Newton's Second Law in Vector Form

Forces $F_1 = (5\mathbf{i}+2\mathbf{j})$ N and $F_2 = (p\mathbf{i}+q\mathbf{j})$ N act on a particle of mass $2$ kg, producing acceleration $\mathbf{a}=(4\mathbf{i}-3\mathbf{j})$ m s$^{-2}$. Find $p$ and $q$.

• $p=3$, $q=-8$

  Correct. $\mathbf{F}_{net}=m\mathbf{a}=2(4\mathbf{i}-3\mathbf{j})=8\mathbf{i}-6\mathbf{j}$. Then $F_1+F_2=F_{net}$: $(5+p)=8 \Rightarrow p=3$, and $(2+q)=-6 \Rightarrow q=-8$.

• B$p=4$, $q=-3$ — taking F2 to be the acceleration vector itself

  This substitutes a directly for F2, skipping both the mass and $F_1$ entirely. Acceleration and an unknown FORCE are different quantities in different units — a is never itself the answer to 'find the unknown force.'

• C$p=-1$, $q=-5$ — using a itself as the resultant force, without multiplying by the mass

  This takes $\mathbf{F}_{net}=\mathbf{a}=4\mathbf{i}-3\mathbf{j}$ (mass never multiplied in), then subtracts $F_1$: $(4-5)\mathbf{i}+(-3-2)\mathbf{j}=-1\mathbf{i}-5\mathbf{j}$. The mass omission happens one line before this, exactly the trap named above.

• D$p=-3$, $q=8$ — subtracting the resultant from F1 instead of F1 from the resultant

  This computes $F_1-F_{net}$ instead of $F_{net}-F_1$, the equation rearranged backwards: $(5-8)\mathbf{i}+(2-(-6))\mathbf{j}=-3\mathbf{i}+8\mathbf{j}$. $F_2$ is what's left of the resultant once $F_1$ is removed from it, not the other way round.

Traps tested: Acceleration mistaken for the unknown force directly · Mass omitted from the vector equation of motion · Equation rearranged backwards

Why is it valid to separately equate the i-coefficients and the j-coefficients on the two sides of a vector equation such as $p\mathbf{i}+q\mathbf{j} = 8\mathbf{i}-6\mathbf{j}$, rather than needing some fresh justification each time it's used?

• Because i and j are independent (perpendicular) directions, so the difference of the two sides can only be the zero vector if BOTH coefficients are zero — no amount of i can produce any j, or vice versa

  Correct — this is exactly the mechanism block's own derivation: $(p-8)\mathbf{i}+(q+6)\mathbf{j}=\mathbf{0}$ forces both coefficients to be zero separately, precisely because i and j point in genuinely different directions.

• BBecause i and j always have magnitude exactly 1, so their coefficients are automatically comparable

  Magnitude-1 length is a real property of unit vectors, but it isn't what licenses equating coefficients — a vector of length 5 in a fixed direction would work exactly the same way. The justification is about DIRECTION (independence), not length.

• CIt isn't always valid — it only works in examples like this one, where both vectors happen to already be given in matching i/j form

  Equating coefficients is a general rule that holds for any equal vectors expressed in i/j form, not a coincidence of this particular example. The mechanism block's derivation applies regardless of where the numbers came from.

• DBecause vector addition and subtraction are commutative

  Commutativity (that $\mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}$) is a real property of vectors, but it has nothing to do with why matching coefficients is valid — that comes specifically from i and j being independent directions, not from the order terms can be added in.

Traps tested: Unit magnitude mistaken for the actual justification · Equate coefficients mistaken for a special case not a general rule · Unrelated vector property cited as the justification

Two forces, $F_1=(3\mathbf{i}-\mathbf{j})$ N and $F_2=(\mathbf{i}+4\mathbf{j})$ N, are the only forces acting on a particle of mass $2$ kg. A student writes $F_1 = m\mathbf{a}$ and solves for a from there. What's wrong?

• Newton's second law needs the RESULTANT of all forces acting, not one of them alone — the student needed $F_1+F_2=m\mathbf{a}$, using $4\mathbf{i}+3\mathbf{j}$ N, not $F_1$ on its own

  Correct. With two forces genuinely both acting, neither one alone is 'the' force $\mathbf{F}=m\mathbf{a}$ refers to — only their sum is.

• BNothing — $F_1$ has the larger magnitude of the two, so it dominates the resultant closely enough to use alone

  There is no 'close enough' shortcut in M1 — Newton's second law is exact, and using one force in place of the true resultant produces the wrong acceleration outright, not an acceptable approximation to the right one.

• C$F_2$ should have been used instead of $F_1$, since it was named second in the question

  The order forces are listed in a question carries no significance for which one (if either) should be used alone — and the real fix isn't swapping which single force is used, it's using their sum.

• DThe mass should be split between the two forces, e.g. 1 kg per force, before applying F=ma to each separately

  Mass is a property of the whole particle, not something to be divided up between the individual forces acting on it. The resultant force acts on the FULL mass — there is no rule that shares mass out across separate forces.

Traps tested: One force assumed to approximate the resultant · Arbitrary forces order mistaken for significance · Mass split between multiple forces

A student's working reads: 'Resultant force = a = 5i + 2j (N), so...' — going on to use this directly as the resultant force for a particle of mass 3 kg. Based on the mark scheme's own general Mechanics-marking principles, what does this cost?

• The M mark for this line — the general marking guidance states omission of mass from an equation of motion is a method error, and writing the resultant force equal to a rather than ma is exactly that

  Correct. "Omission of mass from a resolution is a method error" is verified, general Mechanics-marking guidance (§3 of the research bank) — this is a method error, not a smaller accuracy slip further down the line.

• BOnly the A mark, since the method itself (writing 'resultant force = ...') is still recognisable as an attempt at Newton's second law

  The mark scheme's own general principles are explicit that this SPECIFIC category of omission (the mass) is a method error, not merely an accuracy one — a recognisable ATTEMPT at the right idea doesn't rescue the M mark when a required term is missing from the equation itself.

• CNothing, provided the final numerical answer for any unknown still comes out as a sensible-looking pair of numbers

  A plausible-looking final answer doesn't excuse a method error — the marked-solution above shows exactly this: individually reasonable-looking numbers (p=−5, q=7) that are still wrong, and still M0, because of the missing mass one line earlier.

• DHalf of the marks available for that part, since the method is half right

  M/A/B marks aren't awarded proportionally to how 'close' an error feels — a method mark is either earned (the correct equation, correctly formed) or it isn't. There's no half-credit rule for a required term being present in spirit but missing from the working.

Traps tested: Method error mistaken for an accuracy error · Plausible final answer assumed to excuse a method error · Partial credit assumed proportional to how close the error is

How many distinct WME01 series does WME01-verified-facts.md confirm contain a worked example of Newton's second law in vector form (spec 4.1), in the material reviewed for this lesson?

• One — Jan 2022, Q6

  Correct, and worth knowing plainly rather than assuming otherwise: this is genuinely thinner evidence than most other WME01 topics in this course, which is why this lesson says so directly in its second teach block and its closing flag, rather than presenting one example as if it carried the weight of several.

• BAt least three, the same bar most other WME01 topics in this course clear

  This is the specific overclaim this lesson exists to avoid. Most other WME01 lessons in this course genuinely do clear three or more independently confirmed series — this spec point does not, and inflating the count to match them would misrepresent the actual state of the evidence.

• CNone — this spec point has never been directly tested

  This denies the one real citation this very lesson is built around — Jan 2022 Q6 is a genuine, verified real-paper example, just the only one located in the material reviewed for this course.

• DIt cannot be counted, because the spec guidance only describes what MAY be tested, not what has actually appeared on a real paper

  The guidance text and the real-paper evidence are two different things, and this lesson is careful to distinguish them — but the distinction cuts the other way here: Jan 2022 Q6 is a confirmed REAL past-paper example, not merely a guidance description, so it can and should be counted.

Traps tested: Evidence count inflated to match other topics · Verified example denied despite being cited directly in this lesson · Guidance language mistaken for absence of real evidence