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An explosion problem looks messy: pieces fly off in different directions at different speeds. It is one of the cleanest problems in mechanics, because one vector equation decides almost everything. This post explains why momentum survives an explosion, works three problems with numbers, and lists the slips that cost marks.
Where is this in NCERT and the syllabus?
The 2026-27 reprint of the NCERT Class 11 chapter Systems of Particles and Rotational Motion uses this exact case in Section 6.3. A projectile explodes into fragments midway in the air. The forces leading to the explosion are internal, so they contribute nothing to the motion of the centre of mass, which "continues along the same parabolic trajectory as it would have followed if there were no explosion" (Fig. 6.12). Section 6.4 then states the conservation of total linear momentum when the net external force is zero.
The Syllabus for JEE (Main) 2026 lists the "law of conservation of linear momentum and its applications" in Unit 3 and the centre of mass in Unit 5. The NEET (UG) 2026 syllabus lists the same conservation law, and the JEE (Advanced) 2026 brochure lists "systems of particles; centre of mass and its motion". Check your year's documents.
Why is momentum conserved when gravity is acting?
Strictly, gravity is an external force, so total momentum changes during the flight. But an explosion is short. Compare the two effects for a 3.0 kg shell moving at 20 m/s, with a burst lasting 1 ms (a sample duration):
Momentum of the shell: 3.0 × 20 = 60 kg m/s.
Impulse of gravity during the burst: mgΔt = 3.0 × 10 × 0.001 = 0.03 N s.
That is 0.05% of the momentum. So momentum just before equals momentum just after, to a very good approximation. After the burst, gravity keeps acting on every fragment as usual.
Worked example 1: a bomb at rest bursts into three pieces
A 6.0 kg bomb at rest explodes into three fragments. A 1.0 kg piece flies east at 12 m/s and a 2.0 kg piece flies north at 8.0 m/s. Find the velocity of the 3.0 kg piece and the energy released.
Momentum before is zero, so the three momenta must add to zero.
Take east as x and north as y. p₁ = (12, 0) and p₂ = (0, 16) kg m/s.
p₃ = −(p₁ + p₂) = (−12, −16) kg m/s. Its magnitude is √(144 + 256) = 20 kg m/s.
v₃ = 20 / 3.0 = 6.7 m/s, pointing south of west at tan⁻¹(16/12) = 53° below the west direction.
Energy released = total kinetic energy after = ½(1)(12²) + ½(2)(8²) + ½(3)(6.67²) = 72 + 64 + 66.7 ≈ 203 J.
Worked example 2: a shell bursts at the top of its flight
A 3.0 kg shell is fired at 40 m/s at 60° above level ground (take g = 10 m/s²). At the top of its path it bursts into a 1.0 kg piece that retraces its path back to the gun and a 2.0 kg piece. Where does the heavier piece land?
At the top: u_x = 40 cos 60° = 20 m/s and the vertical velocity is zero.
Time to the top: t = 40 sin 60° / 10 = 3.46 s. Height H = 34.6² / 20 = 60 m. Horizontal position 20 × 3.46 = 69.3 m.
Retracing means the 1.0 kg piece moves at −20 m/s. Momentum: 3.0 × 20 = 1.0 × (−20) + 2.0 × v, so v = 40 m/s.
Both pieces start with zero vertical velocity at 60 m, so both take 3.46 s to land.
The 2.0 kg piece lands at 69.3 + 40 × 3.46 = 208 m from the gun.
Check with NCERT's centre-of-mass idea. Without the burst, range = 40² sin 120° / 10 = 139 m. The centre of mass of the two pieces when both land: (1.0 × 0 + 2.0 × 208) / 3.0 = 139 m. It agrees, and it works only because both pieces land at the same moment.
Energy: kinetic energy just before is ½ × 3.0 × 20² = 600 J; just after it is ½ × 1.0 × 20² + ½ × 2.0 × 40² = 200 + 1600 = 1800 J. The explosion released 1200 J.
If instead one piece had been given a downward velocity, it would land first, the ground would then push on it, and the centre of mass would leave the parabola. Find each landing point separately in that case. Our post on centre of mass problems has another case where the shortcut works.
Common mistakes and how to avoid them
Adding speeds instead of momenta. Momentum is a vector. Write x and y components.
Using energy conservation to find velocities. Kinetic energy increases. Use momentum, and use energy only to find what was released.
Forgetting the zero vertical velocity at the top. A fragment that "retraces its path" has the same speed, opposite horizontal direction, and no vertical velocity at the burst.
Explosions are the reverse of a perfectly inelastic collision: one equation, with kinetic energy changing. Our guide to collisions and the coefficient of restitution covers the forward case. If you want a plan for working through momentum problems in order, Fermi's AI tutor can work out what you already know and build a plan for the session.
A practice routine
Solve two three-fragment problems with the bomb at rest, drawing the momentum triangle each time.
Solve two "burst at the top" problems where both pieces land together, and check with the centre of mass.
Solve one problem where the pieces land at different times, and explain why the shortcut fails.
Find the energy released in every problem.
Try it now: a 4.0 kg shell moving at 10 m/s splits into two equal halves, one of which stops dead. Find the other half's speed and the energy released.
