Total Momentum Before Equals Total Momentum After
When no external impulse acts, the total momentum of a system is the same before and after an explosion, a recoil or a passage of a projectile. Write momentum as a vector sum and solve for the unknown piece.
Momentum Conservation, Recoil and Explosions is the central idea in this question pattern. Identify the condition in the stem before applying the method below.
Explosions from rest
A bomb at rest breaks into parts whose momenta add to zero. A 1 kg bomb splits in the ratio 1 : 1 : 2, so the masses are 0.25, 0.25 and 0.5 kg. If the two equal parts go off perpendicularly at 30 m/s, each has momentum 7.5 kg m/s, their vector sum is 7.5 sqrt(2), and the large part must carry the same momentum the other way: v = 7.5 sqrt(2) / 0.5 = 15 sqrt(2) m/s. A heavy nucleus at rest splitting into fragments with speeds in the ratio 8 : 1 has masses in the ratio 1 : 8, so, because mass is proportional to radius cubed, the radii are in the ratio 1 : 2.
Recoil and the ballistic pendulum
If the recoil is stopped by a spring, (1/2) M V² = (1/2) k d², so V = d sqrt(k/M) and v = (M/m) V = (d/m) sqrt(k M).
For a 10 g bullet at 500 m/s going through a 2 kg block that then rises 10 cm, v1 = sqrt(2 g h) = sqrt(2) m/s and the bullet leaves with v2 = (5 - 2.83)/0.01, about 220 m/s.
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- ✓No external impulse: Σ p before = Σ p after, as vectors
- ✓Explosion from rest: Σ p = 0; masses and speeds are inversely proportional
- ✓Recoil: m v = M V; a spring stopping the gun turns (1/2) M V² into (1/2) k d²
- ✓A block swinging up after a collision: use energy conservation for the swing, momentum for the collision
- ✓Mass is proportional to r³ for fragments of equal density
Do not conserve kinetic energy across an explosion, a recoil or a bullet embedding in a block. Only momentum is conserved in those events; energy is conserved only in the swing or compression that follows.
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