Physics 6A - UC Santa Barbara

Physics 6A

Practice Final (Fall 2010)

solutions

Prepared by Vince Zaccone

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1) A helicopter traveling upwards at 121 m/s drops a package from a height of 500m above the ground.

Assuming free-fall, how long does it take to hit the ground?

This is a straightforward kinematics problem. We can take up to be positive, so the initial velocity is

v0=121 m/s and the acceleration will be –g. Initial height is y0=500m.

Using the basic formula for height in free fall:


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sec28t

)9.4(2

)500)(9.4(4)121(121t

0500t121t9.4

t)8.9(t)121(m5000

gttvyy

2

2

2

s

m21

sm

221

00

2

We can use the quadratic formula and take the positive answer. This is answer c.

This is a projectile problem with a horizontal initial velocity.

Here are the initial and final values that we know:

ga

0v

5.3v

m0y

m5.1y

y

sm

y,0

sm

x,0

0

2) A person skateboarding with a constant speed of 3.5 m/s releases a ball from a height of 1.5m above

the ground. Find the speed of the ball as it hits the ground.


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When the ball hits the ground we will have the same x-component of velocity, but

the y-component will have increased.

sm

y

s

m2y

02

y,02

y

42.5v

)m5.10)(8.9(20v

yyg2vv

2

We need to use the Pythagorean theorem to find the magnitude of the velocity:

sm22

2y

2x

45.6)42.5()5.3(v

vvv

Answer c

sm

sm

00y,0

sm

sm

00x,0

47.11)35sin()20()sin(vv

38.16)35cos()20()cos(vv

We will need to find the components of the initial velocity:

3) A projectile is launched from the origin with an initial speed of 20.0m/s at an angle of 35.0° above

the horizontal. Find the maximum height attained by the projectile.


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m71.6y

)0y)(8.9(2)47.11(0

yyg2vv

max

maxs

m2sm

02

y,02

y

2

The vertical component of velocity will be 0 when it reaches the highest point, so we can use a

kinematics formula to find the maximum height:

Answer b

4) You are standing on a scale in an elevator. When the elevator is at rest the scale reads 750 N. You

press the button for the top floor and the elevator begins to accelerate upward at a constant rate. If

the scale now reads 850 N what is the acceleration of the elevator?

mg

Fscale

Draw a free-body diagram for the person.

The original reading on the scale is the person’s

weight. Divide by 9.8 to get the mass.

Write down Newton’s 2nd law:

2s

m

scale

3.1a

a)kg5.76(N750N850

mamgF

maF


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Answer a

5) Two blocks are connected by a string and are pulled vertically upward by a force F = 90N

applied to the upper block. Find the tension in the string connecting the two blocks.

Draw the force diagram for the system, then use Newton’s 2nd law

twice. The first time is to find the acceleration of the entire

system, the second time just use the forces on the 1kg mass to

find the tension in the string.

2s

m

0

system

2.20a

)a)(kg3(TTN8.9N6.19N90F

This is answer d.


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2 kg

1 kg

F=90N

19.6N

9.8N

T

T

These tensions cancel out because

they are internal forces to the system.

N30T

)2.20)(kg1(N8.9TF2s

mkgblock1

Total mass of system

Weight of

2kg mass

Weight of

1kg mass

6) A jet plane comes in for a downward dive as shown in Figure 3.39. The bottom part of the path is a

quarter circle having a radius of curvature of 350m. According to medical tests, pilots lose consciousness at

an acceleration of 5.5g. At what speed will the pilot black out for this dive?

The given acceleration is 5.5g. This means 5.5 times the

acceleration due to gravity:22 s

m

s

m 9.538.95.5

350mThe path of the airplane is circular, so the given acceleration

must be centripetal (toward the center).

We have a formula for centripetal acceleration:

sm

2

s

m

2

cent

137vm350

v9.53

r

va

2

Answer a

7) A 1000 kg car is driven around a turn of radius 50 m. What is the maximum safe speed of the car if

the coefficient of static friction between the tires and the road is 0.75?

50

mFfriction

The friction force must be directed toward the

center of the circle. Otherwise the car will slide off

the road. If we want the maximum speed, then we

want the maximum static friction force. The road is

flat (not banked), so the Normal force on the car is

just its weight.

N73508.9kg100075.0mgF2s

msmax,static,friction

This friction force is the only force directed toward the center, so it must be the centripetal force:

sm

2

2

lcentripetafriction

2.19v

m50

vkg1000N7350

r

mvFF


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Answer c.

8) A car is traveling at a speed of 40 m/s. The brakes are applied, and a constant force brings the car

to a complete stop in a time of 6.2 seconds. The tires on the car have a diameter of 70 cm.

How many revolutions does each tire make while the car is braking?


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Given: v0=40 m/s; vf=0; Δt=6.2s; diam=70cm→r=0.35m

srevolution562

rad355

radians355s2.64.18s2.63.114tt

4.18s2.6

3.1140

t

3.114m35.0

40

r

v

revrad

2

s

rad21

srad2

21

0

s

radsrad

srads

m0

0

2

2

Answer c.

9) Two blocks of equal mass M are attached by a massless rope, with one block on a frictionless table,

and the other block hanging down below, as shown. When the block on the table is moving in a

circular path at 1 revolution per second, the hanging block is stationary. Find the radius of the circle.


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a)10 cm

b) 25 cm

c) 50 cm

d) 150 cm

Tension in rope = weight of hanging block = Mg

Tension in rope is also the centripetal force on the moving block

Set these equal and solve for R:

22

cent MRR

MvF

use v=Rω

2

2 gRMRMg

we must convert the given speed from

revolutions per second to radians per second srad

revrad

srev 221

Plug in to get R:

cm25m25.0

2

8.9gR

2

srad

s

m

2

2

Answer b.

10) A merry-go-round is initially rotating at a rate of 1 revolution every 8 seconds. It can be treated as

a uniform disk of radius 2 meters and mass 400 kg. A 50 kg child runs toward the merry-go-round

at a speed of 5.0 m/s, jumping on to the rim (tangentially, as shown).

Find the child’s linear speed after jumping onto the merry-go-round.

a) 1.1 m/s

b) 2.3 m/s

c) 5.0 m/s

d) 7.2 m/s


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We can use conservation of angular momentum for this one.

srad

0 785.0rev1

rad2

s8

rev1

Initial angular speed of disk

22

212

21

disk mkg800m2kg400MRI

s

mkgdisk

2

628IL

s

mkg

sm

child

222child

2

500m25kg50mvRL

mkg200m2kg50mRI

s

mkgchilddisktotal

2childdisktotal

2

1128LLL

mkg1000III

sm

srad

child,fsrad

fs

mkgftotalfinal 3.2128.1m2v128.11128IL

2

11) A ballistic pendulum consists of a solid block of titanium with mass 5 kg, suspended from a light

wire. A bullet of mass 5 g is launched toward the block at an unknown speed. The bullet bounces

back at half its original speed, and the block rises to a height of 1.8 cm above its starting point. What

was the initial speed of the bullet?

a) 100 m/s

b) 200 m/s

c) 300 m/s

d) 400 m/s


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v0

Before

Collision

5kg

Highest

Point

1.8cm

5kg½v0 vblock

After

Collision

5kg

Use conservation of momentum for the collision, then

conservation of energy for the swinging to the highest point.

block021

0 vkg5vkg005.0vkg005.0 sm

blocks

m2block2

1 59.0vm018.08.9kg5vkg52

Now we can put this value into our

momentum formula to get the initial

speed of the bullet

sm

021

0 59.0kg5vkg005.0vkg005.0

sm

0 396v

(Round up to get 400m/s)

12) Two cars are moving toward an intersection. Car A is traveling East at 20 m/s, and Car B is

traveling North at 12 m/s. The mass of Car A is 1000 kg and the mass of Car B is 2000 kg. Driver A is

applying mascara to her eyelashes, and driver B is reading a text message, so neither of them slows

down as they approach the intersection. When the cars crash into each other, they stick together.

Find the common velocity of the cars just after the collision.

a) 32.0 m/s at an angle of 45° North of East

b) 12.0 m/s at an angle of 30° North of East

c) 23.3 m/s at an angle of 23° North of East

d) 10.4 m/s at an angle of 50° North of East


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x

y

A

B

20m/s

12m/s

A/Bvfinal

Momentum is conserved in each direction,

so we get two formulas:

sm

y,fy,fbabb

sm

x,fx,fbaaa

8vvmmvm:y

67.6vvmmvm:x

Combine the components with the Pythagorean theorem

to find the final speed, and use tangent to find the angle:

5067.6

8tan

4.10867.6v

sm

sm

sm2

sm2

sm

final

13) The Atwood’s machine system shown is comprised of a block of mass M attached by a massless

rope to block of mass 2M. The rope passes over a solid cylindrical pulley of mass M and radius R,

and the rope does not slip on the pulley.

Find the acceleration of the heavier block. Use g for gravitational acceleration.


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a) 2/7 g

b) 2/5 g

c) 1/2 g

d) 2/3 g

M 2M

M

R

positive torque

T1

T1

T2

T2

2MgMg

We can set up force formulas for the masses,

and a torque formula for the pulley.

gaMaMg2Ma2MaMg

MaTTMRRTRT

Mg2Ma2TaM2gM2T

MaMgTMaTMg

72

21

21

21Ra2

21

21

22

11

Note that acceleration came out negative due to our choice of direction for positive

torque. Make sure the signs in all formulas match up with your choice for positive,

or you will get the wrong answer.

14) A light uniform ladder of length 5m is leaning against a wall so that the top of the ladder is 4m

above the ground and the bottom of the ladder is 3m from the wall, as shown. How high can a person

of mass 150 kg walk up the ladder before the ladder slips? Assume the coefficient of static friction

between the ladder and the ground is 0.6 and that the wall is frictionless.


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a) 1.0 m

b) 2.5 m

c) 3.0 m

d) 4.0 m

4m

5m

3m

dForces on the ladder are shown in blue

(the weight of the ladder is negligible).

We need to find the distance d.

3 formulas we can write down – 2 force

formulas and one torque (use the

ground as the pivot point).

N

fs

mg

Fwall

m4d0dN1470m4N882

0mgm4F0

N1470N0mgN0F

N882N14706.0mgF0Ff0F

53

5d3

w all

y

w allw allsx

ℓ

use similar triangles to find

the lever arm for the weight:

5

d3

m5

d

m3

15) Block A in the figure weighs 60N. The coefficient of static friction between the block and the

tabletop is 0.25. Find the maximum weight, w, of the hanging block so that the system remains at rest.


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Draw the forces on the diagram.

Everything is in equilibrium, so the forces will add up to

zero for each object in the system (essentially we have 3

free body diagrams in the picture).

Block A is on a horizontal table with no vertical forces other

than gravity and normal force, so the normal force must

equal the weight.

This tells us the max. friction force is (.25)(60N)=15N

Since the problem says to find the maximum hanging

weight, we can say Ffriction=15N.

This will be the same as the tension in the horizontal rope,

so T1=15N.

Now look at the hanging weight. The only forces are

gravity and the tension in the vertical rope.

So T2 = w.

Finally, look at the junction where all 3 ropes are tied

together. That point is in equilibrium as well, so we can

write down the forces in each direction and they should

balance out.N15Tw

0w)45sin()45cos(

T

0T)45sin(TF

)45cos(

TT

0T)45cos(TF

1

1

23y

13

13x

N

60N

Ffriction T1 T1

T2

T2

T3

Answer a.

16) Two balls are rolled down a hill. Ball A is a solid sphere with mass M and radius R. Ball B is a

hollow sphere with mass M and radius 2R. Compare the speeds of the balls when they reach the

bottom of the incline.

a) VA = 0.6 VB

b) VA = VB

c) VA = 1.1 VB

d) VA = 1.7 VB


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Moment of Inertia for each ball:

232

B

2

52

A

R2MI

MRI

Use conservation of energy for each ball:

1.1gh

gh

V

V

ghVMghR2MMVMghIMV

ghVMghMRMVMghIMV

56

710

B

A

56

B

2

R2

V2

32

212

B212

BB212

B21

710

A

2

R

V2

52

212

A212

AA212

A21

B

A

17) A box of mass M starts from rest at the top of a frictionless incline of height h. It slides down the hill

and across a horizontal surface, which is also frictionless, except for a rough patch of length h, with

coefficient of kinetic friction 0.25. The box comes into contact with a spring (spring constant = k),

compressing it. The spring then unloads, sending the box back in the opposite direction.


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h

h

Einitial=mgh

When the box slides across the rough patch, energy is lost to friction.

mgh25.0hmghfW kkfriction (each trip)

The spring just changes the direction of the box – energy is conserved while

in contact with the spring. So the total energy lost due to friction is 0.5mgh.

Efinal = 0.5mgh = mghfinal

hfinal = 0.5h

18) A diver tucks her body in mid-flight, reducing her moment of inertia by a factor of 2. What happens

to her angular speed and kinetic energy?

a) Both angular speed and kinetic energy remain the same.

b) Angular speed is doubled, and kinetic energy remains the same.

c) Angular speed is doubled, and kinetic energy is increased by a factor of 4.

d) Both angular speed and kinetic energy are doubled.


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Use conservation of angular momentum.

0ff021

00ff00 2IIII Angular speed is doubled

Plug this in to the kinetic energy formula.

.init,rotK

2002

12002

121

final,rot2002

1.init,rot I22IKIK Kinetic Energy is doubled

19) A uniform marble rolls without slipping down the path shown, starting from rest. Find the minimum

height required for the marble to make it across without falling into the pit.


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a) 23m

b) 14m

c) 30m

d) 35m

36m

25m

45m

h

This is a 2-stage problem. When the marble rolls down the

hill, we can use conservation of energy. Then it’s projectile

motion as it flies across the gap.

m23

8.9

18h

g

vh

mvmgh

r

vmrmvmgh

Imvmgh

EE

2s

m

2

sm

107

2

107

2

107

22

52

212

21

2

212

21

bottomtop

We need to find the speed v

Projectile motion – initial speed is horizontal,

and the marble drops 20m. Find time:

sec2tt8.9m20

gty

2

s

m21

2

21

2

Horizontal distance is 36m:

sm18vsec2vm36

20) A school yard teeter-totter with a total length of 5.2 m and a mass of 36 kg is pivoted at its center.

A child of mass 18-kg sits on one end of the teeter-totter. Where should the parent push downward

with a force of 210 N to balance the teeter totter?

a) 0.5 m from the center

b) 2.2 m from the center

c) 1.3 m from the center

d) 1.9 m from the center


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210 N

5.2 m

176.4 N

352.8 N

d

The torques must balance out.

Measuring from the center, we have:

m2.2d

0dN210m6.2N4.176

The weight of the teeter-totter is at

the center, so it produces no torque.

Physics 6A - UC Santa Barbara

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