DPST1021 Chap.6 Centre of Mass, Momentum and Collisions
Centre of Mass, Momentum and Collisions
Lectures 11 to 13 of Physics 1A move from single bodies to systems. Weeks 4 and 5 of the term introduce the centre of mass of particle collections and extended objects, then build linear momentum, impulse and collisions on top of it.
Everything here belongs to the mechanics block assessed in Test 1, and the ideas return in Lab 3, where gliders and model cars are crashed to compare momentum and kinetic energy before and after impact.
The chapter starts with locating a centre of mass, either by a weighted sum over point masses or by an integral over a continuous body, and shows how a composite shape can be split into simple symmetric pieces.
Newton's second law then applies to the whole system as if all its mass sat at that one point, with only external forces mattering. Writing the law as force equals rate of change of momentum leads straight to the conservation condition, to impulse as the area under a force-time graph, and to a short treatment of rocket propulsion.
Collisions take up the second half.
You decide first whether momentum is conserved during the contact, then whether the bodies stick, and only then whether kinetic energy is conserved. Two-dimensional problems need one momentum equation per axis. A decision diagram, a four-move procedure and solved cases on air-table pucks and a road crash give you a dependable route through collision questions.
What this chapter covers
- 01
Composite-shape centres of mass: split an object into symmetric pieces, place each piece's mass at its own centre, and form the mass-weighted average position.
- 02
System momentum: add the momenta of all parts, equate the total to M times the centre-of-mass velocity, and let only external forces change it.
- 03
Conservation conditions: choose a system boundary that makes contact forces internal, and justify that outside impulse is negligible during a brief collision.
- 04
Impulse and average force: link the change in momentum to the area under a force-time curve and explain why longer contact times reduce peak forces.
- 05
Collision classification: compare kinetic energy before and after to label a collision elastic, inelastic, perfectly inelastic or explosive.
- 06
Two-dimensional collisions: resolve every velocity into components and solve one momentum equation for each axis before recombining magnitude and direction.
Worked example · free
A stationary canister splitting in two
- 1System = both pieces. The ice is frictionless and the burst forces are internal, so total momentum stays at its initial value of zero.
- 1Along the line of motion: 1.5(10) + 2.5v = 0, so v = −6.0 m/s, opposite to the lighter piece.
- 1Kinetic energy afterwards: ½(1.5)(10)² + ½(2.5)(6.0)² = 75 + 45 = 120 J. All of it came from stored energy.
- 1Because P = 0 before and after, vcm = P/M = 0, so the centre of mass stays where the canister was.
Key terms
- Centre of mass
- The mass-weighted average position of a system, moving as if all its mass sat there and every external force acted on it.
- Linear momentum
- The product of mass and velocity, a vector quantity measured in kg m/s.
- Impulse
- The time integral of a force over an interval, equal to the change in momentum it produces and measured in N s.
- Isolated system
- A chosen group of bodies on which the net external force is zero, so that its total momentum cannot change.
- Elastic collision
- A collision that conserves total kinetic energy as well as total momentum.
- Perfectly inelastic collision
- An impact after which the bodies move off joined, sharing one velocity and losing the most kinetic energy that momentum allows.
- Thrust
- The forward force on a rocket from ejecting exhaust backward, equal to exhaust speed times the mass ejection rate.
Centre of Mass, Momentum and Collisions FAQ
Can momentum be conserved while kinetic energy is not?
Yes, and in everyday collisions it is the usual case. Momentum depends only on external forces, which are negligible during a short impact. Kinetic energy has no such protection, since deformation, sound and heating absorb part of it. A car crash conserves momentum almost exactly yet turns much of its kinetic energy into bent panels.
Does the centre of mass have to be inside the object?
No. It is a calculated average position, not a physical part. Rings, horseshoes and bent wires often have it in empty space. High jumpers exploit this: by arching over the bar, their centre of mass can pass beneath the bar while every part of the body clears it.
How do I know whether to include friction in a collision problem?
Compare the contact time with the time scale of the friction. A crash or a bounce is over in a few thousandths of a second, far too short for road friction to rival the enormous contact forces, so momentum just before and just after can be equated. Friction matters afterwards, when you track how far the wreckage slides.
Why do I need one momentum equation per axis in two dimensions?
Conservation holds separately along each independent direction because momentum is a vector. Adding speeds or magnitudes directly ignores the angles and gives wrong results. An x equation and a y equation give two scalar statements, enough for two unknowns such as a final speed and its angle.
Exam move
Begin with centre-of-mass calculations on simple composite shapes until you can set up the weighted sum without hesitation, then try one continuous body by integration to see the same idea with dm. Next, rehearse the argument for conservation out loud: name the system, list the external forces, and say why their impulse is small during contact.
For collisions, follow the decision order every time, momentum first, sticking second, energy last. Draw before and after sketches with axes for every two-dimensional problem and keep components signed. Link the homework set on momentum and collisions to what you observe in the collisions laboratory, where measured speeds let you check both conservation statements.
Finally, watch significant figures: online quiz answers are judged on the number common to all given values.
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