Download File - Mr. Downing Science 20

Science 20 - Unit B Physics
Chapter 1 – motion
B1.1 – 1.3 Review: Average Speed & Velocity


uniform motion
o motion is _____________________________________________________________________________
o
most objects in the natural world move in __________________________________________________
o
motion can only be considered uniform when the object is traveling in ___________________________
_______________________________________.
o
This “ideal” situation – rarely occurs for long periods of time because:
o
forces such as friction or air resistance _____________________________________.
o
most objects _____________________ during its motion, causing an object to ______________
_______________________.
o
when an object is _____________________________, it is still considered to have uniform motion
because it has a _________________________ of ______________
o
average speed (v)
o
represents the _______________________ over time, where
speed
o

d = total distance traveled

t = time elapsed
Rearrange the formula to solve for:

o
v = Δd
Δt
Δd =
∎ Δt=
Speed is a _____________________________________ (hint: Speed
 as such, you do not need to specify direction
 you DO have to include units:
= Scalar)
→ speed is measured in either m/s (metres per second)or km/h (kilometers per hour)

instantaneous speed
o
o
unlike average speed, instantaneous speed is
____________________________________________ Ex.
When you drive to Calgary, your average speed for the trip is
110 km/h
Your instantaneous speed would be when you:
 drive through Red Deer, you glance at the
speedometer and it shows you are traveling at
80km/h
 when you are passing along the highway you see your
speedometer reads 120 km/h
v= d
t
 note the “d” is a specific
distance not a change in
distance (Δd)
 same with the time; “t” is a
specific time, not a change in
time (Δt)

Distance traveled
o Distance traveled can be found in two ways:
o The distance of each leg of the trip can be added up
o

dT = d1 + d2 + d3

dT = ______________________________

dT = ________
OR by rearranging __________________________________________
v = Δd
Δt
Δd= vΔt
Example #1: distance
A driver talks on the phone for 36.0s while she is driving at 100km/h. How far does she travel in this time?
o
If the speed and distance travelled of the moving object is known the time elapsed (gone by) can be
found
Example #2: time elapsed
Standard headlights allow a driver to see up to 60m ahead of the car. If a car is traveling at 120km/h, how many seconds
does a driver have between when they see a person on the side of the road, and when they pass the person?
Conversion factors
 converting between units
o conversion factors allow you to convert from one unit to another

o
ex. 100cm ↔ 1m, 60s ↔ 1min
Hint : the ratio of the value to the unit is ___________________________________________________:

the _________________ unit will have a ______________________number and visa-versa
→ Ex. 100 mm = 10 cm = 0.1 m
→ mm is a smaller unit than cm, but 1000 > 10

Convert the following:
o
245g to kg
o
2.0 mol to mmol
o
3.5hm to m
o
4500 nm to m
o
o
In science most speeds are given in m/s, however in everyday terms, most people use km/h
to convert between m/s and km/h is a bit more difficult because BOTH the distance AND time value
must be converted:
50 m x 1000 m x 1 h
= 50 000 = _________m/s
1h
1 km
3600 s
3600
10 m
1s
o
x
1km
x 3600 s = 36 000 = __________ km/h
1000 m
1h
1000
To simplify…

2 hr to s 

4.5 min to h 

100 km/h to m/s 

18 m/s to km/h 
Example #3:
Use the appropriate conversion factor to determine how many minutes in a week. (Show all of the steps)
 Vector quantities
o scalar quantities simply answer the question “ _______________________”

Ex. ________________________________________________ (remember: Speed = Scalar)
o
vector quantities describe “ __________________ “ AND “___________________________________”
 Ex. _____________________________________________
o
(hint: Velocity = Vector)
the symbols for vector quantities are written with a single-barbed arrow
 Ex.______________________

The difference between distance and displacement
o
Recall, distance is a __________________________ quantity.
o
It measures _______________________ an object travels.
If an object travels with uniform motion 3 m to the right, then turns around and travels 3 m back to where it started from
and then travels and additional 2 m to the left, what distance has the object travelled?
dT= d1 + d2 + d3
dT = 3m + 3m + 2m
dT = 8m
o
__________________________________ is a _____________________________quantity.
o
It measures _________________________________________________________ (the reference point).
If an object travels with uniform motion 3 m to the right, then turns around and travels 3m back past the reference point,
and then travels and additional 2 m to the left, what is the objects displacement?
=
1
+
2+
3
= _______________________________
= ___________
Note: When including a direction the order is always:
value - unit- [direction]
100 km/h [W]

Sign convention
o because direction matters for vector calculations, certain directions are assumed to be either positive
or negative:

_________, _________, _________, ________ are positive.

_________, _________, _________, ________ are negative:
Example #4: sign convention
A trucker drives from Calgary to Edmonton, a distance of 360km, then turns around to make a
delivery in Red Deer.
What is his distance traveled?
What is his displacement?
dT= d1 + d2

=
1
+
2
Velocity ( )
o represents the _________________________ over time, where


=
t
=
o
o
velocity is a vector quantity, you _________________ specify direction
measured in m/s (metres per second)
 Ex. 35 m/s [W]
o The velocity formula is rearranged to solve for displacement (
and time t ) in the same way as the
speed formula.

and
→ You must use ______________________________to solve for _______________and
________________________________ to solve for ____________________, they can
not be used interchangeably.
Example #5: Velocity (
)
A ball rolls 10.0m in 4.00s, bounces against a wall, then rolls 5.0m in 2.00s.
What is the average speed of the ball?
What is the velocity of the ball?
Speed requires distance travelled
Velocity requires displacement
Assignment:
 Practice Problems: 6 & 7 (pg 172)
 1.1 Section Questions #4 (pg 173)
 Practice Problem: 9 a, b & c (pg 175)
& 11 (pg 176)
 1.2 Section Questions 2 & 4 (pg 177 & 178)
 1.3 Section Questions 2-5 (pg 185)
* use practice problem 14 on pg 180 for
guidance if you need help using the
“head-to” tail method”
B1.4 - Using Graphs to Analyze Motion
Using graphs to analyze motion

position-time graphs
o

the slope of a position-time graph allows you to find the rate of travel

The slope of a __________________________ graph = ___________________

The slope of a ________________________________ graph = ________________
Calculating slope:
o

Choose two points that are ___________________

To find the ________________, find the difference in position between point 2 and point 1on
the ____________________

To find the _________________, find the difference in time between point 2 and point 1on the
________________
The formula is:
 slope = rise = ∆y = y2 - y1
run
∆x
x2 - x1
Coordinates are always
given in the format (x,y)
So the coordinates for
point 1 are (0,0) and
point 2 is (10, 100)
Use the graph above to determine the values for points 1 & 2
Example #6: Position-time graphs A cyclist leaves home (Point A) traveling at a constant speed. After 10s, (Point B), she
has traveled 100m. What is the slope of her position-time graph? What information does this give you?
Position-time graph trends
_______________

_________________
_________________
Recall, the __________________ of a distance-time graph is ____________________, so this
information can then be transfer to a speed-time graph.
Plotting speed-time graphs
o
A speed-time graph ALWAYS has:
 the manipulated variable (_______________) on the ____________________.
 the responding variable (_______________) on the __________________.
Example
o Suppose the slopes of the three lines are:
o

Line 1: 5

Line 2: 3

Line 3: 1
The same data on a
speed-time graph would
look like this:
5
1
3
2
3
1
Recall each one is travelling in “uniform motions” (speed_____________________________________________)
Speed-time graphs
o
5
3
1
The ____________________________________________ is always ___________:

On a distance-time graph that mean “zero change in distance”

On a speed-time graph that means “_________________________________”
1
2
3
(aka ______________________________)
o
Although the slope of a uniform motion speed-time graph only allows us to determine the trend, the
_______________________________________ to determine distance traveled
Area on speed-time graphs
o Area = length x width

o
On a speed-time graph the “_____________________” (y-axis) is always __________________ and the
“________________________” (x-axis) is always _____________.
So the formula can be re-written as:

Area = speed x time


This is the same as the distance formula
distance = speed x time
d = v t
Area = length x width
= 5 x 30
= 150
5
Distance travelled, while
travelling at a uniform rate
of 5 m/s for 30 s = 150m
 the area under the line of a speed-time graph = distance
30
Name: _________________________
Chapter 1.4 Graphing Assignment
Show all of your work for calculation questions. Remember the 3 “F’s” Formula- Fill it in- Final Answer
1. Which position–time graph most closely represents an object moving with uniform motion?
Use the graph below to answer questions 2-4
2. Determine the average speed of the
a. motorist
b. pioneer
3.
c. hunters
a. During the trip, did the cyclists stop and rest? Explain how you
identified that area of the graph?
b. If the cyclists stopped for an additional 30 min, determine their
resulting average speed.
4. Who completed the journey with a constant speed? Explain how you identified that area of the graph?
Use the graph below to answer questions 4-6
5. During which interval(s) does the object move with uniform motion?
A. I only
B. II
C. I, II, and III
D. II, and IV
6. During which interval(s) is the automobile accelerating?
A. I
B. I and III C. II and IV
D. IV
7. During which interval does the object move the slowest?
A. I
B. II
C. III
D. IV
6. a. Describe how you would use the graph segments II to determine the average velocity of the object, provide a
formula in your explanation.
b. Describe the motion in:
i. segment III
ii. segment IV
c. Describe the motion of the object for all four segments with a velocity–time graph.
Examine the velocity–time graph of an object travelling with uniform motion to answer the following question
7. a. Describe the motion of the locomotive in terms of:
i. how fast
ii. its direction
iii. the type of motion.
b. Calculate the displacement of the locomotive in the first 6.0 min of the motion represented by the graph.
B 1.5 : Accelerated Motion
Acceleration (
)
o Recall that uniform motion is an object travelling at a __________________________ in only
_____________________.

Most objects in motion will encounter ___________________________________ (like friction) which
will cause them to __________________________ and ______________________________, (a motor)
which will cause them to ________________________.
→
o
o
These object are experiencing ______________________________ ( OR accelerated motion)
Recall that uniform motion is an object travelling at a ______________________ in only ___________________.
Most objects in motion will encounter ______________________________(like friction) which will cause them
to ____________________ and _____________________________, (a motor) which will cause them to
____________________________
 These object are experiencing _________________________________ (accelerated motion)
=
o
Object are experiencing accelerated motion experience a ____________________ ___________ over a
timer period. Where:
Just as you must use the vector

= ____________________________

o
= ____________________________
The equation if expanded to show that change in velocity is a difference in the initial and final rate of
travel, where:

= _________________

o
Δt
quantity displacement with
velocity, you must use the
vector quantity velocity to find
= _________________
Because velocity is a vector, it can change in two ways:
 a ________________________________ (quantity) or a _______________________________.
Change in velocity
speeding up (+ change)
Direction
up, right, N or E (+ direction)
speeding up (+ change)
down, left, S or W (- direction)
slowing down (– change)
up, right, N or E (+ direction)
slowing down (– change)
down, left, S or W (- direction)
Positive or negative acceleration
o
three pieces of information are given by the acceleration of the object:
2
- 9.81 m/s [↓]
positive or negative sign 
tells you if the magnitude of
the acceleration is positive or
negative (recall this can be ___
_________________________
______OR the ________
______________ it is going in)
magnitude 
tells you the _______
____ ______________
(the change in velocity
per second.
direction  this indicates the
trend of the change in velocity.
B/c ( ) is a vector, direction
influences the overall _______
_____ ___ ______________of
the value.
Example #1: positive acceleration
When the traffic light turns from red to green, a car accelerates to a speed of 12.0m/s in 10 seconds. What is the
acceleration of the car?
Example #2: negative acceleration
It takes 10.0s to stop a particular car that was traveling at 15m/s. What is the acceleration of the car?
Example #3: final velocity
A skydiver steps out of a plane and accelerates toward Earth. If the skydiver freefalls for 15.0s what velocity did
reach when her parachute opens?
Example #4:tricky acceleration
A car accelerates from 60km/h to 100km/h in 10.0s. What is the acceleration of the car?
Example #5: initial velocity
A rock is thrown off a cliff and reaches a final velocity of 118m/s after falling for 10s. At what velocity was the rock
thrown downwards?
Assignment:
 Practice Problems 25 (pg 200), 26, 27 b & d, 28 (pg 203)
* Note how velocity-time graphs and the formula used to solve
for acceleration are related
B1.6a: Using Graphs to Calculate Displacement During Accelerated Motion
o
displacement-time graphs
 recall, there are three shapes for a displacement-time graph depending on the kind of motion
____________________
o
____________________
the slope of a position-time graph gives you the __________________________________


____________________
But since we ________________________________________________________________, we transfer
this information to a velocity-time graphs
velocity-time graphs
o a velocity-time graph also has three lines for the three different kinds of motion
NO MOTION
UNIFORM MOTION
ACCELERATED MOTION
o
no motion - the velocity of the object stays at 0m/s as time passes ________________________________
o
uniform motion - the velocity of the objects stays at the same speed as time passes  _________________
o
accelerated motion - the velocity of the object changes at a constant rate as time passes  _____________
__________________
 Note how this looks similar to a graph for uniform motion on a position-time graph…this is b/c the
____________________________________________… its just uniform acceleration vs. uniform motion

Slope calculation
o Since both graphs demonstrate a __________________________________, (a straight line) just like with a
position time graph, we can calculate the slope of the line.
 Recall the slope equation:
slope = rise = change in velocity = (∆v)
run
o
change in time
(∆t)
notice that you already know a formula that looks at the change in velocity over the change in time

o
o
the slope of a velocity-time graph gives you the ________________________________ of the object
the other piece of information that can be found from a velocity-time graph is the displacement.
 Just like with a speed-time graph, this is done by finding the ________________________________
Example #1: determining displacement with a graph
A car left home (Point A) and traveled at 10m/s to a store a few blocks away (Point B). The trip took 45s. How far away
is the store from home?
(2) Draw a vertical line up
from 45s, and shade in the
area under the motion line
(1) Draw a velocity-time
graph for this scenario.
(3) Find the area of the shaded
region.
o area of a rectangle = base x height
o area = (45s)(10m/s)
o area= 450m
 The store is 450m away from home.
o
That was easy, b/c it was travelling with uniform motion, so the area is a rectangle…
 area = length x width
o
What if an object is accelerating?
 area =
Example #2: determining displacement with a graph
An athlete trains by starting from rest and steadily increasing his speed. After two minutes (120s), he is moving at 6m/s.
Sketch a velocity-time graph, and use it to calculate his acceleration and his distance traveled after a minute.
(1) Draw the graph, plot the two
points, draw the line and shade in
the area below the graph.

(2) Use the slope to calculate the
acceleration.
(3) Use the area under the line to
calculate the distance traveled.
slope = rise
run
area of a triangle =
=
velocity-time graphs for objects not starting from rest
o
the examples we’ve looked at so far have all been objects accelerating from rest
 on a graph, that means the line started at the corner of the two axis (0,0)
o
just as often, an object will be traveling at one speed and accelerate to another
 on this graph, the line will still start on the y-axis but not on zero the x-axis
Example #3: determining displacement with a graph when
 0
A car turns off a residential street with a speed limit of 60km/h (17m/s) to a highway with a speed limit of 110km/h
(31m/s). While in the merge lane, it takes the driver 12s to accelerate to the highway speed.
Draw a graph, and use it to find the acceleration of the driver. Find the area under the line to determine the distance of
the merge lane.
(1) Draw the graph, plot the two
points, draw the line and shade in
the area below the graph.
(2) Use the slope to calculate the acceleration.
(3) Use the area under the line to calculate the
distance traveled.
Try This!
 Practice Problems 31 (pg 206)
B1.6b-Using formulas to calculate displacement during accelerated motion

New displacement formulas
o sometimes using graphs is not practical, so displacement is calculated using formulas instead.

we’ve already seen,
but that only works if you have both…
→ a constant velocity (no acceleration)
→ time given to you in the question
o
Since uniform motion is rare in the natural world, there is often a necessity to use these new formulas.
 When you are not given a distance value in a question but you know:
→ ________________________________ ( ______________________________) and a time
→ an _______________________ (or the object starts from rest), but _____________ velocity
→ the _____________________________ of the object and the _________________ interval

New displacement formula #1
o When you know BOTH velocities and time use:
o
Manipulate the formula to solve for:
→
Δt
→
→
Example #4: New displacement formula #1
A car accelerates from 17.0 m/s to 28.0 m/s in 10.0 s. What is the displacement of the car?
∆t =
vi =
vf =
∆d = ?
Try this one!
A car driving at 40.0m/s sees a radar van and slows down to 30.0m/s in 8.00s, just in time to pass the van and avoid a
ticket. How far was the radar van away when the driver spotted it?
∆t =
vi =
vf =
∆d = ?

New displacement formula #2
o when you know initial velocity (or the object starts from rest), but no final velocity AND acceleration
AND time use:
o
Manipulate the formula to solve for:
→
→
Example #5: New displacement formula #2
A ball traveling a 1.75m/s starts rolling down a ramp, which causes it to accelerate at 0.5m/s2 for 1.5s. How long is the
ramp?
vi =
a=
∆t =
∆d = ?
Try this one!
A rock is dropped off a cliff and falls for 7.00s before hitting the ground. How tall is the cliff?
vi =
a=
∆t =
∆d = ?
Assignment:
 Practice Problems: 32 (pg 208), 33 (pg 209) & 34 & 35
(pg 211)
B1.7- Calculating Stopping Distances

Implications of physics principles in driving
o
Headlights
 conventional headlights allow a driver to see about 60m ahead

o
o

→ at highway speeds, this does not give the average driver enough time to react
new HID (high intensity discharge) headlights light up to 100m of the road ahead
→ HID headlights are dangerous for drivers of oncoming traffic because they are very
glaring
Merge lanes
 based on the speed limits of the two roads (e.g. the residential road and the highway)
 by using the displacement formula, engineers can calculate the length of the merge lane needed
to safely accelerate from one speed to another
Yellow traffic lights
 the higher the speed limit on a road leading up to a traffic light, the longer the yellow light will
have to be in order for drivers to safely slow to a stop.
Calculating the stopping distance of a car
o
Stopping distance refers to the distance between a vehicle and the need to come to a stop.
 Stopping distance is made up of two separate distances:
→ Reaction Distance


→ Braking Distance
There are four steps to solving a stopping distance question:
Step 1: Reaction distance
o
reaction time can vary from person to person, and can be affected by distractions such as cell phones,
passengers and loud music,
 average reaction time for drivers is _________
o
reaction distance is how far the car goes during the time between when the ______________
____________________________________________________.
o
Since the vehicle is travelling at a ____________________________ we use the formula velocity times
reaction time.

Example :
→ A car is traveling at 20m/s (about 10km/h over the limit) when the driver sees a
pedestrian in a crosswalk ahead. If the driver has a typical reaction time of 1.50s, how
far does the car travel before the driver reacts?

Step 2 & 3: Braking time & Braking distance
o Braking distance is how far the vehicle travels from the moment the brakes are applied until the car
comes to a complete stop
 a typical deceleration value is ___________________, this allows a vehicle to safely come to a
slow stop
o
Step 2: determine braking time; the amount of __________________________________________,
based on the initial velocity of the car and the rate of deceleration (typically -5.85m/s2)
 Because the vehicle is slowing down, we use the formula for acceleration, and solve for Δt.
But b/c vf is always zero
o
Step 3: braking distance; based on the braking time you calculated, how far the travel until it comes to a
complete stop.
 Since the vehicle is changing speeds, we know BOTH velocities and a braking time, so we use the
formula

Example: If a car decelerates at the typical rate of -5.85m/s2, what distance will the car travel
while slowing from 20m/s to a stop?
Step 2:

Step 3:
Step 4: Stopping distance
o The stopping distance is the total distance the vehicle travels from the beginning of identifying
the need to stop, and actually coming to a complete stop
 the total stopping distance is equal to the reaction distance plus the braking distance
Example #1: Stopping distance
Determine the typical stopping distance for a car traveling at 110 km/h (30.6 m/s). Assume the reaction time of the
driver is 1.50s, and a deceleration of -5.85m/s2
Step 1: Reaction distance
Step 3: Braking distance
Step 2: Braking time
Step 4: Stopping distance
Try This on your own:
A driver, with a typical reaction time of 1.50s attempts to brake on the ice, causing his deceleration rate to reduce to 4.32m/s2, what is the minimum stopping distance if the car is traveling at 20m/s.
Step 1: Reaction distance
Step 3: Braking distance
Step 2: Braking time
Step 4: Stopping distance
Name: _____________________________
Unit B Ch 1.7 & 1.8
A Closer Look at Braking and Determining Stopping Distance
1.
a. The typical reaction time for most drivers is _____________
b. The reaction time is the amount of the time it takes for a driver’s brain to recognize that there is a need to
stop, and for the driver’s foot to reach the brake pedal. The type of motion demonstrated by a vehicle
during the reaction time is: uniform motion / non-uniform motion.
c. The braking time is the time from the instant the driver’s foot hits the brake pedal, to when the vehicle
comes to a complete stop. The type of motion demonstrated by a vehicle during braking time is: uniform
motion / non-uniform motion.
2. A vehicle comes to a stop 10.0 s after the brakes are applied. While the brakes were applied, the vehicle travelled a
distance of 75.0 m. If the vehicle’s acceleration remained constant during the braking, what was the vehicle’s initial
speed?
A. 15.0 km/h
B. 54.0 km/h
C. 108 km/h
D. 111 km/h
3. A vehicle is travelling 105 km/h on a highway when the driver sees a moose in the middle of the lane.
a. Determine the distance travelled by the vehicle while the driver is reacting. Assume the driver’s reaction
time is 1.50 s.
2
b. Determine the braking distance, assuming a deceleration of 5.85 m/s .
c. Determine the stopping distance.
d. Determine a new stopping distance for this vehicle if the driver was impaired and had a reaction time of
3.00s.
e. Explain why it is so dangerous to drive a vehicle while under the influence of drugs or alcohol. Does the
same generalization apply to other impairments to reaction time, such as distractions? Identify some
distractions commonly found in motor vehicles.
f. Explain why posted speed limits represent the maximum speed under ideal conditions and not the speed
you should travel.
Use Ch 1.8 A Closer Look at Braking, pg’s 221-226 in your textbook, to answer the following questions
4. Match each description with the appropriate term listed. Place your answer in the blank space given.
i. force of friction
ii. net force
iii. unit of force
iv. unit of mass
_____ a. a contact force between two surfaces that acts to oppose the motion of one surface past the other
_____ b. the vector sum of all forces acting on an object
_____ c. kilogram
_____ d. newton
5. When winter driving conditions get very icy, sand is often put on the roadway. Explain why this helps drivers brake
and turn with greater safety.
6. a. The drivers of a large transport trucks on the highway sometimes refers to the space immediately in front of the
moving truck as the “no zone.” This is because no other vehicle should ever slip into this space. Explain why the
“no zone” is for the safety of other passenger vehicles as well as for the truck driver.
b. Explain why the “no zone” in front of a large transport truck is longer when the truck is fully loaded.
Newton’s Second Law of Motion states that an object will accelerate, (proportionately
to force applied), in the direction of the net force
7. Which statement correctly describes the relationship among acceleration, net force, and mass?
A. The magnitude of a vehicle’s acceleration increases if the magnitude of the net force increases or if
the mass of the vehicle increases.
B. The magnitude of a vehicle’s acceleration increases if the magnitude of the net force increases or if
the mass of the vehicle decreases.
C. The magnitude of a vehicle’s acceleration increases if the magnitude of the net force decreases or if
the mass of the vehicle increases.
D. The magnitude of a vehicle’s acceleration increases if the magnitude of the net force decreases or if
the mass of the vehicle decreases.
8. A high-performance car initially travelling 97.2 km/h came to a stop in just 2.9 s. The mass of the car and its contents
was 1850 kg. What was the magnitude of the average net force (braking force) on the car while it decelerated?
4
A. 1.7x 10 N
4
B. 1.8 x 10 N
4
C. 4.9x 10 N
4
D. 5.9x 10 N
9. A farmer is on an errand to pick up supplies. The truck and empty flatbed trailer have a combined mass of 2920kg.
The winter driving conditions require extra caution because the midday sun has melted just enough of the packed
snow to add a thin layer of water between the tires and the road. Under these circumstances, the maximum net force
the farmer can expect for braking is about 6500 N. Consider the initial velocity of the vehicle to be in the positive
direction.
a. On the way to pick up supplies, the farmer (driving an empty truck and trailer) approaches an intersection with
an initial velocity of 70.0 km/h. If the farmer has to stop at the intersection, calculate the deceleration of the
truck and trailer.
b. Determine the braking distance for the situation in question 9.a
c. On the way back, the farmer is carrying an additional 1250 kg of supplies on the flatbed trailer. The same
intersection was approached with the same initial velocity. If the braking force remains the same, calculate the
deceleration of the truck and trailer for these circumstances.
d. Determine the braking distance for the situation in question 9.c
e. Given your answers to questions 9.b and 9.d, what driving strategies could the farmer use when approaching
the intersection with the massive load on the trailer?
B1.9 – Newton’s Laws of Motion

Net force
o the vector sum of all forces acting on an object
o in driving, the net force is the force moving the car forward, minus the forces of air resistance, road
resistance and the force applied by the braking system
o because we are dealing with vector quantities signs (+ or – ) of the forces are important

Forces opposing the motion of a car
o force of friction

____________________________________________ of one _________________ past another

o
includes the friction between the tires and the road, and the friction between the brakes and
the wheels
air resistance
 also called “drag force”
 the force resisting the motion of an object ______________________________________

to minimize air resistance, cars are designed to be ____________________________
→ aerodynamic: a design choice to reduce “drag” by changing the surface area of an
object. Ex. A sports car is designed to be aerodynamic; a school bus is not

Forces maintaining the motion of an object
o force: any external action applied on an object.

the _____________________________ is the force
administered to an object to make it move forward

o
o
the larger the applied force, the faster the object will
travel
with ___________________________________, an object that
is moving would __________________ ___________________
Newton’s First Law of Motion
 ______________________________________________
______________________________________________

o
an object at rest will tend to remain at rest
Recall Newton’s Second Law of Motion:
 an object will accelerate, (_________________________________ to force applied), ________
_____________________________________ of the net force.

We use the formula:
For example, once in motion, the space
probe Voyager 2 requires no engines. This
is because there is no air or gravity in
space, therefore, no resistive forces or
applied forces to change its speed. The
probe will simply maintain its speed
indefinitely.

Net force
o To determine the net force, you calculate the _______________________________________ acting on
an object:
o
Notice, (as with deterring displacement) we are _________________________ the values HOWEVER,
because all forces are _______________________________, direction matters and
___________________________________________________.
o
unless stated otherwise:
 we assume that the _____________________________ is the ___________________ direction

we assume that air ________________________________ are in the ______________ direction
Example #1: Newton’s Laws
The engine of a motorcycle supplies an applied force of 1880N east, to overcome the frictional forces of 520N west. The
motorcycle and rider have a combined mass of 245kg. What is the acceleration of the motorcycle?
applied
=
opposing =
net =
m=
* Recall, when the net force is positive, it
indicates that the object is accelerating.
=
Fapplied > Fopposing
Example #2: Newton’s Laws
A car with a mass of 1075kg is traveling on a highway. The engine of the car supplies an applied force of 4800N west to
overcome frictional forces of 4800N east. What is the acceleration of the car?
applied
=
opposing =
net =
m=
=
* Recall, when the net force force is zero, it
indicates that the object is traveling at a
constant speed, NOT that the object isn't
moving; it isn’t accelerating
Fapplied = Fopposing
Example #3: Newton’s Laws
A car with a mass of 995 kg slows down for a traffic light. The frictional forces on the car are 3000N, while the engine
supplies 2800N of applied force. What is the acceleration of the car?
applied
=
opposing =
net =
m=
=
* Recall, when the net force is negative, it
indicates that the object is decelerating
Fapplied < Fopposing
Assignment:
 Ch 1.9 Section Questions 3 & 4 (pg 220)
 Prepare for Ch B1. Exam

Science 20 - unit B
Chapter 2 - Impulse & Momentum
B2.1 – Momentum

Objects in motion tend to stay in motion…changing an object’s motion requires a force to be applied

Recall Newton’s first law of motion: ___________________________________________________
_____________________



since
the amount of effort required to change an object’s motion depends on two factors:
o
_____________________________________________
o
__________________________________________________________________________
A car going 100 km/h has more of a tendency to stay in motion than a tennis ball going 100 km/h.
The more mass and the more velocity an object has, the harder it will be to stop that object.
Practice Example 1: Suggest two reasons why it is harder to stop a semi-truck traveling at 100 km/h than a
small bird traveling at 45 km/h
Momentum

This “quantity of motion” is referred to as momentum.
o Momentum is measured in kg•m/s and is denoted by the letter.
o The formula for momentum is:
o
where



: ____________________________ (kg•m/s )
m : ____________________ (kg)
: ______________________ (m/s)
Rearrange the formula to solve for both mass (m) and velocity ( ).
Practice Problem #2: An airplane has a momentum of 8.3 x 107 kg•m/s [N]. If the airplane is flying at a velocity
of 230 m/s [N], determine its mass.
Try These!!!
 Practice Problems 1-3 (pg 244)
Protective sporting equipment
 Picture the type of equipment that is worn by athletes in three different sports: soccer, hockey and tennis
o a soccer ball has about 3 times the mass as a hockey puck
o a hockey puck can reach velocities of over 160km/h, but a tennis ball can reach velocities of
over 200km/h

You would expect that soccer and tennis players would require more protective
equipment than a hockey player then what makes hockey the most dangerous of
the three sports is the combination of velocity ( ) and mass (m) of the puck;
________________________________________________________
Assignment:
 Complete the Practice Problem #3 in your note package
 Section 2.1 Questions 2-5 & 8 (pg 245)
Practice Problem #3- “Characteristics of an Object in Motion” Lab
o
o
Purpose: You will identify the most significant characteristics of an object in motion.
Materials: Set up the materials as shown in the following diagram.
o
Procedure:
 step 1: Place four marbles in a row inside a horizontal section of tubing. Make sure the marbles
are touching each other.
 step 2: Release a single marble from the 10-cm mark on an inclined portion of tubing so that the
single marble rolls down the tube and strikes the other marbles.
 step 3: Observe how many marbles on the other side are ejected, and measure how far the
Styrofoam cup moves. Record your observations in a table like the one on the right.
o
Analysis Questions:
1. a) What characteristic of the released marble is changed by increasing the height?
b) Describe the effect of increasing the height of the released marble on the other marbles in the
system.
2. a) What characteristic changes when the number of released marbles is doubled from one to
two?
b) Describe the effect of releasing two marbles on the other marbles in the system.
3. Apply your understanding of momentum to explain the behavior of the marbles in the activity.
B2.2 – Change in Momentum

The formula for momentum is

For change to occur in momentum, either the mass or the velocity of an object must change.
o
because the object in question does change its mass mid motion, change in momentum is usually a
result of a change in the object’s velocity

If velocity is changing, _____________________________ must be ______________________________

We can amend the formula for change in momentum to:

To change an object’s motion also requires a force:
since
OR
o So the formula for “force need to change
momentum” can be re-written as:
o
and
o
o

the ∆v’s cancel out
Simplified:
where,

: _________________(N)

:___________________(kg  m/s2)
 ∆t
: ______________________(s)
So there are two formulas to find change in momentum, depending on the information given in the question:
o
If mass and BOTH final and initial velocity
is known use:
o
If force and the time interval is known use :
Practice Problem #1:
A 2.1 kg barn owl flying at a velocity of 15 m/s [E] strikes head-on with a windshield of a car traveling
30 m/s [W].
a) Calculate the change in momentum of the owl.
b) If the time interval for the impact was 6.7 x 10.-3 s determine the force that acted on the owl
c) Predict the effects of the collision on the owl and on the car.
Factors Affecting Change in Momentum

By rearranging the force equation, we get:
o The factors that affect change in momentum are
o the ________________________ on an object,
o the ____________________it takes for that change to occur
In the illustration above, because the truck has mass and velocity, it has momentum. The damage of the collision is
dependent on the amount of force and the time of the impact.
Try These!!!
 Practice Problems 5 & 6 (pg 247), 7 & 8 (pg 251)
Protective sports equipment
 Think back to the sports example:
o what made hockey the most dangerous of the three sports was _________________________________
of the hockey puck, compared to a smaller change in momentum for either the soccer ball or the tennis
ball.
o a collision refers to the impact of one object with another (in this example, the collision of the puck with
the hockey player)
o the greater the change in momentum, _______________________________________________
Vehicular Accident Applications
 Since change in momentum is _________________________________________________, one of those two
things has to be reduced in order to make cars better able to handle collisions
o changing the force:
 the only way to reduce force in
a collision is for the driver to
travel at a slower velocity – this
is _______________________
________________
o changing the time:
 automotive manufacturers have created ‘crumple zones’ in
vehicles.
 crumple zones ______________________
__________________ when it hits another object, therefore
______________________ and minimizing the damage

 ∴ if (change in ) goes
down, goes down
proportionally.
Assignment
 Ch 2.2 Section Questions 2-6 (pg 251)
B2.3 – Impulse

Change in momentum is such an important concept, it receives its own name: impulse(I)
o
\
since

Therefore
too
Where:
 I = __________________(N•s )

= __________________( N)
Notice, impulse and
the change in
momentum are
exactly the same
formula, but they
have different units!
 ∆t = _________________ ( s)
Practice Example 1
A raw egg drops to the floor. If the floor exerts a force of 9.0 N over a time interval of 0.030 s, determine the
impulse required to change the egg’s momentum.

and
o
o
BUT change in momentum can also be found using the formula
which means
This can also be expanded out as:

Where:
 I = ____________________ (N•s )
 m = ____________________( kg)

= __________________ (final & initial) ( m/s)
So even if you are not given a force and time, you can still calculate the impulse of a collision by using the
mass and velocity.
Always collect your variables first, and then use them to figure out which formula best suits the situation.
Practice Example 2
A raw egg with a mass of 0.065 kg falls to the floor. At the moment the egg strikes the floor, it is traveling
4.2 m/s. Assuming that the final velocity of the egg is zero after impact, determine the impulse required to
change the momentum of the egg.
Large Force vs. Large Time Interval
 Think about throwing a baseball straight up into the air, and then catching it in your hands.
o If you hold your hands ‘rigid’ when the ball ________________________________ it will stop almost
instantly (__________________), but _______________________________ (large )), hurting your
hand.
o
If you let your hands “__________” ________________________________ it will take stop over a
greater time period (_______________), but will strike with __________________________, and will
not hurt your hand.
o

Recall the illustration of the truck.
o
o

The impulse to stop the baseball in both cases is the same; however by increasing the time period of the
change in momentum, we therefore decrease this force.
In both illustrations below, the truck’s momentum is the same.
This means that if one of the factors (force or time) is increased,
the other factor has to be decreased.
Observe the consequences on the vehicle.
Large force, short time interval
Smaller force, longer time interval
Try these!!!
 Practice Problems 9 & 11 (pg 254) and 13 (pg 255)
Assignment
 Ch 2.3 Section Questions 2-5, 7, 8 (pg 256)
Lab – Change in Momentum & Forces
Part 1: How crumple zones work
Purpose: The objective of this lab is to calculate the force exerted on a car during a collision, and compare it to
a car that is not designed with a crumple zone.
Procedure
For the first part of this lab, you will need to work with a partner, and each of you will need a separate computer.
1. Partner #1:
a. Go to: http://www.uwsp.edu/physastr/hinaus/Phys203/Crash%20Video.MOV to watch a movie clip
of a mid-sized sedan in an off-centre, front end collision.
b. Useful information: the car is an average mid-sized sedan of unknown make and model. Its initial
speed is 64.4 km/h.
2. Partner #2: Go to http://www.online-stopwatch.com/ .
a. With one of you watching the video, and the other the stopwatch, you will need to record the length
of time of the collision.
b. Start the stopwatch when the car first impacts the wall, and stop it when the video clip ends.
3. This video was shot using high speed film. The video you watched was 30 times slower than the actual
collision. Calculate the actual time of the collision.
4. The mass of the vehicle is not given. Research on the internet what a realistic mass is for an average midsized sedan.
No guessing! Record the website address where you found this information:
5. Summarize your findings in the table below.
Observations/data: (show calculations where necessary)
Variable
Value
Variable
vi
time of collision on video clip
vf
∆t (actual)
Value
m
Analysis:
1. Use the data above to calculate the force required to stop the car.
2. Assuming the same length of time, what force would be required to stop the same vehicle if it was initially
travelling twice as fast? Summarize the relationship between force and the speed of the car.
3. A second crash test is done with a vehicle of the same mass but without crumple zones. The time of the
collision is a quarter of that of the car with crumple zones. Assuming the same momentum, calculate the
force on the car. Summarize the relationship between the time of impact and the force of impact.
Part 2: Breaking an Egg
Objective: compare the force required to break an egg against a hard surface with no “give” and a surface with
more “give”.
Procedure: This lab is messy and will be completed outside. You will need to work in groups of three or four.
Deliberately hitting each other with eggs will result in a zero on the lab.
1. Trial 1: Hard, inelastic surface
a. Stretch out the plastic garbage bag against the brick wall. Place a second garbage bag on the ground
below to catch the egg.
b. Two people will need to hold the plastic tight. Have one of the other team members stand about 5
meters away and throw one egg against the plastic.
2. Trial 2: Hard, elastic surface
a. Move away from the wall and stretch the plastic tight. Again, make sure there is a second garbage
bag on the ground to catch the egg.
b. Stand about the same distance away and throw the egg.
3. Trial 3: Soft, elastic surface
a. This time, hold the bag loosely.
b. Stand about the same distance away and throw the egg.
c. Repeat this trial until you are able to break the egg.
Observations/Analysis:
1. Which trials had the eggs break on the first try?
2. If you were able to break the egg on trial 3, what did you have to change from your first try to be successful?
If you were not successful, what would have helped?
Evaluation/Conclusions:
1. Describe three steps a car company can make to reduce the damage of collisions. Explain using the formulas
for and .
2. Describe two steps you as a driver could take to reduce the damage of collisions. Explain using the formulas
for and
.
B2.4 – Newton’s Third Law and Collisions

Collisions involving a vehicle include three classes of collisions:
o Primary Collision: Vehicle collides with another object
o Secondary Collision: Occupant collides with interior of vehicle
o Tertiary Collision: Organs of occupant collide within the body
Practice Problem #1
The technologies identified in the following table are designed to reduce injury in a motor vehicle collision.
Identify the class for which each technology is designed.
SAFETY TECHNOLOGY
Shock-absorbing bumpers
crumple zones (in the frame)
padded dashboard, steering wheel, etc.
Seat belts
Air bags
CLASS OF COLLISION
Collisions
 Think about what would happen if a car driving down the street loses control and hits a pedestrian.
o The car will apply an enormous force on the pedestrian, causing injury
o The person will also apply a force on the car, causing some minor front end damage to the vehicle (denting
the hood or windshield)
Conservation of momentum
 This concept is summarized in Newton’s third law of motion:
o whenever one object exerts a force on another object, the second object exerts an equal but opposite
force on the first object.
 You may have heard this before: ___________________________________________________
o This can be shown by the equation:

* Note the negative. This indicates that although the
values are the
same, they are in the OPPOSITE direction from one another.
Practice Problem #2
Two people on skates, hold two ends of a rope. Identify the action and reaction forces and the overall direction
of motion, when the person on the right pulls on the rope.
Practice Problem #3
Situation
Action Force
Reaction Force
a passenger collides with an
airbag
a car bumper colliding with
concrete barrier
a hiker pushes against the
ground
a car tire pushes against the
ground
Try This!!!
 Practice Problems #4 in your Note package
Practice Problem #4
In an interaction between a large vehicle and a smaller vehicle, state whether or not the following quantities
must be the same for each vehicle. Circle the correct response, and then justify your answers.
Mass:
Same
Different
Velocity:
Same
Different
Acceleration:
Same
Different
Time:
Same
Different
Change in Momentum:
Same
Different
Force:
Same
Different
Impulse:
Same
Different
B2.5 Conservation of Momentum
Law of Conservation of Momentum

“The total momentum gained by one object is equal to the total momentum lost by the other”
o Just like the “Law of Conservation of Matter” OR “Law of Conservation of Energy”
 Represented by the formula: Σp(before) = Σp(after)
 Where:

Σ = “_____________________” (to add

up)
= _____________________
o So the equation can be expanded out to:

Where:
 m = ________________ (you can assume the mass of the vehicles doesn’t change, so
there’s no need to distinguish between m and m’)

= ________________ (we use
before the collision.


1
and
2
to represent the speed of each vehicle
the ________________ (‘) _______________ to indicate conditions AFTER the
collisions; ’1 and ’2 represent the speed of the vehicles after the collision)
There are three types of collisions:
o
Hit and Rebound
o
Hit and Stick
o
Explosions
 Hit and Rebound
o
the two vehicles ________________________________________________________

in these questions you will be given three of the four velocities (v1, v2, v1’, and v2’ ) and you will
be asked to find the fourth, using this formula:
Hit and Rebound – Example #1
A 3.0kg ball strikes a stationary ball with a mass of 6.0kg. The first ball is moves at 1.5m/s until the collision,
but bounces back going -0.50m/s. Determine the velocity of the second ball after the collision.
m1 =
Especially for
momentum
problems it’s
really important
to identify your
variables, since
there’s so many
of them!
m2 =
1
=
2
=
1’
=
2’
=
 Hit and Stick
o
the two vehicles collide, then travel together as one combined mass
 because the two vehicles are traveling as one unit after the collision, v1’ and v2’ can be simplified
to v’
 the formula for hit and stick collisions is therefore:
HIT AND STICK-Example #2
A 1000-kg car is traveling at 12m/s when it strikes a 2000-kg truck that is stopped at a red light. After the
collision, the two vehicles stick together. What is the speed of the two vehicles after the collision?
m1 =
m2 =
1
=
2
=
’=
 Explosion
o
in an explosion, the mass starts as one contained object (m1 + m2) at rest (
o
after the explosion, the solid object has broken up into two separate masses (m1 and m2), each with
________________________________________
the formula for explosions is:
o
1 and 2
= 0 m/s)
Explosion – Example #3
A 50-g firecracker at rest explodes into two pieces, a 15-g piece which flies to the right at a velocity of 3.50m/s,
and a 35-g piece. Determine the velocity and direction of the 35-g piece.
m1 and 2 =
m1 =
m2 =
1and 2
1’
=
2’
=
=
Assignment
 Complete the Ch 2.4 & 2.5 Assignment:
Newton’s Third Law & Collisions
Name: __________________________________
Ch 2.4 & 2.5 Newton’s Third Law & Collision Assignment
1. In which class(es) of collisions will the occupants of a vehicle likely be injured?
A. Primary
B. Secondary
C. Tertiary
D. Secondary and Tertiary
2. Which statement correctly expresses Newton’s third law of motion?
A. An object will accelerate in the direction of the net force on the object.
B. Without a net force, an object in motion will tend to maintain its velocity and an object at rest will tend to
remain at rest.
C. Whenever one object exerts a force on a second object, the second object exerts an equal but opposite
force on the first object.
D. None of the above
The following diagram shows two objects approaching each other at the same speed.
3. Justify your answer for each of the following. Identify which vehicle will each vehicle will experience the
greatest:
a) Force:
b)
Impulse:
c) Change in momentum:
d) Acceleration:
e) Change in velocity:
f)
Which vehicle would you rather be in during the collision?
Two astronauts in space are attached by a rope. The mass of one astronaut is 100 kg; the mass of the other is 80 kg.
4. If the larger astronaut pulls on the rope with a force of 20 N for 0.20 s, determine the
a) force acting on each astronaut
b) acceleration of each astronaut
m2 = 80 kg
m1 = 100kg
c) final velocity of each astronaut if they both start from rest
d) final momentum of each astronaut if they both start from rest.
e) impulse acting on each astronaut.
5. Imagine you are an astronaut on a spacewalk outside your capsule. You are attached to the capsule with a safety
rope; however, the rope detaches from the capsule and you are floating freely in space. How could you use your
knowledge of Newton’s third law to assist your return?
6. In which type(s) of collision is both the total momentum before and the total momentum after the collision equal
to zero?
A.
Explosion
B.
Hit and stick
C.
Hit and rebound
D.
Hit and stick and explosion
7. In which type(s) of collision is the total momentum before the collision equal to the total momentum after the
collision?
A.
Hit and stick
B.
Hit and rebound
C.
Explosion
D.
All of the above
8. Describe what is meant by a one-dimensional collision.
____________________________________________________________________________________________
____________________________________________________________________________________________
9. A 3500-kg truck travelling 20.0 m/s[E] strikes a 2000-kg, parked minivan. After the collision, the parked minivan is
propelled forward with a velocity of 14.0 m/s[E]. Determine the final velocity of the truck.
10. A 2500-kg van travelling 80.0 km/h[E] strikes a 1500-kg compact car at rest. After the collision, the 1500-kg car
propels forward with a velocity of 60.0 km/h[E]. Determine the final velocity of the van in kilometres per hour.
11. A 20.0-g inflated balloon, initially at rest, is punctured with a nail and explodes into two pieces. If a 12.0-g piece of
balloon explodes to the left with a velocity of 3.00 m/s, calculate the final velocity of the other 8.0-g piece.
12. A 75-kg football player running 1.50 m/s[E] is tackled by another player who is initially at rest. After the collision,
the football players stick together and travel with a combined velocity of 0.80 m/s[E]. Calculate the mass of the
other player.
B2.6 - Work & Energy
Recall from Science 10: Work & Energy
 Work is _____________________________________________________________________________________
 In order for work to be done, there are three conditions that must be satisfied:
o Condition #1:



_______________________________________________________
o
 a push or pull must be applied to the object
 an object that is coasting is not having work done it
Condition #2:
o
▪ _______________________________________________
▪ if the force is not large enough to change the object’s motion, there is no work being done
Condition #3:
▪
______________________________________________________________________________
▪
the movement has to be as a result of the force, so they must be in the same direction
Energy is ______________________________________________
when a certain amount of work is done on an object, that object “gains” that much energy
o Types of Energy:
▪ kinetic (movement)
▪ thermal (heat)
▪ solar / light
▪ electrical (movement of electrons)
▪ potential (energy stored in readiness)
 chemical (stored in chemical bonds)
 gravitational (stored in an object that can fall)
 elastic (as in a stretched elastic or compressed spring)
Mechanical Energy: Em = Ep + Ek

Mechanical energy is the total energy in a system due to its
potential AND kinetic energy
Ep(grav) = Gravitational Potential Energy
o
o
o
if an object is lifted up, the energy acquired is gravitational potential energy
energy due to the position of an object above the Earth’s surface
gravitational potential energy is _______________________________________________________
Ep(grav) = mgh
o
where
▪
▪
▪
▪
Gravitational potential energy is affected
proportionally, by mass AND height.
Ep (grav) = ____________________________________ (J)
m = _________________________(kg)
g = ____________________________________________ (-9.81m/s2)
h = _________________________ (m)

Ek= Kinetic Energy
o if an object is moved, the energy acquired, due to the work done, is kinetic energy.

Kinetic energy is _____________________________________
Ek = ½ mv2

where
Kinetic energy is affected proportionally by mass and
exponentially by the speed at which it is traveling
 EK = _________________________ (J)
 m = ______________ (kg)
 v = ____________________ (m/s)
Conservation of Energy
 in a closed system; where no work or energy is added/ subtracted, the Law of Conservation of Energy applies.
o
this means that _______________________________________________________________________
________________________________________________

in the case of an object that is rising straight up in the air, or falling, there is a transfer between potential and
kinetic energy
o
When the object is at _______________________ (h @ A),
it has __________________________________ energy
o
When the object is _______________________________,
(or just as it leaves the ground) (v @ C), it has
__________________________________ energy
o
Anywhere in the middle of the journey (@B) there is a
combination of potential and kinetic energy
** The sum of the energies at any point in the falling objects journey will remain equal.**
Example
 At the TOP of the diving board, the diver has _______________________
gravitational potential energy, but as he is not falling yet, NO kinetic energy

As the diver FALLS, his ___________________________, so as his potential
energy ____________________________

BUT as he falls, his _________________________ (because gravity causes objects
to accelerate), so his kinetic energy __________________________.

The moment the diver strikes the water’s surface, he is at “ground level” (h =
zero) so he has NO potential energy. However, he has reached his maximum
velocity, so he has _______________________ kinetic energy.
Questions involving a falling or rising object

Because energy is conserved, in a question involving a falling or rising object, the two formulas (Ep and Ek)
can be set equal to each other
Ep(top) = Ek(bottom)
mgh
=
½ mv2
mgh
=
½ mv2
Notice mass is the same, as it is the same object,
and is present on both sides of the equation, so it
will cancel out
gh = ½ v2
Practice Problem #1
If the diving platform is 10m high, what will be the diver’s velocity when he reaches the water?
h=
g = 9.81m/s2 * we can make g positive
because the height refers to
v=
a negative change in
position (down) when the
object falls, so the two
negatives cancel out
Pendulums
 another example of a conversion between Ep and Ek
o at position A, the pendulum is not moving, but is at a maximum height 
_______________________________
o
at position B, the pendulum is moving at a maximum speed,
but is the closest to the ground  _____________ _________
__________________
o
at position C, the pendulum is ___________________________________
Assignment
 Complete Practice Problem 23 (pg 277) Read

Use Example Problem 2.13 (pg 275-276) as a guide
compare the answers you get to the answers for
Example Problem 2.13, explain the condition that
resulted in a difference in the values.
 2.5 Section Questions 2-6 (pg 281)