Cambridge IGCSE Physics (Theory Workbook, Carol Tear) · Exercises 4A–4D, pp.24–29

Physics Workbook — Ch.4 Forces

Graded 2026-06-17 · Sys4Ethan (Claude Opus 4.8 vision, human-verified)
76%
✅ Correct 13 ◐ Partial 4 ✗ Needs fixing 0 ○ Not attempted 0 Total 17
To fix & review (4)Qp26 · Q3 (resultant forces) Qp28 · Q2 (seesaw) Qp28 · Q3 (beam + rope, Fig 4.11) Qp29 · Q2 (metre rule)
Qp24 · Q1 (word puzzle) ✓ Correct forces vocabulary

(a) extension, (c) pivot, (d) lamina, (e) moment, (f) stable/unstable/neutral — all correct. Only (b)'s first blank is hard to read in the photo; double-check the named force that opposes motion.

Qp24 · Q2 (spring vocab) ✓ Correct springs vocabulary

load, elastic, extension/proportional, spring constant, straight/origin — all correct.

Qp25 · Q3 (load–extension graph) ✓ Correct spring constant from a graph

Axes (Load/N, Extension/cm) ✓, limit of proportionality marked ✓, $k=F/e$ ✓, $k=4.0$ N/cm ✓. Tiny fix: write the unit as N/cm or N cm⁻¹, not N/cm⁻¹.

Qp25 · Q1 (4B crossword) ✓ Correct forces crossword

DRAG, RESULTANT, (air) resistance, etc. — readable entries all correct.

Qp26 · Q2 (resultant-force table) ✓ Correct resultant force & motion

All seven ticks correct.

Qp26 · Q3 (resultant forces) ◐ Partial adding forces

(a) $25-12 = 13$ N ✓.

Try this → (b) Looked it up on the clean figure: there are TWO 3 N arrows, both pointing left (negative), and one 5 N pointing right. Add all three with signs ($+5-3-3$). You wrote −2 — did you include BOTH 3 N forces?
Qp26 · Q4 (fill the blanks) ✓ Correct friction & balanced forces

surfaces/motion/heat, zero, direction/speed, circle — all correct.

Qp26 · Q5 (F = ma) ✓ Correct Newton's 2nd law

Ringed $F=ma$ ✓.

Qp27 · Q6 (acceleration) ✓ Correct a = F/m

(a) 0.70, (b) 1.20, (c) 1.94 m/s² — all correct, including the (c) two-mass + friction case.

Qp27 · Q7 (cyclist MCQ) ✓ Correct resultant force then a = F/m

A (0.15 m/s²) with full working ($16-2-3=11$ N, $60+12=72$ kg, $11/72$) — exactly the right habit.

Qp27 · Q8 (circular motion) ✓ Correct what changes centripetal force

mass / radius / speed ✓.

Qp27–28 · Q1 (4C moments) ✓ Correct moment = F × d

All six moments correct: 25.2, 63, 5.4, 0.72, 32.4, 240 N m. Just double-check each direction label against the arrow in its figure.

Qp28 · Q2 (seesaw) ◐ Partial principle of moments

(a) $140\times1.6 = 160d \Rightarrow d = 1.4$ m ✓.

Try this → (b) You described what happens to the 160 N child but didn't answer the question — which WAY does the plank turn when $d$ decreases? (Which side's moment now wins?)
Qp28 · Q3 (beam + rope, Fig 4.11) ◐ Partial moments equilibrium
Try this → Checked the clear figure — the distances are easy to misassign, and your $d=0.53$ m is actually plausible under one reading. Write out your full moment equation (each force × its distance about the pivot, including the 8.0 N rope) so it can be confirmed.
Qp29 · Q1 (mark centre of gravity) ✓ Correct centre of gravity

Crosses placed for the cube, circle and tool.

Qp29 · Q2 (metre rule) ◐ Partial moments on a pivoted rule

(b) distance $50-30 = 20$ cm ✓; (c) weight vector drawn ✓.

Try this → (d) Check where the weight $W$ sits relative to the pivot, then use moments: $W \times (\text{its distance}) = 1.5\,\text{N} \times 20\,\text{cm}$. Your 1.5 N looks like the rule's own weight, not the calculated $W$.
Qp29 · Q3 (non-uniform beam, Fig 4.16) ✓ Correct moments balance

$600 = 450 + 5d \Rightarrow d = 30$ cm — balance set up and solved correctly, working shown.

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