Flashcards · YMA01 · Topic 5

Unit M1: Mechanics 1 flashcards

35 cards on Topic 5, subtopics M1.1 to M1.6. Pick a subtopic or study them all.

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M1.1 Mathematical models in mechanics

Modelling assumption: particle: The object is treated as a mass at a single point, so its size and any rotation are ignored.
Modelling assumptions: light, and inextensible (for a string): Light: negligible mass, so the tension is the same all along the string. Inextensible: does not stretch, so connected objects have the same speed and the same size of acceleration.
Modelling assumptions: smooth and rough surfaces: Smooth: no friction. Rough: friction can act, opposing motion.
Modelling assumptions: lamina, rigid body, rod and uniform: Lamina: flat, negligible thickness. Rigid body: does not bend or deform. Rod: thin and rigid, one-dimensional. Uniform: mass spread evenly, so the weight acts at the middle.
Light smooth pulley: what can you assume?: It has no mass and no friction, so the tension in the string is the same on both sides of the pulley.
How should you state a modelling assumption in an answer?: Pair the assumption with its effect, for example "the string is inextensible, so both particles have the same acceleration". Also state "air resistance ignored" or "g constant" where relevant.

M1.2 Vectors in mechanics

Magnitude and direction of the vector ai + bj: Magnitude √(a² + b²). Angle with i: tan θ = b/a, then place it in the right quadrant from the signs.
Resolve a vector of magnitude R at angle θ above the horizontal: R cos θ i + R sin θ j: adjacent side R cos θ, opposite side R sin θ.
Find the resultant of (6i + 2j), (−3i + 7j) and (4i − j), and its magnitude: Add components: 7i + 8j. Magnitude √(49 + 64) = √113 = 10.6 (3 s.f.).
Position vector of a particle moving with constant velocity v: r = r₀ + vt, where r₀ is the position at t = 0.
A particle is due north of O at some time. What does that tell you?: With i east and j north, its i-component is zero at that time. Solve for t, then find the j-component. "Due east" means the j-component is zero.
Speed compared with velocity: Velocity is a vector (size and direction). Speed is its magnitude: for v = pi + qj, speed = √(p² + q²).

M1.3 Kinematics of a particle moving in a straight line

The five constant acceleration (suvat) equations: v = u + at s = ut + ½at² s = vt − ½at² v² = u² + 2as s = ½(u + v)t They apply only when acceleration is constant.
How do you choose which suvat equation to use?: List s, u, v, a, t. Note the three you know and the one you want, then choose the equation that leaves out the fifth one.
Vertical motion under gravity: value and sign of a: a = −9.8 m s⁻² if up is positive, or +9.8 if down is positive. Keep to one choice. At the highest point v = 0.
Velocity-time graph: what do the gradient and the area give?: Gradient: acceleration. Area under the graph: displacement. Area below the axis is negative displacement; count it as positive for distance.
A car slows uniformly from 28 to 12 m s⁻¹ in 5 s. Find its acceleration and the distance.: a = (12 − 28)/5 = −3.2 m s⁻². s = ½(28 + 12)(5) = 100 m.
A ball lands below its start point. How do you find the time?: Use s = ut + ½at² with s negative (taking up as positive). This gives a quadratic in t. Reject the negative root.

M1.4 Dynamics of a particle moving in a straight line or plane

Newton's second law, weight, and friction on a sliding particle: Resultant force = mass × acceleration: F = ma. Weight W = mg, vertically downwards. Friction on a sliding particle: F = μR, along the surface, opposing the motion.
Particle on a plane inclined at α: components of weight, and acceleration sliding down a rough plane: mg sin α down the slope, mg cos α into it, so R = mg cos α when nothing else acts perpendicular to the plane. Sliding down a rough plane: a = g(sin α − μ cos α).
Connected particles over a smooth pulley: method: Draw each particle with its own forces. Write one F = ma equation for each, taking its direction of motion as positive. Add the equations to remove T, find a, then substitute back for T.
Masses 5 kg and 3 kg hang over a smooth pulley. Find a and T (g = 9.8).: 5g − T = 5a and T − 3g = 3a. Adding: 2g = 8a, so a = 2.45 m s⁻². T = 3(g + a) = 36.75 N.
Momentum, impulse and conservation of momentum: Momentum = mv (N s). Impulse = mv − mu = Ft for a constant force. Direct collision: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, with signed velocities. If the particles join, use one common velocity.
A 0.2 kg ball at 6.5 m s⁻¹ is hit straight back at 2 m s⁻¹. Find the impulse.: Original direction positive: I = 0.2(−2) − 0.2(6.5) = −1.7. Magnitude 1.7 N s, opposite to the original motion.

M1.5 Statics of a particle

Condition for a particle to be in equilibrium, and the method: The resultant force is zero. Resolve in two perpendicular directions and set each total to zero.
Components of a force F at angle θ to the vertical: Vertical component F cos θ; horizontal component F sin θ. (At θ above the horizontal it is the other way round: horizontal F cos θ, vertical F sin θ.)
Friction on a particle at rest compared with limiting equilibrium: At rest: F ≤ μR, and F is whatever balances the other forces. On the point of slipping (limiting): F = μR.
A block of weight 80 N stays at rest under a horizontal pull of 25 N. What can you say about μ?: R = 80 N and F = 25 N. Since F ≤ μR, 25 ≤ 80μ, so μ ≥ 0.3125. The friction force is 25 N.
Tension compared with thrust: Tension pulls along a string or rod. Thrust pushes along a rod. A string can only pull; a rod can do either.
How does a pull at an angle above the horizontal affect the normal reaction?: Its upward component reduces R (R = mg − P sin θ), which reduces the friction available. A downward component increases R.

M1.6 Moments

Moment of a force about a point: Force × perpendicular distance from the point to the line of action of the force. Unit N m. State the sense: clockwise or anticlockwise.
Conditions for a rigid body (such as a rod) to be in equilibrium: The forces balance (resultant force zero), and clockwise moments equal anticlockwise moments about any point.
A beam rests on two supports and is about to tilt about one of them. What do you use?: The reaction at the other support is zero. Take moments about the support it is tilting about.
Where does the weight act for a light, a uniform and a non-uniform rod?: Light: no weight. Uniform: at the midpoint. Non-uniform: at the centre of mass given (or to be found).
A light rod of length 3 m rests on supports at its ends A and B, with 45 N at 1 m from A. Find the reactions.: Moments about B: R_A × 3 = 45 × 2, so R_A = 30 N. Resolving vertically: R_B = 45 − 30 = 15 N.