Transcript Solutions - UCSB CLAS

Physics 6A
Practice Final
Solutions
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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:
2
y  y0  v0t  1
gt
2
1 (9.8 m )  t 2
0  500m  (121 m
)

t

s
2
s2
4.9t2  121t  500  0
121  (121)2  4(4.9)(500 )
t
2(4.9)
t  28 sec
We can use the quadratic formula and take the positive answer. This is answer c.
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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.
This is a projectile problem with a horizontal initial velocity.
Here are the initial and final values that we know:
y 0  1.5m
y  0m
v 0,x  3.5 m
s
v 0,y  0 m
s
ay  g
When the ball hits the ground we will have the same x-component of velocity, but
the y-component will have increased.
v y2  v20,y  2gy  y 0 
v y2  0  2(9.8
m )(0
s2
 1.5m)
v y  5.42 m
s
We need to use the Pythagorean theorem to find the magnitude of the velocity:
v
v2x  v2y
v  (3.5)2  (5.42)2  6.45 m
s
Answer c
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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.
We will need to find the components of the initial velocity:
v0,x  v0  cos(0 )  (20 m
)  cos(35 )  16.38 m
s
s
v0,y  v0  sin( 0 )  (20 m)  sin(35 )  11.47 m
s
s
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:
v y2  v20,y  2gy  y0 
0  (11 .47 m )2  2(9.8
s
m )(y
max
s2
 0)
ymax  6.71m
Answer b
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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?
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.
Fscale
Write down Newton’s 2nd law:
F  ma
Fscale  mg  ma
850N  750N  (76 .5kg)  a
a  1.3
mg
m
s2
Answer a
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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.
Total mass of system
F=90N
2 kg
Fsystem  90N  19 .6N  9.8N
T

T  (3kg)(a)



0
a  20 .2
m
s2
These tensions cancel out because
they are internal forces to the system.
F1kgblock  T  9.8N  (1kg)(20 .2 m2 )
s
T
19.6N
Weight of
2kg mass
T
1 kg
Weight of
1kg mass
T  30N
9.8N
This is answer d.
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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: 5.5 9.8 m   53.9 m


s2 
s2

The path of the airplane is circular, so the given acceleration
must be centripetal (toward the center).
We have a formula for centripetal acceleration:
v2
acent 
r
v2
m
53.9 2 
 v  137 m
s
350m
s
Answer a
350m
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.

Ffriction
50 m
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?

Ffriction,static,max  s  mg  0.751000kg 9.8 m2  7350N
s
This friction force is the only force directed toward the center, so it must be the centripetal force:
Ffriction  Fcentripeta l
mv 2

r

1000kg v 2 
7350N 
50m
v
 19.2 ms
Answer c.
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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?
Given: v0=40 m/s; vf=0; Δt=6.2s; diam=70cm→r=0.35m
40 m
v0
s
0 

 114 .3 rad
s
r
0.35m
rad
 0  114 .3 s


 18 .4 rad
s2
t
6.2s
  0 t 
355rad
rad
2 rev
1
2


6.2s 
t2  114 .3 rad
s
1
2
  18 .4 rad 6.2s2  355 radians


s2 

 56 revolution s
Answer c.
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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.
a)10 cm
b) 25 cm
c) 50 cm
d) 150 cm
use v=Rω
Tension in rope = weight of hanging block = Mg
Tension in rope is also the centripetal force on the moving block Fcent 
Mv 2
 MR2
R
Set these equal and solve for R:
Mg  MR2  R 
g
2
we must convert the given speed from
revolutions per second to radians per second
rad
  1rev
 2 rev
 2 rad
s
s
Plug in to get R:
9.8 m2
g
s
R 2 
 0.25m  25cm
2

2 rad
s


Answer b.
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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
We can use conservation of angular momentum for this one.
0 
1rev 2rad

 0.785 rad
s
8s 1rev
Idisk  21 MR 2 
Ldisk  I 
1
2
Initial angular speed of disk
400kg2m2  800kg  m2
Ichild  mR2  50kg 2m  200kg  m2
2
 
Lchild  mvR  50kg  5 ms 2m  500
kgm2
628 s
kgm2
s
Itotal  Idisk  Ichild  1000kg  m2
L total  Ldisk  Lchild  1128
L final  Itotal  f  1128
kgm2
s
kgm2
s


 f  1.128 rad
 v f ,child  2m 1.128 rad
 2.3 ms
s
s
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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
Before
After
Highest
Collision
v0
Collision
Point
½v0
5kg
vblock
5kg
1.8cm
5kg
Use conservation of momentum for the collision, then
conservation of energy for the swinging to the highest point.
0.005kg  v0   0.005kg   21 v0   5kg  vblock
0.005kg  v0   0.005kg   21 v0   5kg  0.59 ms 
 v0  396 ms
(Round up to get 400m/s)
1
2
5kgvblock 2  5kg9.8 sm2 0.018m  vblock
 0.59 ms
Now we can put this value into our
momentum formula to get the initial
speed of the bullet
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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.
y
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
vfinal
A/B
c) 23.3 m/s at an angle of 23° North of East
20m/s
d) 10.4 m/s at an angle of 50° North of East
Momentum is conserved in each direction,
so we get two formulas:
x : ma v a  ma  mb   v f ,x  v f ,x  6.67 ms
x
A
12m/s
B
y : mb v b  ma  mb   v f ,y  v f ,y  8 ms
Combine the components with the Pythagorean theorem
to find the final speed, and use tangent to find the angle:
v final 
6.67 ms 2  8 ms 2  10.4 ms
tan  
8 ms
6.67 ms
   50
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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.
positive torque
a) 2/7 g
b) 2/5 g
c) 1/2 g
d) 2/3 g
R
M
We can set up force formulas for the masses,
and a torque formula for the pulley.
Mg  T1  Ma  T1  Mg  Ma
T2  2Mg  2Ma  T2  2Ma  2Mg
R 
T1  R  T2  R  1 MR 2  a  T1  T2  1 Ma
2
2
Mg  Ma  2Ma  2Mg  12 Ma  a   27 g
T1
T2
T1
T2
M
Mg
2M
2Mg
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.
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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.
a)
b)
c)
d)
Fwall
1.0 m
2.5 m
3.0 m
4.0 m
5m
N
4m
d
Forces on the ladder are shown in blue
(the weight of the ladder is negligible).
mg
We need to find the distance d.
fs
3 formulas we can write down – 2 force
formulas and one torque (use the
ground as the pivot point).
ℓ
3m
Fx  0  fs  Fwall  0  Fwall  mg  0.6 1470N  882N
Fy  0  N  mg  0  N  1470N
  0
  0  Fwall 4m  mg
3d
5
882N4m  1470N35 d  0  d  4m
use similar triangles to find
the lever arm for the weight:

d
3d


3m 5m
5
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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.
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).
N
T3
Ffriction
T1
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.
T1
T2
60N
T2
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.
Fx  T3  cos(45  )  T1  0
T3 
T1
cos(45  )
Fy  T3  sin( 45  )  T2  0
T1
cos(45  )
 sin( 45  )  w  0
w  T1  15N
Answer a.
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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
Moment of Inertia for each ball:
IA  52 MR 2
IB  32 M2R
2
Use conservation of energy for each ball:
1 MV 2
A
2
1 MV 2
B
2
VA

VB
 21 IA 2A  Mgh  21 MVA2 

1 I 2
2 B B
10 gh
7
6 gh
5
 Mgh 
1 MV 2
B
2

 

2
1 2 MR 2 VA
2 5
R

2R 
1 2M
2 3
 Mgh  VA 
 
2 VB 2
2R
 Mgh  VB 
10
7
gh
6
gh
5
 1 .1
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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.
h
h
Einitial=mgh
When the box slides across the rough patch, energy is lost to friction.
Wfriction  fk  h  kmg  h  0.25mgh (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
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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.
Use conservation of angular momentum.
I0  0  If  f  I0  0 
21 I0  f  f  20
Angular speed is doubled
Plug this in to the kinetic energy formula.
Krot ,init.  21 I0  02  Krot ,final 
  20 2
1 1I
2 2 0




 2 21 I0  02 



 K rot,init . 
Kinetic Energy is doubled
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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.
a)
b)
c)
d)
23m
14m
30m
35m
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.
45m
25m
E top  Ebottom
mgh  21 mv 2  21 I2
mgh 
1
mv 2
2


1 2
mr 2
2 5
36m

v
 
r
2
7
mgh  10
mv 2
h
h
7
10
v2
y  21 gt2
We need to find the speed v
g


2
18 ms
9.8 m2
s
7
10
Projectile motion – initial speed is horizontal,
and the marble drops 20m. Find time:
20m 
1
2
9.8 t
m
s2
2
 t  2 sec
Horizontal distance is 36m:
 23m
36m  v  2 sec  v  18 ms
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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
5.2 m
b) 2.2 m from the center
c) 1.3 m from the center
d
d) 1.9 m from the center
176.4 N
The torques must balance out.
Measuring from the center, we have:
176.4N2.6m  210Nd  0
d  2.2m
210 N
352.8 N
The weight of the teeter-totter is at
the center, so it produces no torque.
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