Flashcards · 9660 · Topic 4

Unit PSM1 – M1: Mechanics (International AS) flashcards

22 cards on Topic 4, subtopics M1.1 to M1.4. Pick a subtopic or study them all.

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Distance compared with displacement, and speed compared with velocity: Distance: total path length, always ≥ 0. Displacement: change in position, + or −. Speed: distance ÷ time, ≥ 0. Velocity: displacement ÷ time, + or −.
The constant acceleration equations: v = u + at s = ut + ½at² s = vt − ½at² s = ½(u + v)t v² = u² + 2as Use the one that leaves out the quantity you neither know nor want.
Displacement-time and velocity-time graphs: what do gradient and area give?: s–t graph: gradient is velocity. v–t graph: gradient is acceleration; area is displacement (below the axis negative; add sizes for distance).
Vertical motion from launch speed u (up positive, g = 9.8): a = −9.8 m/s². At the top v = 0. Maximum height u²/19.6. Time to return to launch level 2u/9.8. Same speed at the same height going up and down.
A train slows uniformly from 25 m/s to 10 m/s in 6 s. Find a and s.: a = (10 − 25)/6 = −2.5 m/s². s = ½(25 + 10) × 6 = 105 m.
A pebble is thrown up at 10 m/s from 3 m above the ground. When does it land?: Up positive, s = −3: −3 = 10t − 4.9t², so 4.9t² − 10t − 3 = 0. t = (10 + √158.8)/9.8 ≈ 2.31 s. Reject the negative root.

Calculus links between s, v and a: v = ds/dt, a = dv/dt = d²s/dt². s = ∫ v dt and v = ∫ a dt, each with a constant from a boundary condition. suvat applies only when a is constant.
Variable acceleration: conditions for "at rest" and for maximum or minimum velocity: At rest: v = 0. Maximum or minimum velocity: a = 0, then test. These usually happen at different times.
How do you find the distance travelled when a particle turns round?: Find when v = 0 in the interval. Find s at the start, each turning time and the end, then add the sizes of the changes. A definite integral of v gives displacement, not distance.
v = t² − 4t, with s = 0 at t = 0. Find the least velocity and the distance travelled in the first 4 s.: a = 2t − 4 = 0 at t = 2, where v = −4 (least). s = t³/3 − 2t², s(4) = −32/3. v < 0 throughout 0 < t < 4, so the distance is 32/3 m.
s = t³ − 6t². Find v and a at t = 1.: v = 3t² − 12t = −9 m/s at t = 1. a = 6t − 12 = −6 m/s² at t = 1.

Newton's three laws of motion: First: zero resultant force means rest or constant velocity. Second: resultant force = ma. Third: when two bodies interact, the forces are equal in size and opposite in direction, acting on different bodies.
Forces in M1: weight, normal reaction, tension, thrust and friction: Weight mg downwards. Normal reaction perpendicular to the surface (found from the vertical forces). Tension pulls; thrust pushes (rods only). Sliding friction F = μR opposes motion. In M1 all forces are horizontal or vertical.
A person of mass 70 kg stands in a lift accelerating upwards at 0.6 m/s². Find the normal reaction.: R − 70g = 70 × 0.6, so R = 70(9.8 + 0.6) = 728 N.
A 6 kg block is pushed by 40 N across a rough floor with μ = 0.3. Find a.: R = 6 × 9.8 = 58.8 N, friction 0.3 × 58.8 = 17.64 N. 40 − 17.64 = 6a, so a = 3.73 m/s² (3 s.f.).
Two particles m₁ > m₂ hang over a smooth pulley. Acceleration and tension?: a = (m₁ − m₂)g/(m₁ + m₂) and T = m₂(g + a). In an exam, write one F = ma equation per particle and show the working.
Car towing a trailer: how do you find the tension in the tow bar?: Whole system: driving force − all resistances = (total mass)a. Trailer alone: T − trailer resistance = (trailer mass)a. With a rigid bar, a negative T means a thrust.

Momentum and impulse: Momentum p = mv, signed in a straight line. Impulse = mv − mu (change in momentum), unit N s; equals Ft for a constant force.
Conservation of momentum for two particles in a straight line: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, with signed velocities. If they coalesce, use (m₁ + m₂)v. If they move apart from rest, the total momentum is zero.
A 5 kg particle at 4 m/s meets a 3 kg particle moving the other way at 2 m/s, and they join. Find the velocity.: 5 × 4 + 3 × (−2) = 8v, so 14 = 8v and v = 1.75 m/s in the direction the 5 kg particle was moving.
A 0.5 kg ball hits a wall at 8 m/s and rebounds at 6 m/s. Find the impulse, and the average force if contact lasts 0.01 s.: Impulse = 0.5 × (6 + 8) = 7 N s away from the wall. Average force = 7/0.01 = 700 N.
Why is momentum not conserved when a ball hits a fixed wall?: The wall gives the ball an impulse, an external force on the ball. In a collision between two particles the impulses on them are equal and opposite, so their total momentum is conserved.