OlympIQ
Read before you practise · 6 min

Internal Forces Never Move the Centre of Mass

The centre of mass of a system moves as if all the mass were concentrated at it and all the external force acted on it. Internal forces, whether attraction, spring or string tension, cannot change its motion.

NSEJS · Collisions and Centre of Mass · 4 linked questions · 100% coverage
Know this first

Motion and force questions become simpler when direction, starting condition, time and the acting forces are marked first.

Distance · Length of the path travelledDisplacement · Straight-line change in positionResultant force · Net force including direction
Study deeper: NCERT Science · Collisions and Centre of Mass

Zero external force means a fixed centre of mass

If the external force is zero,
a_cm = 0, so a centre of mass that starts at rest stays at rest

Two particles initially at rest that attract each other have speeds v and 2v at some instant, but their centre of mass speed is zero, since their momenta are equal and opposite. Two spheres of masses M and 5M with radii R and 2R, released in free space with centres 12R apart, meet when the centres are 3R apart, so together they have covered 9R. Because the centre of mass does not move, the distances covered are inversely proportional to the masses, so the lighter sphere covers 5/6 of 9R, which is 7.5R.

Acceleration of the centre of mass

a_cm = Σ m a / Σ m

For masses 15 kg and 5 kg over a pulley, each has acceleration of magnitude g/2 = 5 m/s², one down and one up, so a_cm = (15(-5) + 5(5))/20 = -2.5 m/s²: the centre of mass accelerates downward at 2.5 m/s² even though the system is internally tied together.

For m and 3m,
a = g/2 and a_cm = [(3m)(g/2) - m(g/2)] / 4m = g/4 downward

OlympIQ exam lens

  • ✓a_cm = F_external / M; internal forces cancel in pairs
  • ✓Zero external force and initial rest: the centre of mass stays at rest
  • ✓Mutually attracting bodies: distances moved are inversely proportional to the masses
  • ✓a_cm = Σ m a / Σ m with signs for direction
OlympIQ trap

Do not set the acceleration of the centre of mass equal to the acceleration of each body. When one body accelerates down and the other up, the centre of mass has a smaller acceleration given by the mass-weighted average with signs.

Ready to try the pattern?

100% coverage across the 4 linked questions.

Practise 4 linked questions
Sign in to practise — your XP, streak and analysis update as you solve.