Title: InkPhysics / Mechanics
On every note of this deck, unless a note says otherwise:
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Front: DEFINITION
Define: Power.
Back: The rate of doing work, equivalently the rate of energy transfer: P = ΔW/Δt.

Front: CHECK YOURSELF
A car brakes uniformly from 24 m s^−1 to rest over 60 m. How long does the braking take?
Show a hint
You know u, v and s. The quickest route uses the equation without a.
Back: Using s = 12(u + v)t: 60 = 12 × t, so t = 5.0 s.
The same answer follows from v = u + at once you find a = −4.8 m s^−2 from v^2 = u^2 + 2as; agreement between routes is the free check the four equations offer.

Front: CHECK YOURSELF
A 2.0 kg trolley moving at 3.0 m s^−1 collides with a stationary 1.0 kg trolley and they couple together. (a) Find their shared speed. (b) Is the collision elastic?
Show a hint
Momentum settles part (a). Part (b) is a kinetic-energy audit.
Back: (a) Momentum before: 2.0 × 3.0 = 6.0 kg m s^−1. After, the same 6.0 is carried by 3.0 kg, so v = 6.0 / 3.0 = 2.0 m s^−1.
(b) Kinetic energy before: ½ × 2.0 × 3.0^2 = 9.0 J. After: ½ × 3.0 × 2.0^2 = 6.0 J. Kinetic energy fell by 3.0 J, so the collision is inelastic, as coupling collisions always are; the 3.0 J became internal energy and sound.

Front: CHECK YOURSELF
A ball rolls off a table 1.25 m high at 4.0 m s^−1. How long is it in the air, and how far from the table does it land? Take g = 9.8 m s^−2.
Show a hint
The fall time comes from the vertical motion alone; the sideways speed is irrelevant to it.
Back: Vertical: s = 12gt^2 with s = 1.25 m gives t^2 = 2 × 1.25 / 9.8 = 0.255, so t = 0.50 s.
Horizontal: constant 4.0 m s^−1 for 0.50 s gives a range of 4.0 × 0.50 = 2.0 m.
Notice the order: the vertical problem produced the time, and the horizontal problem spent it.

Front: DEFINITION
Define: Work.
Back: Energy transferred by a force: the force multiplied by the distance moved in the direction of the force, W = Fs cos θ.

Front: DEFINITION
Define: Centre of mass.
Back: The single point through which an object's weight can be taken to act.

Front: CHECK YOURSELF
A book rests on a table. A student claims the book's weight and the table's normal contact force on it are a Newton's third law pair. Give two reasons the claim is wrong, and state the two genuine pairs.
Show a hint
Check the third law's small print: same type of force, different bodies.
Back: Both forces act on the same body, the book, and they are different types, gravitational and contact. A third-law pair fails on either count alone; this fails on both.
The genuine pairs: the Earth pulls the book down and the book pulls the Earth up (gravitational); the table pushes the book up and the book pushes the table down (contact).
The equality of weight and normal force here is equilibrium, not the third law, and it breaks the moment the book accelerates vertically.

Front: EQUATION — NOT ON THE DATA SHEET
Momentum:
p = ?
Back: p = mv

Front: DEFINITION
Define: Weight.
Back: The gravitational force acting on an object's mass: W = mg, a vector, measured in newtons.

Front: DEFINITION
Define: Couple.
Back: A pair of equal and opposite coplanar forces whose lines of action do not coincide.

Front: DEFINITION
Define: Terminal speed.
Back: The constant speed reached when the resistive forces on a falling object have grown to equal its weight, so the resultant force and acceleration are zero.

Front: CHECK YOURSELF
A car of mass 1200 kg travels at a constant 30 m s^−1 against total resistive forces of 600 N. What power does the engine deliver, and why does the mass not matter?
Show a hint
Constant speed is a statement about the resultant force.
Back: Constant speed means zero acceleration, so the resultant force is zero and the driving force equals the resistive force: 600 N.
P = Fv = 600 × 30 = 18 kW.
The mass is irrelevant because nothing accelerates: mass could only enter through F = ma, and a = 0. It is a distractor, and a deliberate one.

Front: CHECK YOURSELF
Forces of 30 N and 40 N act on a point at right angles to each other. Find the magnitude of the resultant and the angle it makes with the 30 N force.
Show a hint
Tip to tail, then Pythagoras for the size and tangent for the angle.
Back: Magnitude: 30^2 + 40^2 = 2500, and the square root gives 50 N.
Direction: tan θ = 40/30, so θ = 53° to the 30 N force. A 3-4-5 triangle, which is why examiners are so fond of these numbers.

Front: DEFINITION
Define: Principle of conservation of energy.
Back: Energy cannot be created or destroyed, only transferred from one form to another; the total energy of an isolated system is constant.

Front: DEFINITION
Define: Scalar.
Back: A quantity with magnitude only, such as speed, distance, energy or temperature.

Front: DEFINITION
Define: Momentum.
Back: The product of an object's mass and its velocity: p = mv, a vector.

Front: DEFINITION
Define: Mass.
Back: The amount of matter in an object: a scalar, measured in kilograms, unchanged by location.

Front: EQUATION — NOT ON THE DATA SHEET
Weight from mass:
W = ?
Back: W = mg

Front: CHECK YOURSELF
A child of mass 30 kg starts from rest at the top of a slide 2.5 m high and reaches the bottom at 5.0 m s^−1. How much energy was transferred to the surroundings by friction? Take g = 9.8 m s^−2.
Show a hint
Write the budget: potential in, kinetic out, friction takes the difference.
Back: Potential energy given up: ΔE_p = mgΔh = 30 × 9.8 × 2.5 = 735 J.
Kinetic energy gained: E_k = 12mv^2 = ½ × 30 × 5.0^2 = 375 J.
The books must balance, so friction took 735 − 375 = 360 J, now internal energy in the slide, the child and the air.

Front: CHECK YOURSELF
A uniform 4.0 m beam of weight 200 N rests on supports at each end. A 600 N load sits 1.0 m from the left support. Find the force from the left support.
Show a hint
Take moments about the right support, so its unknown force vanishes.
Back: About the right support: anticlockwise, R_left × 4.0. Clockwise: the beam's weight, 200 N at its centre, 2.0 m away, gives 400 N m; the load, 600 N at 3.0 m from the right, gives 1800 N m.
So 4.0 R_left = 400 + 1800 = 2200, giving R_left = 550 N.
The pivot choice did the work: the right support's force never appeared, and one equation held one unknown.

Front: CHECK YOURSELF
An astronaut has a mass of 70 kg. On the Moon, g = 1.62 N kg^−1. What are the astronaut's mass and weight on the Moon?
Show a hint
One of the two quantities never changes.
Back: Mass: 70 kg, exactly as on Earth. Mass is the amount of matter and travels with the astronaut.
Weight: W = mg = 70 × 1.62 = 113 N, about a sixth of the 687 N weight on Earth, because the Moon's field strength is about a sixth of the Earth's.

Front: DEFINITION
Define: Moment of a force.
Back: The force multiplied by the perpendicular distance from the line of action of the force to the pivot.

Front: DEFINITION
Define: Vector.
Back: A quantity with both magnitude and direction, such as velocity, displacement, force or momentum.

Front: DEFINITION
Define: Drag.
Back: The resistive force on an object moving through a fluid, acting opposite to the motion and growing with speed.

Front: CHECK YOURSELF
A skydiver falls at a terminal speed of 55 m s^−1, then opens a parachute and reaches a new terminal speed of 6 m s^−1. Explain, in terms of forces, why the new terminal speed is lower.
Show a hint
The weight has not changed. What has, and what does that do to the speed at which drag can match it?
Back: Opening the parachute greatly increases the area pushing through the air, so at any given speed the drag is far larger than before.
Drag therefore matches the unchanged weight at a much lower speed, and it is at that lower speed that the resultant returns to zero and the fall becomes steady.
Between the two plateaus, drag exceeds weight, the resultant acts upward, and the skydiver decelerates: the steep falling section of the velocity-time graph.

Front: DEFINITION
Define: Elastic collision.
Back: A collision in which kinetic energy is conserved as well as momentum.

Front: DEFINITION
Define: Impulse.
Back: The product of a force and the time for which it acts, FΔt, equal to the change in momentum it produces.