py 211 mid term exam ii - physics.bu.edu

PY 211 Mid Term Exam II

April 3, 2007

Formula Sheet, Useful Data, and Moments of Inertia

Formula Sheet, Useful Data, and Moments of Inertia (Cont.)

Useful Numbers:

G = 6.67 x 10-11 Nm2/kg2

g = 9.8 m/s2

REarth = 6.38 x 103 kmMEarth = 5.97 x 1024 kgRMoon = 1.74 x 103 kmMMoon = 7.35 x 1022 kgEarth-Moon Distance = 384 x 103 km

Problem 1 (Multiple Choice)

_____ Two objects, a spherical ball with mass M and rectangular block with mass 3M, start from

rest on top of a hill. The spherical ball rolls down the hill without slipping. The

rectangular block slides down the hill without friction. Which has the higher linear

velocity of its center-of-mass at the bottom of the hill?

A) The rolling ball

B) The rectangular block

C) Both have the same velocity

D) Not enough information

!

Ei = E f

Ei = mgh" E f = mgh

Rolling: Sliding:

!

E f = KECM + KERolling

!

E f = KECM

B

Problem 1 (Multiple Choice) (Cont.)

_____ The moons of Mars are called Phobos and Deimos and have orbital radii of 9,378 km and

23,459 km respectively. What is the ratio of the period of revolution of Phobos to that of

Deimos?

A) 0.253

B) 0.400

C) 1.58

D) 2.53

E) 4.00

F) 0.158

!

" =2#r

v

FG

=GMm

r2

=mv

2

r

v =GM

r

!

" = 2#r3

GM$ r3 / 2

"1

"2

=r1

r2

%

& '

(

) *

3 / 2

!

"Phobos

"Deimos

=rPhobos

rDeimos

#

$ %

&

' (

3 / 2

=9,378 km

23,497 km

#

$ %

&

' (

3 / 2

) 0.253

A


_____ Compare a ball bouncing elastically off of a wall to a piece of putty of the same mass and

initial velocity hitting the wall and sticking to it. In which case is the impulse to the wall

the largest?

A) The ball hitting the wall

B) The putty hitting the wall

C) They give the same impulse to the wall

D) Not enough information is given

!

I " F # dt$ = %p

Bounce Stick

!

"p = 2pi

!

"p = pi

A


_____ A projectile is launched from Earth’s surface straight up with an initial speed equal to the

escape velocity from Earth, vesc. At an altitude of twice the radius of the Earth (2RE),

neglecting air resistance, what is the projectile’s speed?

A) 2vesc

B) !2vesc

C) vesc/!2

D) vesc/4

E) vesc/2

F) vesc

!

Ei = E f

Ei =mvesc

2

2"GMm

rE

E f =mv

2

2"GMm

2rE

!

mvesc

2

2"GMm

rE

=mv

2

2"GMm

2rE

v2 = v

esc

2 +GM

rE

1" 2( ) = vesc

2"GM

rE

!

vesc

2=2GM

rE

v2

= vesc

2"1

2vesc

2

v = vesc/ 2

C


_____ A figure skater stands on one spot of the ice and spins with her arms extended. As she

pulls her arms in close to her body, her angular velocity changes. Compared to having her

arms extended outwards, with arms pulled in, her new angular momentum:

A) Equals zero

B) Increases

C) Decreases

D) Remains unchanged

E) Not enough information

!

r " = 0#

r L is conserved!$

D

Problem 1 (Multiple Choice) (Cont.)_____ I build a children’s slide to be 3 m high. I can build the slide so that it is either straight or

has curves and bumps. Neglecting friction, how does the shape of the slide affect the time

that it takes to slide down it and the final speed at the bottom?

A) the time required to go down the slide and the final speed are both

independent of the shape

B) the time required to go down the slide depends on the shape, but the final

speed does not

C) the time required to go down the slide and the final speed both depend on the

shape

D) the final speed required to go down the slide depends on the shape, but the

time does not


!

E f = Ei

mgy =mv

2

2" v is unchanged by the shape!

But, the time it takes to slidedown IS affected by the shape!

B


_____ A spring-loaded dart gun shoots a dart straight up into the air and the dart reaches a

maximum height of 24 m. The same dart is shot up a second time with the spring

compressed half as far. How far does the dart go up this time?

A) 96 m

B) 24 m

C) 12 m

D) 6 m


!

Ei = E f

Ei =kx

2

2

E f = mgy

!

y =kx

2

2mg

x"1

2x

y"1

4y

!

y =24 m

4= 6 m

D

Problem 2

2. A 40 kg snowboarder is at the top of an icy 1 m high curved ramp as shown below. The ramp has no friction. At the bottom of the ramp is a flat region of snow with a coefficient of kinetic friction,

k = 0.1. T hree meters away from the bottom of the first ramp is a second icy ramp that the snowboarder can go up. This ramp also has no friction. The snowboarder goes back and forth until eventually coming to a stop. Find the amount of work done by friction each time the snowboarder goes completely across the snow to t he opposite ramp? How many times does the snowboarder cross the flat region? Where does the snowboarder finally stop?

The Math

Skier

!

N

!

f

!

mg

!

N "mg = 0

M = mg

f = µN = µmg

!

W =r F "

r d = µmg#x

W = (0.1)(40 kg)(9.8 m/s2)(3 m)

W =117.6 J

!

Ei = E f

Ei = mgy

Ei = (40 kg)(9.8 m/s2)(1 m)

Ei = 392 J

!

Let n be the number of trips

taken by the skier. Then :

Ei = nW

n =Ei

W= 3.33

Thus the skier makes 3 complete

trips before stopping.

The Math (Cont.)

!

After the 3rd trip, the skier has :

E = Ei" 3W = 39.2 J

!

W" = #KE

39.2 J = µmg#x '

#x '=39.2 J

(0.1)(40 kg)(9.8 m/s2)

#x '=1 m

Thus, the skier stops 1 m from the right

edge of the snowy surface.

Problem 33. A plane is diving toward the ground and then climbs back upwards as shown below. A lift force

acts perpendicular to the plane’s displacement, s = 1.7 x 103 m, which should be taken to be the

same for each case. The engines of the plane exert a thrust as shown which has the same

magnitude during the dive and the climb. The plane’s weight is 5.9 x 104 N. Is the total net work

done by all the forces greater during the dive or the climb? Explain your answer. Find the

difference in the total net work done between the dive and the climb.

The Math

!

W =r F "

r d # = Fdcos$#

The Dive The Climb

!

W =r F lift "

r s +

r F weight "

r s +

r F thrust "

r s

W = 0 + Fweights cos(75o) + Fthrusts

W = (5.9#104 N)(1.7#103 m)cos(75o) + Fthrusts

W = 2.6 #107 J + Fthrusts

!

W =r F lift "

r s +

r F weight "

r s +

r F thrust "

r s

W = 0 + Fweights cos(105o) + Fthrusts

W = (5.9#104 N)(1.7#103 m)cos(105o) + Fthrusts

W = $2.6 #107 J + Fthrusts

More work is done during the fall than the climb.The difference in the work is 5.2 x 107 J (thethrust cancels!)

Problem 4

4. A 75.0 kg stunt man horizontally out of a circus canon with a velocity of 10 m/s. He collides with

a pile of mattresses which are compressed 1.00 m before he is brought to rest. Find the average

force, F , exerted by the mattresses on the stuntman?

!

W" = #KE

W" = F #x

#KE =mv

2

2

F =mv

2

2#x=

(75 kg)(10 m/s)2

2(1.0 m)

F = 3,750 N

Problem 5

5. A billiard ball moving at 5.00 m/s collides elastically with a stationary ball of the same mass.

After the collision the first ball moves at 4.33 m/s at an angle of 30.0 degrees with respect to the

original line of motion. Ignoring friction and any possible rotational motion, find the struck ball's

velocity (direction and magnitude) after the collision.

!

Ei =mv

2

2

px,i = mv

py,i = 0

!

E f =mv

1

2+ mv

2

2

2

px, f = mv1, f cos" + mv

2, f cos#

py, f = mv1, f sin" $mv

2, f sin#=

!

v1, f sin" = v

2, f sin#

!

v2

= v1

2+ v

2

2

!

v = v1, f cos" + v

2, f cos#

!

"

!

v2, f = 25 m

2/s

2 "18.75 m2/s

2

v2, f = 2.5 m/s

sin# =2.17 m/s

2.5 m/s

# = 60o

!

2.17 m/s = v2, f sin"

25 m2/s

2=18.75 m

2/s

2+ v

2, f

2

Problem 6

6. A 60.0 kg person running at an initial speed of 3.25 m/s jumps onto a 120 kg cart that is initially at

rest as shown below. The person slides on the cart's top surface and finally comes to rest relative to

the cart. The coefficient of kinetic friction between the person and the cart is 0.375. Friction

between the cart and the ground can be neglected. (a) Find the final speed of the person and cart

relative to the ground. (b) Find the friction force acting on the person as he is sliding across the top

surface of the cart. (c) Find how long the friction force acts on the person.

The Math

!

All forces are internal, thus :

pf = pi

pi = mvi

pf = (m + M)v

!

v =m

m + Mvi

v =60 kg

60 kg +120 kg3.25 m/s

v =1.08 m/s

Jogger

!

N "mg = 0

N = mg

f = µN = µmg

f = (0.375)(60 kg)(9.8 m/s2)

f = 221 N

!

N

!

mg

!

f

!

"p jogger = Fjogger"t

"t ="p jogger

Fjogger

"t =mvi #mv

µmg=vi # v

µg

"t =3.0 m/s # 1.08 m/s

(0.375)(9.8 m/s2)

"t = 0.33 sec.

Problem 7

7. A string is wound around a uniform disk of radius R and mass M = 2 kg. The disk is released from

rest with the string vertical and its top end tied to a fixed bar as shown below. Find the tension in

the string and the magnitude of the acceleration of the center-of-mass of the disk.

!

T

!

Mg!

R

!

M

The Math

!

r " # = I$ = RT

Fy = macm = mg%T#

!

" =acm

R

I =mR

2

2

T =ma

cm

2

!

macm = mg"macm

2

acm =2

3g

!

T =

m2

3g

2=mg

3

Problem 8

8. A 1 kg point-like mass moves to the right with velocity, v = 2.0 m/s, and strikes a long thin

vertical rod of length L = 1 m and mass M = 3 kg pivoted on a frictionless pin at one end as shown.

The point-like mass sticks to the rod and the combination rotates clockwise. Find the angular

velocity of the combination just after impact of the small mass. (For 5 bonus points, find the

angular velocity of the combination after it has rotated 90°as shown. )

The Math

!

Li = Lf

Li = lmv

Lf = I"

!

I = ml2

+Ml

2

3= m +

M

3

"

# $

%

& ' l

2

!

lmv = m +M

3

"

# $

%

& ' l

2(

( =mv

m +M

3

"

# $

%

& ' l

Bonus:

!

Ei =(m + M /3)l

2" 2

2+ Mg

l

2+ mgl

E f =(m + M /3)l

2" '2

2

"'=2

(m + M /3)l2

(m + M /3)l2" 2

2+ Mg

l

2+ mgl

#

$ %

&

' (

"'= " 2+(2m + M)

(m + M /3)

g

l>"

Problem 99. Two astronauts, each having a mass of 75.0 kg, are connected by a 10.0 m rope of negligible mass.

They are isolated in space, orbiting their center-of-mass with tangential speeds of 5.0 m/s. Treating the astronauts as particles, what are the initial magnitude and direction of the angular momentum of the system? By pulling on the rope, one of the astronauts shortens the distance between them to 5.0 m. What are the astronauts’ new tangential speeds?

!

Li= I" = (2m

d

2

#

$ % &

' (

2

)v

d /2= dmv

Li= (10 m)(75 kg)(5 m/s) = 3,7500 J ) s

The RHR tells us r L points out of the page.

!

Li = Lf

Lf = d'mv'

v'=Li

d'm

v'=3,500 J " s

(5 m)(75 kg)=10 m/s

Problem 10

10. Consider the following table of planetary data. Using these data, find the distance from the center, R, of the planet for a satellite in a synchronous orbit above the equator of Mars. Note that the Earth’s mass is MEarth = 5.97 x 1024 kg.

The Math

To achieve a geosynchronous orbit, the satellite mustmake one revolution about Mars in the same amount oftime it take Mars to rotate once about its own axis.

!

" =2#r

v

F =mv

2

r=GMm

r2

v = GM /r

!

" = 2#r

3

GM

r = GM"

2#

$

% &

'

( )

2$

% & &

'

( ) )

1/ 3

M = MMars

" = Period of Mars Rotation

!

r = (6.67 "10#11 N $m2 $ kg-2)(0.1074)(5.97 "1024 kg)24 hr. (3600 sec./1 hr.) + 37 min. (60 secd /1 min.) + 23 sec.

2%

&

' (

)

* +

2&

' ( (

)

* + +

1/ 3

!

r " 20,417 km

Statistics

Statistics (Cont.)

py 211 mid term exam ii - physics.bu.edu

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