Laws of Motion · Lesson 5 of 14

Conservation of Linear Momentum

“Derive momentum conservation from Newton’s laws and apply it to recoil, separation and collisions.”

Learning Objectives

• Derive the momentum balance for a system of particles. • Identify when external impulse can be neglected. • Solve recoil and two-body separation with signed velocities. • Distinguish momentum conservation from kinetic-energy conservation.

Add the equations for all the parts

Consider two bodies A and B interacting while external forces are absent. On A, the second law gives dp_A/dt = F_A←B. On B it gives dp_B/dt = F_B←A. The third law makes these forces equal and opposite. Adding the equations therefore gives d(p_A + p_B)/dt = 0: the total momentum remains constant, even while each body’s momentum changes.

System momentum balanceLaTeX

P = i \sum p_{i}, \frac{d P}{d t} = \sum F_{external}

Internal interactions redistribute momentum among the bodies but cancel in the system sum.

Conservation conditionLaTeX

J_{external} = 0 ⟹ P_{f} = P_{i}

Zero net external force throughout an interval guarantees conservation throughout. Zero total external impulse guarantees equal endpoint momenta.

For a short collision, the external impulse from gravity or a support may be small compared with the collision impulses. Momentum can then be approximately conserved for the two-body system during the collision. State the interval and direction: horizontal momentum may be conserved even while an external vertical force acts. Momentum conservation is not a claim that every object’s momentum stays fixed.

Recoil: two equal opposite momentum changes

A shell and a freely recoiling gun

Problem
A 0.020 kg shell is fired from a 100 kg gun initially at rest. Take the shell’s launch speed relative to the initial ground frame as 80 m/s. Find recoil speed, neglecting external horizontal impulse.

1.Let the shell’s direction be positive. Initial total momentum is zero.
2.After firing, 0.020 × 80 + 100V = 0, so V = −0.016 m/s.
3.The recoil speed is 0.016 m/s = 1.6 cm/s, opposite to the shell. The heavy gun has a much smaller speed but an equal momentum magnitude.
4.If 80 m/s is instead defined relative to the moving gun, write v_shell − V = 80 along with conservation. Then V = −(0.020 × 80)/(100.020) = −0.0159968 m/s, which rounds to the same quoted value.

Two fragments from an initially stationary object

Two-body separation from restLaTeX

m_{1} v_{1} + m_{2} v_{2} = 0 ⟹ v_{2} = - \frac{m _{1}}{m _{2}} v_{1}

With negligible external impulse, the fragment momentum vectors are equal and opposite. The velocity magnitudes need not be equal.

A nucleus at rest that separates into exactly two fragments must send them in opposite directions under this model. If there were three fragments, the three momentum vectors could close as a triangle rather than all lying on a single line. An initially moving parent also changes the statement: conservation gives the original nonzero total momentum, not necessarily opposite laboratory velocities.

Unequal fragments

Problem
An initially stationary body splits into masses 2 kg and 3 kg. The 2 kg fragment moves at +6 m/s. Find the other velocity.

1.Initial momentum is zero. Write 2 × 6 + 3v = 0.
2.The other fragment has v = −4 m/s.
3.The momenta are +12 and −12 kg m/s. Their speeds differ because the masses differ.

Collisions can conserve momentum without conserving kinetic energy

Momentum is a vector sum and is conserved for an isolated system in both elastic and inelastic collisions. Kinetic energy is a different quantity. It is also conserved in an elastic collision; in an inelastic collision, some initial kinetic energy becomes internal energy, deformation, sound or other forms. Total energy remains conserved, but the kinetic part need not. Momentum conservation alone is usually not enough to determine two unknown final velocities without another condition.

Objects that stick together

Problem
A 2 kg trolley moving at +3 m/s catches a stationary 1 kg trolley and they stick. Find their common velocity.

1.With negligible external horizontal impulse, initial momentum is 2 × 3 + 1 × 0 = 6 kg m/s.
2.The final mass is 3 kg, so 3v = 6 and v = +2 m/s.
3.Initial kinetic energy is ½ × 2 × 3² = 9 J; final kinetic energy is ½ × 3 × 2² = 6 J. Momentum is conserved while 3 J leaves the macroscopic kinetic-energy account.

Choose the system wide enough

A ball rebounding from a wall does not conserve its own momentum: the wall exerts an external impulse on it. Ball plus wall plus Earth is a broader system in which the corresponding internal impulses cancel, with the massive Earth taking an extremely small velocity change. Whether a momentum argument is valid depends on the selected system boundary and external impulse, not on the word “collision” alone.

Conservation and propulsion

A rocket accelerates by ejecting exhaust backward. For a system containing rocket and exhaust, the interaction forces redistribute momentum. For the rocket alone, exhaust crosses the system boundary, so applying F = d(mv)/dt without a momentum-flow term is unsafe. At an instant when thrust is supplied as a force, the rocket’s translational acceleration follows from net force, thrust minus weight and other forces. A full variable-mass rocket derivation belongs to a more detailed momentum balance.

Quick check

Which statement is correct for an isolated inelastic collision?

Key Takeaways

• Internal force pairs cancel in total momentum balance. • Check external impulse over the interval and in the direction of interest. • Recoil and separation conserve total momentum, not each individual momentum. • Elastic and inelastic collisions both conserve momentum when isolated. • A system with mass flowing across its boundary needs careful momentum accounting.

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Newton’s Third Law and System Boundaries

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Forces, Free-Body Diagrams and Equilibrium