TABLE OF CONTENTS 2
CHAPTER 1
3
CHAPTER 2
3
CHAPTER 3
3
CHAPTER 4
4
CHAPTER 5
4
CHAPTER 6
7
CHAPTER 7
7
CHAPTER 8
Velocity & Acceleration Force & Motion Vertical Motion Resolving Forces Friction Connected Particles Work, Energy & Power General Motion in a Straight Line
CIE A-LEVEL MATHEMATICS//9709 Solution:
CIE A-LEVEL MATHEMATICS//9709 Solution: Part (i) Calculating deceleration using Newton’s second law:
1. VELOCITY AND ACCELERATION
0.12 = 0.15 = .. = 0.8 − Calculate at using relevant kinematics equation 0.8 = − = 1.4 − Calculate kinetic energy at = 12 0.151.4 = 0.147 Calculate energy lost: ℎ = 0.1470.072 = 0.075 Calculate speed as leaving using .. formula: 0.075 = 0.15 = 1− Calculate when particle comes to rest: 0.8 = − = 1.25
1.1 Kinematics Equations
= + = + and = = 12 + = +2
1.2 Displacement-Time Graph
•
Gradient = speed
Draw velocity-time graph with data calculated:
1.3 Velocity-Time Graph
•
Gradient = acceleration
•
Area under graph = change in displacement {S12-P42}
Part (ii) Calculate displacement from
to = 4.4 Calculate displacement from to = 1×1.25+ 12 0.81.25 = 0.625 ℎ
Question 7:
0.15 0.12 − 3 2 0.072 − ⃗ ii. Displacement of block from , in the direction is at time . Sketch a displacement-time graph. On graph show values of and when block at and when it comes to rest at .
= 3×2+ 0.82
The small block has mass . The surface is horizontal. The frictional force acting on it is . Block set in motion from with speed . It hits vertical surface at later. Block rebounds from wall directly towards and stops at . The instant that block hits wall it loses of its kinetic energy. The velocity of the block from to direction is at time after it leaves . i. Find values of when the block arrives at and when it leaves . Also find when block comes to rest at . Then sketch a velocity-time graph of the motion of the small block.
Draw displacement-time graph with data calculated:
1.4 Average Velocity •
For an object moving with constant accele ration over a period of time, these quantities are equal: o The average velocity o The mean of initial & final velocities o Velocity when half the time has passed
PAGE 2 OF 8
CIE A-LEVEL MATHEMATICS//9709 1.5 Relative Velocities
CIE A-LEVEL MATHEMATICS//9709 1.5 Relative Velocities
i. ii. iii.
•
•
• •
be the distance travelled by and for = ut+ = ut + If a collision occurs at point + = Let
st
Law of Motion: Object remains at rest or moves with constant velocity unless an external force is applied
Part (ii)
nd
Law of Motion:
=
3. VERTICAL MOTION •
Weight: directly downwards
•
Normal contact force: perpendicular to place of contact
3.1 Common Results of Vertical Motion Finding time taken to reach maximum height by a projectile travelling in vertical motion:
=0
= +
Let and find • The time taken to go up and come back to original position would be double of this Finding maximum height above a launch point use: •
Part (iii) We know when and at same height Find time taken to reach max height for
• •
= 1.25. = + is 0 at max height 0 = 30 10 = 3 Time for above = 3 1.25 = 1.75 seconds
2 = Let = 0 and find
Finding time interval for which a particle is above a given height: and use • Let the height be •
= +
=
4. RESOLVING FORCES
Let • There will be a quadratic equation in • Solve and find the difference between the 2 ’s to find the time interval {S04-P04} Question 7: Particle projected vertically upwards, from horizontal ground, with speed . At same instant projected vertically upwards from tower height , with speed •
30−
10 −
25 = 30 + 12 10 5 + 30 25 = 0 Solve quadratic for = 1 5 reaches tower at = 1 then passes it again when coming down at = 5 Therefore time above tower = 5 1 = 4 seconds Displacement of is , and of is & relationship: = 25 + Create equations for and = 30 + 12 10 = 10 + 12 10 Substitute back into initial equation 30 + 12 10 = 25 + 10 + 12 10 Simple cancelling = 1.25 Find velocities = + = 30 101.25 = 17.5 −− = 10 101.25 = 2.5
2. FORCE AND MOTION
Newton’s 2
Part (i) Substitute given values into displacement equation:
This gives you the time of when the collision occurred Same analysis if motion is vertical
Newton’s 1
Find the time for which is higher than the top of the tower Find velocities of the particles at instant when they are same height Find the time for which is higher than and moving upwards Solution:
•
25
• •
cos90 = sin sin90 = cos
If force makes an angle with a given direction, the effect of the force in that direction is
cos
Forces in equilibrium: resultant = 0 If drawn, forces will form a closed polygon
PAGE 3 OF 8
CIE A-LEVEL MATHEMATICS//9709 Methods of working out forces in equilibrium:
Scenario 2: ring is about to move downwards
CIE A-LEVEL MATHEMATICS//9709 •
Methods of working out forces in equilibrium: o Construct a triangle and work out forces o Resolve forces in and directions; sum of each = 0
Lami’s Theorem: •
Scenario 2: ring is about to move downwards This time friction acts in the opposite direction since friction opposes the direction of motion, thus:
= sin30+ ℎ Using information from before: 0 = sin30 +0.24cos30 20 = 28.3 5.1 Equilibrium
For any set of three forces P,Q and
Force required to keep a particle in equilibrium on a rough plane Max Value Min Value
R in equilibrium
= = sin sin sin 5. FRICTION
Friction = Coefficient of Friction × Normal Contact Force = • •
• •
Friction always acts in the opposite direction of motion
Limiting equilibrium: on the point of moving, friction at max (limiting friction)
The particle is about to move up • Thus friction force acts down the slope •
= +sin
Smooth contact: friction negligible Contact force: o Refers to both and o Horizontal component of Contact force = o Vertical component of Contact force = o Magnitude of Contact force given by the formula:
2
+ = sin
{W12-P43}
= +
{W11-P43} The ring has a mass of . The horizontal rod is rough and the coefficient of friction between ring and rod is . Find the two values of for which the ring is in limiting equilibrium
The particle is about slip down • Thus frictional force acts up the slope •
Question 6:
0.24
Solution: The ring is in limiting equilibrium in two different scenarios; we have to find in both: Scenario 1: ring is about to move upwards
= sin30 ℎ Since the system is in equilibrium, resultant = 0: = cos30 ∴ = 0.24×cos30 Substitute relevant information in to initial equation 0 = sin30 0.24cos3020 = 68.5
Question 6:
0.36
Coefficient of friction is and the particle is in equilibrium. Find the possible values of Solution: The magnitude of friction on particle in both scenarios is the same but acting in opposite directions Calculate the magnitude of friction first:
= 6cos25 ∴ = 0.36×6cos25 Scenario 1: particle is about to move upwards = 6sin25+ = 4.49 Scenario 2: particle is about to move downwards = 6sin25 = 0.578 6. CONNECTED P ARTICLES Newton’s 3
rd
Law of Motion:
For every action, there is an equal and opposite reaction
PAGE 4 OF 8
CIE A-LEVEL MATHEMATICS//9709 {Exemplar Question}
Solution:
CIE A-LEVEL MATHEMATICS//9709 {Exemplar Question} A train pulls two carriages:
Solution:
Diagram showing how to resolve forces:
= 2500
The forward force of the engine is . Find the acceleration and tension in each coupling. The resistance to motion of A, B and C are 200, 150 and 90N respectively. Solution: To find acceleration, regard the system as a single object. The internal s cancel out and give:
2500 200+150+90 = 1900 − ∴ = 1.08 To find , look at C 200 = 1000 2500 200 = 1000×1.08 = 1220 To find , look at A 90 = 400 90 = 400×1.08 = 522
cos40+ cos60 = 5 Resolving forces at A horizontally: sin40 = sin60 Substitute second equation into first: sin60 )cos40+ cos60 = 5 (sin40 Solve to find : = 3.26 Put this value back into first equation to find = 4.40 Resolving forces at vertically:
6.1 Pulleys
•
Equation 1: No backward force
= 2
•
{S12-P41}
Question 6:
∴
Equation 2:
3 = 3
{W05-P04}
Question 3:
has a mass of 0.6 and has a mass of 0.4. The pulley and surface of both sides are smooth. The base of triangle is horizontal. It is given that sin = 0.8.
Initially particles are held at rest on slopes w ith string taut. Particles are released and move along the slope i. Find tension in string. Find acceleration of particles while both are moving. ii. Speed of when it reaches the ground is . When reaches the ground it stops moving. continues moving up slope but does not reach the pulley. Given this, find the time when reaches its maximum height above ground since the instant it was released
2 −
The strings are in equilibrium. The pegs are smooth. All the weights are vertical. Find and
PAGE 5 OF 8
CIE A-LEVEL MATHEMATICS//9709 Solution:
Pulley Case 3
CIE A-LEVEL MATHEMATICS//9709 Pulley Case 3
Solution: Part (i) Effect of weight caused by in direction of slope: Effect of weight where Effect of weight Effect of weight caused by in direction of slope: Effect of weight Body has greater mass than body so when released moves down moves up on their slopes
= sin sin = 0.8 = 4.8 = 0.4×10×0.8 = 3.2 ∴ 4.8 = 0.6 3.2 = 0.4 Solve simultaneous equations: .− = −. = 3.84 . . Substitute back into initial equations to find : 4.83.84 = 0.6 = 1.6 −
Force on pulley =
2cos
Acts: inwards along dotted line which bisects θ
6.3 Two Particles {S10-P43}
Question 7:
Part (ii) Use kinematics equations to find the time which it t ake to reach the ground:
= − and = − = 1.25 .
When reaches the ground, only force acting on its own weight in the direction of slope
is
= 3.2 = 3.2−= 0.4 = 8 Now calculate the time taken for to reach max height This occurs when its final velocity is 0. 8 = 0 2 = 0.25 Now do simple addition to find total time: Total Time = 1.2 5+0.25 = 1.5 6.2 Force Exerted by String on Pulley Pulley Case 1
and are rectangular boxes of identical sizes and are at rest on rough horizontal plane. mass = 200 and mass = 250. If ≤ 3150 boxes remains at rest. If > 3150 boxes move i. Find coefficient of friction between and floor ii. Coefficient of friction between boxes is 0.2. Given that > 3150 and no sliding occurs between boxes. Show that the acceleration of boxes is not greater than 2 − iii. Find the maximum possible value of in the above scenario
Solution: Part (i)
Pulley Case 2
= to max that does not move the boxes = to contact force of both boxes acting on floor ∴ 3150 = ×2000+2500 = 0.7 Find frictional force between and : = 0.2×2000 = 400 Use Newton’s Second Law of Motion to find max acceleration for which boxes do not slide (below 400 = 200 = 2 − has to cause an acceleration of 2− on which will pass on to as they are connected bodies Simply implement Newton’s Second Law of Motion ∴ = 200+2502+3150 3150 The comes from the force required to overcome the friction = 900+ 3150 = 4050 Part (ii)
2
Force on pulley = Acts: downwards
√ 2
Force on pulley = Acts: along dotted line
Part (iii)
PAGE 6 OF 8
CIE A-LEVEL MATHEMATICS//9709 7. W
E
P
=
CIE A-LEVEL MATHEMATICS//9709 7. WORK , ENERGY AND POWER
There is also some work done against resistive force of ; due to law of conservation of energy, this leads us to the main equation:
500
Principle of Conservation of Energy: Energy cannot be created or destroyed, it can only be
.. = ℎ . + ..
changed into other forms
= Kinetic Energy: =
610000 = 315000+500 315000 = 295000 = 590 = 610000 500 500
Work Done:
Gravitational Potential Energy: Power:
= . and =
= ℎ
8. GENERAL MOTION IN A STRAIGHT LINE
7.1 Changes in Energy
=
• • •
•
is the final energy of the object is the initial energy of the object is the energy caused by driving force acting on the object is the energy used up by frictional force or any resistive force
{S05-P04} Question 7: Car travelling on horizontal straight road, mass 1200kg. Power of car engine is and constant. Resistance to motion of car is and constant. Car passes point with speed . Car passes point with speed . Car takes to move from to . i. Find acceleration of the car at ii. Find distance by considering work & energy Solution: Part (i) Use formula for power to find the force at
− 25
20 500 10− 30.5
= 20000 = 10 = 2000 We must take into account the resistance to motion ∴ = = 2000500 = 1500 Use Newton’s Second Law to find acceleration: − 1500 = 1200 = = 1.25 Part (ii) Use power formula to find work done by engine:
= .. .. 20000 = . ..= 610000
There is change in kinetic energy of the car so that means some work done by the e ngine was due to this:
.. = 120010 .. = 120025 ℎ ..= .. .. ℎ ..= 37500060000 = 315000
DIFFERENTIATE
displacement
velocity
acceleration
INTEGRATE
=0
•
Particle at instantaneous rest,
•
Maximum displacement from origin,
•
Maximum velocity,
=0
=0
{W10-P42} Question 7: Particle travels in straight line. It passes point with velocity at time . ’s velocity after leaving given by:
− 5 = 0 = 0.002 0.12 + 1.8 + 5 of is increasing when: 0 < < and > of is decreasing when: < < i. Find the values of and and distance when = ii. Find of when = and sketch velocitytime graph for the motion of
Solution: Part (i) Find stationary points of ; maximum is where and minimum is where
= = = 0.006 0.24 +1.8 Stationary points occur where = 0 ∴ 0.006 0.24+1.8 = 0 Solve for in simple quadratic fashion: = 30 10 Naturally comes before ∴ = 10 = 30 Finding distance by integrating ∴ = ∫ = ∫ 0.002 0.12 + 1.8 +5 = [0.0005 0.04 + 0.9 + 5] = 285
PAGE 7 OF 8
CIE A-LEVEL MATHEMATICS//9709 Part (ii)
20
CIE A-LEVEL MATHEMATICS//9709 Part (ii) Do basic substitution to find
= 0.002 0.12 + 1.8 +5 = 30 = 5 To draw graph find of at using substitution and plot roughly = 13 Graph:
20 seconds has already been = 6.5 2 = 6 = 0.75 ∴ = 2 20+ 12 0.75 20 15 = 2 40 + 150+ 0. 3 75 = 0.375 13 +110 Finally add both to give you = + = 0.375 13 + 110+ 92 = 0.375 13 +202 Since the distance before taken into consideration:
{S13-P42} Question 6: Particle P moves in a straight line. Starts at rest at point and moves towards a point on the line. During first seconds, ’s speed increases to with constant acceleration. During next seconds with constant ’s speed decreases to deceleration. then moves with constant acceleration for seconds reaching point with speed i. Sketch velocity-time graph for ’s motion ii. The displacement of from , at time seconds after leaves , is metres. Shade region of the velocity-time graph representing for a value of where iii. Show that for ,
8
8 − 12 − 2 6 6.5 − 20 ≤ ≤ 26 20 ≤ ≤ 26 = 0.375 13 +202
Solution:
Part (i) and (ii)
Part (ii) First find when since
= 20, this will produce a constant 20 ≤ ≤ 26 = 12 88 + 12 8 + 212 = 92 Finding when 20 ≤ ≤ 26 : = + 12
PAGE 8 OF 8
