Grade 6 Math Circles - ?· Grade 6 Math Circles ... Instead of drawing the square everytime, we use…

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  • Faculty of Mathematics Centre for Education in

    Waterloo, Ontario N2L 3G1 Mathematics and Computing

    Grade 6 Math CirclesFebruary 22 & 23 2016

    Mechanics & Dynamics

    This lesson builds on the Kinematics lesson from February 16/17 2016. It is strongly rec-

    ommended that you read through that lesson first if you are unfamiliar with kinematics.

    Dynamics is the study of forces that act on physical objects to result in motion. Together,

    kinematics and dynamics make up the fundamental basis of physics called mechanics.

    Solving EquationsIn this section, were going to learn to rearrange equations to solve for an unknown. These

    equations will involve four types of operations: addition, subtraction, multiplication, and


    Exercise: fill in the box to make the equality true.

    (a) 5 + 5 = 10

    (b) 3 1 = 2

    (c) 7 + 2 = 9

    (d) 6 3 = 18

    (e) 18/9 = 2

    Instead of drawing the square everytime, we use variables which do the same thing. Vari-

    ables are represented by letters in the alphabet and, just like the squares in the questions

    above, they indicate an open space where we can place any number that will make the equal-

    ity true.

    Example 1:

    In the equation 5m = 10, we say that m = 2 because replacing m with 2 in the equationsatisfies the inequality:

    5m = 10

    5 2 = 10

    10 = 10 which is true!

    We can use any letter we like for the variable.


  • Exercise: Determine the value of each variable. The first one is done for you.

    (a) 5m = 10 = m = 2

    (b) 12 + A = 15 = A = 3

    (c) 3 1 = x = x = 2

    The last one was a lot easier than the other ones. In fact, its clear that solving for a variable

    is easier when the variable is on its own on one side of the equation because all you have to

    do to find its value is perform the correct operations (in this case, 3 1) on the oppositeside of the equality.

    This will be our general technique for solving equations like this, especially as they get more

    complicated: we will try to get the variable on its own side of the equation.

    Example 2: Solve x + 2 = 5.

    To get x on its own, we have to subtract 2 from the left hand side (LHS) of the equation

    because x + 2 2 = x. However: whatever we do to one side of the equation, we have todo to the other side as well so that the equality will continue to hold. Since we subtracted

    2 from the LHS, we have to subtract 2 from the right hand side (RHS):

    x + 2 = 5

    x + 2 2 = 5 2

    x = 5 2

    x = 3

    Does this hold? If I replace x with 3 in my original equation, x + 2 = 5 becomes 3 + 2 = 5

    or 5 = 5, which is true! So x = 3.

    Aside: 5 = 5 is true. If I subtract 1 from the left side: 5 1 = 5 becomes 4 = 5 which is nolonger true unless I also subtract 1 from the right side: 4 = 5 1 becomes 4 = 4, which istrue again!

    Example 3: Solve x 2 = 5To get x on its own on the LHS, we need to add 2: x 2 + 2 = x. So we also have to add 2to the RHS:

    x 2 = 5

    x 2 + 2 = 5 + 2

    x = 5 + 2

    x = 7

    We can also solve equations which involve multiplication and division:


  • Example 4: Let 5n mean 5 n. Solve 5n = 15.

    To get x on its own, I have to divide the LHS by 5 since: 5n/5 = n. So I also have to divide

    the RHS by 5:5n = 15





    n =15


    n = 3

    So n = 3. Does this work? If I replace n with 3 in our original equation, I get that 5n = 15

    becomes 5 3 = 15 which then becomes 15 = 15, which is true!

    Example 5: Solve x/2 = 4

    I have to multiply the LHS by 2 to get x on its own because x2 2 = x. That means I also

    have to multiply the RHS by 2:x

    2= 4


    2 2 = 4 2

    x = 4 2

    x = 8

    Example 6: Solve 5n + 2 = 22.

    Whenever we have an equation involving 2 or more operations, we do any addition or sub-

    traction operations first and any multiplication or division operations last. In this example,

    I subtract 2 from both sides before I divide both sides by 5:

    5n + 2 = 22

    5n + 2 2 = 22 2

    5n = 20





    n =20


    n = 4

    Check if this is correct by replacing n with 4 in our original equation.


  • Exercise: Solve for x. Remember, 5n = 5 n.

    (a) x + 6 = 12 = x = 6

    (b) x 4 = 8 = x = 12

    (c) 4 x = 2 = x = 2

    (d) 3x = 18 = x = 6

    (e) 4x = 16 = x = 4

    (f) x/3 = 4 = x = 12

    (g) x/5 = 5 = x = 25

    (h) 3x + 2 = 20 = x = 6

    (i) 3x 2 = 10 = x = 4

    (j) 6x + 5 = 23 = x = 3

    (k) x/2 4 = 16 = x = 40

    (l) x/2 + 7 = 11 = x = 8

    (m) x/3 3 = 3 = x = 18

    Note that you can always check your answer by replacing the original x in the equation with

    the answer you got to see if LHS = RHS.

    d = vt: Unknowns in the Real WorldIn the velocity section of the Kinematics lesson, we introduced the equation v =


    t. If we

    were finding the average speed of an object, we would say v was the average speed, d was

    the total distance traveled, and t was the time of motion. If we instead wanted the average

    velocity, we said that v was the average velocity, d was the displacement of the object, and

    t again was the total time of motion.

    Multiplying both sides by t, we get a cleaner looking equation: d = vt. We can solve for

    distance, displacement, average velocity, average speed, or time with this equation! Re-

    member, d is displacement as long as v is average velocity and d is total distance

    as long as v is average speed. t is always time.

    As long as we are given two of the values d, t, or v, we can solve for the third value.

    Example 1: My friend Julian leaves his house at 2pm to go to the beach

    90km away. He wants to get there before low tide at 3:15pm,

    when all the good surfin waves are gone. How fast should he

    drive? Give your answer in km/h.

    Begin these questions by writing down what information is pro-

    vided to you by the problem. We are given a distance here: 90

    km (no direction). We are also given a time, which we want

    in hours: from 2pm to 3:15pm. This is 1 hour and 15 minutes.

    15 minutes is 0.25 hours (one quarter of an hour), so the total time that Julian has to get

    to the beach is 1.25 hours. Since 90 km is a distance, we can find the average speed using


  • the relationship d = vt. First, we replace the variables with the information we have:

    d = vt

    90 km = v 1.25 hrs

    We divide each side by 1.25 hrs to get the average speed on its own:

    v =90 km

    1.25 hrs

    v = 72 km/hr

    So Julian has to drive 72 km/h to get to the beach on time.

    We always include the units when making these calculations. The reason is, we want to

    make sure that we are not adding quantities with different units (for example: What is 90

    seconds + 3 km? Nothing! What is 5 m + 10 km? You have to convert to the same units

    before you can add this). Notice we had to do this when we converted 1 hour and 15 minutes

    to 1.25 hours.

    What is dynamics?So far, weve analyzed the movement of objects without considering what causes those move-

    ments. When a pencil rolls off a table without being touched, what caused it to move? Why

    does it tend to fall without provocation? Why doesnt it fly up instead? How come my

    stapler sits on top of the table and doesnt fall through? In my opinion, questions like that

    are much more interesting than the question of how far the pencil fell or how fast it was

    moving. To figure out what made the pencil move, we will now delve into the study of

    dynamics, which focuses on the affects of force on the motion of physical objects.

    Mass & ForceWe use the concept of mass to describe how much matter is inside a physical object. Any-

    thing that you can touch has matter in it. Generally, the bigger an object is, the more mass

    it has, but not always. If you could blow up a balloon to be the same size as an elephant, the

    balloon would still have less matter (or stuff) inside it than the elephant does. Although

    the balloon and the elephant are the same size, we say that the elephant has more mass.

    We often weigh objects to figure out what their mass is. In the above example, the elephant

    would weigh more than the balloon, and that would be one indication that the elephant has

    more mass than the balloon. Having weight is a result of having mass. Weight and mass

    are not the same thing, though. Mass is a scalar quantity measured in kilograms (kg) and

    weight is a consequence of gravity acting on mass (see Mechanics Activity #4).


  • A force is a push or a pull on an object. They can be balanced or unbalanced by other

    forces. Take this example: if you push on a box, then you are applying a force onto that

    box that is causing it to move. We say that the force of your push is unbalanced. If you and

    your friend pushed on the box from opposite directions and each of you is pushing just as

    hard as the other, the box would not move. We say that the force of your push is balanced

    out by the force of her push. If you pushed harder than your friend, then the box would

    move again in the direction that you were pushing. We say that the force of your push is

    unbalanced once again. Gravity is one example of a force. It is pulling you down toward

    the Earth.

    Force is a vector quantity measured in Newtons (N), named for Sir Isaac Newton.

    Newtons Laws of MotionSir Isaac Newton was an English physicist and mathematician who is often credited with

    discovering gr


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