Physics problems go wrong in three places: units, signs and the wrong formula. The way to catch all three is to solve each problem two ways and see whether the answers agree. Here are six examples from the free OpenStax University Physics books, each solved with a calculator and then checked by hand.
Problem: a proton (+1.602 × 10⁻¹⁹ C) and an electron (−1.602 × 10⁻¹⁹ C) are 5.29 × 10⁻¹¹ m apart. Find the force between them.
Calculator: the Coulomb’s law calculator with charges +1 and −1 in elementary charges and r = 5.29e-11 m gives 8.24 × 10⁻⁸ N, attractive.
By hand: F = k|q₁q₂|/r² = 8.99 × 10⁹ × (1.602 × 10⁻¹⁹)² ÷ (5.29 × 10⁻¹¹)². The numerator is 8.99 × 10⁹ × 2.566 × 10⁻³⁸ = 2.307 × 10⁻²⁸; the denominator is 2.798 × 10⁻²¹; the ratio is 8.24 × 10⁻⁸ N. OpenStax prints 8.25 × 10⁻⁸ N; the last digit depends on how the intermediate steps are rounded.
Slips to avoid: squaring the charge but not the distance, and writing a negative force instead of stating “attractive”.
Problem: q₂ = −3e sits at the origin, q₁ = 2e is d = 2.0 × 10⁻⁷ m above it and q₃ = −5e is 2d to its right. Find the net force on q₂.
Calculator: run the pair q₁–q₂ (2 and −3, r = 2.0e-7 m): 3.46 × 10⁻¹⁴ N, attractive, so q₂ is pulled up. Run q₃–q₂ (−5 and −3, r = 4.0e-7 m): 2.16 × 10⁻¹⁴ N, repulsive, so q₂ is pushed left.
By hand: F = (−2.16 × 10⁻¹⁴, +3.46 × 10⁻¹⁴) N. Magnitude √(2.16² + 3.46²) × 10⁻¹⁴ = 4.08 × 10⁻¹⁴ N; direction tan⁻¹(3.46/2.16) = 58° above the −x axis. Both match OpenStax.
Slips to avoid: adding 3.46 and 2.16 as plain numbers (5.62 is wrong) and using d instead of 2d for the second pair.
Problem A: light from air hits a liquid at 30.0° and refracts to 22.0°. What is n₂? Calculator: the Snell’s law calculator in index mode gives 1.33. By hand: n₂ = 1.00 × sin 30.0° ÷ sin 22.0° = 0.500 ÷ 0.375 = 1.33, water.
Problem B: the same 30.0° ray enters diamond (n = 2.419). Calculator: 11.9°. By hand: sin θ₂ = 0.500 ÷ 2.419 = 0.207, so θ₂ = 11.9°.
Slips to avoid: measuring angles from the surface instead of the normal, and a calculator left in radians (sin 30 in radians is −0.988).
Problem: what is the critical angle for light going from polystyrene (n = 1.49) into air? Calculator: enter n₁ = 1.49, n₂ = 1.00 and any angle; the critical angle is 42.2°, and a 50° ray is totally reflected. By hand: θc = sin⁻¹(1.00 ÷ 1.49) = sin⁻¹(0.671) = 42.2°.
Slips to avoid: looking for total internal reflection when light goes into the denser medium (it can’t happen from air into water) and inverting the ratio, which gives an impossible sine above 1.
Problem: a 150 Hz horn on a train moving at 35.0 m/s, with sound at 340 m/s. What does a person beside the track hear as it approaches and as it leaves, and what does the engineer hear?
Calculator: the Doppler effect calculator gives 167.2 Hz approaching, 136.0 Hz receding and 150.0 Hz for the engineer (observer moving with the source).
By hand: approaching, 150 × 340 ÷ (340 − 35) = 150 × 1.115 = 167 Hz; receding, 150 × 340 ÷ 375 = 136 Hz. The shifts, +17 and −14 Hz, are not equal, because the source speed sits in the denominator.
Slips to avoid: flipping the signs (check: approaching must raise the pitch) and treating a moving observer like a moving source.
Problem: how fast must a probe go to escape Earth (M = 5.96 × 10²⁴ kg, R = 6.37 × 10⁶ m), and to escape the Sun starting from Earth’s orbit (M = 1.99 × 10³⁰ kg, r = 1.50 × 10¹¹ m)?
Calculator: the escape velocity calculator’s custom mode, with 5.96e24 kg and 6,370 km, gives 11.18 km/s; with 1.99e30 kg and 150,000,000 km, 42.08 km/s.
By hand: v = √(2GM/r) = √(2 × 6.67 × 10⁻¹¹ × 5.96 × 10²⁴ ÷ 6.37 × 10⁶) = 1.12 × 10⁴ m/s, and √(2 × 6.67 × 10⁻¹¹ × 1.99 × 10³⁰ ÷ 1.50 × 10¹¹) = 4.21 × 10⁴ m/s, matching OpenStax’s 1.12 × 10⁴ and 4.21 × 10⁴ m/s.
Slips to avoid: using kilometers with G in SI units, and dropping the 2 (√(GM/r) is circular orbit speed).
About 8.2 × 10⁻⁸ N, attractive, at 5.29 × 10⁻¹¹ m (OpenStax University Physics Volume 2 Example 5.1 gives 8.25 × 10⁻⁸ N).
Find each pair’s force with Coulomb’s law, split into x and y components, add the components, then take the magnitude and angle of the result.
42.2°, from sin⁻¹(1.00/1.49), per OpenStax University Physics Volume 3 Example 1.4.
The source speed is subtracted from the speed of sound in the denominator when approaching and added when receding, so the two shifts differ: +17 Hz versus −14 Hz for a 150 Hz horn at 35 m/s.
About 11.2 km/s, from v = √(2GM/R) with Earth’s mass and radius.