PHYS 1111 - Summer 2007 - Professor Caillault

Homework Solutions

Chapter 9

3. Picture the Problem: The owner walks slowly toward the northeast while the cat runs eastward and thedog runs northward.Strategy: Sum the momenta of the dog and cat using the component method. Use the knowncomponents of the total momentum to find its magnitude and direction. Let north be in the y direction,east in the x direction.Solution: 1. Use the componentmethod of vector addition to find theowner’s momentum:

rptotal =

rpd +

rpc = md

rvd + mc

rvc =

= 20.0 kg( ) 2.50 m/s y( ) + 5.00 kg( ) 3.00 m/s x( )rptotal = 15.0 kg ⋅m/s( ) x + 50.0 kg ⋅m/s( ) y

2. Divide the owner’s momentum byhis mass toget the components of the owner’svelocity:

rp0 = m0

rv0 =

rptotal

v0 =rptotal

m0

=15.0 kg ⋅m/s( ) x + 50.0 kg ⋅m/s( ) y

70.0 kg

= 0.214 m/s( ) x + 0.714 m/s( ) y

3. Use the known components to findthedirection and magnitude of theowner’s velocity:

θ = tan−1 0.7140.214

= 73.3°

v0 = 0.2143 m/s( )2 + 0.7143 m/s( )2 = 0.746 m/s

Insight: We bent the rules of significant figures a bit in step 3 in order to avoid rounding error. Theowner is moving much slower than either the cat or the dog because of his larger mass.

5. Picture the Problem: The baseball drops straight down, gaining momentum due to the acceleration ofgravity.

Strategy: Determine the speed of the baseball before it hits the ground, then use equation 2-12 to findthe height from which it was dropped.

Solution: 1. Use equation 9-1 tofind the speedof the ball when it lands:

v =pm=

0.780 kg ⋅m/s0.150 kg

= 5.20 m/s

2. (b) Use equation 2-12 to solve for

y0 . Let y = 0 and

v0 = 0 :

v2 = v02 − 2g y − y0( )

y0 =v2

2g=

5.20 m/s( )2

2 9.81 m/s2( )= 1.38 m

Insight: Another way to find the initial height is to use conservation of energy, setting

mgy0 =

12

mv2 and solving for y0.

13. Picture the Problem: The ball rebounds from the floor inthe manner indicated by the figure at right.

Strategy: The impulse is equal to the change in they-component of the momentum of the ball. Use equation 9-6 in the vertical direction to find the impulse.

Solution: Applyequation 9-6 inthe y direction:

I = Δpy = mΔvy

= m v0 cos65° − (−v0 cos65°) = 0.60 kg( ) 5.4 m/s( ) 2cos65°( ) = 2.7 kg ⋅m/s

Insight: There is no impulse in the x direction because the ball does not change its horizontal speed ormomentum.

momentum.

19. Picture the Problem: The two pieces fly in opposite directions at different speeds..

Strategy: As long as there is no friction the total momentum of the two pieces must remain zero, as itwas before the explosion. Combine the conservation of momentum with the given kinetic energy ratioto determine the ratio of the masses. Let

m1 represent the piece with the smaller kinetic energy.

Solution: 1. Set

rp∑ = 0 and solve for

m1 m2 :

p1 + p2 = 0 = m1v1 + m2v2

m1

m2

= −v2

v1

⇒ m1

m2

2

=v2

v1

2

2. Set K2 K1 = 2 :

K2

K1

= 2 =12

m2v22

12

m1v12

⇒ v2

v1

2

= 2m1

m2

3. Combine the expressions from steps1 and 2:

v2

v1

2

=m1

m2

2

= 2m1

m2

⇒ m1

m2

= 2

4. The piece with the smaller kinetic energy has the larger mass.

Insight: The smaller mass carries the larger kinetic energy because kinetic energy increases with thesquare of the velocity but is linear with mass. Its higher speed more than compensates for its smallermass.

21. Picture the Problem: The lumberjack moves to the right while the log moves to the left.

Strategy: As long as there is no friction the total momentum of the lumberjack and the log remainszero, as it was before the lumberjack started trotting. Combine vector addition for relative motion(equation 3-8) with the expression from the conservation of momentum to find

vL, s = speed of

lumberjack relative to the shore. Let vL, log = speed of lumberjack relative to the log, and

vlog, s =

speed of the log relative to the shore.Solution: 1. (a) Write out theequation for relative motion. Letthe log travel in the negativedirection:

rvL, s =

rvL, log +

rv log, s

vL, s = vL, log − vlog, s

vlog, s = vL, log − vL, s

2. Write out the conservation ofmomentumwith respect to the shore:

rp∑ = 0 = mLvL, s − mlogvlog, s

3. Substitute the expression fromstep 1into step 2 and solve for

vL, s

mLvL, s = mlogvlog, s = mlog vL, log − vL, s( )vL, s mL + mlog( ) = mlogvL, log

vL, s =mlogvL, log

mL + mlog( )=

380 kg( ) 2.7 m/s( )85+ 380 kg( )

= 2.2 m/s

4. (b) If the mass of the log had been greater, the lumberjack’s speed relative to the shore would havebeen greater than that found in part (a), because the log would have moved slower in the negativedirection.5. (c) Use the expression fromstep 3 tofind the new speed of thelumberjack:

vL, s =mlogvL, log

mL + mlog( )=

450 kg( ) 2.7 m/s( )85+ 450 kg( )

= 2.3 m/s

lumberjack:Insight: Taking the argument in (b) to its extreme, if the mass of the log equaled the mass of the Earththe lumberjack’s speed would be exactly 2.7 m/s relative to the Earth (and the log). If the mass of thelog were the same as the mass of the lumberjack, the speed of each relative to the Earth would be halfthe lumberjack’s walking speed.

25. Picture the Problem: The initial and final momentumvectors for this collision are depicted at right.

Strategy: Assuming there is no friction between theplayers’ skates and the ice, we can use conservation ofmomentum together with the fact that the players sticktogether after the collision to find the final velocity. Let themotion of player 1 be in the positive x-direction and themotion of player 2 be at an angle of 115° measuredcounterclockwise from the positive x-axis.

Solution: 1. Write out theconservation of momentum andsolve for

rvf :

rp1i +

rp2i =

rpf

mvx( ) + mv cosθx + mv sinθ y( ) = 2mrvf

12

v 1+ cosθ( ) x + 12

v sinθ( ) y =rvf

12

5.25 m/s( ) 1+ cos115°( ) x + 12

5.25 m/s( ) sin115°( ) y =rvf

1.52 m/s( ) x + 2.38 m/s( ) y =rvf

2. Determine the magnitude of rvf :

vf = vfx

2 + vfy2 = 1.52 m/s( )2 + 2.38 m/s( )2 = 2.82 m/s

Insight: The two players slide away from the collision at 57.4° above the positive x axis. Player 2’sinitial momentum provides the y component of the final momentum, but his x momentum in the −x direction is smaller than player 1’s momentum in the x direction, and so the two players have atotal momentum in the positive x direction.

33. Picture the Problem: The apple and orange collide inthe manner depicted in the figure at right. The mass ofthe apple is 0.130 kg and the mass of the orange is 0.160kg.

Strategy: Use conservation of momentum in the ydirection to find they component of the apple’s speed, and conservation ofmomentum in the x direction to find the x component ofthe apple’s speed. The known components can then beused to find the total speed and direction of the appleafter the collision.

Solution: 1. Conservemomentumin the y direction to find

v1f,y :

py∑ = 0 = m1v1f,y − m2v2f,y

v1f,y =m2v2f,y

m1

=0.160 kg( ) 1.03 m/s( )

0.130 kg( )= 1.268 m/s

2. Conserve momentum inthe x direction to find

v1f,x :

px∑ = m1v1i,x + m2v2i,x = m1v1f,x + 0

v1i,x +m2

m1

v2i,x = v1f,x = 1.11 m/s( ) + 0.160 kg0.130 kg

−1.21 m/s( )

= − 0.38 m/s

3. Find the apple’s speed: v1f = v1f,x

2 + v1f,y2 = 1.268 m/s( )2 + − 0.38 m/s( )2 = 1.32 m/s

4. Find the apple’s direction:

θ = tan−1v1f,y

v1f,x

= tan−1 1.27 m/s

− 0.38 m/s

= −73° +180° = 107°

This angle is measured counterclockwise from the positive x axis.Insight: Since the collision is elastic you can also set

Kf =

12

m1v1f2 + 1

2m2v2f

2 = Ki and use the

Ki = 0.197 J given in the example to find

v1f = 1.31 m/s , which is correct to within rounding error.

Note that we bent the rules of significant figures in steps 1 and 3 to avoid such rounding error. Thisapproach leaves an ambiguity in the x component of the apple’s final velocity, however, and you stillneed conservation of momentum in the x direction to resolve it.

43. Picture the Problem: The geometry of the sulfur dioxidemolecule is shown at right.

Strategy: The center of mass of the molecule will liesomewhere along the y axis because it is symmetric in the xdirection. Find

Ycm using equation 9-15. Both oxygen

atoms will be the same vertical distance yO from the origin.

Let m represent the mass of an oxygen atom, ms the sulfur

atom.

Solution: 1. Use equation 9-15 tofind

Ycm

Ycm =my∑

M=

myO + myO + ms ys

2m + ms

=2myO + 0

2m + ms

=2 16 u( ) 0.143 nm( )sin 30°

2 16 u( ) + 32 u= 0.036 nm

2. Recalling that 1 nm = 1×10−9 m, we can write

Xcm , Ycm( ) = 0, 3.6 ×10−11 m( )Insight: If the angle were to decrease from 120° the center of mass would move upward. Forinstance, if the bond angle were only 90°, the center of mass would be located at (0, 5.1×10−11 m).

61. Picture the Problem: The placement of the lead weight moved the center of mass of the tire from acertain location to the geometric center.

Strategy: Let the coordinate system be placed so that the origin is at the geometric center of the wheel.Treat the entire mass of the wheel as a point mass at some small distance d from the origin. Togetherwith the lead weight the center of mass of the system is exactly at zero, so write out equation 9-14 andsolve for d.

Solution: 1. Solveequation 9-14 for d:

Xcm = 0 =mlw xlw + mwhd

mlw + mwh

d = −mlw

mwh

xlw = −0.0502 kg35.5 kg

25.0 cm ×10 mm/cm( ) = − 0.354 mm

2. Before the lead weight was added, the center of mass was 0.354 mm from the center of the wheel.

Insight: Treating the entire wheel as if it were a point mass located at its center of mass is an importantconcept in mechanics. The attempt to prove this concept led Isaac Newton to develop themathematical method we call calculus!