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Summary A level Dynamics unit 1 physics
- Institution
- WJEC
Summary of 1 pages for the course PH1 - Motion, Energy & Charge at WJEC (Dynamics notes)
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Newtons 1st Law Net Force: 2NT DRAG: DERIVING NIL: CENTRE OF GRAVIT
definition: If an object 4N - 2N = 2N - acceleration is not constant. EF-Ma definition: the point w
is moving in constant downwards ☐ ←
- Due to resistive forces. F- AT = Mv-My the entire weight of a
motion or at rest it 4N - Experienced by objects falling, t acts. W-mg
remains this way unless Object being pulled: mg travelling through fluid, vehicles... F- MCV-4) → F- = ma
acted on by resultant force. calculate (h) + (V) components - object collides with air particles t
• Newtons 2nd Law if V-mg A =D then object through the air causing air resistance. FORCES ON A SLOPE: MOMENTS:
Ef- Ma is travelling at constant motion or - Friction occurs due to the tyres vertical = moment = Force ✗ perp
definition: the resultant force is at rest. against surface of road 90-0 Mg COS ⊖ x dista
directly proportional to the rate D α v2 horizontal = f the
normal
of change in momentum of the Object supported Factors affecting drag: o) w mg sin@ pivot
object in the direction of the force. e. g on table • surface area of object
Newtons 3rd Law equal + opposite force • speed of object.
EF-W-D
EXAMPLES OF MOMENTS: PRINCIPL
definition: If object A exerts a of normal and weight MOMENTS
weight DRAG GRAPH TE5N calculate the moment: For an ob
force on object B, object B gradient =
x
exerts an equal but opposite 4 acceleration 30 X is perpendicular to equilibriu
V
force on A. If object in fluid: acceleration is line of action of the sum of clo
0.2m
constant velocity = EF:O U- up thrust decreasing to force (SN) moments
a= 0 w- weight ignritddient-9.81ms-2 terminal To find ×: ✗ = opposite angle point is eq
• : --W
I - , T-mg w velocity sin (301×0-2=0. Im the sum o
2h
Finding Tension: t 9=0, EF:O a 2=0. Im clockwise
{F- T-W PRESSURE in a fluid ÷ about the
NE5YEG INS 2ND LAW
ma- T-W P-① = mg m-density ✗ vol MENT: M=Fd point.
T-math A {F- ma The moment = 5 × 0.1 = 0-SNM EACH =
DENSITY i.In:
1m94
✗ IF = m To determine unknown mass The mass of the
p=m_ (Rgm-3) Pep# =p Ang = phg Apparatus light gataeemefaaf.ge is kept constant
A gradient = M glider of pulley one slotted mass
unknown
• measure mass using a mass balance. mass masses and putting
•
• measure volume with vernier calliper. P=phg = density ✗ height ✗ gravity Force (N) air track mates glider for each read
hanging
• Irregular shapes can be submerged and acceleration
in a known amount of water and MOMENTUM
displaced water = volume of material. • principle of momentum: sum of momenta p-Mv (Rgm5') Rearranging N2L EF-ma to get momentum
e.g before a collision is equal to the sum of → (MMaalltath Mb)bVU_j_mMna,ryatA Mbb {F- ma → {F- DI
250-100 momenta after a collision provided no
= 150m/ resultant force acts ✓ = MA X " " change in momentum = EF ✗ At
100mL 250m/ • conservation of energy: (mat Mb' impulse = change in momentum (P2-
F
Ml → cm³ = ÷ 106 . energy cannot be created or destroyed, it area = impulse/change in mo
150 ÷ × 106 = 1-5×10-4cm³ can only be transferred from one form ↳ calculate by using trapez
to the n act. t triangles + squares (appro
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