Download Name______________ _________Date____________ General

Name_______________________________Period_________Date____________
General Physics – Progress Assessment 1 – Answer Key
Test Date:______________
1. What is the speed of a car that travels 300 miles in 5 hours?
Looking For
Speed
Given
Relationship
Solution
d=300 m
t=5 hrs.
s=d/t
s = 300/5
60 miles per hour
2. A cat runs at 2 m/sec. how far will the cat get in 35 seconds?
Looking For
Distance
Given
Relationship
Solution
v= 2 m/s
t=35 s
d=vt
d = 2 x 35
70 m
3. How long does it take you to run 400 meters if your speed is 4 m/s?
Looking For
Time
Given
Relationship
Solution
d=400 m
v= 4 m/s
t=d/v
t = 400/4
100 s
4. Emma runs around a 400 meter track. What is the distance she travels? What is her
displacement?

Emma travels a distance of 400 meters but her displacement is 0 m because she ends
where she started.
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5. What is acceleration? What is the acceleration of a car that moves from rest to 22 m/s in 5
seconds?
Looking For
Acceleration
Given
Relationship
Solution
v=22 m/s
t=5 s
Acceleration = change
in velocity/time
4.4 m/s2
6. A helicopter’s speed increases from 0 m/s to 25 m/s in 5 seconds. Calculate its
acceleration.
Looking For
acceleration
Given
Relationship
Solution
v=25 m/s
t=5 s
Acceleration = change
in velocity/time
5 m/s2
7. What does a flat line on a position vs. time graph mean?
x
(m)
Not moving.
8. What does the following position vs. time graphs below illustrate about the objects
speed?
x
x
(m)
(m)
t (s)
^constant speed
t (s)
^decreasing speed
t (s)
^no motion
x
(m)
changing speed 
t (s)
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9. Plot the following data on the axes below:
Time (s)
0
1
2
3
4
5
Position (meters)
0
5
15
30
50
80
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10. Does it take more force to stop a bowling ball or a tennis ball? Why?
Bowling ball has more mass thus more inertia (the bowling ball requires a greater force to
stop it.)
11. A force of 500-N propels a 1000-kg train down a track. If a frictional force of 200 N acts
on the train determine the (a) the net force acting on the train and
Looking For
Net force
Given
Relationship
Solution
F2=500 N
F1=200 N
F2-F1
300 N
(b) the acceleration of the train and the direction the train is accelerating.
Looking For
Acceleration
Given
Relationship
Solution
F=300 N
M=1000 kg
a=F/m
0.3 m/s2
12. A 50-N force is applied to a 2-kg box to move it across the floor. A 10-N frictional force
acts on the box. Determine the net force acting on the box? Determine the acceleration
of the box?
10 N
Looking For
Net force
Looking For
Acceleration
50 N
Given
Relationship
Solution
F2=50 N
F1=10 N
F2-F1
40 m/s2
Given
Relationship
Solution
F=40 N
M=2 kg
a=F/m
20 m/s2
13. What is Newton’s 1st law of motion?

An object at rest stays at rest, an object in motion continues in motion at constant
speed in a straight line, unless acted upon by an unbalanced force.
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14. You are traveling in a train and suddenly fall forward. Explain why using one of
Newton’s laws of motion.

According to Newton’s 1st law of motion: the law of inertia. Because of your inertia
you want to continue in motion.
15. What is the definition of a force? Net Force?

A force is a push or a pull.

Net force is the sum of all forces acting upon an object.
16. What is the net force acting on an object that moves with a constant velocity?

0N
17. A car with a mass of 500-kg accelerates at 3 m/s2. Calculate the force acting on the car.
Looking For
Given
m=500-kg
a=3 m/s2
Force
Relationship
Solution
1500 N
F=ma
18. Sam applies a 25-N force to push a 5-kg bowling ball. Sam applies the same 25-N force
to push a 2-kg kickball. Which ball accelerates faster?

The 2-kg ball accelerates faster.
19. What is Newton’s 3rd law of motion?

For every action there is an equal and opposite reaction.
20. According to Newton’s 3rd law of motion if the reaction force is Chris pushing gently on
the wall, what is the reaction force?

The wall pushing on Chris.
21. What is the momentum of a 200 kg motorcycle travelling at 15 m/s?
Looking For
Momentum
Given
Relationship
Solution
m=200 kg
v=15 m/s
P=mv
= 200 x 15
P = 3000 kg m/s
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22. Calculate the mass of a car traveling at 10 m/s that has a momentum of 4,000 kg m/s.
Looking For
Mass
Given
Relationship
Solution
p=4000 kg m/s
v=10 m/s
m=p/v
= 4000/10
400 kg
23. What are the units of momentum? Impulse?

Momentum = Kg m/s

Impulse N·s
24. What is the formula for impulse?

J = F*t
25. What is the difference between an elastic and an inelastic collision?

Elastic collisions are where two objects collide and bounce off each other.

Inelastic collisions are where two objects stick together after the collision.
26. Explain the physics behind padded dashboards.
Padded dashboards increases contact time thus decrease force.
27. A 500-kg car moves at 5 m/s in 2 seconds. Determine the momentum of the car?
Looking For
Given
m=500 kg
v=5 m/s
Relationship
P=mv
Solution
2500 kg m/s
Momentum
Calculate the impulse necessary to stop the car.
Looking For
Given
Relationship
m=500 kg
v=5 m/s
Impulse
Solution
2500 kg m/s
Mv=Ft
Calculate the force acting upon the car as it comes to a stop.
Looking For
Force
Given
m=500 kg
v=5 m/s
t=2 s
Relationship
Solution
1250 N
Mv = Ft
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