Flashcards · 9709 · Topic 4

Mechanics flashcards

29 cards on Topic 4, subtopics 4.1 to 4.5. Pick a subtopic or study them all.

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Limiting vs not limiting: F = μR only at the point of slipping or while sliding.
Components of the weight of a particle on a plane inclined at α: mg sin α down the slope (along the plane). mg cos α into the slope (perpendicular to the plane).
Condition for a particle to be in equilibrium: The resultant force is zero: the sum of the components in every direction is zero (resolve in two perpendicular directions).
Friction: the inequality, and which way it acts: F ≤ μR, with F = μR when limiting or sliding. Friction opposes the motion that would happen (or is happening). A smooth surface has no friction.
A particle is about to slip down a rough plane inclined at α, with no other forces. Find μ.: Perpendicular: R = mg cos α. Along: mg sin α = μR = μmg cos α. So μ = tan α.
Newton’s third law: When two bodies interact, the forces they exert on each other are equal in magnitude and opposite in direction.

Distance vs displacement: Distance adds all motion; displacement is the net change in position.
Speed vs velocity: Velocity has a sign; speed does not.
The constant-acceleration (suvat) equations: v = u + at s = ½(u + v)t s = ut + ½at² v² = u² + 2as Only for constant acceleration.
What do the gradient and area of a displacement–time and a velocity–time graph give?: s–t graph: gradient = velocity. v–t graph: gradient = acceleration; area under the graph = displacement.
Variable acceleration (a function of t): how do you move between s, v and a?: Differentiate: v = ds/dt, a = dv/dt. Integrate: v = ∫ a dt, s = ∫ v dt (with a constant from the initial conditions). Maximum velocity: a = 0. Instantaneous rest: v = 0.
P starts from rest at O with acceleration 2 m s⁻². At the same moment Q passes O at a constant 12 m s⁻¹ in the same direction. When does P catch Q?: P: s = ½ × 2 × t² = t². Q: s = 12t. t² = 12t gives t = 12 s, at 144 m from O.

Momentum vs kinetic energy: Momentum is conserved in impacts; kinetic energy is generally not.
Momentum: definition and unit: Momentum = mass × velocity = mv. Unit: kg m s⁻¹ (equivalently N s). It is a vector, so its sign shows direction.
Principle of conservation of momentum (direct impact in one dimension): Total momentum before = total momentum after: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂. Choose a positive direction first.
Two particles coalesce. What does that mean for the equation?: They join and move with one common velocity v: m₁u₁ + m₂u₂ = (m₁ + m₂)v.
A 0.4 kg ball at 5 m s⁻¹ hits a stationary 0.6 kg ball and rebounds at 1 m s⁻¹. Find the speed of the 0.6 kg ball.: 0.4 × 5 = 0.4 × (−1) + 0.6v, so 2 = −0.4 + 0.6v and v = 4 m s⁻¹. The rebound velocity is −1, not +1.

Mass vs weight: Mass in kg; weight mg in newtons.
Tension vs thrust: A string can only pull; a rod or tow-bar can pull (tension) or push (thrust).
Newton’s second law: Resultant force = mass × acceleration (F = ma), applied in the direction of the acceleration.
How do you solve a connected-particles problem?: 1. Draw a separate force diagram for each particle. 2. Write F = ma for each in its own direction of motion; both have the same acceleration. 3. Add the equations to eliminate the tension. 4. Find the tension from one equation.
Particles of 3 kg and 2 kg hang either side of a smooth pulley (g = 10). Find the acceleration and the tension.: System: 30 − 20 = 5a, so a = 2 m s⁻². 2 kg particle: T − 20 = 2 × 2, so T = 24 N. Check: 30 − 24 = 3 × 2.
Acceleration of a particle sliding down a smooth plane inclined at α: a = g sin α (5 m s⁻² at 30° with g = 10). On a rough plane, the acceleration going up differs from coming down, because friction reverses.

Driving force vs resultant force: P = Fv uses the driving force, not the resultant.
Work done by a constant force F moving through distance d at angle θ to the force: Work done = Fd cos θ, in joules (J).
Kinetic energy and gravitational potential energy: KE = ½mv² GPE = mgh (h is the height gained)
The work–energy principle: Initial energy + work done by driving forces = final energy + work done against resistances (resistance × distance).
Power: the definitions, and what happens at maximum (steady) speed: Power = work done ÷ time, and P = Fv where F is the driving force. At maximum speed a = 0, so the driving force equals the total resistance.
A 50 kg load is raised 6 m at constant speed in 4 s (g = 10). Find the average power.: Work = 50 × 10 × 6 = 3000 J. Power = 3000 ÷ 4 = 750 W.