Chapter 6 Solutions

C.W. Fay

August 23, 2011

College Physics 7th Edition By Wilson, Buffa and Lou. Numbers as followsedition.chapter.number, MC mulitple choice, CQ concept question.

7.6.5, 6.6.12

A 0.150kg baseball traveling with a horizontal speed of 4.50m/s is hit by a bat andthen moves with a speed of 34.7m/s in the opposite direction. What is the changein the ball’s momentum?Given: m = 0.150kg v1 = 4.50m/s v2 = −34.7m/s

∆p = m2v2 −m1v1 = (0.150kg)(−34.7m/s)− (0.150kg)(4.50m/s) = −5.88kgm/s(1)

7.6.6

A 15.0g rubber bullet hits a wall with a speed of 150m/s, if the bullet bounces straightback with a speed of 120m/s, what is the change in momentum of the bullet?Given:: m = 15.0g = 0.0150kg vi = 150m/s vf = −120m/s

∆p = mvf −mvi = (0.0150kg)(−120m/s)− (0.0150kg)(150m/s) = 4.50m/s (2)

7.6.10, 6.6.16

Two runners of mass 70kg and 60kg, respectively, have a total linear momentumof 350kgm/s. The heavier runner is running at 2.0m/s. Determine the possiblevelocities of the lighter runner.

1

Given:m1 = 70kg m2 = 60kg v1 = 2.0m/spTOT = 350kgm/s v2 =?

pTOT = p1 + p2 (3)

p2 = PTOT − p1 (4)

v2 =PTOT −m1v1

m2

= 3.5m/s (5)

or pTOT = p2 − p1 (6)

p2 = PTOT + p1 (7)

v2 =PTOT +m1v1

m2

= 8.2m/s opposite of v1 (8)

7.6.13

A loaded tractor-trailer with a total mass of 5000kg traveling at 3.0km/h hits aloading dock and comes to a stop in 0.64s. What is the magnitude of the averageforce exerted on the truck by the dock?Given:: m = 5000kg vi = 3.0km/h = 0.833m/s vf = 0m/s t = 0.64s

Fave =∆p

∆t=mvf −mvi

∆t=mvi∆t

(9)

=(5000kg)(0.833m/s)

0.64s= 6.5× 103N (10)

7.6.14, 6.6.21

A 2.0kg mud ball drops from rest at a height of 15m. If the impact between the balland the ground lasts 0.50s, what is the average net force exerted by the ball on theground?

Given:m = 2.0kg h = 15m t = 0.50s vf = 0F =?

F∆t = ∆p (11)

(12)

use the conservation of energy to find v,

1

2mv2 = mgh (13)

v =√

2gh (14)

2

Now place it with the impulse equation,

F∆t = mvf −mvi (15)

F =−mvi

∆t= −m

√2gh

∆t= −69N (16)

7.6.20, 6.6.33

An automobile with a linear momentum of 3.0 × 104kgm/s is brought to a stop in5.0s.Given: pi = 3.0× 104kgm/s pf = 0kgm/s ∆t = 5.0s F =?

Impulse = F∆t = ∆p (17)

F =∆p

∆t=

0kgm/s− 3.0× 104kgm/s

5.0s= −6.0× 103N (18)

7.6.21

A pool player imarts an impulse oof 3.2Ns to a stationary 0.25kg cue ball with a cuestick. What is the speed of the ball just after impact?Given:: I = 3.2Ns m = 0.25kg v0 = 0m/s vf =?

I = ∆p = mvf −mvi (19)

vf =I

m=

3.2Ns

kg= 12.8m/s (20)

7.6.26, 6.6.39

A volleyball is traveling toward you. (a) Which action will require a greater forceon the volleyball, your catching the ball or your hitting the ball back? why? (b)A 0.45kg volley ball travels with a horizontal velocity 0f 4.0m/s over the net. Youjump up and hit the ball back with a horizontal velocity of 7.0m/s. If the contacttime is 0.040s, what was the average force on the ball?

Given:m = 0.45kg vi = 4.0m/s vf = −7.0m/s ∆t = 0.040sF =?

a) hitting it backb)

F∆t = ∆p (21)

3

F∆t = mvf −mvi (22)

F = mvf − vi

∆t= (0.45kg)

(4.0m/s)− (−7.0m/s)

0.040s= 1.2× 102N (23)

—–, 6.6.40

A 1.0kg ball is thrown horizontally with a velocity of 15m/s against a wall. If theball rebounds horizontally with a velocity of 13m/s and the contact time is 0.020s,what force is exerted on the ball by the wall?

Given:m = 1.0kg vi = 15m/s vf = −13m/s∆t = 0.020s F =?

∆p = m∆v (24)

F =∆p

∆t=m∆v

∆t=

(1.0)(15− (−13))

0.020(25)

= 1.4× 103Npointingoutofthewall (26)

7.6.29, 6.6.43

A 0.45kg piece of putty is dropped from a height of 2.5m above a flat surface. Whenit hits the surface, the putty comes to rest in 0.30s. What is the average force exertedon the putty by the surface?Given: m = 0.45kg h = 2.5m ∆t = 0.30s F =?

1

2mv2 = mgh (27)

v =√

2gh (28)

F =mv

∆t=m√

2gh

∆t= 10.5N up (29)

7.6.31

An incoming 0.14kg baseball has a speed of 45m/s. The batter hits the ball, givingit a speed of 60m/s. If the contact time is 0.040s, what is the average force of thebat on the ball?Given: m = 0.14kg vi = 45m/s vf = −60m/s t = 0.040s

Fave =∆p

∆t=mvf −mvi

∆t(30)

4

=(0.14kg)(45m/s)− (0.14kg)(−60m/s)

0.040s= 370N (31)

7.6.33, 6.6.44

If the billiard ball in Fig 6.27 is in contact with the rail for 0.010s, what is themagnitude of the average force exerted on the ball?Given: t = 0.010s θ = 60◦ v = 15.0m/s m = 0.20kgSince the force from the bumper is in the y direction the only change in the momentis in the direction.

F =∆p

t=

∆pyt

(32)

=mv cos θ − (−mv cos θ)

t=

2mv cos θ

t(33)

= 3.0× 102N (34)

—-, 6.6.55

A 60kg astronaut floating at rest in space outside a space capsule throws his 0.50kghammer such that it moves with a speed of 10m/s relative to the capsule, whathappens to the astronaut? Given:: m1 = 60kg m2 = 0.50kg vH = 10m/susing the conservation of momentum,

pi = pf (35)

0 = pman − phammer (36)

pman = phammer (37)

m1vm = m2vH (38)

vm =m2vHm1

= 0.083m/s(west) (39)

7.6.38, 6.6.56

In pairs figure-skating competition, a 65kg man and his 45kg female partner standfacing each other on skates on the ice. If they push apart and the woman has avelocity of 1.5m/s eastward, what is the velocity of her partner? (neglect friction)

Given:m1 = 65kg m2 = 45kg v2 = 1.5m/sv1 = pi = 0 pf = 0

pi = pf = 0 = m1v1 −m2v2 (40)

5

v1 =m2v2m1

= 1.0m/s west (41)

7.6.39, 6.6.57

To get off a frozen, frictionless lake, a 65.0kg person takes off a 0.150kg shoe andthrows it horizontally, directly away from the shore with a speed of 2.00m/s. If theperson is 5.00m from the shore, how long doesh he take to reach it?

Given:mm = 60kg ms = 0.50kg vs = 10m/s ∆x = 5.00mt =?

using the conservation of momentum,

pi = pf (42)

0 = pman − pshoe (43)

pman = pshoe (44)

m1vm = msvs (45)

vm =msvsmm

(46)

(47)

since there is no acceleration in the x-direction (frictionless)Given:: ax = 0m/s2 v0 = vm

x = x0 + v0t+ 1/2axt2 (48)

∆x = v0t (49)

t =∆x

vm(50)

=mm∆x

msvs= 1.08× 103s (51)

7.6.41

Consider two sting-suspended balls, both with a mass of 0.15kg. One ball is pulledback in line with the other so it has a vertical height of 10cm, and then is released.(a) What is the speed of the ball just before hitting the stationary one? (b) If thecollision is completely inelastic to what height do the balls swing?Given: m = 0.15kg hi = 0.10mThe problem can be broken into 3 parts. Part I the fall of the ball to just before thecollision, Part II the collision, and Part III the two balls rising to a height of hf . Part

6

I, III can be analyzied by the conservation of energy. Part II by the conservation ofmomentum.Conservation of Energy,

∆K + ∆U = 0 (52)

∆K = −∆U (53)1

2mv2f −

1

2mv2i = mghi −mghf (54)

v2f − v2i = 2mg(h1 − h2) (55)

(56)

Conservation of momentum,

∆p = 0 (57)

pf − pi = 0 (58)

pi = pf (59)

mv = (m+M)V (60)

Now we apply what we know. v the initial velocity from the conservation of momen-tum is the final velocity for the fall of the first ball. V the final velocity from theconservation of momentum for the collision is the initial velocity for the two ballsraising. The velocity of any ball at its highest point is 0 and the height at the lowestpoint is set to 0.

v =√

2ghi = 1.4m/s (61)

V =√

2ghf (62)

m(√

2ghi) = (m+M)(√

2ghf ) (63)

m√

2ghi = 2m√

2ghf (64)

2ghi4

= 2ghf (65)

hf =hi4

= 0.025m (66)

—-, 6.6.61

Two identical cars hit each other and lock bumpers. In each of the following cases,what are the speeds of the cars immediately after coupling bumpers? (a) A car mov-ing with a speed of 90km/h approaches a stationary car; (b) the two cars approach

7

each other with speeds of 90km/h and 120km/h respectively; (c) two cars travel inthe same direction with speeds of 90km/h and 120km/h, respectively.a) Given: m1 = m2 v1 = 90km/h v2 = 0

m1v1 +m2v2 = (m1 +m2)v3 (67)

v3 =(m1v1 +m2v2)

(m1 +m2)(68)

v3 =(mv1)

(2m)=v12

= 45km/h = 12.5m/s (69)

b) Given: m1 = m2 v1 = 90km/h v2 = −120km/h

v3 =(m1v1 +m2v2)

(m1 +m2)(70)

v3 =(mv1 +mv2)

(2m)=v1 + v2

2= −15km/h = −4.17m/s (71)

c) Given: m1 = m2 v1 = 90km/h v2 = 120km/h

v3 =(m1v1 +m2v2)

(m1 +m2)(72)

v3 =(mv1 +mv2)

(2m)=v1 + v2

2= 105km/h = 29.2m/s (73)

7.6.46, 6.6.63

A 10g bullet moving horizontally at 400m/s penetrates a 3.0kg wood block restingon a horizontal surface. If the bullet slows down to 300m/s after emerging from theblock, what is the speed of the block immediately after the bullet emerges?Given: m1 = 10g m2 = 3.0kg v1 = 400m/s v1f = 300m/s

p1 = p2 (74)

m1v1 = m1v1f +m2v2f (75)

v2f =m1v1 −m1v1f

m2

(76)

8

7.6.50

For a movie scene, a 75kg stuntman drops from a tree onto a 50kg sled that is movingonn a frozen lake with a velocity of 10m/s toward the shore. (a) What is the speedof the sled after the suntman is on board? (b) If the sled hits the bank and stops,but the stuntman keeps on going, with what speed does he leave the sled?(neglectfriction)Given:: mm = 75kg ms = 50kg vs = 10m/sa)

msvs = (mm +ms)vf (77)

vf =msvs

(mm +ms)= 4.0m/s (78)

b)

mmvm = (mm +ms)vf (79)

vm =(mm +ms)vf

mm

= 6.7m/s (80)

7.6.52, 6.6.69

A projectile that is fired from a gun has an initial velocity of 90km/h at an angleof 60.0◦ above the horizontal. When the projectile is at the top of its trajectory, aninternal explosion causes it to separate into two fragments of equal mass. One of thefragments falls straight downward as though it has been released from rest. How farfrom the gun does the other fragment land?Given: v = 90.0km/h θ = 60.0◦

We apply the conservation of momentum, the moment before the explosion and weknow that the momentum is pb = mv0x,

pb = pf (81)

mv0x = m2

0 + m2v2x (82)

v2x = 2v0x (83)

We need only concern ourselves with the x-component just before it explodes they-component is equal to zero so the fall time is the same as the rise time. we findthe rise time as,

vytop = v0y − gtrise (84)

trise = =v0yg

=v0 sin θ

g(85)

9

where vytop = 0. Now we write,

x = 3v2o cos θ sin θ

g= 82.8m (86)

=3

2

v2og

sin 2θ (87)

or 3/2 times the range equation.

7.6.53, 6.6.70

A moving shuffleboard puck has a glancing collision with a stationary puck of thesame mass as shown in Fig 6.32. If friction is negligible, what are the speeds of thepucks after the collision?

Given:m1 = m2 = m v10 = v10x = 0.95m/s θ1 = 50◦ θ2 = 40◦

v1oy = 0m/s v20 = 0m/s v1 =? v2 =?

the conservation of momentum in the x-direction,

mv10x = mv1x +mv2x (88)

the conservation of momentum in the y-direction,

mv10y = mv1y +mv2y (89)

to simplify the equations we can write the velocities in terms of v1, v2, θ1, and θ2,

v1x = v1 cos θ1 (90)

v1y = v1 sin θ1 (91)

v2x = v2 cos θ2 (92)

v1y = v2 sin θ2 (93)

Simplifying our equations, we write.

mv10 = mv1 cos θ1 +mv2 cos θ2 (94)

0 = mv1 sin θ1 −mv2 sin θ2 (95)

(96)

Now we have 2 equations and 2 unknowns, so the solution just remains to do somealgebra,

mv1 sin θ1 −mv2 sin θ2 = 0 (97)

v1 sin θ1 = v2 sin θ2 (98)

v1 = v2sin θ2sin θ1

(99)

10

Now put this into the equation for momentum in the x-direction,

v10 = v1 cos θ1 + v2 cos θ2 (100)

v10 = v2sin θ2sin θ1

cos θ1 + v2 cos θ2 (101)

v10 = v2(sin θ2

sin θ1cos θ1 + cos θ2

)(102)

v10 =v2

sin θ1(sin θ2 cos θ1 + cos θ2 sin θ1) (103)

using the trig identity,

sin(θ1 + θ2) = sin θ2 cos θ1 + cos θ2 sin θ1 (104)

We can write,

v10 =v2

sin θ1sin(θ1 + θ2) (105)

since θ1 + θ2 = 90◦ and sin 90◦ = 1,

v10 =v2

sin θ1(106)

v2 = v10 sin θ1 = 0.73m/s (107)

similarly,

v1 = v2sin θ2sin θ1

= v10 sin θ2 = 0.61m/s (108)

7.6.58, 6.6.82

A 4.0kg ball with a velocity of 4.0m/s in the +x-direction collides head-on elasticallywith a stationary 2.0kg ball. What are the velocities of the balls after the collision?Given: m1 = 2m2 v1i = 4.0m/s v2i = 0m/sFind v1f and v2f .This is an elastic collision so energy and momentum are conserved. From the con-servation of momentum,

m1v1i +m2v2i = m1v1f +m2v2f (109)

2m2v1i = 2m2v1f +m2v2f (110)

2v1i = 2v1f + v2f (111)

v1i = v1f +v2f2

(112)

11

v1f = v1i −v2f2

(113)

v21f = v21i − v1iv2f +1

4v22f (114)

From the conservation of energy,

1

2m1v

21i

+1

2m2v

22i

=1

2m1v

21f

+1

2m2v

22f

(115)

m1v21i

= m1v21f

+m2v22f

(116)

2m2v21i

= 2m2v21f

+m2v22f

(117)

v21i = v21f +1

2v22f (118)

v21i = v21i − v1iv2f +1

4v22f +

1

2v22f (119)

v21i − v21i

= −v1iv2f +3

4v22f (120)

v1iv2f =3

4v22f (121)

v1i =3

4v2f (122)

v2f =4

3v1i = 5.33m/s (123)

and,

v1f = v1i −v2f2

= 1.3m/s (124)

7.6.60, 6.6.84

a county fair, two childern ram each other head on while riding on the bumper carsattraction. Jill and her car travling left at 3.50m/s, have a total mass of 325kg. Jackand his car, traveling to the right at 2.00m/s, have a total mass of 290kg.Given: Assuming the collision to be elastic, determine their velocities after thecollision.conservation of moment gives,

m1v1 −m2v2 = −m1v1f +m2v2f (125)

conservation of energy gives,

1

2m1v

21 +

1

2m2v

22 =

1

2m1v

21f +

1

2m2v

22f (126)

12

rewrite equation (??) as,

m1v21 +m2v

22 = m1v

21f +m2v

22f (127)

m1(v21 − v21f ) = m2(v

22f − v22) (128)

m1(v1 − v1f )(v1 + v1f ) = m2(v2f − v2)(v2 + v2f ) (129)

rewrite equation (??) as,

m1(v1 + v1f ) = m2(v2 + v2f ) (130)

Divide equation (??) by equation (??),

m1(v1 − v1f )(v1 + v1f )

m1(v1 + v1f )=

m2(v2f − v2)(v2 + v2f )

m2(v2 + v2f )(131)

v1 − v1f = v2f − v2 (132)

v1 + v2 = v1f + v2f (133)

Solve for v1f ,v1f = v1 + v2 − v2f (134)

Now we can solve for v2f from the conservation of momentum,

m1(v1 + v1 + v2 − v2f ) = m2(v2 + v2f ) (135)

2m1v1 +m1v2 −m1v2f = m2v2 +m2v2f (136)

2m1v1 +m1v2 −m2v2 = m1v2f +m2v2f (137)

v2f =2m1

m1 +m2

v1 +(m1 −m2

m1 +m2

)v2 (138)

v2f = −3.81m/s (139)

Similarly we can solve for v1f ,

v1f =2m2

m1 +m2

v2 +(m2 −m1

m1 +m2

)v1 = 1.69m/s (140)

note: the positive sign means that the initial assignment of a sign to the momentumof mass one final is correct.

13

7.6.63, 6.6.87

A 1.0kg object moving at 10m/s collides perfectly inelastically with a stationary2.0kg object. After the collision the blocks slide up a plane inclined at 37◦. How farup the plane will the combined system travel?Given: m1 = 1.0kg m2 = 2.0kg v0 = 10m/s θ = 37◦

use conservation of momentum to find the velocity at the bottom of the rampand then the conservation of energy to find the distance up the ramp.

conservation of momentum,

m1v1 = m1v2 +m2v2 (141)

v2 =m1

m1 +m2

v1 (142)

(143)

conservation of energy,

1

2(m1 +m2)v

22 = (m1 +m2)gh (144)

h is the change in vertical height of the blocks, and is related to the distance alongthe incline, l, by,

h = l sin θ (145)

this leads to,

1

2(m1 +m2)v

22 = (m1 +m2)gh (146)

1

2(m1 +m2)v

22 = (m1 +m2)gl sin θ (147)

l =v22

2g sin θ(148)

=m2

1v21

(m1 +m2)22g sin θ= 0.94m (149)

—-, 6.6.89

Two balls with masses of 2.0kg and 6.0kg travel toward each other at speeds of 12m/sand 4.0m/s, respectively. If the balls have a head-on, inelastic collision and the 2.0kgball recoils with a speed of 8.0m/s, how much kinetic energy is lost in the collision?

14

Given:m1 = 2.0kg m2 = 6.0kg v1 = 12m/s v2 = −4.0m/sv3 = −8.0m/s

∆p = 0 = m1v1 +m2v2 −m1v3 −m2v4 (150)

v4 =m1v1 +m2v2 −m1v3

m2

=m1

m2

(v − v3) + v2 = 2.67m/s (151)

∆K = (1

2m1v

23 +

1

2m2v

24)− (

1

2m1v

21 + frac12m2v

22) (152)

Ki =1

2m1v

21 + frac12m2v

22 = 192J (153)

Kf =1

2m1v

23 +

1

2m2v

24 = 85J (154)

∆K = 1.1× 102J (155)

7.6.65, 6.6.90

Two balls approach each other as shown in Fig 6.36, where m = 2.0kg, v = 3.0m/s,M = 4.0kg, and V = 5.0m/s. If the balls collide sand stick together at the origin, (a)what are the components of the velocity v of the balls after the collision, and (b) whatis the angle θ? Given: m = 2.0kg v = 3.0m/sx̂ M = 4.0kg V = 5.0m/sThis is an inelastic collision, momentum is conserved, energy is not. Write downmomentum conservation in the x and y directions,

x : mv = (m+M)v′x (156)

y : MV = (m+M)v′y (157)

v′x =mv

(m+M)= 1.0m/s (158)

v′x =MV

(m+M)= 3.3m/s (159)

θ = tan−1(MV/mv) = 73◦ (160)

7.6.73

Show that the fraction of kinetic energy lost in a ballistic pendulum collision is equalto M/(m+M).To find the fraction of energy lost we need to know the energy before and after thecollison. The energy will all be kinetic and as such we need to know the velocity

15

before v and after V the collision.

Fraction of energy lost =1/2mv2 − 1/2(m+M)V 2

1/2mv2(161)

=mv2 − (m+M)V 2

mv2(162)

The conservation of momentum gives.

pi = pf (163)

mv = (m+M)V (164)

V =m

m+Mv (165)

(166)

2 Substituting this into the energy fraction gives,

Fraction of energy lost =mv2 − (m+M)V 2

mv2(167)

=mv2 − (m+M)( m

m+Mv)2

mv2(168)

=mv2

mv2(1− m

m+M) =

m+M −mm+M

=M

m+M(169)

7.6.76, 6.6.106

Find the center of mass of a system composed of thress shperical objects withmasses of 3.0kg, 2.0kg and 4.0kg and centers locaed at (-6.0m,0), (1.0m,0) and(3.0m,0),respectively.

Given:m1 = 3.0kg m2 = 2.0kg m3 = 4.0kg

r1 = (−6.0cm, 0) r2 = (1.0cm, 0) r3 = (3.0cm, 0)by inspection ycm = 0,

xcm =

∑imixi∑imi

(170)

xcm =m1x1 +m2x2 +m3x3

m1 +m2 +m3

= −0.44m (171)

rcm = (−0.44m, 0) (172)

16

7.6.77, 6.6.107

Rework exercise 69, using the concept of the center of mass, and compute ths distancethe other fragment lands from the gun. Given: v = 90.0km/h θ = 60.0◦

At the maximum height the bullet explodes the center of mass travels on the initialparabolic path with an x-compoent of the velocity of v0x. For the bullet fragementto fall straight down it must have a speed v1x = 0 or −vx relative to the center ofmass. This means the other fragment must have a velocity of v2x = 2v0x relativeto the ground (its speed is +vx relative to the center of mass). The time requiredfor the bullet fragments to fall the ground are the same, and are the same as thetime required to reach the maximum height, tmax (assuming the ground is flat). Thismeans the total distance traveled is,

x = v0xtmax + v2xtmax = v0xtmax + 2v0xtmax (173)

= 3v0xtmax = 3vo cos θtmax (174)

To find tmax,

vytop = v0y − gtmax (175)

tmax = =v0yg

=v0 sin θ

g(176)

where vytop = 0, now we write,

x = 3v2o cos θ sin θ

g= 82.8m (177)

=3

2

v2og

sin 2θ (178)

or 3/2 times the range equation.

7.6.79, 6.6.109

A piece of uniform sheet metal measures 25cm by 25cm. If a circular piece with aradius of 5.0cm is cut from the center of the sheet, where is the sheet’s center of massnow?The distribution of the mass as far as its symmetry doesn’t change so the center ofmass is still in the center.

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7.6.80, 6.6.110

Locate the center of mass of the system shown in Fig 6.38 (a) if all the masses areequal; (b) if m4 = m2 = 2m3 = 2m1; (c) if m1 = 1.0kg, m2 = 2.0kg, m3 = 3.0kg,and m4 = 4.0kg.Given: r1 = (0m, 0m) r2 = (0m, 4m) r3 = (4m, 4m) r4 = (4m, 0m)

xcm =

∑imixi∑imi

(179)

xcm =m1x1 +m2x2 +m3x3 +m4x4

m1 +m2 +m3 +m4

(180)

similarly in the y-direction,

ycm =m1y1 +m2y2 +m3y3 +m4y4

m1 +m2 +m3 +m4

(181)

a) m1 = m2 = m3 = m4

rcm = (2.0m, 2.0cm)

b) 2m1 = m2 = 2m3 = m4

rcm = (2.0m, 2.0cm)

c)m1 = 1.0kg, m2 = 2.0kg, m3 = 3.0kg, m4 = 4.0kgrcm = (2.8m, 2.0cm)

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