1v2

2

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€

A

€

v

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?

€

t

€

h

€

u

€

d

€

4kg

€

4kg

€

2kg

€

2kg€

vB = 2gh

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

sec

€

2.8 msec

27

28

€

t

€

h

€

u

€

d

€

4kg

€

4kg

€

2kg

€

2kg

€

2.8 msec

€

p = mv

p =11.2 kg⋅msec

€

v =pm

v =11.2 kg ⋅m

sec

6kgv = 1.86 m

sec

29

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?

€

msec

€

t

€

h

€

u

€

d€

37°

€

70kg€

30kg€

30kg

€

70kg

€

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

sec

€

vboy =pm

vboy =160 Kg⋅m

sec

30kgvboy = 5.3 m

sec

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?

30

€

msec

€

msec

€

b

€

o

€

o

€

m

€

4 msec

€

4 kg

€

2kg

€

60°

€

3 msec

€

2kg

31

€

b

€

o

€

o

€

m

€

4 msec

€

4 kg

€

2kg

€

60°

€

3 msec

€

H

€

16ˆ i Kg⋅msec

€

3ˆ i Kg⋅msec

€

13ˆ i Kg⋅msec

€

V

€

0

€

5.2 ˆ j Kg ⋅msec

€

−5.2ˆ j Kg ⋅msec

€

v =pm

v =14 Kg⋅m

sec

2kgv = 7 m

sec

€

−21.8°

€

2kg

€

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

€

60°

€

θ = tan−1 ji

θ = tan−1 −5.2 j13i

θ = −21.8°

32

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?

€

msec

€

b

€

o

€

o

€

m€

5 msec

€

4.2kg

€

1.4kg

€

90°

€

1.4kg

€

5 msec

€

1.4kg

€

0

€

0

€

H

€

V

€

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 =pm

v =9.9 Kg ⋅m

sec

1.4kgv = 7.07 m

sec

€

K = 12 mv 2 =17.5J

€

K = 12 mv 2 =17.5J

€

K = 12 mv 2 = 35J

€

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

33

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?

€

msec

€

msec

€

b

€

o

€

o

€

m€

5 msec

€

3 kg

€

1 kg

€

90°

€

4 msec

€

1 kg

€

−30°

€

0

€

0

€

H

€

V€

1 kg

€

5 ˆ j Kg⋅msec

€

−2ˆ j Kg⋅msec

€

3.46ˆ i Kg ⋅msec

€

b

€

o

€

o

€

m

€

−3.46ˆ i Kg⋅msec

€

−3ˆ j Kg⋅msec

31

32

€

−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 =pm

v =4.58 Kg⋅m

sec

1.0kgv = 4.58 m

sec , at 221°

36

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?

€

40g

€

50g

€

xCofM =mx∑m∑

0 =40 3( ) + 50 −2( ) + 20 x( )

110g

x =100−120

20= −1m

€

yCofM =my∑m∑

0 =40 4( ) + 50 −6( ) + 20 y( )

110g

y =300 −160

20= 7m

€

20g ⇒ −1, 7( )m

€

20g

37

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? €

msec

€

msec

€

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

38