ls 11 components of vectors

KYS Malaysia Physics AS / 2 S.M.Chappell Lesson 11 2013-2014

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Components of Vectors.


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StarterSolve the following problems

Q1. A ship is pulled by two tugs at 12o to the direction the ship. If each pulls with a force of 6kN. Calculate the resultant force the ship is moved with

Q2. A kite has a mass of 0.5 kg. It is hit by a find with a force of 10N. (Assume g = -9.81 ms-2)

10 N a) Find the weight of the Kite.

b) Using your answer to a) construct a triangle for the forces

c) Using your answers to a) and b) find the tension in the main kite string


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Objectives1) Know the effect of gravity on the movement of . a projectile in terms of components

2) Understand how to resolve projectile problems through resolution of vectors into components

3) Be able to solve projectile problems


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KeywordsAccelerationComponentConstantCosine ruleDisplacementFlight PathHorizontal ComponentMagnitudeMaximumPythagoras theoremSpeedSine ruleTrajectoryVariesVectorsVelocityVertical Component“vutas” equations


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Objects thrown upwards.In these sort of problems it is convention that upwards is a positive direction and downwards a negative direction

Look at the following problemQ1 .A person throws a cricket ball straight up into the air with an initial velocity of 30ms-1. What height will the ball reach from its point of release? (Assume g= -9.81ms -1)Draw a diagram to represent the information

30ms -1

Height 9.81 ms -2

Assign the correct signs to the vectors

-Q. At what point will the ball stop increasing in height?When the upward (+) instantaneous velocity is zero Summarise further information now

u = 30 ms-1

v = 0 ms-1

a = -9.81 ms-2

Height = displacement (s)s = ?

Looking at the data use the appropriate “vutas” equation

v 2 = u2 + 2as v = u +at Eqn 1s = (v + u)t Eqn 2 2s = ut + ½ at2 Eqn 3v2 = 2as + u2 Eqn 4

v 2- u2 = s 2a

0 2- 302 = s2 -9.81

- 900 = s-19.62

s = 45.8 m


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Objects thrown upwards.In these sort of problems it is convention that upwards is a positive direction and downwards a negative direction Look at the following problem

Q2 .A person throws a ball straight up into the air with an initial velocity of 27ms-1, and it goes over a cliff 40m high. How long does it take to reach the top of its trajectory? What is the final velocity the ball will reach the ground at the bottom of the cliff?

Draw a diagram to represent the information

27 ms -1

Height

9.81 ms -2

Assign the correct signs to the vectors

-u = 27 ms-1

v = ms-1

a = -9.81 ms-2

Height = displacement (s)s = - 40 m

v = u +at Eqn 1s = (v + u)t Eqn 2 2s = ut + ½ at2 Eqn 3v2 = 2as + u2 Eqn 4

40 m-

v = u +at Eqn 10 = 27 +(-9.81 x t )

0 -27 = -9.81 x t 0 -27 = t -9.81 t = 2.75 s

Find the time for the object to reach the top of its height

v2 = 2as + u2 Eqn 4Final velocity from full drop

v2 = 2 x -9.81 x -40. + 272

v2 = 1513.8v2 = Ö1513.8.8

v = - 38.91 ms-1


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Components of vectors.Just as several vectors can be made to give a resultant vector the opposite is true. In this that a vector can be split into two perpendicular component vectors. This becomes useful, particularly for projectiles. Imagine the problem you have just solved, except the ball was thrown horizontal. With initial velocity of 30ms-1.Q2. What would the flight path of the ball be?

Down Curved

Q3. Why will it follow an curved down path? Acceleration due to gravity

Constant velocity 30ms-1 (Assuming no drag which is arrows same length common for these type of

questions)

Velocity at each point due to the acceleration due to gravity increasesLonger downward arrows

So there is still a problem. The velocity varies vertically


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Acceleration due to gravity

Tim

e in

mill

isec

onds

Look at the object falling opposite. At each millisecond the position of an identical ball is shown and a line drawn to the axisQ4. What do you notice about the distance the ball moves in the y (downwards) direction ?

Moves the same amount downwards or negative direction

This enables us along with the use of “vutas” equations to solve these types of problemsv =u + at Equation 1. s = (v + u ) t Equation 2.

2s = u t + ½ at 2 Equation 3. v2 = 2as + u2 Equation 4.

Plus the use of the trigonometric formulas

a = b = c .sin a sin b sin c

Sine rule Pythagorus theoremCosine rulea2 = b2 + c2 – 2bc cos A


Solving projectile problems

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Consider the following problem:A cannon fires a steel ball at 83 ms- 1 in the horizontal direction from the top of a cliff 80m high. Find the maximum range of the canon. (g = -9.81ms-2) Draw out the problem

80m

s

Solve vertical function. The time the cannonball will stay in the air s = u t + ½ at 2 Equation 3.

Initial velocity u is zero s = 0 x t + ½ at 2

-80m = ½ -9.81 t 2

-80m x 2 = t 2

-9.81 160/9.81 = t2

16.31 = t2 4.04 s = ts = (v + u ) t Equation 2. 2

Now the horizontal components = (83 + 83 ) 4.04s 2

s = (166 ) 4.04s 2

s = 83 x 4.04s s = 335.32 m

v =u + at Equation 1.s = (v + u ) t Equation 2. 2s = u t + ½ at 2 Equation 3. v2 = 2as + u2 Equation 4.Sine rule a = b = c .sin a sin b sin cCosine rulea2 = b2 + c2 – 2bc cos APythagorus theorem

Or s = v t s = 83 x 4.04 = 335.32 m


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A cannon fires a steel ball at 120 ms- 1 in at an elevation of 30o from the top of a cliff 80m high. Find the maximum range of the canon. (g = -9.81ms-2)

Solving projectile problems 2

120 ms-1Split the balls initial velocity into horizontal and vertical components

30o

120 ms-1 sin q = 0pp hyp sin q x hyp = 0pp

sin 30o x 120ms-1 opp

Vertical component

0.5 x 120ms-1 = opp

Opp = 60 ms-1 vertical component u v

Look at the vertical motion in terms of “vutas” equations.80 m

30o


v2 = 2as + u2

v2 = 2 x -9.81 x -80 + (+60ms-1)2

vv = uv = + 60m s-1 t = av = -9.81m s-2

sv = -80 m

v2 = 1569.6 +3600v2 = 1569.6 +3600v2 = 5169.6 v2 = 5169.6 Övv = -71.89 ms-1

-71.89 ms-1

?

?

Minus because it is the final velocity is downwardsv =u + at Equation 1 v – u = t

a-71.89 – (+60) = t -9.81-71.89 – 60 = t

-9.81-131.89 = t -9.81

t = 13.44 s

1.21 s

Horizontal component vh = uh = ? t = 13.44 s ah = 0 m s-2

sh = ?

cos q = adj . hyp

cos 30o = adj . 120 ms-1

cos q x 120 = adj . 0.866 x 120 = 103.9 ms-1

103.9 ms-1?

s = 103.9 m x 13.44 s s = 1396.4 m

103.9 ms-1

s = v t


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Problem for you to solveA child throws a stone at a 40o angle from the horizontal. If it reaches the sea 100m below, 70m out, 1.4s later. What was the speed it left his hand at ? (Assume g= 9.81 ms -2)

SMC’s tip: draw the problem out

SMC’s tip: write out “vutas” down the side and assign values

vv = ?uv = ?t = 1.4 sav = -9.81 m s-2

sv = ? + -100m

vh = uh uh = vh t = 1.4 s ah = 0 m s-2

sh = 70m

SMC’s tip: Draw out the component vector diagram for the initial vector

40 o

v component

hcomponent

We assume nothing effects the horizontal component we have a displacement and time,


s = vh t

70 m-100m

-9.81 ms-2

40 ou

1.4 s

70 m = vh = 50ms-1 1.4s

50ms-1

cos 40o = 50ms-1

hyp hyp = 50ms-1

Cos 40o

u = 65.3 ms-1

Velocity the stone left the hand at

Horizontal component

Why is the actual velocity it leaves the hand likely to be slightly higher ?For the stone to reach the distance indicated the velocity will have be higher due to air resistance slowing down the horizontal vector for velocity which we assumed was the same value

50ms-1


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Problem for you to solve 2A child throws a stone at a 45o angle from the horizontal. If it reaches the sea 120 m below, 120m sea,170m out, 2.7s later. What was the speed it left his hand at ? (Assume g= -9.81 ms -2)

SMC’s tip: draw the problem out

SMC’s tip: write out “vutas” down the side and assign values

vv = ?uv = ?t = 2.7 sav = -9.81 m s-2

sv = ? + -120m

vh = uh uh = vh t = 2.7 s ah = 0 m s-2

sh = 170m

SMC’s tip: Draw out the component vector diagram for the initial vector

45 o

v component

hcomponent

We assume nothing effects the horizontal component we have a displacement and time,


s = vh t

170 m-120m

-9.81 ms-2

45 ou

2.7 s

170 m = vh = 63 ms-1 2.7s

63 ms-163 ms-1

cos 45o = 63ms-1

hyp hyp = 63ms-1

Cos 45o u = 89.1 ms-1

Velocity the stone left the hand at

Horizontal component

Calculate the height the stone falls from above the cliff. For this vv = 0ms-1

Why? The point where the stones starts its minus velocity due to av

uv sin 45 o = opp 89.1 m

sin 45 o x 89.1= 63 ms-1

0 2 – 632 =s2 x -9.81

v2 = 2as + u2 v2 – u2= s 2a

s= 202.3m

ls 11 components of vectors

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