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Answers · Resultant force

Resultant force — final check key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. FK1 · Final check

Model answer. (a) In P the velocity stays 5.0 m/s towards the east: the same speed along the same straight line. The velocity does not change, so the resultant force is zero; the forces are balanced. (b) In Q the hovercraft moves east and speeds up, so the resultant force is along the motion, towards the east. In R it slows down while moving east, stops, then speeds up towards the west: all the way through, the resultant force is towards the west. (c) In S the speed stays 5.0 m/s, but the direction turns from east through north-east to north and beyond: the resultant force changes only the direction of motion. At the start the hovercraft moves east and turns towards the north, so the force acts towards the north, at right angles to the motion. (d) Towards the west act the air resistance and the pull of the rope: 900 N − 150 N − pull = 400 N, so the rope pulls with 900 N − 150 N − 400 N = 350 N, towards the west. (e) The pupil is wrong. At 4.0 s the velocity is still changing: a moment before, the hovercraft moved east; a moment later, it moves west. A changing velocity needs a resultant force, and here it still acts towards the west. Being at rest for an instant does not mean that the forces are balanced.

A full answer includes

  • (a) Zero: the velocity stays 5.0 m/s east, so the forces are balanced.
  • (b) Q: towards the east, because it speeds up while moving east.
  • (b) R: towards the west, because it slows down while moving east, then speeds up towards the west.
  • (b) I did not give R’s force as towards the east, the way it moves at the start.
  • (c) Only its direction of motion: the speed stays 5.0 m/s.
  • (c) At the start, the force acts towards the north, the way the path turns.
  • (d) Method: 900 N − 150 N − pull of the rope = 400 N.
  • (d) Answer: the rope pulls with 350 N.
  • (d) I did not leave out the air resistance.
  • (e) At 4.0 s the velocity is still changing, from east, through zero, to west.
  • (e) A changing velocity needs a resultant force: it still acts towards the west.
  • (e) The hovercraft is at rest only for an instant.

Also accept

  • (a) zero; (b) Q east, R west; (c) its direction only, towards the north at the start; (d) 350 N; (e) the velocity is still changing, so the resultant force is still towards the west.

Watch for

  • “(c) S has no resultant force, because the speed stays 5.0 m/s.” — The speed is steady, but the direction turns. A change of direction is a change of velocity, so there must be a resultant force.
  • “(d) The rope pulls with 750 N.” — You left out the resultant. The two forces towards the west together must be 400 N smaller than the push of the fan.
  • “(b) In R the resultant force is towards the east, because the hovercraft moves east at the start.” — It moves east but slows down, so the resultant force acts against its motion, towards the west.

2. FK2 · Final check

Model answer. (a) From 0 to 2.0 s the speed rises from 0 to 1.2 m/s in a straight line, so a = Δv ÷ Δt = 1.2 m/s ÷ 2.0 s = 0.60 m/s². (b) F = ma = 0.50 kg × 0.60 m/s² = 0.30 N. The resultant force and the acceleration are always in the same direction, so the force points along the motion, the way the trolley speeds up. (c) a = F ÷ m = 0.10 N ÷ 0.50 kg = 0.20 m/s². The acceleration is in the same direction as the resultant force, backwards, against the motion: the trolley slows down. (d) m = F ÷ a = 0.30 N ÷ 0.40 m/s² = 0.75 kg. (e) (i) On a larger track at the same speed, the path bends less tightly, so a smaller force is needed. (ii) Heavier wagons at the same speed and radius need a bigger force. (f) The two velocities are at right angles: v² = 0.30² + 0.50² = 0.09 + 0.25 = 0.34, so his resultant velocity is √0.34 = 0.58 m/s (at 59° east of north). (g) The velocity and the resultant force have a direction, so they are vectors. The mass, the distance and the time have a size only, so they are scalars.

A full answer includes

  • (a) Method: a = Δv ÷ Δt = 1.2 m/s ÷ 2.0 s.
  • (a) Answer: a = 0.60 m/s².
  • (b) Method: F = ma = 0.50 kg × 0.60 m/s².
  • (b) Answer: F = 0.30 N, in the same direction as the acceleration.
  • (c) Method: a = F ÷ m = 0.10 N ÷ 0.50 kg.
  • (c) Answer: 0.20 m/s², backwards, the same direction as the resultant force.
  • (d) Method: m = F ÷ a = 0.30 N ÷ 0.40 m/s².
  • (d) Answer: m = 0.75 kg.
  • (e)(i) A smaller force.
  • (e)(ii) A bigger force.
  • (f) Method: v² = 0.30² + 0.50² = 0.09 + 0.25 = 0.34.
  • (f) Answer: v = √0.34 = 0.58 m/s.
  • (f) I did not add the velocities to get 0.80 m/s.
  • (g) Vectors: the velocity and the resultant force.
  • (g) I did not choose the mass, the distance or the time: they are scalars.

Also accept

  • (a) 0.60 m/s²; (b) 0.30 N, same direction as the acceleration; (c) 0.20 m/s² backwards; (d) 0.75 kg; (e) (i) smaller; (ii) bigger; (f) 0.58 m/s; (g) the velocity and the resultant force.

Watch for

  • “(c) The acceleration is forwards, because the trolley is still moving forwards.” — The acceleration points the same way as the resultant force, backwards. An acceleration against the motion slows the trolley down.
  • “(d) The mass is 0.12 kg.” — You multiplied the force by the acceleration. m = F ÷ a = 0.30 ÷ 0.40 = 0.75 kg.
  • “(e)(i) A larger track needs a bigger force.” — On a larger circle the path bends less tightly, so at the same speed a smaller force towards the centre is enough.
  • “(f) His resultant velocity is 0.80 m/s.” — The velocities are at right angles, so they combine with Pythagoras’ theorem: √(0.30² + 0.50²) = 0.58 m/s.