Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
Review Problems:
E
1) Which of the following is an accurate statement?
A) The magnitude of a vector can be zero even though one of its components is not zero.
B) It is possible to add a scalar quantity to a vector.
C) Even though two vectors have unequal magnitudes, it is possible that their vector sum
is zero.
D) Rotating a vector about an axis passing through the tip of the vector does not change
the vector.
E) The magnitude of a vector is independent of the coordinate system used.
2)
A
3)
E
What is 0.205!/# , expressed to the proper number of significant figures?
A) 0.348
B) 0.35
C) 0.3
D) 0.3477
The period of a pendulum is the time it takes the pendulum to swing back and forth once. If
the only dimensional quantities that the period depends on are the acceleration of gravity, g,
and the length of the pendulum, ℓ, what combination of g and ℓ must the period be
proportional to? (Acceleration has SI units of m • s-2.)
A) g/ℓ
B) gℓ2
C) gℓ
D)
E)
4)
A
You walk 55m to the north, then turn 60° to your right and walk another 45m. How far are
you from where you originally started?
A) 87 m
B) 50 m
C) 94 m
D) 46 m
Let 𝐴⃗ = 55𝚥̂𝑚; (55 meters North)
-⃗ = 45(𝑐𝑜𝑠(90 − 60)𝚤̂ + 𝑠𝑖𝑛(90 − 60)𝚥̂)𝑚; (45 meters
Let 𝐵
60 degrees east of North)
-⃗ = <45𝑠𝑖𝑛(60)𝚤̂ + =55 + 45𝑐𝑜𝑠(6)>𝚥̂?𝑚
𝑅-⃗ = 𝐴⃗ + 𝐵
-⃗@ = B=45sin(60)>! + =55 + 45cos(60)>! = 86.7m
@R
1
Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
5)
A helicopter is flying horizontally with a speed of 444 m/s over a hill that slopes upward
with a 2% grade (that is, the "rise" is 2% of the "run"). What is the component of the
helicopter's velocity perpendicular to the sloping surface of the hill?
A) 8.9 m/s
B) 220 m/s
C) 435 m/s
D) 444 m/s
6)
As shown in the figure, three force vectors act on an object. The magnitudes of the forces as
shown in the figure are 𝐹$ = 80.0 𝑁, 𝐹! = 60.0𝑁, and 𝐹# = 40.0𝑁, where N is the standard
SI unit of force. The resultant force acting on the object is given by
---⃗
𝐹$ = (80.0𝑐𝑜𝑠(330)𝚤̂ + 80.0𝑠𝑖𝑛(330)𝚥̂) N
---⃗
𝐹! = 60.0𝚥̂ 𝑁
---⃗
𝐹# = −40.0𝚤̂ 𝑁
---⃗! + 𝐹
---⃗#
𝑅-⃗ = ---⃗
𝐹$ + 𝐹
𝑅-⃗ = =(69.3 − 40)𝚤̂ + (−40 + 60)𝚥̂>𝑁
𝑅-⃗ = (29.3𝚤̂ + 20𝚥̂)𝑁
𝑅 = @𝑅-⃗ @ = P29.3! + 20! = 35.5𝑁
A
D
θ = 𝑡𝑎𝑛%$ T
20
U = 34.3&
29.3
A) 180 N at an angle 60.0° with respect to +x-axis.
B) 60.0 N at an angle 90.0° with respect to +x-axis.
C) 20.0 N at an angle 34.3° with respect to +x-axis.
D) 35.5 N at an angle 34.3° with respect to +x-axis.
E) 40.0 N at an angle 60.0° with respect to +x-axis.
7)
B
A teacher sends her students on a treasure hunt. She gives the following instructions:
1. Walk 300 m north.
2. Walk 400 m northwest.
3.Walk 700 m east-southeast and the treasure is buried there.
As all the other students walk off following the instructions, Jane physics student quickly
adds the displacements and walks in a straight line to find the treasure. How far and in what
direction does Jane need to walk?
Use 22.5 degrees south of
A) 187 m in a direction 67.3° north of east
B) 481 m in a direction 40.9° north of east
east for east-southest above
C) 399 m in a direction 52.5° north of east
D) 284 m in a direction 28.2° west of north
E) The treasure position cannot be reached in one straight walk.
2
Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
8)
-⃗ = 3.00𝚤̂ + 4.00𝚥̂. What are the magnitude and
Vector 𝐴⃗ = 1.00ı̂ − 2.00ȷ̂ and vector 𝐵
-⃗?
direction of vector 𝐶⃗ = 𝐴⃗ + 𝐵
A) 7.21 in a direction 33.7° counterclockwise from the positive x axis
B) 6.00 in a direction 63.4° counterclockwise from the positive x axis
C) 4.47 in a direction 6.34° counterclockwise from the positive x axis
D) 4.47 in a direction 26.6° counterclockwise from the positive x axis
E) 7.21 in a direction 56.3° counterclockwise from the positive x axis
9)
-⃗ + 𝐶⃗, where 𝐴⃗ = 1.00𝚤̂ + 4.00𝚥̂ − 1.00𝑘\, 𝐵
-⃗ = 3.00𝚤̂ −
What is the magnitude of 𝐴⃗ + 𝐵
1.00𝚥̂ − 4.00𝑘\ 𝑎𝑛𝑑 ---⃗
𝐶 = −1.00𝚤̂ + 1.00𝚥̂?
A) 7.07
B) 2.00
C) 10.76
D) 6.78
E) 8.12
D
A
A
B
10) If 𝐴⃗ = +4𝚤̂ − 2𝚥̂ − 3𝑘\ 𝑎𝑛𝑑 𝐶⃗ = −4𝚤̂ − 2𝚥̂ − 3𝑘\ which of the following numbers is closest to
the magnitude of 𝐴⃗ − 𝐶⃗?
A) 8
B) 7
C) 9
D) 10
E) 11
11) If an object is accelerating toward a point, then it must be getting closer and closer to that
point.
A) True
B) False
12) The figure shows the position of an object (moving along a straight line) as a function of
time. Assume two significant figures in each number. Which of the following statements
about this object is true over the interval shown?
A
A) The object is accelerating to the left.
B) The object is accelerating to the right.
C) The acceleration of the object is in the same direction as its velocity.
D) The average speed of the object is 1.0 m/s.
3
Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
13) The figure represents the velocity of a particle as it travels along the x-axis. At what value
(or values) of t is the instantaneous acceleration equal to zero?
C
A) t = 0
B) t = 0.5 s and t = 2 s
C) t = 1 s
B
B
14) The position of an object as a function of time is given by x =
- ct, where
and
and x and t are in SI units. What is the instantaneous velocity of the object
when
Note: 𝑥 = 𝑏𝑡 ! − 𝑐 is of the form
$
A) 1.7 m/s
𝑥 = 𝑥' + 𝑣' 𝑡 + ! 𝑎𝑡 ! where
B) 2.1 m/s
𝑥' = 0; 𝑎 = 2𝑏 = 4.0𝑚/𝑠 ! ; and 𝑣' = −𝑐 = −6.7𝑚/𝑠
C) 2.3 m/s
Use: 𝑣 = 𝑣_0 + 𝑎𝑡 = −6.7𝑚/𝑠 + (4.0𝑚/𝑠^2)(2.2𝑠) =
D) 2.7 m/s
2.1 𝑚/𝑠
15) A package is dropped from a helicopter moving upward at
If it takes
before
the package strikes the ground, how high above the ground was the package when it was
released if air resistance is negligible?
A) 810 m
𝑣'( = 15𝑚/𝑠
B) 1000 m
1
C) 1200 m
𝑦 = 𝑦' + 𝑣'( 𝑡 − 𝑔𝑡 !
2
D) 1500 m
1 !
1
𝑦' = 𝑔𝑡 − 𝑣'( 𝑡 = (9.8𝑚/𝑠 ! )(16𝑠)! − (15𝑚/𝑠)(16𝑠)
2
2
= 1,014.4𝑚 = 1000𝑚
4
Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
E
C
A
A
B
16) A runner maintains constant acceleration after starting from rest as she runs a distance of
. The runner's speed at the end of the 60.0 m is 9.00 m/s. How much time did it take
the runner to complete the 60.0 m distance?
A) 6.67 s
B) 15.0 s
C) 9.80 s
D) 10.2 s
E) 13.3 s
17) Two identical stones are dropped from rest and feel no air resistance as they fall. Stone A is
dropped from height h, and stone B is dropped from height 2h. If stone A takes time t to
reach the ground, stone B will take time
A) 4t.
B) 2t.
C) 𝑡√2
D) 𝑡/√2
E) t/2.
18) A ball rolls across a floor with an acceleration of 0.100 m/s2 in a direction opposite to its
velocity. The ball has a velocity of 4.00 m/s after rolling a distance 6.00 m across the floor.
What was the initial speed of the ball?
A) 4.15 m/s
𝑣) = 4.00m/s; 𝑎 = −0.100𝑚/𝑠 ! ; 𝑥) = 6.00𝑚; 𝑥* = 0
B) 5.85 m/s
C) 4.60 m/s
𝑣*! = 𝑣)! − 2𝑎=𝑥) − 𝑥* > = (4𝑚/𝑠)! − 2(−0.100𝑚/𝑠 ! )(6.00𝑚)
D) 5.21 m/s
𝑣* = B𝑣*! = 4.15𝑚/𝑠
E) 3.85 m/s
19) Jan and Len throw identical rocks off a tall building at the same time. The ground near the
building is flat. Jan throws her rock straight downward. Len throws his rock downward and
outward such that the angle between the initial velocity of the rock and the horizon is 30°.
Len throws the rock with a speed twice that of Jan's rock. If air resistance is negligible,
which rock hits the ground first?
A) They hit at the same time.
B) Jan's rock hits first.
C) Len's rock hits first.
D) It is impossible to know from the information given.
20) For general projectile motion, when the projectile is at the highest point of its trajectory,
A) its acceleration is zero.
B) its velocity is perpendicular to the acceleration.
C) its velocity and acceleration are both zero.
D) the horizontal component of its velocity is zero.
E) the horizontal and vertical components of its velocity are zero.
5
Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
21) An electron moves with a constant horizontal velocity of
and no initial
vertical velocity as it enters a deflector inside a TV tube. The electron strikes the screen after
traveling
horizontally and
vertically upward with no horizontal
acceleration. What is the constant vertical acceleration provided by the deflector? (The
effects of gravity can be ignored.)
A) 2.5 × 1014 m/s2
𝑣'+ = 3.0 × 10, m/s; 𝑥* = 0; 𝑥) = 17.0cm; 𝑦' = 0; 𝑦) = 40.0cm;
B) 8.3 × 102 m/s2
𝑎( = ?
C) 1.4 × 104 m/s2
Δ𝑥
D) 1.2 × 1014 m/s2
t=
𝑣'+
1
1
𝑦) = 𝑦' + 𝑣'( + 𝑎𝑡 ! = 0 + 0 + 𝑎𝑡 !
2
2
!
(2)(0.40𝑚)(3 × 10, )!
2𝑦) 2𝑦) 𝑣'+
𝑎= ! =
=
(Δ𝑥)!
(0.17)!
𝑡
A
a = 2.5 × 10$- m/𝑠 !
D
22) A hockey puck slides off the edge of a table with an initial velocity of 28.0 m/s. and
experiences no air resistance. The height of the tabletop above the ground is 2.00 m. What is
the angle below the horizontal of the velocity of the puck just before it hits the ground?
A) 77.2°
𝑣'+ = 28.0𝑚/𝑠
B) 72.6°
𝑣'( = P2𝑔ℎ
C) 12.8°
𝑣'(
D) 12.6°
θ = 𝑡𝑎𝑛%$ T U
𝑣'+
E) 31.8°
6
Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
A
23) A small boat is moving at a velocity of
when it is accelerated by a river current
perpendicular to the initial direction of motion. If the acceleration of the current is 0.750
m/s2, what will be the new velocity of the boat after
A) 25.3 m/s at 82.4° from initial direction of motion
B) 25.3 m/s at 7.59° from initial direction of motion
C) 640.1 m/s at 82.4° from initial direction of motion
D) 640.1 m/s at 7.59° from initial direction of motion
𝑣⃗' = 3.35𝚥̂ m/s; 𝑎⃗ = 0.75𝚤̂ 𝑚/𝑠 ! ; 𝑣⃗ = 𝑣⃗' + 𝑎+ 𝑡; 𝑣⃗ = ((0.750)(33.5)𝚤̂ + 3.35𝚥̂) 𝑚/𝑠
𝑣⃗ = (25.125𝚤̂ + 3.35𝚥̂)𝑚/𝑠
𝑣 = P(25.125)! + (3.35)! = 25.3𝑚/𝑠
θ = 𝑡𝑎𝑛%$
3.35
= 7.6&
25.125
Which is 90& − 7.6& = 82.4& from initial direction of motion
A
24) A hobby rocket reaches a height of
and lands
from the launch point with no
air resistance. What was the angle of launch?
A) 69.0°
𝑣'( = P2𝑔ℎ
B) 67.4°
C) 22.6°
𝑣'( − 𝑣( 2𝑣'(
2𝑔
D) 44.8°
𝑣( = 𝑣'( − gt ⇒ 𝑡 =
=
= 2q
𝑔
𝑔
ℎ
𝑥
𝑥 = 𝑣'+ 𝑡 ⇒ 𝑣'+ =
𝑡
!
𝑣'( 𝑣'( 𝑡 2𝑣'(
4𝑔ℎ 4ℎ
𝑡𝑎𝑛θ =
=
=
=
=
𝑣'+
𝑥
𝑔𝑥
𝑔𝑥
𝑥
4ℎ
θ = 𝑡𝑎𝑛%$
= 69&
𝑥
25) A child throws a ball with an initial speed of
at an angle of 40.0° above the
horizontal. The ball leaves her hand 1.00 m above the ground and experience negligible air
resistance.
(a) What is the magnitude of the ball's velocity just before it hits the ground?
(b) At what angle below the horizontal does the ball approach the ground?
a) 9.14 𝑚/𝑠
b) 48!
7
Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
26) A projectile is fired from point 𝑂 at the edge of a cliff, with initial velocity components of
𝑣'+ = 60.0 𝑚/𝑠 and 𝑣'( = 175 𝑚/𝑠, as shown in the figure. The projectile rises and then
falls into the sea at point P. The time of flight of the projectile is 40.0 𝑠, and it experiences
no appreciable air resistance in flight. What is the height of the cliff?
Height = 840 m
Solution:
Given: 𝑣'+ = 60.0 𝑚/𝑠; 𝑣'( = 175 𝑚/𝑠; 𝑡 = 40.0 𝑠
Want: height of cliff.
The height of the cliff can be found using the following kinematic equation:
1
𝑦 = 𝑦' + 𝑣'( 𝑡 − 𝑔𝑡 !
2
Choose the origin in the coordinate system. I choose to make it the top of the cliff
(i.e. set 𝑦' = 0) and solve for 𝑦, which tells us how far below the base of the cliff the sea is.
1
𝑚
𝑚
𝑦 = 0 + 𝑣'( 𝑡 − 𝑔𝑡 ! = s175 t (40.0 𝑠) − s4.9 ! t (40.0 𝑠)! = −840 𝑚
2
𝑠
𝑠
Therefore, height of cliff = 840 𝑚
8
Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
27) You are arguing over a cell phone while trailing an unmarked police car by 25 m; both your
car and the police car are traveling at 110 km/h. Your argument diverts your attention from
the police car for 2.0 s (long enough for you to look at the phone and yell, “I won’t do
that!”). At the beginning of that 2.0 s, the police officer begins braking suddenly at 5.0 m/s2.
(a) What is the separation between the two cars when your attention finally returns?
Suppose that you take another 0.40 s to realize your danger and begin braking. (b) If you too
brake at 5.0 m/s2, what is your speed when you hit the police car?
a) 15 𝑚
b) 26.26 𝑚/𝑠; 94 𝑘𝑚/ℎ𝑟
Given: 𝑣' = 110𝑘𝑚/ℎ𝑟 = 30.56𝑚/𝑠
𝑥',/01 = 0
𝑥',2&3*45 = 25𝑚
(a)
$
=𝑥2&3*45 − 𝑥(&6 >@78!9 = vs25 + 𝑣',2&3*45 (2.0) + ! =𝑎2&3*45 >(2.0)! t − s0 + 𝑣',(&6 (2.0)tw m
= 15 𝑚
After another 0.4 seconds: =𝑥2&3*45 − 𝑥(&6 >@78!.-9 = 10.6 𝑚
(b) At t = 2.4 seconds you begin braking. Reset the initial conditions of part (b) to the
final conditions of part (a) at t = 2.4 seconds.
;
𝑣',2&3*45
= 𝑣2&3*45 (𝑡 = 2.4𝑠𝑒𝑐) = 30.56𝑚/𝑠 − 5.0𝑚/𝑠 ! (2.4𝑠𝑒𝑐) = 18.56 𝑚/𝑠
;
𝑣',(&6
= 𝑣',(&6 = 30.56𝑚/𝑠
;
𝑥',2&3*45
= 10.6𝑚
Reset clock to 𝑡' = 0
;
;
Set 𝑥2&3*45
− 𝑥(&6
= 0 ⇒ 10.6 = (30.56 − 18.56)𝑡 ⇒ 𝑡 = 0.883𝑠
;
𝑣(&6
@78'.<<#= = =30.56𝑚/𝑠 − (5.0𝑚/𝑠 ! )(0.883𝑠)> = 26m/s = 94km/h
9
Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
28) The position 𝑟⃗ of a particle moving in an xy plane is given by r⃗ = (2.00t # − 5.00t)ı̂ +
(6.00 − 7.00t - )ȷ̂ , with 𝑟⃗ in meters and 𝑡 is seconds. In unit-vector notation, calculate (a) 𝑟⃗,
(b) 𝑣⃗, and (c) 𝑎⃗ for 𝑡 = 2.00 𝑠 (d). What is the angle between the positive direction of the x
axis and a line tangent to the particle’s path at 𝑡 = 2.00 𝑠?
Note: this problem requires calculus and should be ignored.
a)
b)
c)
d)
r⃗ = (6.00ı̂ − 106ȷ̂)m
𝑣⃗(2𝑠𝑒𝑐) = (19.0𝚤̂ − 224𝚥̂)𝑚/𝑠
𝑎⃗(2𝑠𝑒𝑐) = (24.0𝚤̂ − 336𝚥̂)𝑚/𝑠 !
The line tangent to the particles path is simply the velocity vector which in polar
coordinates we get the angle:
v/
−224
θ = tan%$ T U = tan%$ T
U = −85& or + 275&
v>
19
29) A soccer ball is kicked from the ground with an initial speed of 19.5 m/s at an upward angle
of 45°. A player 55 m away in the direction of the kick starts running to meet the ball at that
instant. What must be his average speed if he is to meet the ball just before it hits the
ground?
𝑣⃗?@5 = −5.76m/s
Average speed = 5.76 𝑚/𝑠
10
Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
30) Upon spotting an insect on a twig overhanging water, an
archer fish squirts water drops at the insect to knock it into
the water (see figure). Although the fish sees the insect
along a straight-line path an and angle ϕ and distance 𝑑, a
drop must be launched at a different angle θ' it its
parabolic path is to intersect the insect. If ϕ = 36.0& and
𝑑 = 0.900 𝑚, what launch angle θ' is required for the
drop to be at the top of the parabolic path when it reaches
the insect.
Solution:
First convert the position of the insect into cartesian coordinates:
𝑥*A=547 = 𝑑𝑐𝑜𝑠𝜙; and 𝑦*A=547 = 𝑑𝑠𝑖𝑛𝜙
Use the range and rate equations for projectile motion:
𝑅 = 2𝑥*A=547 =
𝑣'! 𝑠𝑖𝑛2𝜃'
𝑔
and
ℎ = 𝑦*A=547 =
𝑣'! 𝑠𝑖𝑛! 𝜃'
2𝑔
ℎ
𝑠𝑖𝑛! 𝜃'
𝑠𝑖𝑛𝜙
𝑠𝑖𝑛! 𝜃' 𝑠𝑖𝑛𝜙
=
=
⟶
=
= 𝑡𝑎𝑛𝜙
𝑅 2𝑠𝑖𝑛2𝜃' 2𝑐𝑜𝑠𝜙
𝑠𝑖𝑛2𝜃' 𝑐𝑜𝑠𝜙
Use the trig identity (𝑠𝑖𝑛2𝛼 = 2𝑠𝑖𝑛𝛼𝑐𝑜𝑠𝛼) in the above equation to get
𝑠𝑖𝑛! 𝜃'
1
= 𝑡𝑎𝑛𝜃' = 𝑡𝑎𝑛𝜙 ⟶ 𝜃' = 𝑡𝑎𝑛%$ (2𝑡𝑎𝑛𝜙) = 55.5&
2𝑠𝑖𝑛𝜃' 𝑐𝑜𝑠𝜃' 2
11
Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
31) A suspicious-looking man runs as fast as he can along a moving sidewalk from one end to
the other, taking 2.50 s. Then security agents appear, and the man runs as fast as he can back
along the sidewalk to his starting point, taking 10.0 s. What is the ratio of the man’s running
speed to the sidewalk’s speed?
In Case I: The man is moving in the same direction the sidewalk is moving so we can
write:
𝑣!|# = 𝑣!|$ + 𝑣$|# =
𝑥
2.50𝑠
In Case II: The man is moving in the opposite direction the sidewalk is moving so we can
write:
𝑣!|# = 𝑣!|$ − 𝑣$|# =
𝑥
10.0𝑠
Adding the above 2 equations we get:
2𝑣!|$ =
5𝑥
10.0𝑠
Subtracting the second equation from the first equation above we get:
2𝑣$|# =
Therefore:
%!|#
%#|$
&
= ' = 1.67
12
3𝑥
10.0𝑠
Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
32) The graph below represents the position as a function of time of a rocket traveling in a
straight line.
A) What is the rocket’s average velocity between 0 and 4 seconds?
B) At what time(s) is the rocket accelerating?
C) At what time(s) is the rocket stopped?
D) When is the rocket moving in the +x direction? -x direction?
A) 𝑣%&' =
()
(*
=
)( +))
*( +*)
=
,-.+(+0.)
,2+-2
=
,0.
,2
= 12𝑚/𝑠
B) The rocket is accelerating when the position versus time plot is curved. In this
case between 0 and 4 seconds.
C) When the x-t or position versus time plot is flat. This would be between 4 and
7 secs and beyond 9 seconds.
D) The rocket is moving in the +x direction between 0 and 4 seconds. And moving
in the -x direction between 7 and 9 seconds.
13
Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
33) Alice wants to deliver a potato to Bob, who lives 800 meters away. To do this, she builds a
cannon with a muzzle velocity of 150 m/s in her back yard. Assuming the potato lands at the
same height from which it was launched, find the two possible angles she can launch the
potato in order to ensure an accurate delivery. Give your answers to the nearest degree. The
graph below represents the position as a function of time of a rocket traveling in a straight
line.
Given:
|𝑣' | = 150𝑚/𝑠
𝑣',$ = (150𝑚/𝑠, θ$ ) = (150𝑐𝑜𝑠θ$ 𝚤̂ + 150𝑠𝑖𝑛θ$ 𝚥̂)𝑚/𝑠
𝑣',! = (150𝑚/𝑠, θ! ) = (150𝑐𝑜𝑠θ! 𝚤̂ + 150𝑠𝑖𝑛θ! 𝚥̂)𝑚/𝑠
𝑥' = 0
𝑦' = 0
𝑦=𝑡) > = 0
𝑥=𝑡) > = 800𝑚
Find:
θ$ and θ!
$
$
𝑦 = 𝑦' + 𝑣'( 𝑡 − ! 𝑔𝑡 ! = 0 + 𝑣' 𝑠𝑖𝑛θ𝑡 − ! 𝑔𝑡 !
$
At landing 𝑦 = 0 ∴ 0 = s𝑣' 𝑠𝑖𝑛θ − ! 𝑔𝑡t 𝑡 ⟶ 𝑡 =
!@! =*AB
C
𝑥 = 𝑥' + 𝑣'+ 𝑡 = 0 + 𝑣' 𝑐𝑜𝑠θ𝑡 ⟶ 𝑥 = 𝑣' 𝑐𝑜𝑠θ𝑡
Substitute expression for t derived using the equation of motion for y
⟶ 𝑥 = 𝑣' 𝑐𝑜𝑠θ s
!@! =*AB
@!" (!=*AB4&=B)
C
C
t=
Use trig identity: 2𝑠𝑖𝑛θ𝑐𝑜𝑠θ = 𝑠𝑖𝑛(2θ) ⟶ 𝑥 =
∴ 𝑠𝑖𝑛(2θ) =
C<''F
@!"
@!" =*A(!B)
C
= 800𝑚 at landing
= 0.3484 ⟶ 2θ = 𝑠𝑖𝑛%$ (0.3484)
Calculator gives 2θ = 20.39& but remember 𝑠𝑖𝑛(𝑥) = 𝑠𝑖𝑛(180 − 𝑥) therefore
2θ = 180 − 20.39 = 159.61 is also valid.
∴ 2θ = 20.39& , 159.61& are both valid
Finally divide both angle by 2 to obtain θ
θ$ = 79.8& ≈ 80& and θ! = 10.2& ≈ 10&
14
Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
34) A cannonball is launched from the top of a flat
mountainside, which slopes upwards at 35 degrees
with respect to the horizontal. The cannonball
exits the cannon moving at 150 m/s, at an angle of
55 degrees with respect to the horizontal.
A) How much time (in seconds) elapses
between the moment the cannonball is fired
and the moment it lands on the
mountainside.
B) How far down the mountainside (in meters)
does the cannonball land?
Givens:
𝑣_𝑜 = 150𝑚/𝑠
𝑦' = 0
𝑥' = 0
𝑣+' = 𝑣' 𝑐𝑜𝑠55&
𝑣(' = 𝑣' 𝑠𝑖𝑛55&
Consider the diagram above and recognize that when the cannon ball lands it must hit the slope
and the relationship between y and is 𝑦 = −𝑥𝑡𝑎𝑛(35& ), which gives one of the three equations
you need to solve this problem.
Next write down the equations of motion:
$
$
𝑥 = 𝑥' + 𝑣+' 𝑡 + ! 𝑎+ 𝑡 ! = 𝑣' 𝑐𝑜𝑠(55& )𝑡;
$
𝑦 = 𝑦' + 𝑣(' + ! 𝑎( 𝑡 ! = 𝑣' 𝑠𝑖𝑛(55& )𝑡 − ! 𝑔𝑡 !
Substituting the expression for the slope of the mountainside (𝑦 = −𝑥𝑡𝑎𝑛(35& )) into the
kinematic equation for 𝑦 we get:
$
−𝑥𝑡𝑎𝑛(35& ) = 𝑣' 𝑠𝑖𝑛(55& )𝑡 − ! 𝑔𝑡 !
Next substitute the kinematic equation for 𝑥 into the above to get:
$
−𝑣' 𝑐𝑜𝑠(55& )𝑡𝑎𝑛(35& )𝑡 = 𝑣' 𝑠𝑖𝑛(55& )𝑡 − ! 𝑔𝑡 ! and then divide through by 𝑡.
$
⟶ −𝑣' 𝑐𝑜𝑠(55& )𝑡𝑎𝑛(35& ) = 𝑣' 𝑠𝑖𝑛(55& ) − ! 𝑔𝑡 and now solve for 𝑡.
𝑡=
!@! G=*A(HH# )I4&=(HH# )7?A(#H# )J
C
= 37.37 𝑠 ⟶ 𝑡 ≈ 37𝑠
Now that we have the time, we can solve for 𝑑 (refer to figure above):
+
d = =*A(HH# ) =
@! 4&=(HH# )7
=*A(HH# )
= 𝑣' t =
($H'F/=)4&=G(HH# )J(#K.#K=)
=*A(HH# )
15
= 3,925𝑚 ⟶ 𝑑 ≈ 3,900𝑚
Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
35) Your friend started riding her bike to get to her physics exam, but right after she left, you
realized that she forgot to take her calculator. You grab the calculator and get on your bike,
riding as fast as you can to catch up with her, along a long, flat, straight road. Your friend is
riding her bike at a constant speed of 5.0 m/s. At the moment when you are 12 meters
behind her, you toss the calculator toward your friend and it makes a perfect landing in the
basket that's mounted on her bicycle. In your reference frame, you threw the calculator with
a speed of 10 m/s (relative to you) and at an angle of 45 degrees above the horizontal. How
fast was your bicycle moving (relative to the ground) at the moment when you threw the
calculator? (Assume that the starting height above the ground for your throw is the same as
the height where the calculator lands in the basket.)
Solution: This problem is problem is very similar to the Indiana Jones Problem solved at the
beginning of the lab lecture this week.
First lets visualize the problem and to do this we will first define some subscripts.
𝑔 − ground frame of reference, 𝑢 − your frame of reference, 𝑓 − your friend,
𝑐 − calculator
List the givens:
𝑣4|6,' = 10
𝑚
𝑚
𝑐𝑜𝑠(45& )𝚤̂ + 10 𝑠𝑖𝑛(45& )𝚥̂
𝑠
𝑠
𝑣)|C = 5.0
𝑚
𝑠
We need to find the velocity of your friend relative to you such that when the calculator lands
your friend and the calculator are in the same position or 𝑥4 = 𝑥) and 𝑦4 = 𝑦) .
In your frame of reference the position of the calculator and your friend are given by the
following kinematic equations: Note to simplify the subscripts the following equations are in
your frame of reference.
𝑥4 = 0 + (10 𝑚/𝑠)𝑐𝑜𝑠(45& )𝑡;
1
𝑦4 = 0 + (10 𝑚/𝑠)𝑠𝑖𝑛(45& )𝑡 − 𝑔𝑡 !
2
16
Physics 44-28794-28797 Spring 2023 Midterm Exam #1 Review Worksheet Answers
𝑥) = 12 𝑚 + 𝑣) 𝑡;
𝑦) = 0
Solve for 𝑦4 = 𝑦) = 0 to find the time the calculator is in the air.
0 = (10 𝑚/𝑠)𝑠𝑖𝑛(45& )𝑡 − (4.9 𝑚/𝑠 ! )𝑡 ! ⟶ 𝑡 =
(10 𝑚/𝑠)𝑠𝑖𝑛(45& )
= 1.443 𝑚/𝑠 !
4.9 𝑚/𝑠 !
Now armed with the time we can find the friends velocity relative to you by setting 𝑥) = 𝑥4 .
& )𝑡
12𝑚 + 𝑣) 𝑡 = (10 𝑚/𝑠)𝑐𝑜𝑠(45
= −1.245 𝑚/𝑠
(10 𝑚/𝑠)𝑐𝑜𝑠(45& )(1.443 𝑠) − 12 𝑚
⟶ 𝑣) =
1.443\ s
𝑣)|6 = −1.245 𝑚/𝑠
or.
𝑣6|) = 1.245 𝑚/𝑠
Finally, using relative velocity and knowing the velocity of the friend relative to the ground we
can find the velocity of you relative to the ground.
𝑣6|C = 𝑣6|) + 𝑣)|C = 1.245 𝑚/𝑠 + 5.0 𝑚/𝑠 = 6.245 𝑚/𝑠 ≈ 6.2 𝑚/𝑠
17