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A-Level Kinematics, uniformly accelerated motion, momentum and Newton's laws

What the A-Level syllabus expects for Kinematics, uniformly accelerated motion, momentum and Newton's laws, and how to practise it.

What the syllabus expects

How it's examined

Questions on this topic most often ask you to find, explain, state. About 6% of the past-paper style questions in Rae's bank for this subject sit in this topic.

Worked examples

Example 1 (2 marks)

(a) Explain what an inertial frame of reference is, and state how such a frame is moving.

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An inertial frame of reference is one in which Newton's first law holds: a body with no resultant force acting on it stays at rest or continues to move in a straight line at constant velocity. Such a frame is not accelerating; it moves with constant velocity (constant speed in a straight line).

Example 2 (3 marks)

At the instant a fully submerged diver has zero horizontal velocity, the viscous drag on him is 950 N upward, the upthrust is 740 N, and his mass is 78 kg. Find the size and direction of his acceleration.

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Principle: Newton's second law, resultant force F = ma, with the forces first combined as vectors. Take vertically upwards as the positive direction. Step 1 - list every vertical force acting on the diver at that instant. Upthrust from the water, U = 740 N, acting upwards (upthrust always acts upwards). Viscous drag, D = 950 N, acting upwards as stated (drag opposes motion, so the diver must be moving downwards at this instant - consistent with his horizontal velocity being zero). Weight, W = mg, acting downwards, where m = 78 kg and g = 9.81 N kg⁻¹: W = 78 kg × 9.81 N kg⁻¹ W = 765 N. Step 2 - find the resultant by adding the upward forces and subtracting the downward one. F = U + D − W F = 740 N + 950 N − 765 N F = 925 N The answer is positive, so the resultant is 925 N acting vertically upwards. Step 3 - apply Newton's second law. Rearrange F = ma by dividing both sides by m: a = F/m a = 925 N / 78 kg a = 11.9 m s⁻² a ≈ 12 m s⁻². Step 4 - direction. Acceleration is always in the same direction as the resultant force, so the acceleration is 12 m s⁻² directed vertically upwards (the diver is moving downwards but slowing down).

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