# Quest 2 Review Vectors & 2d kinematics KEY ```Version 001 – Quest 2 Review Vectors &amp;amp; 2d kinematics – tubman – (Phys1A2122p3)
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1
~ =C
~ +D
~
2. P
~ =C
~ +B
~
3. P
~ =A
~ −B
~ correct
4. P
001 (part 1 of 2) 10.0 points
~
Express the vector R
C
R
A P
D
B
~
~
~
~ the edges of a
in terms of A, B, C, and D,
parallelogram.
~ =D
~ −A
~
1. R
~ =A
~ −D
~
2. R
~ =A
~ −B
~
3. R
~ =C
~ +B
~
4. R
~ =B
~ +D
~
5. R
~ =B
~ +A
~ correct
6. R
~ =A
~ +D
~
7. R
~ =A
~ −C
~
8. R
~ =B
~ −A
~
9. R
~ =C
~ +D
~
10. R
Explanation:
Apply the parallelogram rule of addition:
~ and B;
~ the
join the tails of the two vectors A
resultant vector is the diagonal of a parallel~ and B
~ as two of its
ogram formed with A
sides.
002 (part 2 of 2) 10.0 points
~ in terms of A,
~ B,
~ C,
~ and
Express the vector P
~
D,
~ =B
~ −A
~
1. P
~ =C
~ −A
~
5. P
~ =B
~ +D
~
6. P
~ =B
~ +A
~
7. P
~ =D
~ −A
~
8. P
~ =A
~ −D
~
9. P
~ =A
~ +D
~
10. P
Explanation:
By the triangle method of addition
~ +P
~ =A
~
B
~ =A
~ −B
~ .
P
003 10.0 points
Resultant Displacement 05 109772
A person walks 34 m East and then walks
31 m at an angle 30◦ North of East.
What is the magnitude of the total displacement?
1. 76.6011
2. 60.2479
3. 56.1532
4. 70.9834
5. 62.79
6. 67.4562
7. 58.459
8. 57.5253
9. 66.7191
10. 56.8649
Explanation:
Let :
Ra = 34 m ,
Rb = 31 m ,
θ = 30◦ .
and
Version 001 – Quest 2 Review Vectors &amp;amp; 2d kinematics – tubman – (Phys1A2122p3)
The total displacement is the vector sum of
the two displacements, so
Rx = Ra + Rb cos θ = 34 m + (31 m) cos 30◦
= 60.8468 m and
Ry = Rb sin θ = (31 m) sin 30◦ = 15.5 m .
The final displacement is
R=
=
q
q
Rx2 + Ry2
(60.8468 m)2 + (15.5 m)2
= 62.79 m .
004 (part 1 of 2) 10.0 points
Blown Off Course 168934
A ship is expecting to travel to its home port
510 km due East. Before the ship starts to
travel, a severe storm comes up and blows the
ship 490 km due South.
How far is the ship from its home port?
1. 776.981
2. 643.506
3. 672.012
4. 806.04
5. 514.782
6. 797.559
7. 477.074
8. 707.248
9. 766.094
10. 757.694
Explanation:
Let : dship = 510 km and
dwind = 490 km .
Choose a coordinate system with the positive x-axis representing 0◦ and the positive
y-axis representing 90◦ .
2
N
Ship
W ind
R
θ
E
The displacements are at right angles to
each other, so
q
~
kRk = d2ship + d2wind
q
= (510 km)2 + (490 km)2
= 707.248 km .
005 (part 2 of 2) 10.0 points
At what angle North of East must the ship
travel to reach its destination? Let East be 0◦
and North 90◦ .
1. 32.7352
2. 25.8664
3. 28.8427
4. 23.9625
5. 30.1137
6. 25.0675
7. 26.5651
8. 21.6444
9. 43.8543
10. 34.8545
Explanation:
The ship was blown to the South, so it
must travel to the North and East to reach its
destination.
If θ is the angle its destination makes with
East, then W is the side opposite θ and S is
W
490 km
θ = arctan
= arctan
S
510 km
= 43.8543◦ .
006 (part 1 of 2) 10.0 points
Version 001 – Quest 2 Review Vectors &amp;amp; 2d kinematics – tubman – (Phys1A2122p3)
Big Bertha 36992
During World War I, the Germans had a gun
called Big Bertha that was used to shell Paris.
The shell had an initial speed of 2.15 km/s at
an initial inclination of 22.7◦ to the horizontal.
The acceleration of gravity is 9.8 m/s2 .
How far away did the shell hit?
1. 559.786
2. 408.362
3. 283.289
4. 85.7919
5. 117.531
6. 186.477
7. 24.8054
8. 641.356
9. 483.685
10. 335.851
Explanation:
The range R is given by
v02
sin(2 θ0)
g
(2150 m/s)2
sin 45.4◦
=
9.8 m/s2
= 335.851 km .
R=
007 (part 2 of 2) 10.0 points
How long was it in the air?
1. 319.039
2. 142.012
3. 96.466
4. 169.326
5. 351.341
6. 220.215
7. 264.121
8. 104.188
9. 232.938
10. 493.495
Explanation:
The time in the air is
R
t=
v0x
R
=
v0 cos θ0
3
3.35851 &times; 105 m
(2150 m/s) cos 22.7◦
= 169.326 s .
=
008 (part 1 of 4) 10.0 points
Horizontal Projection 40319
A ball is thrown horizontally from the top of
a building 47.7 m high. The ball strikes the
ground at a point 108 m from the base of the
building.
The acceleration of gravity is 9.8 m/s2 .
Find the time the ball is in motion.
1. 2.39046
2. 1.96396
3. 3.17516
4. 3.12005
5. 2.79212
6. 2.28125
7. 2.82482
8. 2.10926
9. 1.98463
10. 2.13331
Explanation:
In this solution, we take the origin of the x
and y axes at the base of the building. The
positive y direction is taken to point upward,
so that the vertical acceleration is −g. The
time the ball is in flight is found from
y = y0 + vy0 t +
1
a y t2
2
1
0 = y 0 + 0 − g t2
2
r
2 y0
t=
g
s
2 (47.7 m)
=
9.8 m/s2
= 3.12005 s .
009 (part 2 of 4) 10.0 points
Find the initial velocity of the ball.
1. 38.0337
Version 001 – Quest 2 Review Vectors &amp;amp; 2d kinematics – tubman – (Phys1A2122p3)
2. 34.6149
3. 39.5346
4. 61.3605
5. 10.9751
6. 31.0983
7. 29.6678
8. 15.041
9. 11.2234
10. 42.8439
4
1. -31.085
2. -25.5859
3. -30.5765
4. -28.4169
5. -33.0708
6. -29.1657
7. -17.542
8. -25.8905
9. -16.6829
10. -18.783
Explanation:
There is no acceleration in the horizontal direction. Therefore, we have
x = x0 + vx0 t
x = 0 + vx0 t
x
vx0 =
t
108 m
=
3.12005 s
= 34.6149 m/s .
010 (part 3 of 4) 10.0 points
Find the x component of its velocity just before it strikes the ground.
1. 35.9132
2. 29.6678
3. 10.9751
4. 21.3483
5. 34.6149
6. 38.515
7. 19.5451
8. 56.0463
9. 42.8439
10. 58.5836
Explanation:
The y component of the velocity just before
the ball strikes the ground is
vy = vy0 + ay t
= 0−gt
= −(9.8 m/s2 ) (3.12005 s)
= −30.5765 m/s .
012 (part 1 of 3) 10.0 points
AP B 1993 MC 64 66 84565
A ball is thrown and follows the parabolic
path shown. Point Q is the highest point on
the path and points P and R are the same
height above the ground.
Q
P
R
Explanation:
Since there is no acceleration in the horizontal
direction, the x component of the velocity is
constant. So
vx = vx0 = 34.6149 m/s .
011 (part 4 of 4) 10.0 points
Find the y component of its velocity just before it strikes the ground.
How do the speeds of the ball at the three
points compare? Air friction is negligible.
1. k~vQ k &lt; k~vR k &lt; k~vP k
2. k~vQ k &lt; k~vP k = k~vR k correct
Version 001 – Quest 2 Review Vectors &amp;amp; 2d kinematics – tubman – (Phys1A2122p3)
3. k~vP k = k~vR k &lt; k~vQ k
5
9.
4. k~vR k &lt; k~vQ k &lt; k~vP k
5. k~vP k &lt; k~vQ k &lt; k~vR k
Explanation:
The speed of the ball in the x-direction is
constant. Because of gravitational acceleration, the speed in the y-direction is zero at
point Q. Since points P and R are located at
the same point above ground, by symmetry
they have the same vertical speed component
(though they do not have the same velocity),
so vQ &lt; vP = vR .
Explanation:
Since air friction is negligible, the only acceleration on the ball after being thrown is
that due to gravity, which acts straight down.
014 (part 3 of 3) 10.0 points
Which diagram best indicates the direction of
the net force, if any, on the ball at point Q?
1.
2.
013 (part 2 of 3) 10.0 points
Which diagram best indicates the direction of
the acceleration, if any, on the ball at point
R?
correct
3.
4.
1.
5.
2.
6.
3.
4.
correct
5.
6.
7. The ball is in free fall and there is no
acceleration at any point on its path.
8.
7.
8.
9. The ball is in free-fall and there is no
acceleration at any point on its path.
Explanation:
By Newton’s second law, the force is in the
direction of the acceleration (downward).
015 (part 1 of 3) 10.0 points
AP B 1994 FR 1 short 95850
Version 001 – Quest 2 Review Vectors &amp;amp; 2d kinematics – tubman – (Phys1A2122p3)
A ball of mass 0.6 kg, initially at rest, is
kicked directly toward a fence from a point
20 m away, as shown below.
The velocity of the ball as it leaves the
kicker’s foot is 17 m/s at angle of 57 ◦ above
the horizontal. The top of the fence is 7 m
high. The ball hits nothing while in flight and
air resistance is negligible.
The acceleration due to gravity is 9.8 m/s2 .
m/
s
b
b
b
b
b
b
b
b
b
b
b
b
b
17
b
b
b
b
7m
◦
b
57
b
b
20 m
Determine the time it takes for the ball to
reach the plane of the fence.
1. 1.49296
2. 1.86943
3. 1.70916
4. 1.37341
5. 1.55884
6. 1.51383
7. 1.79945
8. 1.35642
9. 2.16009
10. 2.05111
Explanation:
Let : θ = 57 ◦ and
d = 20 m .
The horizontal component of the velocity is
constant, so
vhoriz = v0 cos θ
= (17 m/s) cos 57 ◦
= 9.25886 m/s .
The horizontal motion defines the time of
flight:
vhoriz t = d
t=
6
d
vhoriz
20 m
=
9.25886 m/s
= 2.16009 s .
b
016 (part 2 of 3) 10.0 points
How far above the top of fence will the ball
pass? Consider the diameter of the ball to be
negligible.
1. 0.207175
2. 1.42074
3. 1.01814
4. 0.962449
5. 0.437954
6. 0.305372
7. 0.403716
8. 0.933905
9. 0.81545
10. 0.706199
Explanation:
The vertical component of the initial velocity is
vvert = v0 sin θ
= (17 m/s) sin 57 ◦
= 14.2574 m/s .
The height of the ball during its flight is
given by
1
g t2
2
= (14.2574 m/s) (2.16009 s)
1
− (9.8 m/s2 ) (2.16009 s)2
2
= 7.93391 m .
y = vvert t −
Therefore, the distance that the ball passes
above the fence is
∆y = (7.93391 m) − (7 m) = 0.933905 m .
017 (part 3 of 3) 10.0 points
Version 001 – Quest 2 Review Vectors &amp;amp; 2d kinematics – tubman – (Phys1A2122p3)
What is the vertical component of the velocity
when the ball reaches the plane of the fence?
1. -6.455
2. -6.71698
3. -6.59675
4. -7.18939
5. -2.87927
6. -4.20553
7. -6.9115
8. -4.59697
9. -6.01041
10. -7.03111
Explanation:
Thus the the vertical component of the velocity when the ball reaches the plane of the
fence is
vvert = v0 sin θ − g t
= (17 m/s) sin 57 ◦
− (9.8 m/s2 ) (2.16009 s)
= −6.9115 m/s .
Vertical Component of Velocity (m/s)
This is verified by analyzing the graph below
Vertical Velocity vs Time
25
20
15
10
5
0
−5
−10
−15
−20
−25
0
0.5
1.0
1.5
Time (s)
2.0
2.5
7
```