momentum practice solutions

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Version 001 – Momentum Practice – Ross – (20691)

This print-out should have 31 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.

Animal Rescue Plane

001 (part 1 of 2) 0.0 pointsAn animal-rescue plane flying due east at17 m/s drops a bale of hay from an altitude of 67 m .

The acceleration due to gravity is9.81 m/s2 .

If the bale of hay weighs 200 N , what is themomentum of the bale the moment it strikesthe ground?

Correct answer: 816.396 kg m/s.

Explanation:

Let : v plane = 17 m/s ,

h = 67 m , and

W = 200 N .

v plane

v⊥

θ

67 m

The bale strikes the ground at an angle, soits final velocity v will have both a horizontaland a vertical component, and

v2 = v2⊥ + v2

h

Vertically, conservation of energy gives us

E kf = E po

1

2 m v2

⊥ = m g h

v2⊥ = 2 g h

The horizontal component of the bale is v, thspeed of the plane, so the speed of the bale ait strikes the ground is

v =

v2

⊥ + v2

plane =

2 g h + v2

plane

The momentum of the bale is thus

p = m v = W

g

2 g h + v2

plane

= 200 N

9.81 m/s2

×

2 (9.81 m/s2) (67 m) + (17 m/s)2

= 816.396 kg m/s .

002 (part 2 of 2) 0.0 pointsAt what angle of inclination will the balstrike? Answer between −180◦ and +180◦.

Correct answer: 64.8791◦.

Explanation:When the bale strikes the ground, the ver

tical component is the side opposite the angl

of inclination and the horizontal componenis the side adjacent, so

tan θ = v⊥v plane

θ = arctan

v⊥v plane

= arctan

√ 2 g h

v plane

= arctan 2 (9.81 m/s2) (67 m)

17 m/s

= 64.8791◦ .

keywords:

Conceptual 06 01003 (part 1 of 4) 0.0 points




Calculate the momentum for a 0.2 kg riflebullet traveling 100 m/s.

Correct answer: 20 kg · m/s.

Explanation:

Let : m = 0.2 kg and

v = 100 m/s .

p = m v

= (0.2 kg)(100 m/s) = 20 kg · m/s .

004 (part 2 of 4) 0.0 points

What momentum does a 1100 kg automobiletraveling 0.3 m/s (a few miles per hour) have?


Explanation:

Let : m = 1100 kg , and

v = 0.3 m/s .

p = (1100 kg)(0.3 m/s) = 330 kg · m/s .

005 (part 3 of 4) 0.0 pointsWhat momentum does a 40 kg person running11 m/s (a fast sprint) have?


Explanation:

Let : m = 40 kg , and

v = 11 m/s .

p = (40 kg)(11 m/s) = 440 kg · m/s .

006 (part 4 of 4) 0.0 points

What momentum does a 18000 kg truck traveling 0.04 m/s (a slow roll) have?


Explanation:

Let : m = 18000 kg , and

v = 0.04 m/s .

p = (18000 kg)(0.04 m/s) = 720 kg · m/s .

Hewitt CP9 07 R32007 0.0 points

Can momenta cancel? Can kinetic energie

cancel?

1. Neither can cancel.

2. They both can cancel.

3. Momenta cannot cancel; kinetic energiecan.

4. Momenta can cancel; kinetic energies cannot cancel. correct

Explanation:Momentum is a vector quantity, so the to

tal momentum of two moving objects can bless than the momentum of either one aloneKinetic energy is a scalar and always positiveso the total kinetic energy of two moving ob

jects is always greater than the kinetic energof either one alone.

Hewitt CP9 07 R33008 0.0 points

If a moving object doubles its speed, howmuch more momentum does it have? Howmuch more kinetic energy?

1. Momentum won’t change; kinetic energdoubles.

2. Both will double.

3. Both will remain the same.




4. Unable to determine

5. Momentum doubles; kinetic energy won’tchange.

6. Momentum doubles; kinetic energy willincrease by four times. correct

Explanation:Momentum is defined as m v and kinetic

energy is defined as 1

2 m v2; so momentum

doubles and kinetic energy increases by fourtimes when the speed doubles.

AP M 1993 MC 16009 0.0 points

A balloon of mass M is floating motionless inthe air. A person of mass less than M is ona rope ladder hanging from the balloon. Theperson begins to climb the ladder at a uniformspeed v relative to the ground.

How does the balloon move relative to theground?

1. The balloon does not move.

2. Down with speed v

3. Up with speed v

4. Up with a speed less than v

5. Down with a speed less than v correct

Explanation:Let the mass of the person be m.Total momentum is conserved (because the

exterior forces on the system are balanced),especially the component in the vertical di-

rection.When the person begins to move, we have

m v + M vM

= 0 ,

=⇒ vM = − m

M v < v

=⇒ |vM | =

m

M v < v,

since m < M =⇒ m

M < 1.

Thus the balloon moves in the oppositdirection.

Cannon Recoil010 (part 1 of 2) 0.0 points

A revolutionary war cannon, with a mass o

2100 kg, fires a 17.7 kg ball horizontally. Thcannonball has a speed of 141 m/s after it haleft the barrel. The cannon carriage is on flat platform and is free to roll horizontally.

What is the speed of the cannon immediately after it was fired?

Correct answer: 1.18843 m/s.

Explanation:

Let : m = 17.7 kg ,M = 2100 kg , and

v = 141 m/s .

The cannon’s velocity immediately after iwas fired is found by using conservation omomentum along the horizontal direction:

M V + m v = 0

⇒ −V =

m

M v

where M is the mass of the cannon, V is thvelocity of the cannon, m is the mass of thcannon ball and v is the velocity of the cannonball. Thus, the cannon’s speed is

|V | = m

M |v|

= 17.7 kg

2100 kg (141 m/s)

= 1.18843 m/s .

011 (part 2 of 2) 0.0 pointsThe same explosive charge is used, so the totaenergy of the cannon plus cannonball systemremains the same.

Disregarding friction, how much faste

would the ball travel if the cannon wermounted rigidly and all other parameters remained the same?





Explanation:By knowing the speeds of the cannon and

the cannon ball, we can find out the totalkinetic energy available to the system

K net = 1

2 m v2 +

1

2 M V 2 .

This is the same amount of energy availableas when the cannon is fixed. Let v′ be thespeed of the cannon ball when the cannon isheld fixed. Then,

1

2 m v′2 =

1

2 (m v2 + M V 2 ) .

⇒ v′ =

v2 +

M

m V 2

= v

1 +

m

M

= (141 m/s)

1 +

17.7 kg

2100 kg

= 141.593 m/s .

Thus, the velocity difference is

v′ − v = 141.593 m/s − 141 m/s

= 0.592967 m/s .

Jump Up012 0.0 points

Bill (mass m) plants both feet solidly on theground and then jumps straight up with ve-

locity →v .

The earth (mass M ) then has velocity

1. V Earth

= + →v man .

2. V Earth

= +

m

M

→vman

.

3. V Earth

= +

M

m

→vman

.

4. V Earth

= −

m

M

→vman

.

5. V Earth

= −m

M

→vman

. correct

6. V Earth

= −

M

m

→vman

.

7. V Earth

= − →vman

.

8. V Earth

= +

m

M

→vman

.

Explanation:The momentum is conserved. We have

m →v man + M

→

V Earth

= 0

So→

V Earth

= −m

M

→vman

.

Attack Helicopter013 (part 1 of 2) 0.0 points

An attack helicopter is equipped with a 20mm cannon that fires 158 g shells in thforward direction with a muzzle speed o1160 m/s. The fully loaded helicopter haa mass of 2820 kg. A burst of 140 shells ifired in a 1.66 s interval.

What is the resulting average force on thhelicopter?

Correct answer: 15457.4 N.Explanation:

The impulse imparted to the shells equalthe change in momentum:

F av ∆t = ∆mv

The mass change is

∆m = n m = (140 shells) (158 g) = 22.12 kg

so the average force is

F = v ∆m

t

= (1160 m/s)(22.12 kg)

1.66 s

= 15457.4 N .

Since the velocity of the shells is much greatethan the velocity of the helicopter, there is nneed to use relative velocity.




014 (part 2 of 2) 0.0 pointsBy what amount is its forward speed reduced?


Explanation:

From conservation of momentum ∆ p = 0,so

∆v = ∆m v

M

= (22.12 kg) (1160 m/s)

2820 kg

= 9.09901 m/s ,

Force on a Golf Ball

015 0.0 pointsA golf ball (m = 72.6 g) is struck a blow thatmakes an angle of 35.4◦ with the horizontal.The drive lands 284 m away on a flat fairway.

The acceleration of gravity is 9.8 m/s2 .If the golf club and ball are in contact for

10.2 ms, what is the average force of impact?Neglect air resistance.

Correct answer: 386.399 N.

Explanation:

Let : g = 9.8 m/s2 ,

m = 72.6 g ,

θ = 35.4◦ , and

t = 10.2 ms .

Note that the range of the golf ball is given by

L = v2 sin2θ

g ,

so the initial velocity of the ball is

v0 =

L g

sin(2 θ) =

(284 m) (9.8 m/s2)

sin(2 × 35.4◦)

= 54.2875 m/s .

The average force exerted is the change in itsmomentum over the time of contact

F = m v0

t

= (72.6 g) (0.001 kg/g) (54.2875 m/s)

(10.2 ms) (0.001 s/ms)

= 386.399 N .

Hammer and Nail

016 0.0 pointsA(n) 3.6 lb hammer head, traveling at 6.7 ft/strikes a nail and is brought to a stop i0.00078 s.

The acceleration of gravity is 32 ft/s2 .What force did the nail receive?

Correct answer: 966.346 lb.

Explanation:

Let : W = 3.6 lb ,

vo = 6.7 ft/s , and

t = 0.00078 s .

Impulse is the change in momentum:

F ∆t = m ∆v = m (vf − vo) .

The hammer’s final velocity is 0, so

F t = W g

(0 − vo)

Thus the force the hammer felt was

F = −W vo

g t

(it had to be a negative force to stop it) andby Newton’s Third Law of Motion, the forcthe nail received was

F = W vo

g t =

(3.6 lb)(6.7 ft/s)

(32 ft/s2) (0.00078 s)

= 966.346 lb .

Rebounding Ball 02017 0.0 points

A 2.3 kg steel ball strikes a massive wall a6.67 m/s at an angle of 31.5◦ with the plane o




the wall. It bounces off with the same speedand angle.

x

y6 .6 7 m

/ s

6. 6 7 m

/ s

31.5◦

31.5◦

If the ball is in contact with the wall for0.244 s , what is the magnitude of averageforce exerted on the ball by the wall?


Explanation:

Let : m = 2.3 kg ,

v = 6.67 m/s ,

θ = 31.5◦ , and

t = 0.244 s .

x

yvf

m

vim

θ

θ

F ∆t = ∆ p .

Only the component of the ball’s velocityperpendicular to the wall will change. Thisvelocity component before hitting the wall is

v⊥ = v cos θ = (6.67 m/s)cos31.5◦

= 5.68711 m/s .

After hitting the wall, this component is−5.68711 m/s, because the rebound angle isalso 31.5◦ . The change in momentum duringcontact with the wall is therefore

∆ p = m vf − m v0 = m (−v⊥) − m v⊥= −2 m v⊥ = −2 (2.3 kg) (5.68711 m/s)

= −26.1607 kg · m/s ,

so the average force on ball is

F =

∆ p

t

= 26.1607 kg · m/s

0.244 s = 107.216 N

Velocity Change in Space018 0.0 pointsSmall rockets are used to make small adjustments in the speed of satellites. One suchrocket has a thrust of 51 N.

If it is fired to change the velocity of a(n74000 kg space craft by 59 cm/s, how lonshould it be fired?

Correct answer: 856.078 s.

Explanation:

Impulse is defined by

I = F t = m ∆v

so

t = m ∆v

F

Dimensional analysis: for t

kg · cm/s

N · 1 m

100cm =

kg · m/s

kg

·m/s2

= s

AP M 1998 MC 13019 0.0 points

A disc of mass m moves horizontally to thright with speed v on a table with negligiblfriction when it collides with a second disc omass 2 m. The second disc is moving horizon

tally to the right with speed v

6 at the momen

of impact.

m 2 m

v

v

6 before

The two discs stick together upon impact.

3 m

vf

after




What is the speed of the composite bodyimmediately after the collision?

1. None of these is correct.

2. vf = 2

3 v

3. vf = 5

21 v

4. vf = 4

9 v correct

5. vf = 5

9 v

6. vf = 3

10 v

7. vf = 1

3 v

8. vf = 11

35 v

9. vf = 2

5 v

10. vf = 7

15 v

Explanation:The total momentum of the system is con-

served because there is no exterior force. Sowhat we have is

(m + 2 m) vf = m v + 2 m v6

3 m vf = m v

1 +

2

6

3 vf =

6

6 +

2

6

v

3 vf = 8

6 v

vf = 8

(6) (3) v

= 818

v , so

= 4

9 v .

keywords:

Inelastic Collision 08020 0.0 points

If all three collision in the figure are totallyinelastic, which cause(s) the most damage?

m v

brick

wall

I

m 2mv 0.5v

II

m 0.5mv 2v

III

1. III correct

2. I, II

3. II, III

4. II

5. all three

6. I, III

7. I

Explanation:The right car in III loses more kinetic en

ergy in the collision than the right car in II o

the wall (which has zero kinetic energy) in ISince any amount of kinetic energy lost goeinto deforming the cars, the most damage occurs in III.

Water Fills a Bucket021 0.0 points

Water (ρ = 1000 kg/m3) falls without splashing at a rate of 0.253 L/s from a height o62.3 m into a 1.05 kg bucket on a scale.

The acceleration of gravity is 9.8 m/s2 .

If the bucket is originally empty, what doethe scale read after 3.56 s ?


Explanation:From conservation of energy, the velocity i

m g h = 1

2 m v2

v =

2 g h




=

2 (9.8 m/s2)(62.3 m)

= 34.944 m/s .

u is the volume rate of the flow, so the massrate of the flow is

um = ρ u

= (1000 kg/m3)(0.253 L/s)

1 m

100 cm

3

× 1000 cm3

1 L= 0.253 kg/s .

The force exerted by the water hitting thebucket is

F w = m v

t

= um t vt

= um v

= (0.253 kg/s)(34.944 m/s)

= 8.84082 N .

Thus the overall force on the scale is

F = F w +

M bucket + um t

g

= 8.84082 N

+1.05 kg + (0.253 kg/s)(3.56 s)×(9.8 m/s2)

= 27.9575 N .

Circular Hole in a Square022 0.0 points

A circular hole of diameter 16.3 cm is cutout of a uniform square of sheet metal havingsides 32.6 cm, as in the figure.

d

l

l

What is the distance between the center of mass and the center of the square?

Correct answer: 2.81602 cm.

Explanation:Choose the center of the square as the ori

gin. Let the radius of the circle be r. Usinthe formula for center of mass, we have

xcm = mi ximi

and likewise in y direction. Notice that thside of the square has length 4 r. Thinking othe absence of the circle as being a negativmass element there, we have, assuming masdensity of ρ,

xcm = ρ (4 r)2 (0) − r ρ π r2

ρ (4 r)2 − ρ π r2 =

−π

16 − π

r

The calculation is the same in y directionTherefore, the distance from the center of thsquare is

s =√

2

π

16 − π

r .

Serway CP 08 08023 0.0 points

A water molecule consists of an oxygen atom

with two hydrogen atoms bound to it. Thangle between the two bonds is 106◦.

y

x

53◦

53◦

O−

H+

H+

0 . 1 2 7

n m

0 . 1 2 7

n m

If each bond is 0.127 nm long, how far fromthe oxygen atom is the center of mass of thmolecule? Take the mass of an oxygen atomto be 16 times the mass of a hydrogen atom.

Correct answer: 0.00849227 nm.

Explanation:




Given : θ = 106◦ ,

L = 0.127 nm , and

mO = 16 mH .

Fromxcm =

mi xi

mi

if we choose the x-axis to start from the oxy-gen atom and so that the hydrogen atoms aresymmetrical with respect to it, then

ycm = 0

xcm = 2 mH L cos θ

2

2 mH + 16 mH

= 2 mH L cos θ

2

18 mH

= L cos θ

2

9

= (0.127 nm) cos 53◦

9

= 0.00849227 nm .

Tipler PSE5 12 20

024 (part 1 of 2) 0.0 pointsA square plate is produced by welding to-gether four smaller square plates, each of sidea. The weight of each of the four plates isshown in the figure.

x

y

90 N

20 N 80 N

80 N

(0, 0) (2a, 0)

(0, 2a)(2a, 2a)

Find the x-coordinate of the center of grav-ity (as a multiple of a).

Correct answer: 1.09259 a .

Explanation:

Let : x1 = x2 = 1

2 a ,

x3 = x4 = 3

2 a ,

W 1 = 90 N ,

W 2 = 20 N ,

W 3 = 80 N , and

W 4 = 80 N .

y

xW 1

W 2 W 3

W 4

a

2

3 a

2

a

2

3 a

2

The total weight is

W = W 1 + W 2 + W 3 + W 4 = 270 N .

Applying the definition of center of gravity

xcg = i W i xi

W 1 + W 2 + W 3 + W 4

=(W 1 + W 2)

a

2 + (W 3 + W 4)

3 a

2W

= (90 N + 20 N) + 3 (80 N + 80 N)

2 (270 N) a

= 1.09259 a .

025 (part 2 of 2) 0.0 pointsFind the y-coordinate of the center of gravit(as a multiple of a).

Correct answer: 0.87037 a .

Explanation:

Let : y1 = y4 = 1

2 a and

y2 = y3 = 3

2 a .



Version 001 – Momentum Practice – Ross – (20691) 1

ycg =

i W i yi

W 1 + W 2 + W 3 + W 4

=(W 1 + W 4)

a

2 + (W 2 + W 3)

3 a

2W

= (90 N + 80 N) + 3 (20 N + 80 N)

2 (270 N)

a

= 0.87037 a .

Bucket of Water026 0.0 points

The bottom and the top of a bucket haveradii 25 cm and 39 cm, respectively. Thebucket is 32 cm high and filled with water.

Where is the center of gravity relative to thecenter of the bottom of the bucket? Ignorethe weight of the bucket itself.

Correct answer: 18.2967 cm.

Explanation:

Let : rb = 25 cm ,

rt = 39 cm , and

h = 32 cm .

The center of gravity lies above the centerof the bottom. Consider a disk of water atheight y above the bottom. Its radius is

r = rb + y rt − rb

h ,

its area is

A = π r2 = π rb + yrt − rb

h 2

= π

rb + y

∆r

h

2

,

its volume is

V = A dy = dy π

rb + y

∆r

h

2

,

and its mass is

m = ρ V = ρ dy π

rb + y

∆r

h

2

.

The total mass of the water is

M = h

y=0 dm = h

y=0 ρ π

rb + y

∆r

h2

dy

= ρ π

r2

b h + h ∆r rb + 1

3 h ∆r2

.

The height of the center of gravity is

yCG =

h0

y dm

M =

ρ

M

h

0

π r2 y dy

= π ρ

M

h

0

rb + y

∆r

h

2

y dy

=12

r2b h2 + 2

3 ∆r rb h2 + 1

4 ∆r2 h2

r2b h + ∆r rb h +

1

3 ∆r2 h

Since1

2 r2

b h2 + 2

3 ∆r rb h2 +

1

4 ∆r2 h2

= 1

2 (25 cm)2 (32 cm)2

+ 2

3 (14 cm) (25 cm) (32 cm)2

+ 14 (14 cm)2 (32 cm)2

= 6.09109 × 105 cm4 and

r2b h + ∆r rb h +

1

3 ∆r2 h

= (25 cm)2 (32 cm)

+ (14 cm) (25 cm) (32 cm

+ 1

3 (14 cm)2 (32 cm)

= 33290.7 cm3 , then

yCG = 6.09109 × 105 cm4

33290.7 cm3 = 18.2967 cm .

Serway PSE 12 08027 0.0 points

Pat builds a track for his model car out owood. The track is 9 cm wide (along the coordinate), 1 m high (along the y coordinate) and 3.4 m long (along the x coordinatstarting from x = 0).




The runway is cut such that it forms partof the left-hand side of a parabola, see figurebelow.

x

1 m

y

3.4 m

y = 1

(11.56 m) [x − (3.4 m)]2

9 cm

Locate the horizontal position of the centerof gravity of this track.

Correct answer: 0.85 m.

Explanation:

Let : a = 3.4 m ,

b = a2 = 11.56 m ,

z = 9 cm , and

σ = surface density .Basic Concepts:

rCM =

r ρ dV ρ dV

=

r ρ dV

M

x

1 m

y

3.4 m

y = 1

(11.56 m) [x − (3.4 m)]2

9 cm

x

dx

Solution: Let σ represent the mass-perface area. A vertical strip at position x, wit

width dx and height (x − a)2

b , has mass

dm = σ (x − a)2

b dx .

The total mass is

M =

dm =

3.4 m

x=0

σ (x − a)2

b dx

= σ

b

3.4 m

0 m

(x2 − 2 a x + b) dx

= σ

b

x3

3 − a x2 + b x

3.4 m

0 m

.

The x coordinate of the center of gravity is

xcg = 1

M

x dm

= 1

M

3.4 m

x=0

σ x (x − a)2 dx

b

= σ

b M

3.4 m

x=0

x (x − a)2 dx

= σ

b M

3.4 m

0 m

(x3 − 2 a x2 + b x) dx

= σ

b M x4

4 − 2 a x3

3 + b x2

2

3.4 m

0 m

=

x4

4 − 2 a x3

3 +

b x2

2x3

3 − a x2 + b x

3.4 m

0 m

= 11.1361 m2

13.1013 m= 0.85 m .

Chain Falling off Table028 (part 1 of 2) 0.0 points

Given: A uniform flexible chain whose masis 7.6 kg and length is 4 m. A table whose topis frictionless.

Initially you are holding the chain at resand one-half of the length of the chain is hunover the edge of the table. When you let loosof the chain it falls downward.




The acceleration of gravity is 9.8 m/s2 .

a

4 m

2 .

4 m

Radius of tableis negligible

compared to thelength of chain

Mass of chainis 7.6 kg .

Find the acceleration a of the chain when

the length of the chain hanging vertically is2.4 m .

Correct answer: 5.88 m/s2.

Explanation:Note: The initial condition does not enter

into the consideration for the acceleration.

Let : g = 9.8 m/s2 ,

L = 4 m ,

ℓ = 2.4 m , andm = 7.6 kg .

The linear density of the chain is

λ = m

L =

7.6 kg

4 m = 1.9 kg/m .

F = ℓ λ g

cm

The free body diagram in the vertical direction gives

F y = ℓ λ g = L λ a .

Therefore

a = ℓL g (1

= 2.4 m

4 m (9.8 m/s2)

= 5.88 m/s2 .

029 (part 2 of 2) 0.0 pointsFind the magnitude of the velocity of the othe chain when 2.4 m of the chain is hangingvertically.


Explanation:The change in kinetic energy is

∆K = 1

2 m v2

= 1

2 λ L v2 . (2

Let ℓi = L

2 and ℓf = ℓ.

Using the table top as the origin of thy-coordinate and down as the positive y direction

ycm =

mon table

0

+ mhanging

ℓ

2

mon table + mhanging

ycmi =

L − L

2

λ 0 +

L

2 λ

L

4

λ L

ycmf = (L − ℓ) λ 0 + ℓ λ

ℓ

2

λ L

The vertical center of mass difference ∆ycm i

∆ycm = ycmf − ycmi

=λ ℓ

ℓ

2 − λ

L

2

L

4λ L

= 1

8 L [4 ℓ2 − L2] . (3




The change in potential energy is

∆U = λ L g ∆ycm

= 1

8 λ g [4 ℓ2 − L2] . (4)

From conservation of energy ∆K = ∆U , Eq.2 and Eq. 4, we have

1

2 λ L v2 =

1

8 λ g (4 ℓ2 − L2)

v2 = g

4 L [4 ℓ2 − L2] . (5)

Therefore

v =

g4 L

[4 ℓ2 − L2] (6)

=

(9.8 m/s2)

4 (4 m) [4(2.4 m)2 − (4 m)2]

= 2.07654 m/s .

Alternative: You can use the kinematicexpression and remember that the center of mass accelerates at g [not Eq. 1], since the

acceleration of gravity is not a function of mass.

v2 = 2 a ∆ycm

= 2 g 1

8 L [4 ℓ2 − L2]

= g

4 L [4 ℓ2 − L2] , (7)

where ∆ycm is obtained from Eq. 3.Note: Equation 7 is identical to Eq. 5.

Figuring Physics 01030 0.0 points

Consider the pair of identical blocks about tobe simultaneously released from rest. BlockA is completely free, and block B is attachedto one end of a massive chain, the other endheld as shown. When dropped, both blockshit the floor below — a vertical distance equalto the length of the chain.

A B

Which block hits first?

1. Block A

2. The density of the chain must be given todetermine which block hits the floor first.

3. Both blocks hit the floor simultaneously.

4. Block B correct

Explanation:Block B hits the floor first. Notice tha

the race isn’t between blocks A and B, bubetween A and the end part of the B pluchain system, which isn’t in free fall becausone end is fastened to the post. So it doesn’have to accelerate at g like block A. ThB plus chain center of mass, initially closeto the floor, accelerates at less than g. Buacceleration of its “free” end increases in fallsurpassing g — like the tip of a falling polaccelerates more than g when it rotates to thground.

What a surprise this is to bungee jumperto be falling faster than g. So whereas the onldownward force on block A is due to gravityblock B is additionally pulled downward by




the chain and gravity.Alternative: The distance the center of

mass of Block A falls is greater than the dis-tance the center of mass of Block B plus thechain. Therefore it will take Block A longerto fall.

See Paul Hewitt’s books for more examplesof “Figuring Physics”.

Vertical Chain Drop031 0.0 points

A chain whose mass is 8 kg and length is3 m is shown below in the figure.

The acceleration of gravity is 9.8 m/s2 .Consider a chain held at one end such that

the bottom end of the chain just touches thefloor. The top end of the chain is released.

3 m

Chain isfallingto thefloor

9.8 m/s2

Mass of chain is

8 kg .

What is the force exerted by the chain onthe floor just as the last link of the chain landson the floor?

Correct answer: 235.2 N.Explanation:

Let : m = 8 kg ,

L = 3 m ,

λ = m

L = 2.66667 kg/m ,

g = 9.8 m/s2 , and

x = length of chain on the floor .

L

x

L−x

dx

Length of

chain onthe floor

is x

N

g v

Mass of chain is

m .

Since the mass of the chain on the floor im = λ x , the normal force is

N ≡ d p

dt

= m d v

dt + v

d m

dt

= λ x g + v d m

dt

= m

L x g + v

d m

dt , (1

where m is the mass of the chain and L is thlength of the chain.

The first term is the weight of that part othe chain already on the floor.

From energy conservation

1

2 m v2 = m g x

v2 = 2 g x . (2

The second term of Eq. 1 is the force exerteby the link just landing and being brought trest. If dm is the mass and dx is the lengtof a link (dm = λ dx) and v is the speed withwhich it strikes the floor, we have

v d m

dt = v

λ dx

dt




= v m

L

d x

dt

= m

L v2 , using Eq. 2

= m

L (2 g x) , (3)

where x is the original height above the floorof the link just landing.

Substituting v d m

dt from Eq. 3 into Eq. 1,

the normal force becomes

N = m

L x g + 2

m

L x g

= 3 m

L x g

x=L

= 3 m g (4)

= 3 (8 kg) (9.8 m/s2)

= 3 (78.4 N)

= 235.2 N ,

where x = L just as the last link lands on thetable.

It is useful to drop a chain into your handand notice that the force when the last linktouches your hand is much greater than theforce required to just hold the stationary chainin your hand.

Insight: Suppose you have a length of string of length ∆x held at a height L abovethe floor with the same linear density as

the chain; i.e., λ = m

L . When the string is

dropped from a height x it will impact the

floor for a period of time ∆t = ∆x

v =

∆x√ 2 g x

.

The impulse I ≡ ∆ p (change in momentum)will be I = F ∆t = ∆ p = ∆m v = λ ∆x v .

Therefore the force exerted on the floor willbe

F = I

∆t =

∆m v∆x

v

= λ ∆x v

∆x

v

= m

L v2 , from Eq. 2

= m

L 2 g L , where x = L

= 2 m g , (3′)

which is the same as Eq. 3.

momentum practice solutions

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