Unit B Review Questions
1.
Quantity
Symbol
Scalar or Vector
Most Common Unit(s)
distance
scalar
m
km
time
scalar
s
h
scalar
m/s
km/h
position
vector
m
km
displacement
vector
m
km
average
velocity
vector
m/s
km/h
acceleration
vector
m/s2
net force
vector
N
scalar
kg
vector
kg•m/s
vector
N•s
kg•m/s
scalar
J
kg•m2/s2
average
speed
v
mass
m
momentum
impulse
impulse
kinetic
energy
Ek
2.
Application
This equation can be used to determine the . .
Equation
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average velocity of a glacier’s retreat
displacement of a vehicle on a highway during a driver’s
reaction time
time a vehicle takes to reach an object illuminated at the end
of the headlights’ beam
acceleration of a cart down a ramp
final velocity of an object dropped from rest for a given period
of time
time for a vehicle to come to a stop, given an initial velocity
and a value for deceleration
displacement of a car in an acceleration lane
braking distance for a vehicle, given the initial velocity and the
time needed for stopping
initial velocity of an object that is thrown into the air, given the
maximum height and the time to reach that height
time for an object thrown into the air to reach its highest point,
given the initial velocity, the maximum height, and the travel
time
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maximum height of a gymnast
length of time for a diver to reach the water given
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force exerted on a passenger in a motor vehicle accident
acceleration that results when a net force acts on a known
mass
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momentum of a hockey puck
velocity of a motorcycle
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force exerted by a bird colliding with a windshield
force exerted by a puck on a goalie if the time of the impact is
made larger through the use of protective equipment
force exerted on a passenger in a vehicle during an accident
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force of the air bag exerted on the passenger
force of a car on a truck in a collision
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velocity of a curling rock after it has been struck head-on by
another curling rock with a known velocity (example of a hitand-rebound collision)
combined velocity of a car and truck after a truck rear-ends a
stationary car at an intersection, given both the masses and
the initial velocity of the truck (example of a hit-and-stick
collision)
velocity of the second piece from a stationary firecracker, given
the masses and the velocity of the first piece (example of an
explosion)
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W = F•d
3. a.
b.
c.
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kinetic energy of a vehicle before and after a collision, given
the mass of the vehicle and the speed of the vehicle before and
after
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work done raising an egg test dummy to the release height
(Investigation: Designing and Testing a Helmet)
According to Newton’s first law of motion, the driver would tend to maintain her velocity
and would continue moving forward even though the car had come to a stop.
There are several technologies that would help reduce the woman’s injuries. The list
includes seat belts, air bags, a collapsible steering wheel, and a crumple zone in the
frame of the vehicle.
The seat belt would provide an external net force to decelerate the driver with greater
safety than if the net force was provided by a collision between the driver and the
steering wheel. The seat belt allows the force to act over a larger time interval, so the
size of the force can be reduced. A collision with the interior of the car means that the
woman will be acted upon by the same impulse, but the time of the interaction is much
d.
4. a.
less, resulting in a larger net force.
Since a pregnant woman carries her unborn child in her abdomen, the seat belt must be
designed to exert forces on the woman without harming the unborn child. Similarly, the
steering wheel must be properly positioned so that minimal forces will be exerted on the
pregnant woman’s abdomen in the event of a crash.
Let east be the positive direction.
The woman’s momentum is 1.06 × 103 kg•m/s [E].
b.
The driver’s momentum after the collision is 0.
c.
The impulse required to decelerate the driver is 1.06 × 103 N•s [W].
d.
The force required to stop the driver’s forward motion is 1.18 × 103 N [W].
e.
f.
5. a.
The force required to stop the driver in 0.090 s is 1.18 × 104 N [W].
The computer image on the far right best shows a collision that reduces the forces due to
the longer stopping time. The computer image in the centre shows the larger forces
acting (ten times larger) because the stopping time is so small (ten times smaller).
Truck
The displacement of the truck while reacting is 25.0 m [N].
Car
b.
The displacement of the car while reacting is 30.0 m [N].
The magnitude of these displacements is called the reaction distance.
6. Since solving for time involves dividing velocity by acceleration and since the question is asking
for distance (a scalar), the vector notation is dropped.
a. Truck
Car
vi = +20.0 m/s
vf = 0
a = –5.85 m/s2
First, determine the time.
Next, determine braking distance.
First, determine the time.
Next, determine braking distance.
The car travelled 34.2 m while braking.
The truck travelled 23.7 m while braking.
b. Each of these values is called the braking distance.
7. a. Truck
Car
The truck’s stopping distance is 48.7 m.
The car’s stopping distance is 64.2 m.
b.
8. a.
The sports car’s displacement is 56.0 m.
b.
c.
This driver is in a situation where neither option is safe. If the driver attempts to brake,
the vehicle will not stop until it is at least one car length into the intersection. If the
driver attempts to continue driving through the intersection, the vehicle will still be
approaching the intersection at high speed when the light turns red. This situation is
caused by the driver approaching the intersection so quickly that no safe options were
available. Defensive drivers avoid this situation by driving at an appropriate speed and by
using clues, such as the “Don’t Walk” signal for pedestrians, to anticipate that the traffic
light is about to turn yellow.
9.
10.
11. The motion of the skateboard on Ramp A is accelerated motion because the best-fit line is a
curve. The fact that this curve gets steeper as time goes on indicates that the skateboard is
speeding up.
12. The motion of the skateboard on Ramp B is accelerated motion because the best-fit line is a
curve. The fact that this curve gets less steep as time goes on indicates that the skateboard
is slowing down.
13.
14.
15. Ramp A
Ramp B
16. For Ramp A, the skateboard is rolling downhill, speeding up as it goes.
For Ramp B, the skateboard is rolling uphill, slowing down as it goes. Because the magnitude
of the acceleration is slightly greater for Ramp B, this ramp must be a little steeper.
17. a. The type of motion depicted by the graph for the truck is uniform motion because the
graph is a straight line.
b. The type of motion depicted by the graph for the car is decelerated motion because the
graph is a curved line.
c. Since the graphs cross at 2.00 s, this means that both vehicles are at the same position.
A collision likely occurred.
18. a. Let east be the positive direction.
m1 = 1200 kg
m2 = 900 kg
After the collision, the vehicles had a velocity of 14.2 m/s [E].
b.
The change in the momentum of the truck is 6.27 × 103 kg•m/s [W].
c.
The impulse on the truck is 6.27 × 103 N•s [W].
d.
The force on the truck is 4.83 × 103 N [W].
19. Let east be the positive direction.
a.
The final velocity of the car is 17.1 m/s [E].
b.
The change in momentum of the truck is 8.88 × 103 kg•m/s [W].
c.
The change in momentum of the car is 8.88 × 103 kg•m/s [E].
d. Yes, the law of conservation of momentum states that the magnitudes of the change in
momentums must be the same. However, one object is losing momentum and the other object
is gaining momentum.
e.
The impulse of the truck on the car is 8.88 × 103 N•s [E].
f.
The impulse of the car on the truck is 8.88 × 103 N•s [W].
g. Yes, the impulses have the same magnitude but opposite directions.
h.
The force of the truck on the car is 6.83 × 103 N [E].
i.
The force of the car on the truck is 6.83 × 103 N [W].
j. Yes, the magnitudes of the forces acting on both vehicles are the same but point in opposite
directions. This is stated in Newton’s third law.
k. F = 6830.769 231 N
W = Fd
d = 0.300 m
= (6830.769 231 N)(0.300 m)
W=?
= 2.05 × 103 J
The work done to compress the vehicle is 2.05 × 103 J.
l.
To do this work, 2.05 × 103 J of energy is required.
20. No, forensic engineers do not determine who was at fault in a collision. Forensic engineers collect
and analyze data to try to determine exactly what happened.
21. Although answers may vary, the following chart shows typical student responses.
ACCIDENT RECONSTRUCTION
Data Collected by Forensic
Engineer at Scene of
Collision
Science Terms and
Concepts Used in
Analyzing Data
Information About
Collision Derived
from Data
length of skid marks
left on roadway
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displacement
negative acceleration
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the initial speed of the vehicle
that created the skid marks prior
to the collision
amount of crumpling
or deformation of
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displacement
kinetic energy
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the initial speed of the vehicles
metal parts
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crumple energy
law of conservation
of energy
rest positions of the
vehicles at the scene
of a collision
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momentum
conservation of
momentum
Science 20 © 2006, Alberta Education
prior to the collision
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displacement
the point of impact
the initial speed of the vehicles
prior to the collision