CCommons

Physics problem solving is mostly checking your answer before you know it's wrong

Dimensions, limiting cases and symmetry. Three boring checks that catch more mistakes than rereading ever did.

Neel Chatterjee12th Sample

4 Oct 2026 Β· 4 min read

Photo: Andrey Korzun Β· CC BY-SA 4.0 Β· via Wikimedia Commons

In chess, the commonest way to lose is not a bad plan. It's a good-looking move you didn't check. You see a clean tactic, play it, and only after your opponent replies do you notice the one square you forgot to look at. Strong players aren't smarter at finding moves. They have a habit of one extra look before letting go of the piece.

Physics problem solving has the same disease. I derive an expression, circle it, move on. And later I find out in the answer key that I was off by a factor of two, or that I'd squared the wrong thing. For a long time I thought the fix was "be more careful". That never works. The fix is a blunder check, a small routine that takes twenty seconds. I have three.

Check 1: dimensions

Every term you add must have the same dimensions, and your final answer must have the dimensions of what was asked. For a pendulum:

T=2Ο€LgT = 2\pi\sqrt{\frac{L}{g}}

Lg\dfrac{L}{g} has dimensions mmΒ sβˆ’2=s2\dfrac{\text{m}}{\text{m s}^{-2}} = \text{s}^2, so the square root is in seconds. βœ“. If you had misremembered it as g/L\sqrt{g/L}, you'd get sβˆ’1\text{s}^{-1} and know immediately that it's wrong.

Dimensions catch: adding unlike quantities, forgetting to square or take a root, a gg that fell to the wrong side of a fraction. They don't catch: pure numbers like 2, Ο€\pi, 12\tfrac12 and trigonometric factors.

Check 2: limiting cases

Push each variable to an extreme where you already know the answer. This is the most powerful check and nobody teaches it.

Atwood machine, two masses m1>m2m_1 > m_2 over a light frictionless pulley. A derivation gives

a=g m1βˆ’m2m1+m2,T=2m1m2m1+m2 ga = g\,\frac{m_1 - m_2}{m_1 + m_2}, \qquad T = \frac{2m_1m_2}{m_1+m_2}\,g

Now test it:

  • m1=m2m_1 = m_2: a=0a = 0. Balanced, nothing moves. βœ“
  • m2β†’0m_2 \to 0: aβ†’ga \to g. m1m_1 is in free fall, nothing pulling back. βœ“
  • m2β†’0m_2 \to 0: Tβ†’0T \to 0. A free-falling mass has no rope tension. βœ“
  • m1=m2=mm_1 = m_2 = m: T=mgT = mg. The rope holds up one weight. βœ“
  • Swap the labels m1↔m2m_1 \leftrightarrow m_2: aa changes sign. Motion reverses. βœ“

Five checks in under a minute, and I haven't looked at the answer key.

Here's one that caught a real mistake. In a mock I wrote the projectile range as

R=u2sin⁑θgR = \frac{u^2\sin\theta}{g}

Dimensions: u2/gu^2/g is a length. βœ“ Looks fine. Now limits: at ΞΈ=0\theta = 0 we get R=0R = 0, correct (no height, lands immediately). But at ΞΈ=90∘\theta = 90^\circ it gives R=u2/gR = u^2/g, and a stone thrown straight up lands at its starting point, so RR should be 00. βœ—. The correct formula has sin⁑2ΞΈ\sin 2\theta:

R=u2sin⁑2θgR = \frac{u^2\sin 2\theta}{g}

which is 00 at both ends and has a maximum at 45∘45^\circ. That wrong answer would have passed my dimension check. The limit check killed it.

Check 3: symmetry and conservation

Ask what the problem should not depend on.

  • A block slides down a smooth track of unknown shape from height hh. Speed at the bottom? You don't need the shape. Energy conservation gives v=2ghv = \sqrt{2gh} for any smooth path. If your answer depends on the path, you made it harder than it is.
  • Two identical masses collide head-on elastically. They exchange velocities. If your answer doesn't do that, something's off.

I think of conservation laws as endgame tables in chess. In an endgame with few pieces, the result is already known, so you skip the whole middle and read the answer. When a conservation law applies, you skip the differential equation and read the answer.

Check 4 (the free one): is the number believable?

If someone jumps with a launch speed of 3 m/s, the height is u22g=920β‰ˆ0.45\dfrac{u^2}{2g} = \dfrac{9}{20} \approx 0.45 m. That's a plausible human jump. If my answer is 45 m, I've lost a factor of 100 somewhere.

What it looks like in practice

CheckTimeCatchesMisses
Dimensions10 swrong powers, mixed-up quantitiesnumerical factors, sin⁑\sin vs cos⁑\cos
Limiting cases20–30 swrong trig factors, wrong signs, wrong denominatorserrors that vanish at the limits you tried
Symmetry / conservation15 sanswers that depend on things they shouldn'tnot much, but needs insight
Sanity of size5 sfactors of 10, 100small errors

None of them proves your answer is right. They only prove it isn't wrong in these ways. But "not wrong in four ways" is a long distance from "I circled it and hoped".

I didn't expect this habit to change my scores much. It did. Mostly because my mistakes were rarely about not knowing physics. They were about not looking once more. In chess I play slower now too, and I lose fewer pieces to my own hurry.

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11 responses

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  • [+]Harsh Gupta12thSampleΒ· 15d ago

    neel posts, chess analogy appears. it's like a law of nature at this point. but yes the projectile range catch is a good one, I've written sin ΞΈ instead of sin 2ΞΈ in an actual test

  • [+]Arjun Nair12thSampleΒ· 15d ago

    Agree with all of it. One thing worth adding to the dimensions check: it also tells you what combinations are possible. If you know the answer can only depend on LL and gg, dimensions force T∝L/gT \propto \sqrt{L/g} up to a constant. You can get the form of a result without deriving anything. It's a very good trick for unfamiliar problems.

  • [+]Sara Fernandes11thSampleΒ· 15d ago

    I'm PCB but we do projectile motion in 11 as well. For football penalty practice my coach says 45 degrees gives the farthest kick. Is that always true? Or is that only in physics problems with no air?

    • [+]Neel Chatterjee12thSampleΒ· 15d agoAuthor

      In the idealised case (no air resistance, ball lands at the same height it was kicked from), 45∘45^\circ gives the farthest range. If the ball lands lower than it started, say you throw from a cliff, the best angle is slightly less than 45∘45^\circ. And with air drag it's also lower in practice. For a football kicked on level ground it's something like 30 to 40 degrees in reality, but I haven't measured that, so take it as "less than 45".

  • [+]Rohan Sharma11thSampleΒ· 15d ago

    for the Atwood machine with a heavy pulley the answer is the same right? since the pulley just rotates in place

    • [+]Arjun Nair12thSampleΒ· 15d ago

      Not quite. A pulley with mass has rotational inertia, so some of the energy goes into spinning it, and the acceleration is smaller. With moment of inertia II and radius rr: a=(m1βˆ’m2) gm1+m2+I/r2a = \frac{(m_1 - m_2)\,g}{m_1 + m_2 + I/r^2} Note the limit check works here too. Let Iβ†’0I \to 0 (light pulley) and you get Neel's formula back.

      • [+]Rohan Sharma11thSampleΒ· 15d ago

        oh nice, so the limit check is a way of checking the more complicated formula against the simple one. makes sense. thanks

  • [+]Aman YadavDropperSampleΒ· 14d ago

    Good habit. In the exam hall I do dimensions on every second step. Costs nothing. Saved me 4 marks in my last test.

  • [+]Nisha Verma12thSampleΒ· 14d ago

    Not JEE but the "is the number believable" check works for NEET numericals too. I once got the speed of a car as 3000 m/s and just moved on. Next test I checked. The answer was 30. Missed a decimal. Never again.

  • [+]Ishaan Bose11thSampleΒ· 14d ago

    the table at the end is clean. for chemistry I think the equivalent of limiting cases is "what happens when concentration goes to zero / infinity". I never used it, I'll try

    • [+]Tanvi Deshmukh12thSampleΒ· 14d ago

      yes!! I use it for pH questions. If I get pH 8 for 10⁻⁸ M HCl, I know it's wrong, because an acid can never make water basic. Water's own H⁺ matters at that dilution, and the real answer comes out just under 7 (about 6.98). Great check!

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