Saturday, 31 March 2012

Forces, movement, shape and momentum

(as momentum is for Single Science, I won't be talking about it unless you want smth on it, anything for Single Science is labelled SS in front of the specification point.)

1.8 express a force as a push or pull of one body on another

A force is a push or pull that one object exerts on another
3 things a force can do to an object:

  • change its speed
  • change is shape
  • change its direction
1.9 identify various types of force (for example gravitational, electrostatic etc.)

various types of force:
  • push/pull (contact force)
  • tension-the pull at both ends of a stretch spring, string or rope
  • compression
  • thrust/upthrust
  • load
  • effort
  • *weight/gravitational
  • *electrical/electrostatic
  • *magnetic
*the last three forces are non-contact, they can act without touching an object

SS 1.10 distinguish between vector and scalar quantities 
Scalars have magnitude (size), Vectors have both magnitude and direction. 
Scalars
Vectors
Speed
Velocity
Distance
Displacement
Mass
Weight/tension/compression à forces
Temperature
Thrust
Energy
Drag
Charge
Upthrust
Volume
Acceleration
Area
Field strength – magnetic/electrical/gravitational


SS 1.11 appreciate the vector nature of a force
Forces have magnitude and direction, it is therefore a vector. For example, weight is a force with magnitude, and it acts downwards. It is a vector, and you would normally use an arrow pointing downwards with its magnitude ("xN" N being the unit-Newtons) to represent it in a diagram.

SS 1.12 add forces that act along a line
The forces act along a line, meaning they are collinear. You can add the forces together like scalars. For instance if two forces are acting on a box, both pushing towards the right side, one with a force of 4N and the other with 6N, then 4N+6N=10N (resultant force). So a total of 10N is acting on the box, pushing it to the right. 

However, if one of the forces was acting to the left with say, 2N and the other to the right with 7N, then the resultant is 5N to the right. The 2N to the left cancels out 2N from the right.  You can think of the force acting to the left as a negative value, like the reverse direction, so 7N + (-2N) = 5N. It's good to use diagrams to help you out. 

1.13 understand that friction is a force that opposes motion
Friction is a force that always opposes motion between 2 surfaces in contact.

1.14 recall and use the relationship between unbalanced force, mass and acceleration:

force=mass x acceleration
F= m.a

1.15 recall and use the relationship between weight, mass and g:

weight= mass x g
W= m x g
g=gravitational field strength, this is the gravitational force exerted per unit mass at a point in the field, it is a vector quantity. unit= N/kg
on earth, we consider g as 10N/kg, or g = 9.8 m/sbut we use g = 10 m/sthis is how an object in the Earth's gravitational field would accelerate if it was free fall-without friction (this doesn't happen in the real world). 
( to explain how we get to this:
a= F/m = mg/m = g ) 

1.16 describe the forces acting on falling objects and explain why falling objects reach a terminal velocity

2 forces act on falling objects: 
  • Earth's gravity-Weight (W): has direction (vector quantity), pulls object downwards towards the centre of the Earth
  • air resistance/drag (D)--upwards force, pushes object upwards
Air resistance increases with speed. When an object first starts falling, their weight is greater than air resistance. (W>D) Hence it falls faster, accelerating towards the ground. However, as it gains speed, the air resistance increases until it eventually equals the object's weight. (W=D) Acceleration then becomes zero (a=0) and the object will have reached terminal velocity. 

1.17 describe the factors affecting vehicle stopping distance including speed, mass, road condition and reaction time

vehicle stopping distance= *( reaction time (a.k.a thinking time) x constant speed ) + breaking distance 
vehicle stopping distance= thinking distance + braking distance
*d=s x t, so it becomes thinking distance
thinking distance: how far the car travels at constant speed before the driver reacts by applying the car brakes
braking distance: distance travelled by the car as it decelerates to a stop


  • as speed increases, stopping distance increases
  • as mass increases, force needed to stop car increases (F=ma)
  • dry weather=more friction, rainy=less friction, as water acts as lubricant
  • as reaction time increases, stopping distance increases (If driver was drunk, their reaction would be slower, so reaction time increases, hence stopping distance increases)
SS 1.18 recall and use the relationship between momentum, mass and velocity:
momentum = mass x velocity
p= m x v

SS 1.19 use the ideas of momentum to explain safety features

SS 1.20 use the conservation of momentum to calculate the mass, velocity or momentum of objects

SS 1.21 use the relationship between force, change in momentum and time taken:
force= change in momentum/time taken

SS 1.22 understand Newton's third law
Newton's Third Law of Motion:
For every action, there is an equal and opposite reaction, and these forces act on mutually opposite bodies. 
i.e. forces occur in pairs.
Fab = -Fab

E.g. when an object is falling, not only is weight acting on it downwards, but there is also air resistance pushing it upwards -- the forces act in opposite directions on the same object.
Or for instance, when a car is driving, the driving force pushes the car forwards, but there is also friction acting in the opposite direction, if the driving force wasn't bigger than friction, the car would be slowing down...

1.23 recall and use the relationship between moment of a force and its distance from the pivot

moment= force x perpendicular distance from the pivot
moment= F x d

1.24 recall that the weight of a body acts through its centre of gravity

SS 1.25 recall and use the principle of moments for a simple system of parallel forces acting in one plane

SS 1.26 understand that the upward forces on a light beam, supported at its ends, vary with the position of a heavy object placed on the beam

1.27 describe how extension varies with applied force for helical springs, metal wires and rubber bands
they all obey Hooke's Law up to their elastic limit
-thanks to a correction by an anonymous person (see comments), it apparently should be:
helical springs and metal wires do obey Hooke's Law, but rubber bands do NOT follow Hooke's Law  and the extension is NOT directly proportional to the force causing it. 
(y)

1.28 recall that the initial linear region of a force-extension graph is associated with Hooke's Law

Hooke's Law: The extension is directly proportional to the stretching force. 


 

Hooke's Law only applies to the straight part of the graph-initial linear region, up to the limit of proportionality. 
Elastic limit-if a spring is taken beyond this limit, it won't return to its old shape=deformed.

1.29 associate elastic behaviour with the ability of a material to recover its original shape after the forces causing deformation have been removed


Deformation from the limit of proportionality to the Elastic limit is still reversible; but 
Beyond the elastic limit, deformation is irreversible=permanent deformation or plastic deformation. 




20 comments:

  1. which one is paper 1 and which is paper 2 for sciences? (single or double award?)

    ReplyDelete
  2. I post stuff for paper 1 which is Double Award, and the spec points listed are for double award too. :)

    ReplyDelete
    Replies
    1. struggling with astronomy! please help! :)

      Delete
    2. alright, see if the new 'astronomy' post helps. :) i did it according to specification so that's really all you need to know, i hope.

      Delete
  3. thank u, ya the spec only has a few points but gotta noe them too D:

    ReplyDelete
  4. for 1.17, its "braking" not "breaking"

    ReplyDelete
  5. What does the specification point, '1.12 add forces that act along a line' mean?

    ReplyDelete
    Replies
    1. It just means the forces act along the same line, so they are collinear, and therefore can be added as scalars. e.g. if two forces acting towards the right, one with a force of 3N and the other with 5N, then 3N + 5N = 8N (Resultant force)
      But if one was acting to the left with 3N and the other to the right with 5N, then the resultant is 2N to the right. (You can think of the force acting towards the left as a negative value-the reverse direction if you like.)
      :)

      Delete
  6. This has been so helpful, but I was wondering what grade you got?

    ReplyDelete
    Replies
    1. I did Single Science and got an A*. :)

      Delete
  7. well then why u bother commenting. she has spent so long on all of these spec points, id like to see you do better idiot.

    ReplyDelete
  8. i love you for this

    ReplyDelete
  9. 1.27 is supposed to be describe experiments on how to... help please!

    ReplyDelete
    Replies
    1. Hey, just thought I'd mention that she specifically said that this blog is tailored to the 2009 specification - and only in the 2011 specification does it say we have to describe experiments.. if you have a student book/revision guide it explains experiments really well. :)

      However, she did also say that 'they all obey Hooke's Law up to their elastic limit' which is incorrect. The first two do, but it is vital that you know rubber bands do NOT follow Hooke's Law at all and the extension is NOT directly proportional to the force causing it. :)

      (Also, I want to say thanks again for this amazing blog...I was the person asking about mole ratios last week and decided to check out your posts for physics and biology too! :D)

      Delete
    2. ooh haha I see, thanks for your commment! and yes, seems like at least one person has realised that my blog is tailored to the old spec..! i'll add your correction to the post btw, thanks for it. :)

      and you're very welcome, :D

      Delete
  10. Thank you so so so so much hahaha

    ReplyDelete
  11. This really helped me type up revision notes for my mocks and my exams. really helpful but on some of the single science points such as 1.25 I didnt quite understand this and the fact that it had no extra explanation didnt help me. thank you though really good :)

    ReplyDelete
  12. can u please explian 1.26

    ReplyDelete

Please tick a box showing your reaction, and any feedback is appreciated. :)

Note: This blog will no longer be updated as I finished IGCSEs in 2012. Sorry! :( If you are interested in buying IB notes though, please contact me. :)