Busy. Please wait.
Log in with Clever

show password
Forgot Password?

Don't have an account?  Sign up 
Sign up using Clever

Username is available taken
show password

Make sure to remember your password. If you forget it there is no way for StudyStack to send you a reset link. You would need to create a new account.
Your email address is only used to allow you to reset your password. See our Privacy Policy and Terms of Service.

Already a StudyStack user? Log In

Reset Password
Enter the associated with your account, and we'll email you a link to reset your password.
Didn't know it?
click below
Knew it?
click below
Don't Know
Remaining cards (0)
Embed Code - If you would like this activity on your web page, copy the script below and paste it into your web page.

  Normal Size     Small Size show me how


The displacement from A to B is the vector AB (arrow overhead), while the displacement from B to A is the vector BA (arrow overhead), thus AB* -1.
If the displacement from A to B is f, while the displacement from B to C is g, the displacement from A to C is f+g.
A vector quantity has both magnitude and direction.
A scalar quantity only has magnitude.
As distance isn't affected by d[] direction, direction is a scalar.
Distance in a particular direction is called displacement, hence displacement is a vector.
The displacement from O to P is OP. The distance from O to P is the [] of the vector OP magnitude--.
Speed is not affected by the [] of movement direction--, speed is a scalar.
Speed in a particular direction is called velocity, thus velocity is a vector.
The speed of an object is its velocity's magnitude.
For an object moving with constant velocity: velocity = displacement/time, thus displacement = velocity x time.
In kinematics, one studies an object's motion without worrying about why it's moving in a particular way. In the graphs, we don't take into account the object's m[] mass or any forces that may act on the object; generally things are simpified further by only considering motion in a straight line, and taking any acceleration to be constant.
In terms of value, distance can only be either positive or zero.
Generally, with graphs, the independent variable tend to be shown on the [] axis horizontal, while on the vertical axis, what tends to be shown is the dependent variable.
speed = distance/time; in terms of value, time is only ever taken to be 0 or positive, thus, in terms of value, speed can only be positive or zero.
In terms of value, displacement along a straight line may be positive, negative or zero.
Displacement is measured from the starting point.
acceleration = change in speed/time taken
A negative acceleration is sometimes called retardation
In terms of value, velocity along a straight line may be positive, negative or zero.
The area under a velocity/time graph represents displacement. Area below the horizontal axis counts as negative displacement. If there is some area above the horizontal axis (A), and some area below (B), then the final displacement equals A+B (if B is taken to be negative).
The area under an acceleration/time graph represents a [] in v[] change in velocity.
A particle comes to rest when it's final velocity equals 0.
The pulling force that a planet exerts on every object is called gravity. On Earth, if we ignore the effect of air resistance, the acceleration of a falling object due to gravity is 9.8 m/s². {On another planet, the acceleration due to that planet's gravity may different to ours' depending on the mass of the planet.}
Throughout this unit, the model used for calculations assumes: air resistance is small enough to be ignore, and the acceleration due to gravity is the same for every object regardless of its weight and that it is c[] constant--.
Projecting an object downwards is much like dropping it, except its initial velocity isn't 0.
When an object is projected upwards, it is easier to chose [] as the positive direction upwards--; the acceleration due to gravity is then given a [] sign negative--.
If asked to find the greatest height reached by an object - find the point where the final velocity (v) equals 0.
If an object is projected upwards with speed x m/s from a height of h metres above the ground, to calculated the time taken for the object to reach the ground (i.e. after it's reached its top height), in the suvat equations: s=[], u=[] and a=[] s=-h, u=x, a=-9.8.
Force is measured in Newtons.
Weight is a type of f[] force.
The weight of an object with mass of m kg is 9.8m N: this is the force that is exerted on it by the Earth.
Tension is a type of f[] force.
Newton's first law: if an object is stationary, then it will be [] unless it is acted upon by a [] stationary--force--; if an object is moving, unless it's acted upon by a force, it will continue to move at the same [] in a [] line speed--straight--.
A common way of simplifying problems in mechanics is to represent objects as a dot (/particle); forces and their directions are represented with labelled arrows.
If there is only one force acting on an object, then the object would move, in the direction of the force. If there is a force (A) acting on an object, but the object doesn't move, then there is another force (B) with the [] size and in the [] direction same--opposite--; if the other force (B) was smaller than A, then the object would [] in the [] of A move--direction--(and vice-versa) ; if B wasn't opposite A, then the object would move in the direction of [] at an [] A--angle--.
If an object is on a suface, then the surface exerts an [] force on the object upwards--, called the normal reaction force (so called as it and the surface are perpendicular).
If an object is released from a height: it will accelerate downwards because, acting on it is only one force - its weight (i.e. in the model). Newton's first law tells us that the apple must move; but it can't move with constant velocity as this only happens when there is no force, so the object must accelerate.
An object's horizontal component of velocity is the force that determines its [] velocity horizontal--; an object's vertical component of velocity is the force that determines its [] velocity vertical--.
If a projected object has constant horizontal component of velocity of h: its path of movement when its constant vertical component of velocity (v) is 0 is a straight line; when v is a constant amount greater than 0 downwards, the path of movement is a [] line tending [] straight--downwards--, and if v (going downwards) increases, the path of movement is a straight line sloping downwards, but at a [] angle steeper--. If throughout, no [] is [] horizontally, then the ball will, horizontally, continue to have the same speed force--acting--.
If the downward component of velocity is constant, then the object will move in a straight line, however, projected objects have curved paths as the downward component of the velocity [] at a [] rate increases--constant-- (remember Newton's first law - the force acting is the weight);if no force is acting horizontally, the horizontal componenent of velocity remains constant, but the vertical component [] with [] increases with time at a [] rate due to [] constant--gravity--.
In the standard model for mechanics, a string which suspends a weight is taken to be light and inextensible; 'light' means it has no weight and 'inextensible' means that it won't stretch. A string may be used to apply, to an object, some force; the force is the [] in the string and is an example of a [] force tension--pulling--.
Let there be an object, weight - w N, suspended by a string, with tension T N. By Newton's 1st law, if an object remains stationary, there is no net force: the downward weight is balanced by the [] [] upward tension (i.e. T = w).
In the standard model for mechanics, a pully is taken to be light and frictionless. Let there be two weights (w1 N and w2 N), one either side of a pulley; let the string have tension T N one either side (the same tension on both sides). If there's equal weight on both sides, then the objects will be stationary. If the weights are unequal, then the heavier object goes [] and the lighter one goes [] down--up--; through that knowledge, one can deduce information about the value of T, often using inequalities.
[#A43#] (1,2 ss) A and B, attached by strings, hang in a vertical line below a fixed point C: find the tension in the upper (T1 N) and lower (T2 N) string, assuming they are light and inextensible. A has weight 15 N and T2 acts [] on A upward; B has weight 10 N, T2 acts [] on B and T1 acts [] on B downwards--upwards--. From Newton's first Law, there must be no net force acting on A (as it is stationary), thus T2 = 15; additionally, there must be no net force acting on B, thus T1 = [] + [] = [] 10+15=25.
If A N pulls downwards on a stationary pulley with tension T N on either side, T = A/2 (A = 2T).
When an object is acted upon by force a resistance force may increase to match the pulling force, to stop the object moving, but at some point, the resistance force can't increase further; increasing the other force beyond that point causes the object to move.
An object is pulled witha horizontal force of P N and the corresponding resistance force is R N. The value of R when P = A N and the object remains stationary is A; the value of R when P = A N and the object moves at constant speed is A. The value of R when P = A N and the object gains speed is R where R is (using inequalities) 0<R<A. The value of R when P = A N and the object slows down is R where R is (using inequalities) R>A.
With Newton's second law, doubling the net force produces double the acceleration, as the mass remains constant.
Applying the same force to an object of twice the mass produces [] of the acceleration half--. Note: 2 * 1/2 = 1: both sides have been multiplied by the same amount.
The acceleration due to gravity on Earth is denoted by g [#A44#].
The weight of an object is the [] that gives it an acceleration of [] force--g--. Using F = ma with a = g, the weight of an object of mass m grams is mg N.
On a forces diagram, write acceleration as [] arrows double--.
Let there be a jumper (J) of mass m kg, thus weight mg N. If J is attached to a string, there is tension in the string (T N). mg is [], while, as the string stretches, T [] constant--increases--. The net downward force (F) is (mg - T)N (=ma). While mg-T > 0, J continues to [], but at a [] rate (in comparison to when there was no tension) accelerate--reduced--. When mg-T=0: J moves [], but there is no [] downwards--acceleration--(J moves with constant velocity). When mg-T<0, J [] in the direction that's [] accelerates--upwards--; the upward acceleration increases with T, thus if mg-T was > 0 previously, J has to first [], then move upwards stop--; the upwards acceleration is greatest at the [] point, when J is momentarily [] lowest point--stationary--.
When an object moves over a rough surface the [] force must be taken into account resistance (to be given)--.
Newton's 3rd law tells us that every action has an equal and opposite reaction; it applies whether or not the object moves. If action and reaction are equal but opposite, they don't [] each other [] cancel--out--; that is because the action and reaction forces act on [] objects different.
A book of mass m kg lies on the floor of a lift, thelift accelerates upwards at a rate of a m/s². What force does the book exert on the floor of the lift? The book accelerates upwards at a m/s². Vertically, the forces acting on the book are its [] and the [] force R N from the [] weight--upward--floor--. Find R, using ma = R - mg (upwards is the positive direction here). As R is the force that the floor exerts on the book, the force that the book exerts on the floor is R. Note: the weight of the book is not a force that the book is exerting on the floor, but a force is exerted on the book by the planet; if the lift accelerated downwards, adapt by taking the positive direct to be downwards.
Two objects of mass X kg and Y kg (pulled by a force of F N) are connected by a light inextensible string on a smooth horizontal surface. The equal and opposite force that each object exerts on the other is the [] in the [] tension (T N)--string--. If we want to find the acceleration of the system and the tension in the string, we can consider each object seperately: in the X kg object: the equation for the force is T = Xa, while for the Y kg object: the equation for force is F-T = Ya. The equations should be treated as simulaeneous equations, and used to find the value of [] and [] a (the acceleration) and T (the tension) - thus is the answer.
If a string is on either side of a pulley and on each side has tension T N, then the total force that the string exerts on the pulley is 2T N.
A force has both magnitude and direction, thus is a [] quantity vector--. Multiple forces combine to act as a [] force called the [] force single--resultant--. The resultant force is found by [] the forces as [] adding--vectors--. Any force can be represented by two perpendicular components (i.e. i and j).
F = ma is a formula for motion in a straight line. The good news is that the formula still works in two or even three dimensions: to make this work, F and a are given as vectors and m is a scalar. You may also use the constant acceleration formulae.
If two forces (A and B, B<A) act on the same line on an object, then the resultant force is (A-B)N; the particle will behave as if the only force acting on it is the resultant force, so it will [] in that direction accelerate--.
If there are two forces that do not act along the same line, then they can be taken to be two adjacent sides of a parallelogram; the resultant force, R N, can be represented by the parallelogram's diagonal. The value of R can be found using the cosine rule, and the angle that the resultant makes with the horizontal can be found using the sine rule. Note: if the two primary forces are perpendicular, the resultant is represented by the diagonal of a rectangle.
If you have a triangle with hypothenuse 'h', opposite side 'o', adjacent side 'a' and angle θ between p and o: o = hsinθ, a = hcosθ. Thus if you have a resultant force, you can find its c[] components, that is, the two [] forces that it's a resultant of perpendicular--.
When more than one force act on an object, it is often useful to resolve them in a particular direction. This means that you find the [] of the forces in that [] components--direction-- and add them together. Components in the opposite direction are taken to be negative. Adding components in this way gives what we call the [] [] in the given direction algebraic sum--. You can write R (-->) to show that you intend to resolve the forces in the direction of the arrow.
We often use [] to represent a resultant force R--. Its horizontal and vertical components may be written, respectively, RH and RV (H,V ss) - these can be found by [] horizontally and vertically, respectively resolving--.
RV and RH (V,H ss) may be shown as the sides of a right-angled triangle or as the resultant force.
A particle is in equilibrium if the resultant force acting on it is zero. To solve problems on the equilibrium of a particle, use the fact that the reolved forces, horizontal and vertically, or in any two perpendicular directions, amount to zero. Through this, one can find the value of unknowns in the equilibrium.
If a particle of mass m is suspended in eqilibrium by a light inextensible string, then the tension must balance the weight (i.e. T=mg N). If the particle is them pulled to one side by a horizontal force of F N, the particle remains in equilibrium: you can find the angle that the string makes with the vertical and the new tension in the string by creating a [] with the forces triangle--(i.e. a triangle of forces). The hypothenuse would be [] N, and the two legs would be the [] and the [] force T--weight--horizontal force; θ is the angle between T N and the weight; using Pythagoras' theorem and trigonometry, T N and θ may be found.
In mechanics, a device that we often use to deliver a pushing force is a rod. The pushing force is called the [] in the rod thrust. In the model, the rod is light and rigid - it has not weight and it doesn't compress.
While the tension in a string acts [] []lly from each end inwards equally--, the thrust in a rod acts [] []lly from each end outwards equally--[#A45#]. It is also possible for a rod to be under [], which makes it act like a string tension--. You can [] with a rod as well as push pull--, however a string cannot be under thrust (you can't push something with a piece of string)!
A particle pushed by a rod (inclined at x° to the horizontal) with a force of FN; the particle has mass m, thus weight mg N [#A49#]; the particle moves horizontally with constant speed. Since the head moves in a straight line with constant speed, we know from Newton's 1st law that it must be in equilibrium. What is the force exerted on the particle by the floor? One approach to finding the force is to use the [] of forces triangle of forces: the three sides of the triangle are FN, mgN and RN (the sought force) [#A50#]; using the sine rule and the cosine rule, R and θ can be found. An alternative approach would be to find the two [] []s of the required force perpendicular components-- by [] []lly and []lly resolving horizontally and vertically; you could the find the resultant force.
If a particle is on a smooth horizontal plane, the two forces acting on the particle are its weight (W N) and the normal reaction from the plane (R N). If the plane, is inclined at an angle of θ: the angle between the particle's weight and the plane is 90-θ, so the angle between W and R is θ. In [#A46#] the component of weight normal to the plane is Wcosθ, thus R = Wcosθ; the component of weight acting down the plane is Wsinθ. If θ (the inclination of the plane) increases, cosθ reduces, making the component of weight onto the plane, balanced by R, smaller; additionally, if θ (the inclination of the plane) increases, sinθ [], making the component of weight acting down the plane [] increases--greater--.
Braking force on a car opposes its motion, thus in a diagram, would be represented with an arrow that's facing in a direction [] to the direction of movement opposite--.
If asked to find the acceleration of an item in a system or the acceleration of a system, and you don't know, for example, the tension, but can find two equations containing the T, then treat them as simultaeneous equations.
If a particle is on a smooth inclined plane and a light rigid rod pushes the particle with a horizontal thrust, it exerts a force of F N (in terms of direction) horizontally and vertically. Note: if the plane is inclined at θ, then the amount of degrees between the rod and the plane is θ; the component of the thrust acting on the plane is [] N, and the component of thrust acting along the plane is [] N Fsinθ--Fcosθ--.
Generally, an object on a rough surface, with force acting on it, may not move, as friction is acting on it; when the force increases, friction increases until it reaches its maximum value. When the pushing force is equal to the maximum value for the friction, the object will be in limiting equilibrium; the box is then 'on the point of slipping'.
Pressing an object harder onto the surface means that a [] force will be needed to overcome friction larger--; increasing the downward force, thus the normal reaction force and the [] value of friction will be [] maximum--increased--; (let 'fm [m ss] be the maximum value of the friction and R the normal reaction force): as R and fm vary, fm/R remains constant, thus they are in direct [] to each other proportion--.
Let Fm (m ss) be the maximum value of friction and R the normal reaction force: the value of Fm/R is called the coefficient of friction; it is usually represented with μ. The connection between Fm, R and μ may be written as Fm = μR. In general, if an object remains stationary, then the force of friction, F, is given by F =< μR.
The value of μ (the coefficient of friction) depends on the []s in contact surfaces--; a combination of [] sufaces has a relatively high value of μ, possibly even greater than 1 rough--; a combination of smoother surfaces has a relatively [] value of μ low--, possibly as low as 0.04.
If an object is on an inclined, and is prevented from sliding down by a pushing force of P N acting up the plane. In the scenario, friction acts [] the plane up--. The minimum value of P occurs when friction acts up the plane at its maximum value (i.e. when the object is overcome the friction and would go down if there wasn't the pushing force).
What happens to the force due to friction once an object is moving? For mechanics 1, we use a simplified model in which: the force of friction remains constant, regardless of speed; the size of the force of friction is the same as the [] force of static friction maximum--(i.e. F = μR). Note: as friction is a force, F = ma, may be used with it.
Sometimes in questions involving dynamic friction: some elements (such as mass) are not told, in which case, just represent them using algebra. Q) A object slides along a horizontal floor. The coefficient of friction is 0.6. What is the acceleration of the object? A) Let 'g' represent the constant of gravitational acceleration. Resolving vertically, R = mg. Horizontally the only force is the friction (F). F equals [] and [], thus [] = [] ma and μR--ma = μR; you can substitute μ for 0.6 and [] for [] R--mg--. The result is 0.6mg = ma, hence acceleration equals 0.6g = 0.6 x 9.8 = 5.88.
If a string is being pulled, the pulling force and the tension are equal.
The difference between the two forces becomes more significant as the angle made with the horizontal is increased.
If the acceleration is defined in an equation, where the other side is a constant (e.g. a = gsin40-μgcos40), then the acceleration is constant, thus you can use the suvat equations.
If there are two objects (A and B) moving in opposite directions at the same speed, where B has greater mass than A (A is heading rightward, while B is heading leftward): if they collide, A [] direction switches--, while B moves [] at a [] speed left--reduced--. However, A would be able to affect B enough to make it switch direction if it approached it with great enough speed.
The [] and [] of an object are used to define its momentum mass--velocity--.
Momementum is a useful quantity to help us work out what happens in a c[] collision.
If an object has mass m kg and velocity v m/s in some direction, then its momentum is given by momentum = mv Ns. Note: product of the unit of mass (kg) and the units of velocity is kgms^-1, and through F = ma, we see 1 N = 1 kgms^-2, thus 1 Ns = kgms^-1.
Does momentum have direction? Yes. Does momentum have size? Yes. Thus, it's a [] quantity vector--.
The direction of an object's momentum is the same as the direction of its velocity. Note: with momentum, choose one direction of velocity to be positive; momentum in the opposite direction will be taken as negative.
If object A has momentum X Ns and object B has momentum Y Ns, then their total momentum is (X+Y) Ns.
If two objects collide and there are no external forces: total momentum before collision = total momentum after collision: this is known as the 'principle of conservation of momentum'. Note: the principle can work in the presence of external forces provided that the resultant external force on each object is 0.
If two object coalesce: they [] [] and move as [] stick together--one--.
Questions about momentum using vectors are treated in almost exactly the same way as one-dimensional question. The differences are: each velocity is given as a (usually a column) vector; the speed of an object is the velocity vector's magnitude.
The change in momentum is called impulse and is measured in Ns. Impulse is a [] quantity vector--.
The direction of the impulse on an object is the same as the direction of the force that produced it.
When two objects collide, they exert an impulse on each other of [] magnitude and in [] directions equal--opposite--.
Where F is a force, t is time, u is initial velocity and v is final velocity, the formula for impulse is Ft = mv - mu.
The magnitude of an impulse is its modulus.
The impulse exerted on object A by object B is the change in momentum in object A.
A light horizontal beam will be in [] if the forces acting upon it act along the same line equilibrium--; if the forces don't, a [] effect may be created turning-- (and the beam would no longer be in equilibrium). The greater the distance [] the forces, the greater the turning effect distance between--.
When considering the equilibrium of a large object, we need to take into account, the [] effect of all forces turning--.
The turning effect of a force about some point depends on: the force's [] and [] size and direction--, and the [] distance from the point to the line of action of the [] perpendicular--force--.
Where d m is the perpendicular distance to the line of action of the force, and the force is F N, then the moment of the force about P is Fd Nm [#A47#].
If there are multiple moments about a point (P), they can be combine to make a single [] moment resultant--. For example: if you take clockwise as the positive direction, and there is a clockwise moment of A and an anticlockwise moment of B: the total moment of forces about P is (A-B) Nm clockwise.
With moments: in place of a light rod, one may have a uniform beam of given mass. When taking moments, the weight of a uniform bean acts at its midpoint.
For a 'larger object' (i.e. a beam) to be in equilibrium, the [] force acting on it must be zero resultant-- and the resultant [] about any point must be zero moment--. This is useful to, for example, find the force that supports exert on a beam.
Taking moment about A can be shown in writing as M(A) and is likely to be followed by a colon (e.g. M(A): X = Y).
Any object which is projected from a point is a projectile. The only force acting during the flight is the weight, meaning that the only acceleration which acts on a projectile is v[] vertical.
If a projectile hits the ground at the same height it was projected from: its path is s[] symmetrical.
A projectile moves in two directions simultaeneously: we resolve the velocity [] and [] vertically and horizontally.
(x,y ss) With projectiles: vertically: initially velocity = [], velocity at time t later = [], acceleration = [], displacement = [] uy--vy--g (=9.8m/s²)--sy--; horizontally: displacement = [], initial velocity = [], acceleration = [] sx--ux--0--; as the horizontal acceleration is 0, the velocity at time t later can be written (instead of vx) ux.
An object is projected at v m/s at a direction of θ above the horizontal, resolving horizontally - ux = vcosθ, resolving vertically - uy = vsinθ.
You can use the suvat equations with projectiles if the type (vertical or horizontal) specific quanities (e.g sx [x ss]) are all of the same type.
If any number of objects are propelled (straight or at a decline) or dropped (from the same height), they will fall at an [] rate equal-- (because the acceleration affecting them is the same); the horizontal speed doesn't affect the rate at which they fall.
If velocity is constant, displacement = velocity x time.
If the horizontal velocity of an object is x and the vertical velocity is y: as a vector, the velocity is (x, y) [in a column vector], the speed is thus √(x²+y²).
The time of flight for a projectile is determined by its [] motion vertical--.
The time of flight for a projectile to return to the level of projection is t = 2usinθ/g (u is the initial velocity, θ is the angle of projection and g is the acceleration due to gravity).
The horizontal distance that a projectile travels before it lands is its range; range = [...] x [...] horizontal component of velocity x time of flight.
The 'time of flight' is the time a particle takes to move from its point of projection to the point where it [] the [] [] strikes the horizontal plane--.
If you know the initial velocity of a projectile, then you can easily find an expression for both the horizontal and vertical [] at time t displacement--: sx [x ss] and sy [y ss] respectively. You can then eliminate [] to produce a single equation connecting sx and sy t--; this equation is called a trajectory equation. However, sx is usually just called [] and sy is usually called [] x--y--.
[#A42#] With constant acceleration: where u is the initial veclocity in m/s, v is the final velocity in m/s after t seconds, a is the acceleration in ms^-2 and s is the displacement of the particle in m after t seconds: a = v-u/t, thus v= u+at, thus v² = u² + 2as; s = [] = [] (u+v/2)t = ut + at²/2. Note: the decorum units of measurement are given, but they are obviously adaptable, and also work where u, v, a and s are vectors.
A vector has no vertical displacement when the j component equals 0.
A support of a beam will exert a [] on the beam force, and an object on the beam will exerts its [] on it weight--.
A vector quantity has both magnitude and direction.
A scalar has only magnitude.
Distance is not affected by d[] direction, thus it's a scalar.
Newton's second law: F = ma. 'F' is the [] [] acting on the object net force-- (when there is more than one force, they combine to give the [] overall force net--); 'm' is the object's []; 'a' is the object's mass--acceleration-- - the acceleration and the net force have the same direction. The force is measured in [], the mass is measured in [] and the acceleration is measured in [...] Newtons (N)--kilograms (kg)--metres per second per second (ms^-2). F newtons is the force required to give a mass of m kg an acceleration of a m/s².
Created by: Toluo
Popular Standardized Tests sets




Use these flashcards to help memorize information. Look at the large card and try to recall what is on the other side. Then click the card to flip it. If you knew the answer, click the green Know box. Otherwise, click the red Don't know box.

When you've placed seven or more cards in the Don't know box, click "retry" to try those cards again.

If you've accidentally put the card in the wrong box, just click on the card to take it out of the box.

You can also use your keyboard to move the cards as follows:

If you are logged in to your account, this website will remember which cards you know and don't know so that they are in the same box the next time you log in.

When you need a break, try one of the other activities listed below the flashcards like Matching, Snowman, or Hungry Bug. Although it may feel like you're playing a game, your brain is still making more connections with the information to help you out.

To see how well you know the information, try the Quiz or Test activity.

Pass complete!
"Know" box contains:
Time elapsed:
restart all cards