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Pearson Edexcel A Level Maths - Mechanics Summary Sheets £14.49   Add to cart

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Pearson Edexcel A Level Maths - Mechanics Summary Sheets

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Make Mechanics easier with these comprehensive, student-friendly summary sheets tailored to the Pearson Edexcel A Level Maths syllabus. Designed to simplify complex concepts, these notes are perfect for quick revision and in-depth understanding. What’s included: - Clear and concise explanat...

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  • September 10, 2024
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A LEVEL MATHS - MECHANICS REVISION NOTES

1 KINEMATICS East
 Distance - a scalar quantity with no direction
= 160 m
 Displacement - a vector quantity – measured
from the starting position
= 40 m (East of starting point)
 Position - a vector quantity – distance from a fixed origin

AVERAGE SPEED = 𝑇𝑜𝑡𝑎𝑙 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
AVERAGE VELOCITY=
𝑇𝑜𝑡𝑎𝑙 𝑇𝑖𝑚𝑒 𝑇𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛


USING Position-Time and Velocity-Time GRAPHS

Position- time graph Velocity – time graph




stationary Moving back
Velocity ms-1


towards the
starting point




Gradient = velocity


Time (s)


VELOCITY TIME GRAPH
Gradient = acceleration

DISPLACEMENT – AREA
CONSTANT ACCELERATION UNDER THE GRAPH
(Straight Line)
VELOCITY = 0


Moving in the opposite
Velocity ms-1




direction – back towards the
starting position




1

, EQUATIONS FOR CONSTANT ACCELERATION -

s: displacement (m) u : initial velocity (ms-1) v : final velocity (ms-1) a : acceleration (ms-2)

t = time (s)
v = u + at v2 = u2 + 2as

s = ½(u + v)t s = ut + ½at2 s = vt - ½at2
 Acceleration due to gravity is 9.8 ms-2 (unless given in the question)
 Negative Acceleration means retardation/deceleration
 You may need to show how the equations can be derived from the graph

A car starts from rest and reaches a speed of
15 ms-1 after travelling 25m with constant
A ball is thrown vertically upwards with a speed
acceleration. Assuming the acceleration remains
of 12 ms-1 from a height of 1.5 m. Calculate the
constant, how much further will the car travel
maximum height reached by the ball.
the next 4 seconds?
u = 12 ms-1
u = 0 ms v = 15 ms s = 25 m
-1 -1
a = -9.8 ms-2
At maximum height v = 0
v2 = u2 + 2as 152 = 2a × 25
v2 = u2 + 2as
a = 4.5 ms-2
0 = 144 - 2×9.8×s
s = 7.35 m

u = 15 ms-1 Maximum height = 1.5 + 7.35
t=4 = 8.85 m
a = 4.5 s = ut + ½ at2
s = 15 × 4 + ½ × 4.5 × 16
= 96 m
2 FORCES and ASSUMPTIONS

KEY FORCES
W : weight (mg = mass × 9.8)
R : reaction (normal reaction – at right angles to the point of contact)
F : friction (acts in a direction opposite to that in which the object is moving or is on the point
of moving)
T : Tension


ASSUMPTIONS
 Objects are modelled as masses concentrated at a single point so no rotational forces.
 Strings are inextensible (inelastic) so any stretch can be disregarded
 Strings and rods are light (no mass) so weight can be disregarded
 Pulleys are smooth so no frictional force at the pully needs to be considered.




2

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