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Collision questions look like they need a new formula for every case: elastic, perfectly inelastic, a ball bouncing off the floor, a bullet in a block. They do not. Two equations, one for momentum and one for how bouncy the impact is, solve nearly all of them. This post shows you which equation to trust in each case, works through the same collision three ways, and lists the mistakes that cost marks.
What is conserved in a collision, and what is not?
The NCERT Class 11 chapter Work, Energy and Power puts it plainly: in all collisions the total linear momentum is conserved. The reason is Newton's third law. During the brief impact, the two bodies push on each other with equal and opposite forces, so whatever momentum one gains, the other loses.
Kinetic energy is different. NCERT calls a collision elastic if the total kinetic energy after the collision equals the total before it, and completely inelastic if the bodies move together afterwards. Most real collisions are in between, and some kinetic energy turns into heat, sound and deformation.
Both ideas are in the syllabus. The JEE (Main) 2026 syllabus lists "elastic and inelastic collisions in one and two dimensions", and the JEE (Advanced) 2026 brochure lists impulse, elastic and inelastic collisions, and collisions of point masses with rigid bodies. Check the syllabus for your own exam year.
Why do students find these problems hard?
It is rarely the algebra. Singh and Rosengrant (2003) built a 25-question test of energy and momentum concepts and interviewed students who took it. They found that most introductory students had difficulty interpreting the basic principles qualitatively and applying them to physical situations. In practice that shows up as using energy conservation in a collision that loses energy, or forgetting that momentum is a vector with a sign.
What does the coefficient of restitution mean?
For a head-on collision, the coefficient of restitution is
e = (speed of separation) / (speed of approach) = (v₂ − v₁) / (u₁ − u₂)
where u are velocities before and v after, all measured along the line joining the centres. For an elastic collision e = 1, for a perfectly inelastic one e = 0, and real collisions fall in between. NCERT does not name e, but it turns the "in between" case into something you can calculate.
Worked example: one collision, three ways
A 2 kg ball moving at 3 m/s hits a 1 kg ball at rest, head-on. Take the direction of motion as positive.
Momentum before: 2 × 3 = 6 kg m/s. Kinetic energy before: ½ × 2 × 3² = 9 J.
Case 1: elastic (e = 1). Momentum: 2v₁ + v₂ = 6. Restitution: v₂ − v₁ = 3. Adding the second equation to the first gives 3v₁ = 3, so v₁ = 1 m/s and v₂ = 4 m/s. Check the energy: ½ × 2 × 1² + ½ × 1 × 4² = 1 + 8 = 9 J, unchanged.
Case 2: perfectly inelastic (e = 0). The balls move together: 3v = 6, so v = 2 m/s. Kinetic energy after: ½ × 3 × 2² = 6 J. One-third of the energy, 3 J, is lost.
Case 3: partly elastic (e = 0.5). Momentum: 2v₁ + v₂ = 6. Restitution: v₂ − v₁ = 0.5 × 3 = 1.5. Substituting v₂ = v₁ + 1.5 gives 3v₁ = 4.5, so v₁ = 1.5 m/s and v₂ = 3.0 m/s. Kinetic energy after: ½ × 2 × 1.5² + ½ × 1 × 3² = 2.25 + 4.5 = 6.75 J, lost 2.25 J.
Look at the pattern. The energy left, 9 J, 6.75 J and 6 J, falls as e falls, and the momentum is 6 kg m/s every time. That is a good check on any answer.
A bounce off the floor. The floor does not move, so e = (rebound speed) / (impact speed). A ball dropped from 1.25 m with e = 0.8 hits at √(2 × 9.8 × 1.25) ≈ 4.95 m/s and leaves at about 3.96 m/s. Since height goes as speed squared, it rises to e² × 1.25 = 0.64 × 1.25 = 0.80 m.
Which mistakes cost the most marks?
Conserving kinetic energy by default. Unless the question says elastic, or gives e = 1, kinetic energy is not conserved. Our work-energy theorem guide covers where energy methods do apply.
Dropping signs. A ball that rebounds has a negative velocity. Pick a positive direction, write it on your diagram and keep every velocity as a signed number.
Using e with speeds along the wrong line. In an oblique collision, apply restitution only to the velocity components along the line joining the centres. Components at right angles to that line do not change for smooth spheres.
Ignoring an outside impulse. Momentum is conserved only if no large external force acts during the impact. A ball hitting a block fixed to a wall, or a pendulum bob whose string jerks taut, needs more care.
Forgetting the centre of mass frame. In the centre of mass frame the total momentum is zero, which makes many two-body problems quicker. Our post on centre of mass problems shows how.
If you know these rules but still end up with the wrong final velocity, compare your working line by line with a full solution. The Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you can see where a sign went wrong.
A practice routine for this chapter
Day 1: Five one-dimensional collisions. For each, write the momentum equation and the restitution equation before solving.
Day 2: Three bounce problems, including total distance travelled by a ball bouncing many times (a geometric series in e²).
Day 3: Three oblique collisions of equal spheres, resolving along and across the line of centres.
Day 4: A mixed set with bullets, blocks and springs, where the collision and what happens after it are separate stages. More traps from this part of the syllabus are in our post on common mistakes in JEE mechanics.
Tonight, solve the worked example above with e = 0.25 and check that the kinetic energy lands between 6 J and 6.75 J.
