Name
Bell
Honors Physics 1st Semester Review
1. The drawing below shows a moving object which has been photographed using a strobe flash.
The object starts at rest and moves to the right. The
dots show the location of the object each second.
a. Plot the position vs. time for the object on the graph
to the right.
b. Write a mathematical equation (for this graph) to
describe the motion of the object. Symbols and units must be correct for full credit (Hint: it
should be in the form of y = mx + b)
x = (2m/s)t + 0m
c. Explain the meaning of the slope and the significance of the y-intercept.
slope = velocity of the object. Y-intercept tells you the initial position of the object.
d. Define displacement and then find the displacement on the graph above for 2s – 4s.
displacement is the straight line distance between where you start and where you finish; it is
also the change in position of the object: xf - xi = 8m – 4m = 4m.
For questions 2 – 11 draw the graph and select the letter combination for the graph that most closely
represents the motion of the needed answer from the provided Graph Answer Key. This will give
you practice with selecting the appropriate graph for the final exam.
Draw the position vs. time, velocity vs. time graph, and acceleration vs. time graph created by
walking toward a motion sensor at a constant velocity. Assume you started at a positive initial
position and you are moving toward a position of zero, (moving in the negative direction).
2.
3.
x
4.
v
a
t
t
1
t
Honors Physics Semester 1 Review
Draw the motion maps for the above scenario and select the letter combination for the motion map
that most closely represents the motion of the needed answer from the provided Motion Map Answer
Key
5.
6.
Draw the position vs. time, velocity vs. time graph, and acceleration vs. time graph for an object
dropped from rest from the top of a building. Assume the ground is a vertical position of zero.
7.
8.
9.
a
v
x
t
t
t
Draw the position vs. time, velocity vs. time graph, and acceleration vs. time graph to represent the
motion of an object thrown straight up with an initial position of zero. Just consider the object’s
motion until it reaches its final height or highest point.
10.
11.
12.
v
x
a
t
t
t
13. What does the area under the velocity versus time curve represent?
The displacement of the object….
Consider the position vs. time graph for objects A and B below on the left.
14. How does the motion of object A differ from that of object B?
A moves with a constant velocity, and B starts off fast and slows
down, eventually stopping sometime after 5 seconds.
Draw the general shapes of the velocity graphs produced by object
A and object B. Label each.
v
B
A
t
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Honors Physics Semester 1 Review
15. In the chart below, you will qualitatively compare values for objects A and B then provide a
brief explanation for each answer.
Comparison:
is A > B, A < B, or A = B
A=B
a. Displacement at t = 5s
b. Average velocity from 0s - 5s
= (total displacement)/ (time)
c. Instantaneous velocity at 5s
A=B
A>B
B>A
d. Magnitude ( or size) of the
acceleration at t = 5s
How do you know?
they both start and finish at the
same position, therefore they
have the same displacement.
they both have the same
displacement over the same
time, therefore the same average
velocity. (V = Δx/Δt).
line A is steeper at a time of 5
seconds than the tangent line to
the curve for B at the same time
A has zero acceleration since its
velocity is constant and B’s
slope is changing indicating that
the velocity is changing, hence
a non-zero acceleration and
larger magnitude / size.
16. A 80kg snowboarder starts from rest on the top of a snowy hill.
What is the snowboarder’s final velocity after traveling 20m down
the hill with at a constant acceleration of +1m/s/s?
vf2 = vi2 + 2ax
vf2 = (0m/s)2 + 2(1m/s/s)(20m)
vf2 = + 40m/s
vf = + 6.3m/s
17. Draw a quantitative velocity versus time graph for the
snowboarder’s motion. Label each axis.
a. After 10 seconds, what is the snowboarder’s
displacement? This can be found either graphically or
algebraically (with equations).
10
5
x = ½ at2 + vit
x = ½ (1m/s/s)(10s)2 + (0 m/s)(10s)
x = 50 m
OR solve using area under the curve
d = ½ bh
d = ½ (10s)(10m/s)
d = ½ (100m)
d = 50m
10
10
b. Write a mathematical equation (for this graph) to describe the motion of the object.
Symbols and units must be correct for full credit.
v = (1m/s/s)t + 0m/s
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Honors Physics Semester 1 Review
18. A 65kg girl steps off of a high dive 10m above the surface of the water. Assume there is no air
resistance. You will need to remember the value for the acceleration caused by gravity!
a. How long is she in the air? Hint: Find the time she is being accelerated by gravity alone.
x = ½ at2 + vit
m = ½ (-10 m/s/s/)t2 + 0
Remember that the acceleration caused
2 s2 = t 2
by gravity is always -10m/s/s regardless
t = 1.414 s
of the object’s mass.
b. What is the girl’s final velocity right before she hits the water?
vf2 = vi2 + 2ax
vf2 = 0 + 2(-10m/s/s)m)
vf2 = 200m2/s2
vf = 14.1 m/s
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Honors Physics Semester 1 Review
For questions 20-23our students are pushing each with +100N on a 2000kg vehicle. Assume that
there is a frictional force of 200N between the truck and the road.
20. Draw a force diagram for the truck AND find the truck’s
acceleration. Hint: 1st find the sum of the forces (∑F) on the
car.
Fx = Fp + Fp + Fp + Fp – f
FN
ma = 4(Fp) – f
Ff
Fp
(2000kg)(a) = 4(100N) – (200N)
(2000kg)(a) = 200N
a = 0.1m/s/s
Fg
Fy = FN – Fg = 0N
OR… a = F/m = (400N + (-200N)) / 2000kg = 0.1m/s/s
21. How long will it take to travel a distance of 50 m?
x = vot + ½ at2
m = 0 + ½ (0.1m/s2)t2
100m/0.1m/s2 = t2
t = 32s
22. What is the truck’s final velocity at the end of the 50m if it started from rest?
vf2 = vi2 + 2ax
vf2 = 02 + 2(0.1m/s/s)(50m)
vf2 = 10m2/s2
vf = 3.16 m/s
23. If four 70kg students sit on top of the vehicle, what is the size of the normal force of the ground
pushing back upon the car?
Ff
Fp
Fy = FN – Fg - Fg - Fg - Fg - Fg = 0N
FN = Fg + Fg + Fg + Fg + Fg
FN = mg + mg + mg + mg + mg
FN = (2000kg)(10N/kg)+ (70kg)(10N/kg)+ (70kg)(10N/kg)+ (70kg)(10N/kg)+ (70kg)(10N/kg)
Fgc
Fgs
Fgs
Fgs
Fgs
FN = 20,000N+700N+700N+700N+700N
FN = 22,800N
For each question (24 - 28), one or more features of the system have been changed. You are to
indicate what effect the change will have on the acceleration compared to that in question #17
part (a).
Use the following answer key.
a. The acceleration will be greater.
b. The acceleration will be less.
c. The acceleration remains the same.
d. It's not possible to tell.
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Honors Physics Semester 1 Review
___B__24. The mass of the vehicle is increased by adding 4 students in the backseat. (If the mass
increases what effect does that have on the acceleration?)
___A__25. Five students now push each with a 100N force.
___C__ 26. The 1st vehicle is replaced with a 4000kg truck and each of the four students now push
with 200N of force. We will also assume the frictional force is twice as big, -400N.
___B__ 27. The students push at a downward angle of 45 degrees instead of completely horizontal.
___A__ 28. The vehicle is pushed down a small incline.
Means F = 0 N
29. A 100kg person traveling downward on an elevator at a constant speed.
Draw a qualitative force diagram for the man while moving at a constant
negative velocity.
FN
Fg
30. If the elevator is traveling at a constant -4m/s what is the size of the normal force? Does the
person feel heavier, lighter, or normal?
Since the person is traveling at a CONSTANT velocity, the normal
force equals the force of the earth (Fg) so the person feels normal
FN
31. Draw a qualitative force diagram for the same 100kg person accelerating
upward (when the elevator begins to move from the 1st to the 2nd floor). Is
the normal force larger, smaller or the same size as Fg? Since the
elevator is accelerating upwards, there is a net force in the upward
direction. The upward force to the man is supplied by the elevator floor
that is experiencing an upward net force and therefore an upward
acceleration. This causes the man to feel “heavier” than he normally would,
so FN > Fg
Fg
32. Does the person, as described in part “31”, feel heavier, lighter, or normal in weight? Why?
Whoops, see above.
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Honors Physics Semester 1 Review
34. Draw and calculate the x and y components (Fx and Fy) for the following forces at an angle.
To solve the following problems, either use SOH-CAH-TOA, or the equations you can use as
long as the angle  touches the x-axis…
10N
Fy
8N
Fy
25°
60°
Fx
Fx = Fcos
Fy = Fsin



Fx = Fcos
Fy = Fsin
Fx = 10Ncos
Fy = 10sin 


Fx = 8cos Fy = 8sin
Fx = 9.06N
Fy = 4.22N
Fx = 4N
Fy = 6.93N
35. A person pushes to the right with a force of 200N on a child hanging on to a rope swing. If the
person on the rope swing is not moving and the rope makes a 60 degree angle with the horizontal…
Fx
FT
a. Label the forces in the force diagram to
the right for the child hanging on the
rope swing.
FTy
b. What is the size of the force of tension in the rope?
60˚
FTx
Fp = +200N
Fg
Solution A
Fx = Fp – FTx = 0N
Fx = 200N – FTx = 0N
FTx = 200N
Solution B

Looking at the force diagram we can see the Fp
= FTx = 200N

FTx = (FT)cos
200N= (FT)cos
 FT
FT = 400N
cos  
A FTx


H FT
cos 60  
Once you know the value of FTx or FTy
then you can use SOH-CAH-TOA or
the equations that are reliable if theta
touches the x-axis to back track and
find FT as you see above…
200 N

FT
FT  cos 60  200 N 
FT  200 N cos 60  400 N 
c. What is the mass of the child on the rope?
Fy = FTy – Fg = 0N
Remember, FTy = FTsin as long as
touches the x-axis
FTy – Fg = 0N
FTy = Fg
FTsin = Fg
400Nsin = Fg
346.4N = Fg
Alternate Solution:

Looking at the force diagram we can see the
Fg = FTy 

sin  
7
O FTy


H FT
Honors Physics Semester 1 Review

if Fg = 346.4N then mass = 34.6kg
sin 60 
FTy
400 N

FT y  400 N  sin 60 
FTy  346.4 N  Fg 
if Fg = 346.4N then mass = 34.6kg
36. An 80kg snowboarder starts from rest on the top of a snowy hill. The incline makes a 30° angle
with the horizontal.
a. Draw a force diagram for the snowboarder.
FN
Ff
OR
FN
Fgx
Ff
Fgy
60°
30°
Fg
Fgy
Fgx
Fg
30°
b. How big is the component of gravity pulling the snowboarder
down the hill, parallel to the surface? (Fgx =?)
Solution A
Solution B

The component of gravity pulling the
Use the force diagram above and to the
snow boader down the hill is equal to
right.
the “x-component” of gravity
O Fgx
sin  

Fgx= Fgcos60

H
Fg
Fgx= mgcos60
Fgx= (80kg)(10n/kg)cos60 = 400N
F
sin 30  gx 
800 N
Fgx  800 N  sin 30 
Fgx  400 N 
c. How big is the component of gravity pulling the snowboarder into the hill, perpendicular to
the surface? (Fgy = ?)
d.
Solution A
Solution B

Fgy= Fgsin60
Use the force diagram above and to the
Fgy= mgsin60
right.
Fgy=(80kg)(10N/kg)sin60
A Fgy
cos  

Fgy= 692.8N

H
8
Fg
Honors Physics Semester 1 Review
cos 30 
Fgy
800 N

Fgy  800 N  cos 30 
Fgy  692.8N 
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Honors Physics Semester 1 Review