Mechanics - motion, forces, energy and equilibrium
Mechanics
The mechanics questions in the Myanmar matriculation paper test whether you can move between describing motion (equations, graphs) and explaining it (resultant force, momentum, energy). Expect multi-step numericals: a body accelerating under two or three forces, a vertical throw, a braking car, a conical pendulum or satellite orbit, and a beam balanced on two supports.
Core rules
- Uniform acceleration: v = u + at, s = ut + ½at², v² = u² + 2as. On a velocity–time graph the gradient is acceleration and the area is displacement.
- Newton's second law applies to the resultant force: ΣF = ma. Weight W = mg acts at the centre of gravity.
- Momentum is conserved when no external resultant force acts: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂, for both elastic and inelastic collisions.
- Circular motion requires a centripetal resultant F = mv²/r = mω²r towards the centre. It is supplied by friction, tension, gravity or a normal reaction - it is never an extra force added on top.
- Work done by the resultant force equals the change in kinetic energy; power = Fv.
- For equilibrium, ΣF = 0 and Σmoments about any point = 0, using the perpendicular distance from the pivot.
Worked example
A 1200 kg car accelerates uniformly from rest to 20 m/s in 8 s against a constant resistance of 400 N. Find the driving force.
Acceleration a = (20 − 0)/8 = 2.5 m/s². Resultant force needed = ma = 1200 × 2.5 = 3000 N. Driving force = 3000 + 400 = 3400 N.
Traps
Mixing mass (kg) with weight (N); applying v = u + at when acceleration is not uniform; leaving the resistance force out of the resultant; drawing centripetal force as an extra arrow on a free-body diagram; using a sloping length instead of the perpendicular distance in a moment; changing the sign convention for g partway through a vertical motion problem.