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Physics 1

Momentum, centre of mass and collisions

Impulse and momentum, the centre of mass and its motion, conservation of momentum, elastic and inelastic collisions in one dimension, the ballistic pendulum and the rocket equation.

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When two bodies collide, the forces involved are huge and last a few milliseconds: there is no way of knowing them instant by instant. Yet the outcome of a collision is easy to calculate, thanks to a quantity that never changes in an isolated system: momentum. In this lesson we define it, introduce the centre of mass and solve collisions in one dimension.

Momentum and impulse

The momentum of a body is its mass times its velocity:

    \[ \vec p = m\,\vec v \]

Newton himself wrote the second law in this form, which is more general than \vec F = m\vec a because it also holds when the mass changes:

    \[ \vec F = \frac{d\vec p}{dt} \]

Integrating over time gives the impulse-momentum theorem: the impulse of a force, its integral over time, equals the change in momentum.

    \[ \vec J = \int_{t_1}^{t_2}\vec F\,dt = \Delta\vec p \]

This is why airbags and helmets work. The change in momentum in a crash is fixed (from mv to zero); stretching the stopping time from 5 to 50 milliseconds makes the average force F = \Delta p/\Delta t ten times smaller.

Systems of particles and the centre of mass

For a system of N bodies the centre of mass is defined as

    \[ \vec r_{cm} = \frac{\sum_i m_i\,\vec r_i}{\sum_i m_i} = \frac{1}{M}\sum_i m_i\,\vec r_i \]

It is an average of the positions weighted by mass. Differentiating, the total momentum of the system is \vec P = \sum m_i\vec v_i = M\,\vec v_{cm}.

Proof: the motion of the centre of mass

Each body feels forces from outside the system and internal forces from the other bodies of the system. Adding the law d\vec p_i/dt = \vec F_i over all bodies:

    \[ \frac{d\vec P}{dt} = \sum_i\vec F_i^{\,ext} + \sum_i\sum_{j\ne i}\vec F_{ij} \]

where \vec F_{ij} is the force of body j on body i. By the third law \vec F_{ij} = -\vec F_{ji}: in the double sum the internal forces cancel in pairs. What is left is

    \[ \frac{d\vec P}{dt} = M\,\vec a_{cm} = \vec F^{\,ext} \]

The centre of mass moves like a single point carrying the whole mass of the system, acted on only by the external forces. A firework exploding in flight: the fragments fly in every direction, but their centre of mass carries on along the same parabola.

Conservation of momentum

If the net external force is zero (an isolated system), then d\vec P/dt = 0:

    \[ \vec P = \sum_i m_i\,\vec v_i = \text{constant} \]

The principle holds even when the internal forces are huge and unknown, and it is more general than conservation of mechanical energy, which often fails in collisions. During a short collision it can be used even with external forces such as weight, because their impulse over a few milliseconds is negligible compared with that of the impact forces.

Collisions in one dimension

In every collision between isolated bodies momentum is conserved. We distinguish:

  • elastic collision: kinetic energy is conserved too (steel balls, gas molecules);
  • inelastic collision: part of the kinetic energy becomes deformation and heat;
  • perfectly inelastic collision: the bodies stick together and move on as one.
Elastic collision in one dimension: mass m1 hits mass m2 at rest
Before the collision (t_1) mass m_1 moves at v_1 towards m_2 at rest; after it (t_2), if m_1 \lt m_2, m_1 bounces back and m_2 moves forward.

Proof: velocities after an elastic collision

Let m_2 be initially at rest. Conservation of momentum and of kinetic energy:

    \[ m_1 v_1 = m_1 v_1' + m_2 v_2' \]

    \[ \tfrac12 m_1 v_1^2 = \tfrac12 m_1 v_1'^2 + \tfrac12 m_2 v_2'^2 \]

Rewrite both equations grouping the m_1 terms:

    \[ m_1(v_1 - v_1') = m_2 v_2',\qquad m_1(v_1 - v_1')(v_1 + v_1') = m_2 v_2'^2 \]

Dividing the second by the first (with v_1' \ne v_1, otherwise there is no collision) gives v_1 + v_1' = v_2': in an elastic collision the relative velocity is reversed. Substituting into the first equation:

    \[ v_1' = \frac{m_1 - m_2}{m_1 + m_2}\,v_1,\qquad v_2' = \frac{2m_1}{m_1 + m_2}\,v_1 \]

Three cases worth remembering:

  • equal masses: v_1' = 0, v_2' = v_1. The velocities are swapped, as in Newton’s cradle or a head-on shot in billiards;
  • m_1 \ll m_2 (a ball against a wall): v_1' \approx -v_1, the ball bounces back at the same speed;
  • m_1 \gg m_2 (a truck hitting a ball): v_2' \approx 2v_1, the ball flies off at twice the truck’s speed.

Perfectly inelastic collision

The two bodies move on with the same velocity V. From momentum alone, with m_2 at rest:

    \[ V = \frac{m_1}{m_1 + m_2}\,v_1 \]

The kinetic energy lost is the fraction m_2/(m_1 + m_2) of the initial one: it is largest when the target is much heavier than the projectile.

Example: the ballistic pendulum. A 10 g bullet lodges in a 2 kg wooden block hanging from a string, which rises by 5 cm. After the collision energy is conserved: V = \sqrt{2gh} = \sqrt{2\cdot 9.81\cdot 0.05} \approx 0.99 m/s. During the collision momentum is conserved: v_1 = V\,(m_1 + m_2)/m_1 = 0.99\cdot 2.01/0.01 \approx 199 m/s. Be careful not to use energy conservation across the collision: more than 99% of the bullet’s energy becomes heat and deformation.

Variable-mass systems: the rocket

A rocket ejects gas at speed u relative to itself. In a time dt its mass changes by dm \lt 0 and, by conservation of momentum in empty space, m\,dv = -u\,dm. Integrating from the initial mass m_0 to the final mass m_f gives Tsiolkovsky’s equation:

    \[ \Delta v = u\,\ln\frac{m_0}{m_f} \]

The logarithm explains why rockets are almost all fuel: for a velocity change of 9 km/s, what it takes to reach low orbit, with u = 3 km/s you need m_0/m_f = e^3 \approx 20. Payload and structure are only 5% of the lift-off mass, which is why multi-stage rockets are used.

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