10.1. Model: We will use the particle model for the bullet (B) and the bowling ball (BB).Visualize:

Solve: For the bullet,

K m vB B B2 kg m/s J= = =1

2

1

20 01 500 12502( . )( )

For the bowling ball,

K m vBB BB BB kg m s J= = =1

2

1

210 10 5002 2( )( / )

Thus, the bullet has the larger kinetic energy.Assess: Kinetic energy depends not only on mass but also on the square of the velocity. The above calculationshows this dependence. Although the mass of the bullet is 1000 times smaller than the mass of the bowling ball, itsspeed is 50 times larger.

10.2. Model: Model the hiker as a particle.Visualize:

The origin of the coordinate system chosen for this problem is at sea level so that the hiker’s position in DeathValley is y0 8 5= − . .mSolve: The hiker’s change in potential energy from the bottom of Death Valley to the top of Mt. Whitney is

∆U U U mgy mgy mg y y= − = − = −

= − − = ×gf gi f i f i

2kg ( m/s m m J

( )

( ) . )[ ( )] .65 9 8 4420 85 2 87 106

Assess: Note that ∆U is independent of the origin of the coordinate system.

10.3. Model: Model the compact car (C) and the truck (T) as particles.Visualize:

Solve: For the kinetic energy of the compact car and the kinetic energy of the truck to be equal,

K K m v m v vm

mvC T C C T T C

T

CT

kg

kgkm/hr km/hr= ⇒ = ⇒ = = =1

2

1

2

20 000

100025 1122 2 ,

( )

Assess: A smaller mass needs a greater velocity for its kinetic energy to be the same as that of a larger mass.

10.4. Model: Model the oxygen and the helium atoms as particles.Visualize: We denote the oxygen and helium atoms by O and He, respectively. Note that the oxygen atom is fourtimes heavier than the helium atom, so m mO He= 4 .Solve: The energy conservation equation K KO He= is

1

2

1

24 2 02 2 2 2m v m v m v m v

v

vO O He He He O He HeHe

O

= ⇒ = ⇒ =( ) .

Assess: The result v vHe O= 2 , combined with the fact that m mHe O= 14 , is a consequence of the way kinetic

energy is defined: it is directly proportional to the mass and to the square of the speed.

10.5. Model: Model the car (C) as a particle. This is an example of free fall, and therefore the sum of kineticand potential energy does not change as the car falls.Visualize:

Solve: (a) The kinetic energy of the car is

K m vC C C kg m/s J= = = ×1

2

1

21500 30 6 75 102 2 5( )( ) .

(b) Let us relabel KC as Kf and place our coordinate system at yf = 0 m so that the car’s potential energy Ugf iszero, its velocity is vf, and its kinetic energy is Kf. At position yi, v Ki i m/s or J= =0 0 , and the only energy thecar has is U mgygi i= . Since the sum K + Ug is unchanged by motion, K U K Uf gf i gi+ = + . This means

K mgy K mgy K K mgy

yK K

mg

f f i i f i i

ii

2

J - 0 J

kg m / s m

+ = + ⇒ + = +

⇒ = − = × =

0

6 75 10

1500 9 845 9

5( ) ( . )

( )( . ).f

(c) From part (b),

yK K

mg

mv mv

mg

v v

gif i

f if i= − =

−=

−( )( )1

2

1

22

2 22 2

Free fall does not depend upon the mass.

10.6. Model: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does notchange as the ball rises and falls.Visualize:

The figure shows a ball’s before-and-after pictorial representation for the three situations in parts (a), (b) and (c).Solve: The quantity K + Ug is the same during free fall: K U K Uf gf i gi+ = + . We have

(a) 1

2

1

2

2 10 0 2 9 8 5 10

12

1 02

0

1 02

12 2 2 2

mv mgy mv mgy

y v v g

+ = +

⇒ = −( ) = − × =[( / ) ( / ) ]/( . / ) .m s m s m s m

5.1 m is therefore the maximum height of the ball above the window. This is 25.1 m above the ground.

(b) 1

2

1

222

2 02

0mv mgy mv mgy+ = +

Since y y2 0 0= = , we get for the magnitudes v v2 0 10= = m / s.

(c) 1

2

1

22 2 23

23 0

20 3

23 0

20 3

202

0 3mv mgy mv mgy v gy v gy v v g y y+ = + ⇒ + = + ⇒ = + −( )

⇒ = + − − =v32 2 210 2 9 8 0 20 492( / ) ( . / )[ ( )]m s m s m m m / s2 2

This means the magnitude of v3 is equal to 22.2 m/s.Assess: Note that the ball’s speed as it passes the window on its way down is the same as the speed with which itwas tossed up, but in the opposite direction.

10.7. Model: This is a problem of free fall. The sum of the kinetic and gravitational potential energy for theball, considered as a particle, does not change during its motion.Visualize:

The figure shows the ball’s before-and-after pictorial representation for the two situations described in parts (a) and (b).Solve: The quantity K + Ug is the same during free fall. Thus, K U K Uf gf i gi+ = + .

(a) 1

2

1

22

0 2 9 8 10 166 6 12 9

12

1 02

0 02

12

1 0

02 2

0

mv mgy mv mgy v v g y y

v v

+ = + ⇒ = + −

⇒ = + − = ⇒ =

( )

( / ) ( . / )( ) . / . /m s m s m 1.5 m m s m s2 2 2

(b) 1

2

1

22

166 6 2 9 8 1 5 14 0

22

2 02

0 22

02

0 2

22 2

2

mv mgy mv mgy v v g y y

v v

+ = + ⇒ = + −

⇒ = + − ⇒ =

( )

. / ( . / )( . ) . /m s m s m 0 m m s2 2

Assess: An increase in speed from 12.9 m/s to 14.0 m/s as the ball falls through a distance of 1.5 m is reasonable.Also, note that mass does not appear in the calculations that involve free fall.

10.8. Model: Model the puck as a particle. Since the ramp is frictionless, the sum of the puck’s kinetic andgravitational potential energy does not change during its sliding motion.Visualize:

Solve: The quantity K + Ug is the same at the top of the ramp as it was at the bottom. The energy conservationequation K U K Uf gf i gi+ = + is

1

2

1

22

0 2 9 8 1 026 20 11 4 48

2 2 2 2

2

mv mgy mv mgy v v g y y

v v

f f i i i f f i

i2 2 2 2

im s m s m 0 m m s m s

+ = + ⇒ = + −

⇒ = + − = ⇒ =

( )

( / ) ( . / )( . ) . / . /

Assess: An initial push with a speed of 4.48 m/s ≈ 10 mph to cover a distance of 3.0 m up a 20° ramp seemsreasonable.

10.9. Model: Model the skateboarder as a particle. Assuming that the track offers no rolling friction, the sumof the skateboarder’s kinetic and gravitational potential energy does not change during his rolling motion.Visualize:

The vertical displacement of the skateboarder is equal to the radius of the track.Solve: The quantity K + Ug is the same at the upper edge of the quarter-pipe track as it was at the bottom. Theenergy conservation equation K U K Uf gf i gi+ = + is

1

2

1

22

0 2 9 8 3 0 58 8 7 67

2 2 2 2

2

mv mgy mv mgy v v g y y

v v

f f i i i f f i

i2 2

im s m s m 0 m m s m s

+ = + ⇒ = + −

= + − = ⇒ =

( )

( / ) ( . / )( . ) . / . /

Assess: Note that we did not need to know the skateboarder’s mass, as is the case with free-fall motion.

10.10. Model: Model the ball as a particle undergoing rolling motion with zero rolling friction. The sum of theball’s kinetic and gravitational potential energy, therefore, does not change during the rolling motion.Visualize:

Solve: Since the quantity K + Ug does not change during rolling motion, the energy conservation equations apply.For the linear segment the energy conservation equation K U K U0 0 1 1+ = +g g is

1

2 12

1mv mgy+ = 1

2 02

0mv mgy+ ⇒ + = + ⇒ =1

2m

1

2m s

1

2m v mg m mgy mv mgy1

2 20 1

200 0( ) ( / )

For the parabolic part of the track, K U K U1 1 2 2+ = +g g is

1

2

1

2

1

20

1

2

1

212

1 22

2 12 2

2 12

2mv mgy mv mgy mv mg mgy mv mgy+ = + ⇒ + = + ⇒ =( ) ( /m m 0 m s)

Since from the linear segment we have 12 1

20 0 2 2 0 1 0m v mgy mgy mgy y y= = = =, we get or m. . Thus, the ball rolls

up to exactly the same height as it started from.Assess: Note that this result is independent of the shape of the path followed by the ball, provided there is norolling friction. This result is an important consequence of energy conservation.

10.11. Model: In the absence of frictional and air-drag effects, the sum of the kinetic and gravitationalpotential energy does not change as the pendulum swings from one side to the other.Visualize:

The figure shows the pendulum’s before-and-after pictorial representation for the two situations described in parts(a) and (b).Solve: (a) The quantity K + Ug is the same at the lowest point of the trajectory as it was at the highest point. Thus,K U K U1 1 0 0+ = +g g means

1

2

1

22 2

2 0 0 2 2

12

1 02

0 12

1 02

0

12 2

0 1 0

mv mgy mv mgy v gy v gy

v g gy v gy

+ = + ⇒ + = +

⇒ + = + ⇒ =( ) ( / )m m s

From the pictorial representation, we find that y L L0 30= − °cos . Thus,

v gL1 2 1 30 2 9 8 0 75 1 30 1 403= − ° = − ° =( cos ) ( . )( . )( cos ) .m/s m m/s 2

(b) Since the quantity K + Ug does not change, K U K Ug g2 2 1 1+ = + . We have

1

2

1

22

1 403 0 2 9 8 0 100

22

2 12

1 2 12

22

22 2

mv mgy mv mgy y v v g

y

+ = + ⇒ = −( )⇒ = − × =[( . ) ( ) ]/( . ) . m/s m/s m/s m2

Since y L L2 = − cosθ , we obtain

cos( . ) ( . )

( . ). cos ( . )θ θ= − = − = ⇒ = = °−L y

L2 10 75 0 10

0 750 8667 0 8667 30

m m

m

That is, the pendulum swings to the other side by 30° .Assess: The swing angle is the same on either side of the rest position. This result is a consequence of the factthat the sum of the kinetic and gravitational potential energy does not change. This is shown as well in the energybar chart in the figure.

10.12. Model: Model the child and swing as a particle, and assume the chain to be massless. In the absence offrictional and air-drag effects, the sum of the kinetic and gravitational potential energy does not change during theswing’s motion.Visualize:

Solve: The quantity K + Ug is the same at the highest point of the swing as it is at the lowest point. That is,K U K U0 0 1 1+ = +g g . It is clear from this equation that maximum kinetic energy occurs where the gravitational

potential energy is the least. This is the case at the lowest position of the swing. At this position, the speed of theswing and child will also be maximum. The above equation is

1

2

1

22

0 2 0 2

02

0 12

1 12

02

0 1

12 2

0 1 0

mv mgy mv mgy v v g y y

v g y v gy

+ = + ⇒ = + −

⇒ = + − ⇒ =

( )

( / ) ( )m s m

We see from the pictorial representation that

y L L

v gy

0

1 0

45 3 0 3 0 45 0 879

2 2 9 8 0 879 4 15

= − ° = − ° =

⇒ = = =

cos ( . ) ( . )cos .

( . / )( . ) . /

m m m

m s m m s2

Assess: We did not need to know the swing’s or the child’s mass. Also, a maximum speed of 4.15 m/s isreasonable.

10.13. Model: Model the car as a particle with zero rolling friction. The sum of the kinetic and gravitationalpotential energy, therefore, does not change during the car’s motion.Visualize:

Solve: (a) The initial energy of the car is

K U mv mgy0 0 02

02 41

2

1

21500 10 0 7 50 10+ = + = = ×g kg m/s J( )( . ) .

The energy of the car at the top of the hill is

K U K mgy K K1 1 1 1 1 141500 9 8 5 7 35 10+ = + = + = + ×g

2kg m s m J( )( . / )( ) .

If the car just wants to make it to the top, then K1 = 0. In other words, an energy of 7.35 × 104 J is needed to get tothe top. Since this energy is less than the available energy of 7.50 × 104 J, the car can make it to the top.(b) The conservation of energy equation K U K U0 0 2 2+ = +g g is

7 50 101

27 50 10 1500 1500 9 8 5 0

14 1

422

24

22

2

. . ( ) ( )( . / )( . )

. /

× = + ⇒ × = + −

⇒ =

J J1

2kg kg m s m

m s

2mv mgy v

v

Assess: A higher speed on the other side of the hill is reasonable because the car has increased its kinetic energyand lowered its potential energy compared to its starting values.

10.14. Model: Assume an ideal spring that obeys Hooke’s law.Visualize:

Solve: (a) The spring force on the 2.0 kg mass is F k ysp .= − ∆ Notice that ∆y is negative, so Fsp is positive. This

force is equal to mg, because the 2.0 kg mass is at rest. We have .− =k y mg∆ Solving for k:

k mg y= − = − − − − =( / ) ( . )( . / )/( . ( . ))∆ 2 0 9 8 0 15 0 10 3922kg m s m m N/m

(b) Again using − =k y mg∆ :

∆y mg k

y y y y

= − = −′ − = − ⇒ ′ = − = − − = − = −

/ ( . )( . / )/( / )

. . . . .

3 0 9 8 392

0 075 0 075 0 10 0 175 17 5

kg m s N m

m m m 0.075 m m cm

2

e e

The length of the spring is 17.5 cm when a mass of 3.0 kg is attached to the spring. The position of the end of thespring is negative because it is below the origin, but length must be a positive number.

10.15. Model: Assume that the spring is ideal and obeys Hooke’s law. We also model the 5.0 kg mass as a particle.Visualize: We will use the subscript s for the scale and sp for the spring.

Solve: (a) The scale reads the upward force Fs on m that it applies to the mass. Newton’s second law gives

( ) ( . )( . / )F F w F w mgyon m s on m s on m

2kg m s N∑ = − = ⇒ = = = =0 5 0 9 8 49

(b) In this case, the force is

( )

( )/ ( )/ .

F F F w k y mg

k mg y

yon m s on m sp N = 0

N N 20 N m N/m

∑ = + − = ⇒ + −

⇒ = − = − =

0 20

20 49 0 02 1450

∆

∆(c) In this case, the force is

( )

/ ( )/( / ) .

F F w k y mg

y mg k

yon m sp

N N m m = 3.4 cm

∑ = − = ⇒ − =

⇒ = = =

0 0

49 1450 0 0338

∆

∆

10.16. Model: Assume that the spring is ideal and obeys Hooke’s law. Also model the sled as a particle.

Visualize: The only horizontal force acting on the sled is rFsp .

Solve: Applying Newton’s second law to the sled gives

( )F F ma k y max x xon sled sp= = ⇒ =∑ ∆

⇒ = = =a k x mx ∆ / ( )( . )/ .150 0 20 20 1 50N/m m kg m/s2

10.17. Model: Assume that the spring is ideal and obeys Hooke’s law.Visualize: According to Hooke’s law, the spring force acting on a mass (m) attached to the end of a spring isgiven as F k xsp ,= ∆ where ∆x is the change in length of the spring. If the mass m is at rest, then Fsp is also equal to

the weight w mg= .Solve: We have .spF k x mg= =∆ We want a 0.10 kg mass to give ∆x = 0 010. m. This means

k mg x= = =/ ( . )( . / )/( . ) /∆ 0 10 9 8 0 010 98kg N m m N m

10.18. Model: Model the student (S) as a particle and the spring as obeying Hooke’s law.Visualize:

Solve: According to Newton’s second law the force on the student is

( )

( )( . / . / )

F F w ma

F w ma mg ma

y y

y y

on S spring on S

spring on S2 2 kg m s m s N

= − =

⇒ = + = + = + =∑

60 9 8 3 0 768

Since , N. This means N N m m.spring on S S on springF F k y k y y= = = = =∆ ∆ ∆768 768 2500 0 307( )/( / ) .

10.19. Model: Assume an ideal spring that obeys Hooke’s law.Solve: The elastic potential energy of a spring is defined as U k ss = 1

22( ) ,∆ where ∆s is the magnitude of the

stretching or compression relative to the unstretched or uncompressed length. We have cm 0.20 m∆s = =20and N/mk = 500 . This means

U k ss N m m J= = =1

2

1

2500 0 20 102( ) ( / )( . )∆

Assess: Since ∆s is squared, Us is positive for a spring that is either compressed or stretched. Us is zero when thespring is in its equilibrium position.

10.20. Model: Assume an ideal spring that obeys Hooke’s law.Solve: The elastic potential energy of a spring is defined as U k ss = 1

22( ) ,∆ where ∆s is the magnitude of the

stretching or compression relative to the unstretched or uncompressed length. ∆Us = 0 when the spring is at itsequilibrium length and ∆s = 0. We have J and N/msU k= =200 1000 . Solving for ∆s:

∆s U k= = =2 2 200 0 632s / J / 1000 N/m m( ) .

10.21. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, so the mechanical energyK + Us is conserved. Also model the book as a particle.Visualize:

The figure shows a before-and-after pictorial representation. The compressed spring will push on the book until thespring has returned to its equilibrium length. We put the origin of our coordinate system at the equilibrium positionof the free end of the spring. The energy bar chart shows that the potential energy of the compressed spring isentirely transformed into the kinetic energy of the book.Solve: The conservation of energy equation K U K U2 2 1+ = +s s1 is

1

2

1

2

1

2

1

222

22

12

12mv k x x mv k x x+ − = + −( ) ( )e e

Using x2 = xe = 0 m and v1 = 0 m/s, this simplifies to

1

2

1

20

1250 0 040

0 5002 002

21

22

12 2

mv k x vkx

m= − ⇒ = = =( )

( )( . )

( . ). m

N/m m

kg m/s

Assess: This problem can not be solved using constant-acceleration kinematic equations. The acceleration is not aconstant in this problem, since the spring force, given as F k xs = − ∆ , is directly proportional to ∆x or | |.x x− e

10.22. Model: Assume an ideal spring that obeys Hooke’s law. Since there is no friction, the mechanicalenergy K + Us is conserved. Also, model the block as a particle.Visualize:

The figure shows a before-and-after pictorial representation. We have put the origin of our coordinate system at theequilibrium position of the free end of the spring. This gives us x1 = xe = 0 cm and x2 = 2.0 cm.Solve: The conservation of energy equation K U K U2 2 1 1+ = +s s is

1

2

1

2

1

2

1

222

22

12

12mv k x x mv k x x+ − = + −( ) ( )e e

Using v2 = 0 m/s, x1 = xe = 0 m, and x2 − xe = 0.02 m, we get

1

2

1

222

12

2 1k x x mv x x xm

kv( ) ( )− = ⇒ = − =e e∆

That is, the compression is directly proportional to the velocity v1. When the block collides with the spring withtwice the earlier velocity (2v1), the compression will also be doubled to 2(x2 − xe) = 2(2.0 cm) = 4.0 cm.Assess: This problem shows the power of using energy conservation over using Newton’s laws in solvingproblems involving nonconstant acceleration.

10.23. Model: Model the grocery cart as a particle and the spring as an ideal that obeys Hooke’s law. We willalso assume zero rolling friction during the compression of the spring, so that mechanical energy is conserved.Visualize:

The figure shows a before-and-after pictorial representation. The “before” situation is when the cart hits the springin its equilibrium position. We put the origin of our coordinate system at this equilibrium position of the free end ofthe spring. This give x1 = xe = 0 and (x2 − xe) = 60 cm.Solve: The conservation of energy equation K U K U2 2 1 1+ = +s s is

1

2

1

2

1

2

1

222

22

12

12mv k x x mv k x x+ − = + −( ) ( )e e

Using v2 = 0 m/s, (x2 − xe) = 0.60 m, and x1 = xe = 0 m gives:

1

2

1

2

250

100 60 3 002

212

1 2k x x m v vk

mx x( ) ( ) ( . ) . /− = ⇒ = − = =e e

N/m

kgm m s

10.24. Model: Model the jet plane as a particle, and the spring as an ideal that obeys Hooke’s law. We willalso assume zero rolling friction during the stretching of the spring, so that mechanical energy is conserved.Visualize:

The figure shows a before-and-after pictorial representation. The “before” situation occurs just as the jet planelands on the aircraft carrier and the spring is in its equilibrium position. We put the origin of our coordinate systemat the right free end of the spring. This gives x x1 0= =e m. Since the spring stretches 30 m to stop the plane,x x2 30− =e m.Solve: The conservation of energy equation K U K U2 2 1 1+ = +s s for the spring-jet plane system is

1

2

1

2

1

2

1

222

22

12

12mv k x x mv k x x+ − = + −( ) ( )e e

Using v x x x x2 1 20 0 30= = = − =m/s m, and m yieldse e,

1

2

1

2

60 000

15 00030 602

212

1 2 1k x x mv vk

mx x( ) ( )

,

,( )− = ⇒ = − = =e

N/m

kgm m/s

Assess: A landing speed of 60 m/s or ≈ 120 mph is reasonable.

10.25. Model: We assume this is a one-dimensional collision that obeys the conservation laws of momentumand mechanical energy.Visualize:

Note that momentum conservation alone is not sufficient to solve this problem because the two final velocities( ) (v vx xf f and )1 2 are unknowns and can not be determined from one equation.Solve: Momentum conservation:

Energy conservation:

i i f f

i i f f

m v m v m v m v

m v m v m v m v

x x x x

x x x x

1 1 2 2 1 1 2 2

1 12

2 22

1 12

2 221

2

1

2

1

2

1

2

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

+ = +

+ = +

These two equations can be solved for ( ) (v vx xf f and )1 2 , as shown by Equations 10.39 through 10.43, to give

( ) ( ) ( . / ) . /

( ) ( )( )

( . / ) . /

vm m

m mv

vm

m mv

x x

x x

f i

f i

g 20 g

g 20 gm s m s

g

g 20 gm s m s

11 2

1 21

21

1 21

50

502 0 0 857

2 2 50

502 0 2 86

= −+

= −+

=

=+

=+

=

Assess: These velocities are of a reasonable magnitude. Since both these velocities are positive, both balls movealong the +x direction.

10.26. Model: This is the case of a perfectly inelastic collision. Momentum is conserved because no externalforce acts on the system (clay + block). We also represent our system as a particle.Visualize:

Solve: (a) The conservation of momentum equation pfx = pix is

( ) ( ) ( )m m v m v m vx x x1 2 1 1 2 2+ = +f i i

Using ( ) ( ) ,v v vx xi i and 1 0 2 0= = we get

vm

m mv v v vx x x xf i i i

kg

kg 0.05 kg =

+=

+= =1

1 21 1 1 0

0 05

10 0476 0 0476( )

.

( )( ) . ( ) .

(b) The initial and final kinetic energies are given by

K m v m v v v

K m m v v v

x x

x

i i i

f f

kg kg m/s kg

kg 0.050 kg

= + = + ( ) =

= + = + =

1

2

1

2

1

20 050

1

21 0 0 025

1

2

1

21 0 0476 0 00119

1 12

2 22

02 2

02

1 22 2

02

02

( ) ( ) ( . ) ( ) ( . )

( ) ( )( . ) .

The percent of energy lost = −

× = −

× =K K

Ki f

i

% % %100 10 00119

0 025100 95 2

.

..

10.27. Model: In this case of a one-dimensional collision, the momentum conservation law is obeyed whetherthe collision is perfectly elastic or perfectly inelastic.Visualize:

Solve: In the case of a perfectly elastic collision, the two velocities ( ) (v vx xf f and )1 2 can be determined by combiningthe conservation equations of momentum and mechanical energy. By contrast, a perfectly inelastic collision involvesonly one final velocity vfx and can be determined from just the momentum conservation equation.(a) Momentum conservation:

Energy conservation:

i i f f

i i f f

m v m v m v m v

m v m v m v m v

x x x x

x x x x

1 1 2 2 1 1 2 2

1 12

2 22

1 12

2 221

2

1

2

1

2

1

2

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

+ = +

+ = +

These two equations can be solved as shown in Equations 10.39 through 10.43:

( ) ( )( ) ( )

( ) ( )( / ) . /

( ) ( )( )

( ) ( )( / ) . /

vm m

m mv

vm

m mv

x x

x x

f i

f i

g g

g gm s m s

g

100 g gm s m s

11 2

1 21

21

1 21

100 300

100 30010 5 0

2 2 100

30010 5 0

= −+

= −+

= −

=+

=+

= +

(b) For the inelastic collision, both balls travel with the same final speed v xf . The momentum conservationequation isf ip px x=

( ) ( ) ( )

( ) .

m m v m v m v

v

x x x

x

1 2 1 1 2 2

10010 0 2 5

+ = +

⇒ =+

+ =

f i i

f

g

100 g 300 g m/s m/s m/s

Assess: In the case of the perfectly elastic collision, the two balls bounce off each other with a speed of 5.0 m/s.In the case of the perfectly inelastic collision, the balls stick together and move together at 2.5 m/s.

10.28. Model: This is a case of a perfectly elastic collision between a proton and a carbon atom. The collisionobeys the momentum as well as the energy conservation law.Visualize:

Solve: Momentum conservation:

Energy conservation:

P i P C i C P f P C f C

P i P C i C P f P C f C

m v m v m v m v

m v m v m v m v

x x x x

x x x x

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

+ = +

+ = +1

2

1

2

1

2

1

22 2 2 2

These two equations can be solved, as described in the text through Equations 10.39 to 10.43:

( ) ( ) ( . ) .

( ) ( ) ( . )

vm m

m mm m v

m m

m m

vm

m mv

m

m m

x x

x x

f PP C

P CP C i P

P P

P P

f CP

P Ci P

P

P P

m/s m/s

m/s

= −+

+ = −+

× = − ×

=+

=+

×

12

122 0 10 1 69 10

2 2

122 0 10

7 7

7 == ×3 08 106. m/s

After the elastic collision the proton rebounds at 1 69 107. × m/s and the carbon atom moves forward at 3 08 106. × m/s.

10.29. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant.Visualize:

The particle is released from rest at x = 1.0 m. That is, K = 0 at x = 1.0 m. Since the total energy is given by E = K + U,we can draw a horizontal total energy (TE) line through the point of intersection of the potential energy curve (PE)and the x = 1.0 m line. The distance from the PE curve to the TE line is the particle’s kinetic energy. These valuesare transformed as the position changes, causing the particle to speed up or slow down, but the sum K + U does notchange.Solve: (a) We have E = 5.0 J and this energy is a constant. For x < 1.0, U > 5.0 J and, therefore, K must benegative to keep E the same (note that K = E − U or K = 5.0 J − U). Since negative kinetic energy is unphysical, theparticle can not move to the left. That is, the particle will move to the right of x = 1.0 m.(b) The expression for the kinetic energy is E − U. This means the particle has maximum speed or maximumkinetic energy when U is minimum. This happens at x = 2.0 m. Thus,

K E U mv vmmax min max max( . ) ( . ) . .

( . ) ./= − = − = = ⇒ = = =5 0 1 0 4 0

1

24 0

2 4 0 8 0202J J J J

J J

0.020 kgm s

The particle possesses this speed at x = 2.0 m.(c) The total energy (TE) line intersects the potential energy (PE) curve at x = 1.0 m and x = 6.0 m. These are theturning points of the motion.

10.30. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant.Visualize:

The particle with a mass of 500 g is released from rest at A. That is, K = 0 at A. Since E = K + U = 0 J + U, we candraw a horizontal TE line through U = 5.0 J. The distance from the PE curve to the TE line is the particle’s kineticenergy. These values are transformed as the position changes, causing the particle to speed up or slow down, butthe sum K + U does not change.Solve: The kinetic energy is given by E − U, so we have

1

222mv E U v E U m= − ⇒ = −( )/

Using U U UB C DJ, J, and J= = =2 0 3 0 0. . , we get

v v

v

B C

D

J 2.0 J kg m/s J 3.0 J kg m/s

J 0 J kg m/s

= − = = − =

= − =

2 5 0 0 5 3 46 2 5 0 0 5 2 83

2 5 0 0 5 4 47

( . )/ . . ( . )/ . .

( . )/ . .

10.31. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant.Visualize:

Since the particle oscillates between x = 2.0 mm and x = 8.0 mm, the speed of the particle is zero at these points.That is, for these values of x, E = U = 5.0 J, which defines the total energy (TE) line. The distance from thepotential energy (PE) curve to the TE line is the particle’s kinetic energy. These values are transformed as theposition changes, but the sum K + U does not change.Solve: The equation for total energy E = U + K means K = E − U, so that K is maximum when U is minimum.We have

K mv U

v U m

max max min

max min

.

( . )/ ( . )/ . .

= = −

⇒ = − = − =

1

25 0

2 5 0 2 5 0 0 002 63 2

2 J

J J 1.0 J kg m/s

10.32. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant.Visualize:

For the speed of the particle at A that is needed to reach B to be a minimum, the particle’s kinetic energy as itreaches the top must be zero. Similarly, the minimum speed at B for the particle to reach A obtains when theparticle just makes it to the top with zero kinetic energy.Solve: (a) The energy equation K U K UA A top top+ = + is

1

20

2 2 5 7 75

2mv U U

v U U m

A A top

A top A

J

J 2 J 0.100 kg m/s

+ = +

⇒ = − = − =( )/ ( )/ .

(b) To go from point B to point A, K U K UB B top top+ = + is

1

20

2 2 5 10 0

2mv U U

v U U m

B B top

B top B

J

J 0 J 0.100 kg m/s

+ = +

⇒ = − = − =( )/ ( )/ .

Assess: The particle requires a higher kinetic energy to reach A from B than to reach B from A.

10.33. Model: Model your vehicle as a particle. Assume zero rolling friction, so that the sum of your kineticand gravitational potential energy does not change as the vehicle coasts down the hill.Visualize:

The figure shows a before-and-after pictorial representation. Note that neither the shape of the hill nor the angle ofthe downward slope is given, since these are not needed to solve the problem. All we need is the change inpotential energy as you and your vehicle descend to the bottom of the hill. Also note that

35 35000 9 722km / hr m / 3600 s m s= =( ) . /Solve: Using yf = 0 and the equation K U K Ui gi f gf+ = + we get

1

2

1

22

2 9 722 2 9 8 15 19 7 71

2 2 2 2

2 2

mv mgy mv mgy v gy v

v v gy

i i f f i i f

f i i

m s m s m m s km hr

+ = + ⇒ + =

⇒ = + = + = =( . / ) ( . / )( ) . / /

You are driving over the speed limit. Yes, you will get a ticket.Assess: A speed of 19.7 m/s or 71 km/hr at the bottom of the hill, when your speed at the top of the hill was35 km/s, is reasonable. From the energy bar chart, we see that the initial potential energy is completely transformedinto the final kinetic energy.

10.34. Model: This is case of free-fall, so the sum of the kinetic and gravitational potential energy does notchange as the cannon ball falls.Visualize:

The figure shows a before-and-after pictorial representation. To express the gravitational potential energy, we putthe origin of our coordinate system on the ground below the fortress.Solve: Using yf = 0 and the equation K U K Ui gi f gf+ = + we get

1

2

1

22

2 80 2 9 8 10 81 2

2 2 2 2

2 2

mv mgy mv mgy v gy v

v v gy

i i f f i i f

f i i2

m/s m/s m m/s

+ = + ⇒ + =

= + = + =( ) ( . )( ) .

Assess: Note that we did not need to use the tilt angle of the cannon, because kinetic energy is a scalar. Also notethat using the energy conservation equation, we can find only the magnitude of the final velocity, not the directionof the velocity vector.

10.35. Model: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does notchange as the rock is thrown. Model the Frisbee and the rock as particles.Visualize:

The coordinate system is put on the ground for this system, so that yf m= 16 . The rock’s final velocity vf must beat least 5.0 m/s to dislodge the Frisbee.Solve: (a) The energy consrevation equation for the rock K U K Uf gf i gi+ = + is

1

2

1

22 2mv mgy mv mgyf f i i+ = +

This equation involves only the velocity magnitudes and not the angle at which the rock is to be thrown to dislodgethe Frisbee. This equation is true for all angles that will take the rock to the Frisbee.(b) Using the above equation we get

v gy v gy v v g y yf f i i i f f i2m s m s m 2 m m s2 2 2 22 2 2 5 0 2 9 8 16 17 3+ = + = + − = + − = ( ) ( . / ) ( . / )( ) . /

Assess: Kinetic energy is defined as K mv= 12

2 and is a scalar quantity. Scalar quantities do not have directionalproperties.

10.36. Model: For the marble spinning around the inside of a smooth pipe, the sum of the kinetic and gravita-tional potential energy does not change.Visualize:

We use a coordinate system with the origin at the bottom of the pipe, that is, y1 = 0. The radius (R) of the pipe is10 cm, and therefore ytop = y2 = 2R = 0.20 m. At an arbitrary angle θ, measured from the bottom of the circle, y =R − R cosθ.Solve: (a) The energy conservation equation K U K U2 2 1 1+ = +g g is

⇒ + = +

⇒ = + − = + − =

1

2

1

2

2 3 0 2 9 8 0 2 25

22

2 12

1

2 12

1 22

mv mgy mv mgy

v v g y y( ) ( . / ) ( . / )( ) . /m s m s m 0.20 m m s2

(b) Expressing the energy conservation equation as a function of θ :

K U K U mv mgy mv

v v gy v gR

( ) ( ) ( ) ( )

( ) ( cos )

θ θ θ θ

θ θ θ

+ = + ⇒ + = +

⇒ ( ) = − = − −

g g J1 12

12

12

12

1

2

1

20

2 2 1

Using v1 = 3.0 m/s, g = 9.8 m/s2, and R = 0.10 m, we get v( ) . ( cos )θ θ= − −9 1 96 1 (m/s)(c) The accompanying figure shows a graph of v for a complete revolution ( ).0 360° ≤ ≤ °θ

Assess: Beginning with a speed of 3.0 m/s at the bottom, the marble’s potential energy increases and kineticenergy decreases as it gets toward the top of the circle. At the top, its speed is 2.25 m/s. This is reasonable sincesome of the kinetic energy has been transformed into the marble’s potential energy.

10.37. Model: Assume that the rubber band behaves similar to a spring. Also, model the rock as a particle.Visualize:

Please refer to Figure P10.37.Solve: (a) The rubber band is stretched to the left since a positive spring force on the rock due to the rubber bandresults from a negative displacement of the rock. That is, ( ) ,F kxxsp = − where x is the rock’s displacement from the

equilibrium position due to the spring force Fsp.(b) Since the Fsp versus x graph is linear with a negative slope and can be expressed as F kxsp = − , the rubber band

obeys Hooke’s law.(c) From the graph, | | |∆ ∆F xsp N for | cm. Thus,= =20 10

kF

x= = =

| |

| |/

∆∆

sp N

0.10 mN m

20200

(d) The conservation of energy equation K U K Uf sf i si+ = + for the rock is

1

2

1

2

1

2

1

2

1

2

1

20

1

20

1

2

200

0 050 30 19 0

2 2 2 2 2 2 2 2mv kx mv kx mv k m kx

vk

mx

f f i i f i

f i

m m/s

N/m

kgm m/s

+ = + ⇒ + = +

= = =

( ) ( )

.( . ) .

Assess: Note that xi is ∆x, which is the displacement relative to the equilibrium position, and xf is the equilibriumposition of the rubber band, which is equal to zero.

10.38. Model: Model the block as a particle and the springs as ideal springs obeying Hooke’s law. There is nofriction, hence the mechanical energy K + Us is conserved.Visualize:

Note that xf = xe and xi − xe = ∆x. The before-and-after pictorial representations show that we put the origin of thecoordinate system at the equilibrium position of the free end of the springs.Solve: The conservation of energy equation K U K Uf sf i si+ = + for the single spring is

1

2

1

2

1

2

1

22 2 2 2mv k x x mv k x xf f e i i e+ − = + −( ) ( )

Using the value for vf given in the problem, we get1

20 0

1

2

1

2

1

202 2

02 2mv k x mv k x+ = + ⇒ = J J ( ) ( )∆ ∆

Conservation of energy for the two-spring case:1

20 0

1

2

1

22 2 2mV k x x k x xf i e i e J J+ = + − + −( ) ( )

1

22 2mV k xf = ( )∆

Using the result of the single-spring case,1

222

02mV mv V vf f 0= ⇒ =

Assess: The block separates from the spring at the equilibrium position of the spring.

10.39. Model: Model the block as a particle and the springs as ideal springs obeying Hooke’s law. There is nofriction, hence the mechanical energy K + Us is conserved.Visualize:

The springs in both cases have the same compression ∆x. We put the origin of the coordinate system at theequilibrium position of the free end of the spring for the single-spring case (a), and at the free end of the twoconnected springs for the two-spring case (b).Solve: The conservation of energy for the single-spring case:

K U K U mv k x x mv k x xf sf i si f f e i i e+ = + ⇒ + − = + −1

2

1

2

1

2

1

22 2 2 2( ) ( )

Using xf = xe = 0 m, vi = 0 m/s, and vf = v0, this equation simplifies to1

2

1

22 2mv k x0 = ( )∆

Conservation of energy in the case of the two springs in series, where each spring compresses by ∆x /2, is

K U K U mV mv k x k xf sf i si f i 2+ = + ⇒ + = + +1

20

1

2

1

2

1

222 2 2 2( / ) ( / )∆ ∆

Using xf = xe′ = 0 m and vi = 0 m/s, we get

1

2

1

2

1

22 2mV k xf =

( )∆

Comparing the two results we see that Vf = v0 2/ .Assess: The block pushes on the spring until the spring has returned to its equilibrium length.

10.40. Model: Model the ball and earth as particles. An elastic collision between the ball and earth conservesboth momentum and mechanical energy.Visualize:

The before-and-after pictorial representation of the collision is shown in the figure. The ball as it is dropped from aheight of 10 m has zero velocity. Its speed just before it hits earth can be found using kinematics. We can thenapply the conservation laws to find earth’s recoil velocity.Solve: (a) To get the speed of the ball at collision:

( ) ( ) ( ) ( ) ( . )( . )

( ) ( . )( . ) .

v v g y y

v

12

02

1 0

1

2 0 2 9 8 10 0

2 9 8 10 0 14 0

B B B2

B2

m/s m/s m

m/s m m/s

= − − = − −

⇒ = =In an elastic collision, the velocity of the object that is struck is

( ) ( )( . )

.( . ) . /v

m

m mv2 1 24

242 2 0 50

5 98 1014 0 2 33 10E

B

E BB

kg

kgm / s m s=

+=

×= × −

(b) The time it would take earth to move 1.0 mm at this speed is given by

speeddistance

velocity

m

2.33 10 m/ss years= =

×= × = ×

−

−

104 29 10 1 36 10

3

2420 13. .

10.41. Model: Model the marble and the steel ball as particles. We will assume an elastic collision between themarble and the ball, and apply the conservation of momentum and the conservation of energy equations. We willalso assume zero rolling friction between the marble and the incline.Visualize:

This is a two-part problem. In the first part, we will apply the conservation of energy equation to find the marble’sspeed as it exits onto a horizontal surface. We have put the origin of our coordinate system on the horizontalsurface just where the marble exits the incline. In the second part, we will consider the elastic collision between themarble and the steel ball.Solve: The conservation of energy equation K U K U1 1 0 0+ = +g g gives us:

1

2

1

212

1 02

0m v m gy m v m gyM M M M M M( ) ( )+ = +

Using ) m / s and m we get M M M( , ( ) ( ) .v y v gy v gy0 1 12

0 1 00 01

22= = = ⇒ = When the marble collides with the

steel ball, the elastic collision gives the ball velocity

( ) ( )vm

m mv

m

m mgy2 1 0

2 22S

M

M SM

M

M S

=+

=+

Solving for y0 gives

yg

m m

mv0 2

21

2 20 258 25 8= +

= =M S

MS m cm( ) . .

10.42. Model: Model the two packages as particles. Momentum is conserved in both inelastic and elasticcollisions. Kinetic energy is conserved only in an elastic collision.Visualize:

Solve: For a package with mass m the conservation of energy equation is

K U K U m v mgy m v mgym m1 1 0 0 12

1 02

0

1

2

1

2+ = + ⇒ + = +g g ( ) ( )

Using ) m / s and m0( ,v ym = =0 01

1

22 2 9 8 3 0 7 6681

20 1 0

2m v mgy v gym m( ) ( ) ( . / )( . ) . /= ⇒ = = =m s m m s

(a) For the perfectly inelastic collision the conservation of momentum equation is

p p m m v m v m vx x m m mf i 3 2= ⇒ + = +( )( ) ( ) ( )( )2 22 1 1

Using ) m / s we get2( ,v m1 0=( ) ( ) / . /v vm m2 1 3 2 563 m s= =

(b) For the elastic collision, the mass m package rebounds with velocity

( ) ( ) . .vm m

m mvm m3 1

2

2

1

37 668 2 56= −

+= − ( ) = − m / s m / s

The negative sign with (v3)m shows that the package with mass m rebounds and goes to the position y4. We candetermine y4 by applying the conservation of energy equation as follows. For a package of mass m:

K U K U m v mgy m v mgym mf gf i gi+ = + ⇒ + = +1

2

1

242

4 32

3( ) ( )

Using ) m s m, and ) m/s( . / , ( ,v y vm m3 3 42 55 0 0= − = = we get

mgy m y42

4

1

22 56 33 3= − ⇒ =( . / ) .m s cm

10.43. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, and therefore themechanical energy K + Us + Ug is conserved.Visualize:

The figure shows a before-and-after pictorial representation. We have chosen to place the origin of the coordinatesystem at the position where the ice cube has compressed the spring 10.0 cm. That is, y0 = 0.Solve: The energy conservation equation K U U K U U2 2 2 0 0 0+ + = + +s g s g is

1

2

1

2

1

2

1

222 2

0 02 2

0mv k x x mgy mv k x x mgy+ − + = + − +( ) ( )e e e

Using m / s m and m / sv y v2 0 00 0 0= = =, , ,

mgy k x x yk x x

mg22

2

2 2

2

1

2 2

25 0 10

2 0 050 9 825 5= − ⇒ = − = =( )

( ) ( / )( . )

( . )( . / ).e

e N m m

kg m s cm

The distance traveled along the incline is y2 51 0/ sin 30 cm.° = .Assess: The net effect of the launch is to transform the potential energy stored in the spring into gravitationalpotential energy. The block has kinetic energy as it comes off the spring, but we did not need to know this energyto solve the problem.

10.44. Model: Assume an ideal spring that obeys Hooke’s law. Since this is a free fall problem, the mechanicalenergy K U U+ +g s is conserved. Also, model the safe as a particle.

Visualize:

We have chosen to place the origin of our coordinate system at the free end of the spring which is neither stretchednor compressed. The safe gains kinetic energy as it falls. The energy is then converted into elastic potential energyas the safe compresses the spring. The only two forces are gravity and the spring force, which are bothconservative, so energy is conserved throughout the process. This means that the initial energy—as the safe isreleased—equals the final energy—when the safe is at rest and the spring is fully compressed.Solve: The conservation of energy equation K U U K U U1 1 1 0 0 0+ + = + +g s g s is

1

2

1

2

1

2

1

212

1 12

02

02mv mg y y k y y mv mg y y k y y+ − + − = + − + −( ) ( ) ( ) ( )e e e e e

Using v v1 0 0= = m/s and ye = 0 m, the above equation simplifies to

mgy ky mgy1 12

0

1

2+ =

⇒ = − = − −−

= ×kmg y y

y

2 2 1000 9 8 2 0 0 5

0 51 96 100 1

12

5( ) ( )( . )( . ( . ))

( . ).

kg m/s m m

mN/m.

2

2

Assess: By equating energy at these two points, we do not need to find how fast the safe was moving when it hitthe spring.

10.45. Model: Assume an ideal spring that obeys Hooke’s law. The mechanical energy K + Us + Ug is conservedduring the launch of the ball.Visualize:

This is a two-part problem. In the first part, we use projectile equations to find the ball’s velocity v2 as it leaves thespring. This will yield the ball’s kinetic energy as it leaves the spring.Solve: Using the equations of kinematics,

x x v t t a t t v t

v t t v

y y v t t a t t

x x

y y

3 2 2 3 2 3 22

2 3

2 3 3 2

3 2 2 3 2 3 22

1

25 0 30 0 0

30 5 0 5 0 30

1

2

1 5

= + − + − ⇒ = + ° − +

° = ⇒ = °

= + − + −

−

( ) ( ) . ( cos )( )

( cos ) . ( . cos )

( ) ( )

.

m 0 m s m

m m/

mm s m s s2= + ° − + −0 30 01

29 8 02 3 3

2( sin )( ) ( . / )( )v t t

Substituting the value for t3, ( . ) ( sin ).

cos. /

.

cos− = °

°

− ( ) °

1 5 305 0

304 9

5 0

3022

2

2

2

mm

m sm

vv v

⇒ − = + − ⇒ =( . ) ( . ).

.1 5 2 887163 33

6 10222 2m m m/s

vv

The conservation of energy equation K U U K U U2 2 2 1 1 1+ + = + +s g s g is

1

2

1

20

1

2

1

222 2

2 12 2

1mv k mgy mv k s mgy+ + = + +( ) ( ) m ∆

Using y2 = 0 m, v1 = 0 m/s, ∆s = 0.2 m, and y1 = −(∆s)sin 30º, we get

1

20 0 0

1

230 2 30

0 2 0 02 6 102 2 0 02 9 8 0 2 0 5 19 6

22 2 2

22

2 2 2

mv k s mg s s k mv mg s

k k

+ + = + − ° = + °

= + ⇒ =

J J J

m kg m s kg m s N/m

( ) ( )sin ( ) ( )sin

( . ) ( . )( . / ) ( . )( . / )( . )( . ) .

∆ ∆ ∆ ∆

Assess: Note that y s1 30= − °( )sin∆ is with a minus sign and hence the gravitational potential energy mgy1 is− °mg s( )sin .∆ 30

10.46. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction and hence the mechanicalenergy K + Us + Ug is conserved.Visualize:

Solve: (a) When releasing the block suddenly, K U U K U U2 2 2 1 1 1+ + = + +s g s g

1

2

1

2

1

2

1

222

22

2 12

12

1mv k y y mgy mv k y y mgy+ − + = + − +( ) ( )e e

Using v2 = 0 m/s, v1 = 0 m/s, and y1 = ye, we get

01

2490 0 0 245

245 5 0 9 8 0 20

22

2 1 22

1 2

1 22

1 2 1 2

J N/m J J N/m

N/m kg m/s m

1 1

2

+ − + = + + ⇒ − = −

⇒ − = − ⇒ − =

( )( ) ( )( ) ( )

( )( ) ( . )( . )( ) ( ) .

y y mgy mgy y y mg y y

y y y y y y

(b) When lowering the block gently until it rests on the spring, the block reaches a point of static equilibrium.

F k y mg ymg

knet

kg m s

N mm= − = ⇒ = = =∆ ∆0

5 0 9 8

4900 10

2( . )( . / )

/.

(c) In part (b), at a point 0.10 m down, the forces balance. But in part (a) the block has kinetic energy as it reaches0.10 m. So the block continues on past the equilibrium point until all the gravitational potential energy is stored inthe spring.

10.47. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, and thus the mechanical energyK + Us + Ug is conserved.Visualize:

We place the origin of our coordinate system at the spring’s compressed position y1 = 0. The rock leaves the springwith velocity v2 as the spring reaches its equilibrium position.Solve: (a) The conservation of mechanical energy equation is

K U U K U U mv k y y mgy mv k y y mgy2 2 2 1 1 1 22

22

2 12

12

1

1

2

1

2

1

2

1

2+ + = + + + − + = + − +s g s g e e ( ) ( )

Using y2 = ye, y1 = 0 m, and v1 = 0 m/s, this simplifies to

1

20 0

1

20

1

20 4 0 4 9 8 0 30

1

21000 0 30 14 8

22

2 12

22 2

2

mv mgy k y y

v v

+ + = + − +

+ = ⇒ =

J J

kg kg m/s m N/m m m/s

e

2

( )

( . ) ( . )( . )( . ) ( )( . ) .

(b) Let us use the conservation of mechanical energy equation once again to find the highest position (y3) of therock where its speed (v3) is zero:

K U K U mv mgy mv mgy

g y y v y yv

g

3 3 2 2 32

3 22

2

3 2 22

3 222 2

1

2

1

2

01

2 2

14 8

2 9 811 2

+ = + ⇒ + = +

⇒ + − = ⇒ − = = =

g g

2 m/s

m/sm( ) ( )

( . )

( . ).

If we assume the spring’s length to be 0.5 m, then the distance between ground and fruit is 11.2 m + 0.5 m = 11.7 m.This is much smaller than the distance of 15 m between fruit and ground. So, the rock does not reach the fruit.

10.48. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, so the mechanical energyK + Ug + Us is conserved.Visualize:

Place the origin of the coordinate system at the end of the unstretched spring, making ye = 0 m.Solve: The clay is in static equilibrium while resting in the pan. The net force on it is zero. We can start by usingthis to find the spring constant.

F w k y y ky mg kmg

ysp e

2 kg)(9.8 m / s

m N / m= ⇒ − − = − = ⇒ = − = −

−=( )

( . )

..1 1

1

0 10

0 109 8

Now apply conservation of energy. Initially, the spring is unstretched and the clay ball is at the ceiling. At the end,the spring has maximum stretch and the clay is instantaneously at rest. Thus

K U U K U U mv mgy ky mv mgy2 2 2 0 0 012 2

22

12 2

2 12 0

20 0+ + = + + ⇒ + + = + +( ) ( ) ( ) ( )g s g s J

Since v0 = 0 m/s and v2 = 0 m/s, this equation becomes

mgy ky mgy ymg

ky

mgy

k

y y

212 2

20 2

22

0

22

2

2 20

0 20 0 10 0

+ = ⇒ + − =

+ − =. .

The numerical values were found using known values of m, g, k, and y0. The two solutions to this quadraticequation are y2 = 0.231 m and y2 = –0.432 m. The point we’re looking for is below the origin, so we need thenegative root. The distance of the pan from the ceiling is

L = |y2| + 50 cm = 93.2 cm

10.49. Model: Assume an ideal spring that obeys Hooke’s law. Also assume zero rolling friction between theroller coaster and the track, and a particle model for the roller coaster. Since no friction is involved, the mechanicalenergy K + Us + Ug is conserved.Visualize:

We have chosen to place the origin of the coordinate system on the end of the spring that is compressed andtouches the roller coaster car.Solve: (a) The energy conservation equation for the car going to the top of the hill is

K U U K U U mv mgy k x x mv mgy k x x2 2 2 0 0 0 22

22

02

0 021

2

1

2

1

2

1

2+ + = + + + + − = + + −g s g s e e e ( ) ( )

Noting that m m/s m/s and | m0 ey v v x x= = = − =0 0 0 2 02 1 0, , , | . , we obtain

0 0 0 01

22 0

2

2 0

2 400 9 8 10

2 019 600

22

22 2

J J J J m

m

kg m/s m

m N/m

2

+ + = + +

⇒ = = =

mgy k

kmgy

( . )

( . )

( )( . )( )

( . ),

We now increase this value for k by 10% for safety, giving a value of 21,600 N/m.(b) The energy conservation equation K U U K U U3 3 3 0 0 0+ + = + +g s g s is

1

2

1

2

1

2

1

232

32

02

0 02mv mgy k x x mv mgy k x x+ + − = + + −( ) ( )e e e

Using m, m / s m and m, we getey v y x x3 0 0 05 0 0 0 2 0= − = = − =. , , | | .

1

25 0 0 0 0

1

21

2400 400 9 8 5 0

1

22 156 10 2 0

17 7

32

02

32 4 2

3

mv mg k x x

v

v

+ − + = + + −

− = ×

⇒ =

( . ) ( )

( ) ( )( . )( . ) ( . )( . )

.

m J J J

kg kg m/s m N/m m

m/s

e

2

10.50. Model: Since there is no friction, the sum of the kinetic and gravitational potential energy does notchange. Model Julie as a particle.Visualize:

We place the coordinate system at the bottom of the ramp directly below Julie’s starting position. From geometry,Julie launches off the end of the ramp at a 30° angle.

Solve: Energy conservation: K U K U1 1 0 0+ = +g g ⇒ + = +1

2

1

212

1 02

0mv mgy mv mgy

Using v y y0 0 10 25 3= = = m/s m, and m,, the above equation simplifies to

1

22 2 9 8 25 20 771

21 0 1 0 1mv mgy mgy v g y y+ = ⇒ = − = − =( ) ( . )( ) .m/s m 3 m m/s2

We can now use kinematic equations to find the touchdown point from the base of the ramp. First we’ll considerthe vertical motion:

y y v t t a t t v t t t t

t t t t

y y2 1 1 2 1 2 12

1 2 1 2 12

2 12

2 1

1

20 3 30

1

29 8

20 77 30

4 9

3

= + − + − = + − + − −

− − − −

( ) ( ) ( sin )( ) ( . )( )

( )( . )sin

( . )( )

( )

(

m m m / s

m/s

m/s

m

4.9 m/s

2

2 2

°

°))

( ) ( . )( ) ( . ) ( ) .

=

− − − − = ⇒ − =

0

2 119 0 6122 0 2 3772 12

2 1 2 1t t t t t ts s s2

For the horizontal motion:

x x v t t a t tx x2 1 1 2 1 2 121

2= + − + −( ) ( )

x x v t t2 1 1 2 130 0 20 77 30 2 377 42 7− = − + = =( cos )( ) ( . )(cos )( . ) .° ° m m/s s m

Assess: Note that we did not have to make use of the information about the circular arc at the bottom that carriesJulie through a 90° turn.

10.51. Model: We assume the spring to be ideal and to obey Hooke’s law. We also treat the block (B) and theball (b) as particles. In the case of an elastic collision, both the momentum and kinetic energy equations apply. Onthe other hand, for a perfectly inelastic collision only the equation of momentum conservation is valid.Visualize:

Place the origin of the coordinate system on the block that is attached to one end of the spring. The before-and-afterpictorial representations of the elastic and perfectly inelastic collision are shown in figures (a) and (b), respectively.Solve: (a) For an elastic collision, the ball’s rebound velocity is

( ) ( ) ( . ) .vm m

m mvf b

b B

b Bi b

g

120 g m/s m/s= −

+= − = −80

5 0 3 33

The ball’s speed is 3.33 m/s.(b) An elastic collision gives the block speed

( ) ( ) ( . ) .vm

m mvf B

B

b Bi b

g

120 g m/s m/s=

+= =2 40

5 0 1 667

To find the maximum compression of the spring, we use the conservation equation of mechanical energy for theblock + spring system. That is :s sK U K U1 1 0 0+ = +

1

2

1

2

1

2

1

20 0

0 100 1 667 20 11 8

21 0

2 20 0

21 0

2 2

1 02

m v k x x m v k x x k x x m v

x x

B BB f B B f B f

kg m/s N/m cm

( ) ( ) ( ) ( ) ( ) ( )

( ) ( . )( . ) /( ) .

′ + − = + − + − = +

− = =

(c) Momentum conservation p pf i= for the perfectly inelastic collision means

( ) ( ) ( )

( . . ) ( . )( . ) .f

m m v m v m v

v vB B f b i b B i B

fkg kg kg m/s m/v m/s

+ = ++ = + ⇒ =0 100 0 020 0 020 5 0 0 0 833

The maximum compression in this case can now be obtained using the conservation of energy equation K U1 1+ =s

K U0 0+ s :

0 1 2 1 2 0

0 120

200 833 0 0645 6 45

2 2 J / / J

kg

N/mm/s m cm

B b f

B bf

+ = + +

⇒ = + = = =

( ) ( ) ( )( )

.( . ) . .

k x m m v

xm m

kv

∆

∆

10.52. Model: Assume an ideal spring that obeys Hooke’s law. We treat the bullet and the block in the particlemodel. For a perfectly inelastic collision, the momentum is conserved. Furthermore, since there is no friction, themechanical energy of the system (bullet + block + spring) is conserved.Visualize:

We place the origin of our coordinate system at the end of the spring that is not anchored to the wall.Solve: (a) Momentum conservation for perfectly inelastic collision states p pf i= . This means

( ) ( ) ( ) ( )m M v m v M v m M v mv vm

m Mv+ = + ⇒ + = + ⇒ =

+

f i m i M f B f B kg m/s0

where we have used vB for the initial speed of the bullet. The mechanical energy conservation equation K U1 1+ =s

K Ue se+ as the bullet embedded block compresses the spring is:

1

2

1

2

1

2

1

2

01

2

1

20

21

2 2 2

22

22

2

m v k x x m M v k x x

kd m Mm

m Mv v

m M kd

m

f( ) ( ) ( )( ) ( )

( )( )

′ + − = + + −

+ = ++

+ ⇒ = +

e f e e

B B J J

(b) Using the above formula with m = 5.0 g, M = 2.0 kg, k = 50 N/m, and d = 10 cm,

vB kg kg N/m m / m/s= + =( . . )( )( . ) ( . )0 005 2 0 50 0 10 0 005 2002 2

(c) The fraction of energy lost is1

2

1

21

2

1 1

1 10 005

0 0050 9975

2 2

2

2 2mv m M v

mv

m M

m

v

v

m M

m

m

m M

m

m M

B f

B

f

B

+

kg

kg 2.0 kg

− += −

= − ++

= −+

= −+

=

( )

.

( . ).

That is, during the perfectly inelastic collision 99.75% of the bullet’s energy is lost. The energy is dissipated insidethe block. Although it is common to say, “The energy is lost to heat,” in the next chapter we’ll see that it is moreaccurate to say, “The energy is transformed to thermal energy.”

10.53. Model: Because the track is frictionless, the sum of the kinetic and gravitational potential energy doesnot change during the car’s motion.Visualize:

We place the origin of the coordinate system at the ground level directly below the car’s starting position. This is atwo-part problem. If we first find the maximum speed at the top of the hill (point C), we can use energyconservation to find the maximum initial height.Solve: Because the its motion is circular, at the top of the hill the car has a downward-pointing centripetal accel-

eration ra mv r jc = −( / ) ˆ.2 Newton’s second law at the top of the hill is

( ) ( )F n w n mg m amv

Rn mg

mv

Rm g

v

Ry y y ynet c= + = − = = − ⇒ = − = −

2 2 2

If m / s v n mg= =0 , as expected in static equilibrium. As v increases, n gets smaller—the apparent weight of thecar and passengers decreases as they go over the top. But n has to remain positive for the car to be on the track, sothe maximum speed vmax occurs when n → 0 . We see that v gRmax .= Now we can use energy conservation torelate point C to the starting height:

K U K U mv mgy mv mgy mgR mgR mghC C i i C C i J+ = + ⇒ + = + ⇒ + = +1

2

1

2

1

202

12

max ,

where we used vC = vmax and yC = R. Solving for hmax gives h Rmax .= 32

(b) If R = 10 m, then hmax = 15 m.

10.54. Model: This is a two-part problem. In the first part, we will find the critical velocity for the ball to goover the top of the peg without the string going slack. Using the energy conservation equation, we will then obtainthe gravitational potential energy that gets transformed into the critical kinetic energy of the ball, thus determiningthe angle θ.Visualize:

We place the origin of our coordinate system on the peg. This choice will provide a reference to measure gravitationalpotential energy. For θ to be minimum, the ball will just go over the top of the peg.Solve: The two forces in the free-body force diagram provide the centripetal acceleration at the top of the circle.Newton’s second law at this point is

w Tmv

r+ =

2

where T is the tension in the string. The critical speed vc at which the string goes slack is found when T → 0. Inthis case,

mgmv

rv gr gL= ⇒ = =C

2

C2 3

The ball should have kinetic energy at least equal to

1

2

1

2 32mv mg

Lc =

for the ball to go over the top of the peg. We will now use the conservation of mechanical energy equation to getthe minimum angle θ. The equation for the conservation of energy is

K U K U mv mgy mv mgyf gf i gi f2

f i2

i+ = + ⇒ + = +1

2

1

2

Using and the above value for , we getf c f i c2v v y L v v= = =, , ,1

3 0

1

2 3 3 2mg

Lmg

Lmgy y

L+ = ⇒ =i i

That is, the ball is a vertical distance 12 L above the peg’s location or a distance of

2

3 2 6

L L L−

=

below the point of suspension of the pendulum, as shown in the figure on the right. Thus,

cos/

.θ θ= = ⇒ = °L

L

6 1

680 4

10.55. Model: This is a two-part problem. In the first part, we will find the critical velocity for the block to goover the top of the loop without falling off. Since there is no friction, the sum of the kinetic and gravitationalpotential energy is conserved during the block’s motion. We will use this conservation equation in the second partto find the minimum height the block must start from to make it around the loop.Visualize:

We place the origin of our coordinate system directly below the block’s starting position on the frictionless track.Solve: The free-body diagram on the block implies

w nmv

R+ = c

2

For the block to just stay on track, n = 0 . Thus the critical velocity vc is

w mgmv

Rv gR= = ⇒ =c

c

22

The block needs kinetic energy 12 mvc

2 = 12 mgR to go over the top of the loop. We can now use the conservation of

mechanical energy equation to find the minimum height h.

K U K U mv mgy mv mgyf gf i gi f f i i+ = + ⇒ + = +1

2

1

22 2

Using m/sf c f iv v gR y R v= = = =, , ,2 0 and yi = h, we obtain

1

22 0 2 5gR g R gh h R+ = + ⇒ =( ) .

10.56. Model: Model Lisa (L) and the bobsled (B) as particles. We will assume the ramp to be frictionless, sothat the mechanical energy of the system (Lisa + bobsled + spring) is conserved. Furthermore, during the collision,as Lisa leaps onto the bobsled, the momentum of the Lisa + bobsled system is conserved. We will also assume thespring to be an ideal one that obeys Hooke’s law.Visualize:

We place the origin of our coordinate system directly below the bobsled’s initial position.Solve: (a) Momentum conservation in Lisa’s collision with bobsled states p p1 0= , or

( ) ( ) ( ) ( ) ( )

( ) ( / ) . /

m m v m v m v m m v m v

vm

m mv

L B L L B B L B L L

L

L BL

kg

40 kg 20 kgm s m s

+ = + ⇒ + = +

⇒ =+

=+

=

1 0 0 1 0

1 0

0

4012 8 0

The energy conservation equation: K U U K U U2 2 2 1 1 1+ + = + +s g s g is

1

2

1

2

1

2

1

222

22

2 12 2

1( ) ( ) ( ) ( ) ( ) ( )m m v k x x m m gy m m v k x x m m gyL B e L B L B e e L B+ + − + + = + + − + +

Using v2 = 0 m/s, k = 2000 N/m, y2 = 0 m, y1 = (50 m) sin 20° = 17.1 m, v1 = 8.0 m/s, and (mL + mB) = 60 kg, we get

01

22000 0

1

260 8 0 0 60 9 8 17 12

2 2J N / m J kg m / s J kg m / s me2+ − + = + +( )( ) ( )( . ) ( )( . )( . )x x

Solving this equation yields ) m.e( .x x2 3 46− =(b) As long as the ice is slippery enough to be considered frictionless, we know from conservation of mechanicalenergy that the speed at the bottom depends only on the vertical descent ∆y. Only the ramp’s height h is important,not its shape or angle.

10.57. Model: We can divide this problem into two parts. First, we have an elastic collision between the 20 gball (m) and the 100 g ball (M). Second, the 100 g ball swings up as a pendulum.Visualize:

The figure shows three distinct moments of time: the time before the collision, the time after the collision butbefore the two balls move, and the time the 100 g ball reaches its highest point. We place the origin of ourcoordinate system on the 100 g ball when it is hanging motionless.Solve: For a perfectly elastic collision, the ball moves forward with speed

( ) ( ) ( )vm

m mv vM

m

m Mm m1 0 0

2 1

3=

+=

In the second part, the sum of the kinetic and gravitational potential energy is conserved as the 100 g ball swings upafter the collision. That is, K U K U2 2 1 1+ = +g g . We have

1

2

1

222

2 12

1M v Mgy M v MgyM M( ) ( )+ = +

Using ) J ) m and ( , (( )

, , cos ,v vv

y y L LM Mm

2 10

1 203

0= = = = − θ the energy equation simplifies to

g L Lv

v L

m

m

( cos )( )

( ) ( cos ) ( ( . ( cos .

− =

⇒ = − = − ° =

θ

θ

1

2 9

18 1 1 0 1 50 7 94

02

0 g 18 9.8 m/s ) m) ) m/s2

10.58. Model: Model the balls as particles. We will use the Galilean transformation of velocities (Equation10.44) to analyze the problem of elastic collisions. We will transform velocities from the lab frame S to a frame S′ inwhich one ball is at rest. This allows us to apply Equations 10.43 to the case of a perfectly elastic collision in S′, findthe final velocities of the balls in S′, and then transform these velocities back to the lab frame S.Visualize: Let S′ be the frame of the 200-g ball. Denoting masses as m1 = 100 g and m2 = 200 g, the initialvelocities in the S frame are ( )v xi m/s1 4= and ( )v xi m/s.2 3= −

Figure (a) shows the before-collision situation as seen in frame S, and figure (b) shows the before-collisionsituation as seen in frame S′. The after-collision velocities in S′ are shown in figure (c), and figure (d) indicatesvelocities in S after they have been transformed to frame S from S′.Solve: (a) In the S frame, ( )v xi m/s1 4= and ( )v xi m/s2 3= − . S′ is the reference frame of the 200-g ball, so V =–3 m/s. The velocities of the two balls in this frame can be obtained using the Galilean transformation of velocitiesv′ = v – V.

So,( ) ( )v v Vx xi i m/s ( m/s) m/s′ = − = − − =1 1 4 3 7 ( ) ( )v v Vx xi i m/s ( m/s) m/s′ = − = − − − =2 2 3 3 0

Figure (b) shows the “before” situation, where ball 2 is at rest.Now we can use Equations 10.43 to find the after-collision velocities in frame S′:

( ) ( ) (

( ) ( ) (

vm m

m mv

vm

m mv

x x

x x

f i

f i

g 200 g

g 200 gm/s) m/s

(100) g

g 200 gm/s) m/s

′ = −+

′ = −+

= −

′ =+

′ =+

=

11 2

1 21

21

1 21

100

1007

7

3

2 2

1007

14

3

Finally, we need to apply the reverse Galilean transformation v = v′ + V, with the same V, to transform the after-collision velocities back to the lab frame S:

( ) ( ) .

( ) ( ) .

v v V

v v V

x x

x x

f f

f f

m/s m/s m/s

m/s m/s m/s

1 1

2 2

7

33 5 33

14

33 1 67

= ′ + = − − = −

= ′ + = − =

Figure (d) shows the “after” situation in the lab frame. The 100 g ball is moving left at 5.33 m/s; the 200 g ball ismoving right at 1.67 m/s.

(b) The momentum conservation equation p px xf i= for a perfectly inelastic collision is

( ) ( ) ( )

( . ) ( . )( . ) ( . )( . ) .

m m v m v m v

v vx x x

x x

1 2 1 1 2 2

0 100 0 100 4 0 0 200 3 0 0 667

+ = ++ = + − ⇒ = −

f i i

f fkg 0.200 kg kg m/s kg m/s m/s

Both balls are moving left at 0.667 m/s.

10.59. Model: Model the balls as particles. We will use the Galilean transformation of velocities (Equation10.44) to analyze the problem of elastic collisions. We will transform velocities from the lab frame S to a frame S′ inwhich one ball is at rest. This allows us to apply Equations 10.43 to a perfectly elastic collision in S′. After finding thefinal velocities of the balls in S′, we can then transform these velocities back to the lab frame S.Visualize: Let S′ be the frame of the 400-g ball. Denoting masses as m1 = 100 g and m2 = 400 g, the initialvelocities in the S frame are ( ) .v xi m/s1 4 0= + and ( ) .v xi m/s.2 1 0= +

Figures (a) and (b) show the before-collision situations in frames S and S′, respectively. The after-collision velocitiesin S′ are shown in figure (c). Figure (d) indicates velocities in S after they have been transformed to S from S′.Solve: In frame S, ( ) .v xi m/s1 4 0= and ( ) .v xi m/s.2 1 0= Because S′ is the reference frame of the 400-g ball, V =1.0 m/s. The velocities of the two balls in this frame can be obtained using the Galilean transformation of velocitiesv′ = v – V.

So,( ) ( ) . . .v v Vx xi i m/s m/s m/s′ = − = − =1 1 4 0 1 0 3 0 ( ) ( ) . .v v Vx xi i m/s m/s m/s′ = − = − =2 2 1 0 1 0 0

Figure (b) shows the “before” situation in frame S′ where the ball 2 is at rest.Now we can use Equations 10.43 to find the after-collision velocities in frame S′:

( ) ( ) ( . .

( ) ( ) ( . .

vm m

m mv

vm

m mv

x x

x x

f i

f i

g 400 g

g 400 gm/s) m/s

(100 g)

g 400 gm/s) m/s

′ = −+

′ = −+

= −

′ =+

′ =+

=

11 2

1 21

21

1 21

100

1003 0 1 80

2 2

1003 0 1 20

Finally, we need to apply the reverse Galilean transformation v = v′ + V, with the same V, to transform the after-collision velocities back to the lab frame S.

( ) ( ) . . .

( ) ( ) . . .

v v V

v v Vx x

x x

f f

f f

m/s m/s m/s

m/s m/s m/s1 1

2 2

1 80 1 0 0 80

1 20 1 0 2 20

= ′ + = − + = −= ′ + = + =

Figure (d) shows the “after” situation in frame S. The 100 g ball moves left at 0.80 m/s, the 400 g ball right at 2.20 m/s.Assess: The magnitudes of the after-collision velocities are similar to the magnitudes of the before-collision velocities.

10.60. Model: Use the model of the conservation of mechanical energy.Visualize:

Solve: (a) The turning points occur where the total energy line crosses the potential energy curve. For E = 12 J,this occurs at the points x = 1 m and x = 7 m.(b) The equation for kinetic energy K = E − U gives the distance between the potential energy curve and totalenergy line. U = 8 J at x = 2 m, so K = 12 J − 8 J = 4 J. The speed corresponding to this kinetic energy is

vK

m= = =2 2 4

4 0( )

.J

0.5 kgm/s

(c) Maximum speed occurs for minimum U. This occurs at x = 1 m and x = 4 m, where U = 0 J and K = 12 J. Thespeed at these two points is

vK

m= = =2 2 12

6 93( )

.J

0.5 kgm/s

(d) The particle leaves x = 1 m with v = 6.93 m/s. It gradually slows down, reaching x = 2 m with a speed of 4.00 m/s.After x = 2 m, it speeds up again, returning to a speed of 6.93 m/s as it crosses x = 4 m. Then it slows again, cominginstantaneously to a halt (v = 0 m/s) at the x = 7 m turning point. Now it will reverse direction and move back to theleft.(e) If the particle has E = 4 J it cannot cross the 8 J potential energy “mountain” in the center. It can either oscillateback and forth over the range 1.0 m ≤ x ≤ 1.5 m or over the range 3 m ≤ x ≤ 5 m.

10.61. Model: We will use the conservation of mechanical energy.Visualize:

The potential energy (U) of the nitrogen atom as a function of z exhibits a double-minimum behavior; the twominima correspond to the nitrogen atom’s position on both sides of the plane containing the three hydrogens.Solve: (a) At room temperature, the total energy line is below the “hill” in the center of the potential energy curve.That is, the nitrogen atom does not have sufficient energy to pass from one side of the molecule to the other. There’s astable equilibrium position on either side of the hydrogen-atom plane at the points where U = 0. Since E > 0, thenitrogen atom will be on one side of the plane and will make small vibrations back and forth along the z-axis—that is,toward and away from the hydrogen-atom plane. In the figure above, the atom oscillates between points A and B.(b) The total energy line is now well above the “hill,” and the turning points of the nitrogen atom’s motion (wherethe total energy line crosses the potential curve) are at points C and D. In other words, the atom oscillates from oneside of the H3 plane to the other. It slows down a little as it passes through the plane of hydrogen atoms, but it hassufficient energy to get through.

10.62. Model: The potential energy of two nucleons interacting via the strong force is

U U e x x= − −0 1 0[ ]/

where x is the distance between the centers of the two nucleons, x0152 0 10= × −. m, and U0

116 0 10= × −. J.Visualize: Nucleons are protons and neutrons, and they are held together in the nucleus by a force called thestrong force. This force exists between nucleons at very small separations.Solve: (a)

(b) For x U= × = ×− −5 0 10 55 1 1015 12. .m, J. This energy is represented by a total energy line.

(c) Due to conservation of total energy, the potential energy when x = × −5 0 10 15. J is transformed into kineticenergy until x = twice the radius = × −1 0 10 15. m. At this separation, u = × −15 7 10 12. J. Thus,

1

2

1

215 7 10 55 1 102 2 12 12mv mv v+ + × = × ⇒ ×− −. . J J = 1.53 10 m/s8

Assess: A speed of 1.53 × 108 m/s is approximately 0.5 c where c is the speed of light. This speed is understandablefor the present model.

10.63. A 2.5 kg ball is thrown upward at a speed of 4.0 m/s from a height of 82 cm above a vertical spring. Whenthe ball comes down it lands on and compresses the spring. If the spring has a spring constant of k = 600 N/m, byhow much is it compressed?

10.64. (a) A 1500 kg auto coasts up a 10.0 m high hill and reaches the top with a speed of 5.0 m/s. What initialspeed must the auto have had at the bottom of the hill?(b)

We place the origin of our coordinate system at the bottom of the hill.(c) The solution of the equation is vi = 14.9 m/s. This is approximately 30 mph and is a reasonable value for thespeed at the bottom of the hill.

10.65. (a) A spring gun is compressed 15 cm to launch a 200-g ball on a horizontal, frictionless surface. The ball has aspeed of 2.0 m/s as it loses contact with the spring. Find the spring constant of the gun.(b)

We place the origin of our coordinate system on the free end of the spring in the equilibrium position. Because thesurface is frictionless, the mechanical energy for the system (ball + spring) is conserved.(c) The conservation of energy equation is

K U K U

mv k m k

k

k

f sf i si

m m/s m

kg m/s m

N/m

+ = +

+ = + −

= −=

1

2

1

20

1

20

1

20 15

0 200 2 0 0 15

35 6

12 2 2 2

2 2

( ) ( ) ( . )

( . )( . ) ( . )

.

10.66. (a) A 100 g lump of clay traveling at 3.0 m/s strikes and sticks to a 200 g lump of clay at rest on africtionless surface. The combined lumps smash into a horizontal spring with k = 3.0 N/m. The other end of thespring is firmly anchored to a fixed post on the surface. How far will the spring compress?(b)

(c) Solving the conservation of momentum equation we get v1x = 1.0 m/s. Substituting this value into the conservationof energy equation yields ∆x2 31 6= . cm.

10.67. (a) A spring with spring constant 400 N/m is anchored at the bottom of a frictionless 30° incline. A 500-gblock is pressed against the spring, compressing the spring by 10 cm, then released. What is the speed with whichthe block is launched up the incline?(b) The origin is placed at the end of the uncompressed spring. This is point from which the block is launched asthe spring expands.

(c) Solving the energy conservation equation, we get vf = 2.65 m/s.

10.68. Model: Choose yourself + spring + earth as the system. There are no forces from outside this system, soit is an isolated system. The interaction forces within the system are the spring force of the bungee cord and thegravitational force. These are both conservative forces, so mechanical energy is conserved.Visualize:

We can equate the system’s initial energy, as you step off the bridge, to its final energy when you reach the lowestpoint. We do not need to compute your speed at the point where the cord starts to stretch. We do, however, need tonote that the end of the unstretched cord is at y y U k y y0 1 2

12 2 0

230= − = = −m 70 m, so s ( ) . Also note that U1s = 0,since the cord is not stretched. The energy conservation equation is

K U U K U U mgy k y y mgy2 2 2 1 1 1 2 2 02

101

20 0+ + = + + ⇒ + + − = + +g s g s J J J( )

Multiply out the square of the binomial and rearrange:

mgy ky ky y ky mgy

ymg

ky y y

mgy

ky y

2 22

0 2 02

1

22

0 2 02 1

22

2

1

2

1

2

22

2100 8 980 0

+ − + =

⇒ + −

+ −

= − + =.

This is a quadratic equation with roots 89.9 m and 10.9 m. The first is not physically meaningful because it is aheight above the point where the cord started to stretch. So we find that your distance from the water when thebungee cord stops stretching is 10.9 m.

10.69. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, hence the mechanical energyK + Ug + Us is conserved.Visualize:

We have chosen to place the origin of the coordinate system at the point of maximum compression. We will uselengths along the ramp with the variable s rather than x.Solve: (a) The conservation of energy equation K U U K U U2 2 2 1 1 1+ + = + +g s g s is

1

2

1

2

1

20

1

20 0

1

2

1

20 4 0 30 0

1

2250 10 9 8 4 0

22

22

12

12

2 2 2

2

mv mgy k s mv mgy k

m mg k s m mg s

s

+ + = + +

+ + = + + ° +

= +

( ) ( )

( ) ( ) ( ) ( ) ( . )sin

( )( ) ( )( . )( .

∆

∆ ∆

∆

m

m/s m m/s m J

N/m kg m/s m2 ∆∆s)1

2

This gives the quadratic equation:

( )( ) ( )

.

125 49 196 0

1 46

2N/m kg m/s kg m /s

m and 1.07 m (unphysical)

2 2 2∆ ∆∆

s s

s

− ⋅ − ⋅ =⇒ = −

The maximum compression is 1.46 m.(b) We will now apply the conservation of mechanical energy to a point where the vertical position is y and theblock’s velocity is v. We place the origin of our coordinate system on the free end of the spring when the spring isneither compressed nor stretched.

1

2

1

2

1

2

1

20

1

230

1

20 4 0 30 0

2 212

12

2 2

mv mgy k s mv mgy k

mv mg s k s mg

+ + = + +

+ − ° + = + ° +

( ) ( )

( sin ) ( ) ( . sin )

∆

∆ ∆

m

J m J

1

230

1

230 4 0 02 2k s mg s mv mg( ) ( sin ) sin ( . )∆ ∆− ° + − ° =m

To find the compression where v is maximum, take the derivative of this equation with respect to ∆s:

1

22 30

1

22 0 0k s mg m v

dv

d s( ) ( sin )∆

∆− ° + − =

Since at the maximum, we havedv

d s∆= 0

∆s mg k= ° = =( sin )/ ( )( . )( . )/( ) .30 10 9 8 0 5 250 19 62kg m/s N/m cm

10.70. Model: Assume an ideal, massless spring that obeys Hooke’s law. Let us also assume that the cannon(C) fires balls (B) horizontally and that the spring is directly behind the cannon to absorb all motion.Visualize:

The before-and-after pictorial representation is shown, with the origin of the coordinate system located at thespring’s free end when the spring is neither compressed nor stretched. This free end of the spring is just behind thecannon.Solve: The momentum conservation equation p px xf i= is

m v m v m v m vx x x xB f B C f C B i B C i C( ) ( ) ( ) ( )+ = +

Since the initial momentum is zero,

( ) ( ) ( ) ( )vm

mv v vx x x xf B

C

Bf C f C f C

kg

10 kg= − = −

= −20020

The mechanical energy conservation equation for the cannon + spring K U K Uf sf i si+ = + is

1

2

1

2

1

20 0

1

2

1

2

20 000

2000 5 5 0

2 2 2 2 2m v k x m v k x m v

vk

mx

x

x

( ) ( ) ( ) ( ) ( )

( )( , )

( . ) .

f C i C f C

f C

J J

N/m

kgm m/s

+ = + ⇒ + =

⇒ = ± = ± = ±

∆ ∆

∆

To make this velocity physically correct, we retain the minus sign with (vfx)C. Substituting into the momentum con-servation equation yields:

( ) ( . )v xf B m/s m/s= − − =20 5 0 100

10.71. Model: Assume an ideal spring that obeys Hooke’s law. There is no friction, hence the mechanicalenergy K + Ug + Us is conserved.Visualize:

We have chosen to place the origin of the coordinate system on the free end of the spring that is neither stretchednor compressed, that is, at the equilibrium position of the end of the unstretched spring. The bullet’s mass is m andthe block’s mass is M.Solve: (a) The energy conservation equation K U U K U U2 2 2 1 1 1+ + = + +s g s g becomes

12

12

12

122

22

22 1

21

21( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )m M v k y y m M g y y m M v k y y m M g y y+ + − + + − = + + − − + −e e e e

Noting v2 = 0 m/s, we can rewrite the above equation as

k y m M g y y m M v k y( ) ( ) ( ) ( ) ( )∆ ∆ ∆ ∆22

2 1 12

122+ + + = + +

Let us express v1 in terms of the bullet’s initial speed vB by using the momentum conservation equation pf = pi

which is ( ) .m M v mv Mv+ = +1 B block Since vblock = 0 m/s, we have

vm

m Mv1 =

+

B

We can also find the magnitude of y1 from the equilibrium condition k y y Mg( ) .1 − =e

∆yMg

k1 =

With these substitutions for v1 and ∆y1, the energy conservation equation simplifies to

k y m M g y ym v

m Mk

Mg

k

vm M

mg y y

m M

m

M g

kk

m M

my

( ) ( ) ( )( )

( )( )

( )

∆ ∆ ∆

∆ ∆ ∆

22

1 2

2 2 2

22

1 2 2

2 2

2 22

2

2

+ + + =+

+

⇒ = +

+ − + + +

B

B

We still need to include the spring’s maximum compression (d) into this equation. We assume that d = ∆y1 + ∆y2,that is, maximum compression is measured from the initial position (y1) of the block. Thus, using ∆ ∆y d y2 1= − =( / )d Mg k− , we have

vm M

mgd

m M

m

M g

kk

m M

md Mg kB = +

− +

+ +

−

2

2

2

2 2

22

1 2

( / )

(b) Using m = 0.010 kg, M = 2.0 kg, k = 50 N/m, and d = 0.45 m,

v

g

B2 kg

kgm/s m kg

kg

0.010 kgm/s N/m

N/m kkg

m 2.0 kg) m/s N

=

−

+ − ×

22 010

0 0109 8 0 45 2 010

2 09 8 50

50 2 0101

0 0100 45 9 8 50

2

2

2

2 2

22

.

.( . )( . ) ( . )

.( . ) /( )

( )( . )( . )

[ . ( ( . )/ //m

m/sB

]

2

453⇒ =v

Assess: This is a reasonable speed for the bullet.

10.72. Model: This is a collision between two objects, and momentum is conserved in the collision. Inaddition, because the interaction force is a spring force and the surface is frictionless, energy is also conserved.Visualize:

Let part 1 refer to the time before the collision starts, part 2 refer to the instant when the spring is at maximumcompression, and part 3 refer to the time after the collision. Notice that just for an instant, when the spring is atmaximum compression, the two blocks are moving side by side and have equal velocities: vA2 = vB2 = v2. This is animportant observation.Solve: First relate part 1 to part 2. Conservation of energy is

1

2

1

2

1

2

1

2

1

2

1

212

22

22 2

22 2m v m v m v k x m m v k xA A A A B B A B= + + = + +( ) ( ) ( )max max∆ ∆

where ∆xmax is the spring’s compression. Ug is not in the equation because there are no elevation changes. Also notethat K2 is the sum of the kinetic energies of all moving objects. Both v2 and ∆xmax are unknowns. Now add theconservation of momentum:

m v m v m v m m v vm v

m mA A A A B B A BA A

A B

m/s1 2 2 2 21 2 667= + = + ⇒ =

+=( ) .

Substitute this result for v2 into the energy equation to find:1

2

1

2

1

2

0 046

21

222

12

22

k x m v m m v

xm v m m v

k

( ) ( )

( ).

max

max

∆

∆

= − +

⇒ = − + = =

A A A B

A A A B m 4.6 cm

Notice how both conservation laws were needed to solve this problem.(b) Again, both energy and momentum are conserved between “before” and “after.” Energy is

1

2

1

2

1

216

1

212

32

32

32

32m v m v m v v vA A A A B B A B= + ⇒ = +

The spring is no longer compressed, so the energies are purely kinetic. Momentum is

m v m v m v v vA A A A B B A B1 3 3 3 38 2= + ⇒ = +

We have two equations in two unknowns. From the momentum equation, we can write v vB A3 32 4= −( ) and use thisin the energy equation to obtain:

161

24 4 3 16 32 3 16 16 03

23

2 2

32

3 32

3= + ⋅ −( ) = − + ⇒ − + =v v v v v vA A A A A A

This is a quadratic equation for vA3 with roots vA3 = (4 m/s, 1.333 m/s). Using vB3 = 2(4 − vA3), these two roots givevB3 = (0 m/s, 5.333 m/s). The first pair of roots corresponds to a “collision” in which A misses B, so each keeps itsinitial speed. That’s not the situation here. We want the second pair of roots, from which we learn that the blocks’speeds after the collision are v vA Bm/s and m/s.3 31 33 5 33= =. .

10.73. Model: Model the balls as particles, and assume a perfectly elastic collision. After the collision is over,the balls swing out as pendulums. The sum of the kinetic energy and gravitational potential energy does not changeas the balls swing out.Visualize:

In the pictorial representation we have identified before-and-after quantities for both the collision and the pendulumswing. We have chosen to place the origin of the coordinate system at a point where the two balls at rest barelytouch each other.Solve: As the ball with mass m1, whose string makes an angle θi1 with the vertical, swings through its equilibriumposition, it lowers its gravitational energy from m gy m g L L1 0 1 1= −( cos )θ i to zero. This change in potential energytransforms into a change in kinetic energy. That is,

m g L L m v v gL1 1 1 1 12

1 1 1

1

22 1( cos ) ( ) ( ) ( cos )− = ⇒ = −θ θi i

Similarly, ( ) ( cos ). , ( . ( ) .v gL v v1 2 2 1 2 1 1 1 22 1 45 2 396= − = = ° = =θ θ θi i i Using we get ) m/s Both balls are moving

at the point where they have an elastic collision. Since our analysis of elastic collisions was for a situation in whichball 2 is initially at rest, we need to use the Galilean transformation to change to a frame S′ in which ball 2 is at rest.Ball 2 is at rest in a frame that moves with ball 2, so choose S′ to have V = –2.396 m/s, with the minus sign becausethis frame (like ball 2) is moving to the left. In this frame, ball 1 has velocity ( ′v1 )1 = (v1)1 – V = 2.396 m/s + 2.396m/s = 4.792 m/s and ball 2 is at rest. The elastic collision causes the balls to move with velocities

( ) ( ) – ( . .

( ) ( ) ( .

′ = −+

′ = −

′ =+

′ =

vm m

m mv

vm

m mv

2 11 2

1 21 1

2 21

1 21 1

1

34 792 597

2 2

32 396

m/s ) = 1 m/s

m/s ) = 3.194 m/s

We can now use v = v´ + V to transform these back into the laboratory frame:( ) . . .

( ) . . .

v

v2 1

2 2

1 597 2 396 3 99

3 195 2 396 0 799

= − − = −= − =

m / s m / s m / s

m / s m / s m / s

Having determined the velocities of the two balls after collision, we will once again use the conservation equationK U K Uf gf i gi+ = + for each ball to solve for the θf1 and θf2.

1

21

1

201 3 1

21 1 1 2 1

2m v m gL m v( ) ( cos ) ( )+ − = +θ f J

Using ( ) ,v3 1 0= this equation simplifies to

gLgL

( cos ) ( ) cos ( ) .11

21

1

279 31

21

21− = − ⇒ = − − ⇒ = °θ θ θf f f3.99 m/s 3.99 m/s

The 100 g ball rebounds to 79.3°. Similarly, for the other ball:1

21

1

202 3 2

22 2 2 2 2

2m v m gL m v( ) ( cos ) ( )+ − = +θ f J

Using ( ) ,v3 2 0= this equation becomes

cos ( . ) .θ θf f22

211

20 799 14 7= −

⇒ = °gL

The 200 g ball rebounds to 14.7°.

10.74. Model: Model the sled as a particle. Because there is no friction, the sum of the kinetic andgravitational potential energy is conserved during motion.Visualize:

Place the origin of the coordinate system at the center of the hemisphere. Then y0 = R and, from geometry, y1 = Rcos .φSolve: The energy conservation equation K U K U1 1 0 0+ = + is

1

2

1

2

1

22 11

21 0

20 1

21mv mgy mv mgy mv mgR mgR v gR+ = + ⇒ + = ⇒ = −cos ( cos )φ φ

(b) If the sled is on the hill, it is moving in a circle and the r-component of rFnet has to point to the center with

magnitude F mv Rnet = 2 / . Eventually the speed gets so large that there is not enough force to keep it in a circulartrajectory, and that is the point where it flies off the hill. Consider the sled at angle φ . Establish an r-axis pointingtoward the center of the circle, as we usually do for circular motion problems. Newton’s second law along this axisrequires:

( ) cos cos

cos cos

F w n mg n mamv

R

n mgmv

Rm g

v

R

r rnet = − = − = =

⇒ = − = −

φ φ

φ φ

2

2 2

The normal force decreases as v increases. But n can’t be negative, so the fastest speed at which the sled stays onthe hill is the speed vmax that makes n → 0. We can see that v gRmax cos .= φ(c) We now know the sled’s speed at angle φ, and we know the maximum speed it can have while remaining onthe hill. The angle at which v reaches vmax is the angle φmax at which the sled will fly off the hill. Combining the twoexpressions for v1 and vmax gives:

2 1 2 1

2

3

2

348 21

gR gR R R( cos ) cos ( cos ) cos

cos cos .

max max

max max

− = ⇒ − =

⇒ = ⇒ =

= °−

φ φ φ φ

φ φ