Motion of the centre of mass and the momentum of a system · Physics Class 11–12 | StemPy Lab
Course 6.1·Lesson 2 of 3
Motion of the centre of mass and the momentum of a system
The centre of mass of a system moves as if all the mass were there and all the external forces acted on it: Macm=Fext
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Try the code and the simulations beside the lesson.
The motion of the centre of mass
Add up Newton's second law for every particle of a system. The internal forces come in third-law pairs and cancel, so only the external forces are left:
Macm=Fext
The centre of mass moves like a single particle of mass M pushed by the external forces alone.
Momentum of a system
The total momentum is the sum of the parts' momenta, and it equals
P=Mvcm
With no external force, P is conserved, and the centre of mass moves at a steady velocity (or stays at rest), whatever happens inside.
A shell that bursts
A shell bursts at the top of its flight. The burst is internal, and gravity is unchanged, so the centre of mass of the pieces carries on along the same parabola.
The path of a shell in a vertical plane. The shell goes up along a solid blue curve to its highest point, where it bursts. A dashed blue curve shows where the centre of mass would go, landing at the red cross. Part A falls straight down from the top and lands at half the range. Part B moves on along a flatter green curve and lands at one and a half times the range. The centre of mass lands exactly between the two parts, at the range.
Try different bursts.
You should find that the other piece lands farther away the bigger the share that drops, but the centre of mass always lands at the same place, the shell's own range.
Watch out. This holds only while all the pieces are in the air. Once one piece lands, the ground pushes on it, and that is an external force.
A worked example
A 60kg diver stands on a 240kg boat at rest, and dives off the back at 2.5m/s.
Momentum before. Zero: everything is at rest.
After.60kg×2.5m/s+240kgv=0, so the boat moves forward at v=0.625m/s.
The centre of mass. The momentum is 60kg×2.5m/s+240kg×(−0.625m/s)=0, so vcm=0: it stays still, as it must.
Think first. Ask whether any external force acts along the direction you care about. If none does, the centre of mass does not move along it.
Practise with new numbers
Try it in Python
The program finds where the second piece lands for bursts of different shares.
Remember these
Macm=Fext: internal forces cannot move the centre of mass.
P=Mvcm; with no external force, vcm stays the same.
A burst shell's centre of mass keeps to the old path while all the pieces fly.
Check yourself
Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.
A 3.0kg cart moves east at 4.0m/s and a 1.0kg cart is at rest. What is the velocity of their centre of mass?
Can you move your own centre of mass by pulling on your belt?
A cat and a dog sit at the two ends of a trolley on smooth ice, and walk toward each other. What does the trolley do?
Why does a gun recoil, but the centre of mass of gun and bullet does not move when it fires?
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