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Many JEE problems give you kinetic energy and ask for momentum, or the other way round. Students often go through the speed every time: find v, then use it. That works, but it is slow and opens the door to slips. One relation, K = p²/2m, does the job in a line. This post derives it, shows four problem types where it saves time, and lists the traps.
Where does K = p²/2m come from?
Kinetic energy is K = ½mv², as in the NCERT Class 11 chapter on work, energy and power, and momentum is p = mv. Substitute v = p/m:
K = ½m(p/m)² = p²/2m, and so p = √(2mK).
Two consequences follow at once:
Same momentum: K is inversely proportional to m. The lighter body has more kinetic energy.
Same kinetic energy: p is proportional to √m. The heavier body has more momentum.
Kinetic energy, momentum and the de Broglie relation are all in the JEE (Main) 2026 syllabus, so this relation links three units.
Type 1: percentage changes
If K rises by 44%, by how much does p rise? p is proportional to √K, so p becomes √1.44 = 1.2 times its value: a 20% rise.
If p rises by 50%, K becomes 1.5² = 2.25 times its value: a 125% rise, not 100%.
Type 2: an explosion
A 3 kg shell at rest explodes into pieces of 1 kg and 2 kg, releasing 300 J of kinetic energy. Find each piece's kinetic energy.
Momentum is conserved and started at zero, so the two pieces have equal and opposite momenta of the same size p. With equal p, K is inversely proportional to m, so K₁ : K₂ = 2 : 1. Hence K₁ = 200 J and K₂ = 100 J.
Check: p = √(2 × 1 × 200) = 20 kg m/s and √(2 × 2 × 100) = 20 kg m/s. The speeds are 20 m/s and 10 m/s.
Type 3: same force, same time or same distance?
Two pucks of mass m and 4m start from rest on a smooth surface and are pushed by the same constant force.
Same time: the impulse is the same, so both gain the same momentum p. Then K = p²/2m gives the lighter puck four times the kinetic energy.
Same distance: the work is the same, so both gain the same kinetic energy K. Then p = √(2mK) gives the heavier puck twice the momentum.
This is exactly the comparison that tripped up students in a well-known study. Lawson and McDermott (1987) asked undergraduates to compare the changes in momentum and kinetic energy of two dry-ice pucks moved by the same constant force, and found that most could not relate the formulas they had learned to the motion they watched. The fix is to ask first: is the force acting for the same time (impulse-momentum) or over the same distance (work-energy)?
Type 4: de Broglie wavelength from kinetic energy
The NCERT Class 12 chapter on dual nature of radiation and matter gives the de Broglie relation λ = h/p. With p = √(2mK):
λ = h/√(2mK).
For an electron with K = 100 eV: K = 100 × 1.602 × 10⁻¹⁹ J, so p = √(2 × 9.109 × 10⁻³¹ × 1.602 × 10⁻¹⁷) ≈ 5.40 × 10⁻²⁴ kg m/s, and λ = 6.626 × 10⁻³⁴ / 5.40 × 10⁻²⁴ ≈ 1.23 × 10⁻¹⁰ m, about 0.123 nm.
An electron and a proton with the same kinetic energy: λ is proportional to 1/√m, so the electron's wavelength is √(m_p/m_e) ≈ √1836 ≈ 43 times longer. Our post on de Broglie wavelength problems has more of these.
Which mistakes cost marks?
Treating momentum as a scalar in explosions. The pieces have equal and opposite momenta; the total is zero, not 2p.
Doubling K when p doubles. K goes up four times.
Using eV directly in SI formulas. Convert to joules first.
Assuming kinetic energy is conserved in collisions. Only momentum is conserved in every collision; see our post on collisions and momentum.
If you keep reaching for v when you do not need it, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer, so you can compare your route with a shorter one.
A practice routine
Memorise both forms: K = p²/2m and p = √(2mK).
Each day for a week, solve one problem of each type above, without finding v unless asked.
Before each problem, write "same p", "same K" or "neither". This one line decides the method.
Mix in one collision problem and one photoelectric or matter-wave problem, so you practise choosing.
Now try this: a 0.5 kg ball and a 2 kg ball have the same momentum. Find the ratio of their kinetic energies (answer: 4 : 1).
