Press Next (or Play) to walk through each story one small step at a time. The numbers come last.
1. Two friends on skates pull a rope
Read the story as text
Two friends stand still on smooth ice, 10 meters apart. One is big (60 kg), one is smaller (40 kg).
Each holds an end of a rope. They start pulling the rope hand over hand, so they slide toward each other.
The ice is so slick that it pushes on them only straight up, never sideways. Nobody else touches them.
They will meet somewhere on the ice. But where? In the middle? Closer to the big skater? Closer to the small one?
The question this topic answers: when the parts of a group push and pull on each other, what
happens to the group as a whole?
2. Where is the balance point?
3. Inside push or outside push?
Quick check. Jo and her little brother Kai stand still on smooth ice and push off each other. They glide apart. What happens to their balance point (center of mass)?
Another example: Walking on a rolling cart
Another example: Balancing on a board
4. Who is in the system?
5. The firework's balance point
6. Pulling the sled
Quick check. Two kids on a sled (50 kg in all) on smooth ice shove each other with 100 N. A friend pulls the sled forward with 25 N. How fast does the center of mass speed up?
Another example: Theo pushes from inside the cart
7. Check yourself
Think of your answer first, then tap to see it.
a) Two skaters stand still on smooth ice and push off each other. Does the center of mass of the pair move?
Show answer
No. The pushes are inside the system, so they cancel. The pair started at rest, so its center of mass stays put.
b) A ball falls toward Earth. If the system is just the ball, is gravity an inside or an outside force?
Show answer
Outside. Earth is not part of the system. If the system were ball + Earth, gravity would be an inside force.
c) A 3 kg ball sits at x = 0 and a 1 kg ball sits at x = 4 m. Where is their center of mass?
Show answer
xcm = (3 kg × 0 + 1 kg × 4 m) ÷ 4 kg = 1 m, closer to the heavier ball.
Already know this?
Three quick questions. Get all three right on the first try and you can skip ahead to the simulation.
1. Two skaters, 70 kg and 35 kg, stand still on smooth ice and pull a rope between them until they meet. Where do they meet?
2. You sit in a wagon and push forward on its front edge. The system is you and the wagon. Why does the wagon not speed up?
3. A 2 kg ball sits at x = 0 and a 6 kg ball sits at x = 4 m. Where is their center of mass?
The idea in plain words
A system is whatever you choose to study. You draw an imaginary line around it. It can be one
skater, the other skater, or both skaters together. Picking the system is always the first step.
Internal forces act between two things that are both inside your system: the rope pulling on
skater A and the rope pulling on skater B, if the system is "both skaters and the rope".
External forces come from something outside the system: Earth's gravity, the ice pushing up,
a friend giving a push.
The center of mass is the average position of all the mass, where heavier parts count more.
For two objects on a line:
xcm = (m1x1 + m2x2) / (m1 + m2)
Trap: putting the center of mass halfway between the objects. Instead: weight each position by its mass; with different masses the answer sits closer to the heavier object.
For a uniform, symmetric object (a ball, a meter stick, a brick) the center of mass is at its geometric center.
The center of mass does not have to be inside the material: a ring's center of mass is in the empty middle.
The big rule: only external forces can change how the center of mass moves.
acm = Fnet, external / Mtotal
Trap: thinking pushes between parts of the system (skaters pulling a rope, you pushing the dashboard) can move the system. Instead: internal forces come in equal, opposite pairs and cancel; only an outside force can change acm.
Internal forces always come in equal and opposite pairs (you will see why in 2.3), so inside the system they
cancel. The skaters can pull as hard as they like: if nothing outside pushes them sideways, their center of mass
stays exactly where it was. That is also why you can treat a whole car, or a whole person, as a single dot at its
center of mass when you only care about how it moves as a whole.
Worked numbers: where do the skaters meet?
Read the steps as text
Put skater A (60 kg) at x = 0 and skater B (40 kg) at x = 10 m. Positive is to the right. Skater A pulls the
rope with 120 N.
Choose the system: both skaters (and the light rope).
Why: then the rope forces are internal and cancel, which makes the question easy.
Find the center of mass at the start: xcm = (60 kg × 0 m + 40 kg × 10 m) / (60 kg + 40 kg) = 400 / 100 = 4.0 m.
Why: heavier parts count more, so the center is closer to the 60 kg skater.
List the external forces: gravity (down) and the ice (up). They cancel. No sideways external force.
Why: only external forces can move the center of mass.
So acm = 0 / 100 kg = 0. The center of mass started at rest, so it stays at x = 4.0 m.
Why: zero acceleration and zero starting velocity means no motion.
They meet at the center of mass: x = 4.0 m, which is 4 m from the big skater and 6 m from the small one.
Why: when they meet they are at the same spot, and that spot must be the center of mass.
Check with forces on each skater. The rope pulls each with 120 N. aA = 120 N / 60 kg = 2.0 m/s² (right),
aB = 120 N / 40 kg = 3.0 m/s² (left). They close the 10 m gap at 2 + 3 = 5 m/s², so 10 = ½(5)t², t = 2.0 s.
In 2.0 s skater A moves ½(2.0)(2.0)² = 4.0 m. Same answer.
Why: the system view and the one-object-at-a-time view must agree. The system view was much quicker.
Lab: skaters on the ice
Change the masses, the rope pull and the gap. Then try an outside push: a friend pushes skater A
(positive is to the right). Watch the dashed CM marker and the center of mass graphs. The faint marker
shows where the center of mass started.
Use the buttons to pick the system for the free body diagram. Notice that the rope forces appear on one skater's
diagram but disappear from the "both skaters" diagram.
Both skaters: the rope forces are internal, so they are not on this diagram.
Center of mass finder: three masses on a 1 m stick
The stick itself is very light. Move the masses and change their size. The triangle shows the balance point, the
center of mass.
Examples
Example 1: two blocks on a line basic
A 2.0 kg block sits at x = 0 and a 6.0 kg block sits at x = 4.0 m. Where is the center of mass?
Show solutionRead the steps as text
xcm = (2.0 × 0 + 6.0 × 4.0) / (2.0 + 6.0) = 24 / 8.0 = 3.0 m.
Why: weight each position by its mass.
Check: 3.0 m is closer to the 6.0 kg block. Good, the heavier block pulls the center toward itself.
Example 2: walking in a canoe medium
A 70 kg person stands at the back of a 30 kg canoe that floats at rest. The person walks 3.0 m forward
(measured along the canoe). Ignore water drag. How far does the canoe move, and which way?
Show solutionRead the steps as text
System: person + canoe. Sideways external forces: none (we ignore drag).
Why: the push between feet and canoe is internal.
So the center of mass stays put: 70 × dperson + 30 × dcanoe = 0, where d is each one's move relative to the water (forward positive).
Why: the center of mass cannot move, so the weighted moves add to zero.
The person moves 3.0 m relative to the canoe: dperson = dcanoe + 3.0.
The canoe moves 2.1 m backward. The person moves 3.0 − 2.1 = 0.9 m forward relative to the water.
Check: 70 × 0.9 = 63 and 30 × 2.1 = 63. Equal and opposite, so the center did not move.
Example 3: pushing two boxes AP
On a frictionless floor, a 2.0 kg box touches a 4.0 kg box. You push the 2.0 kg box with 30 N to the right, so
both boxes speed up together. (a) Find the acceleration of the center of mass. (b) Find the force the 2.0 kg box
exerts on the 4.0 kg box. (c) Explain why your push is not 30 N on the 4.0 kg box.
Show solutionRead the steps as text
(a) System: both boxes. External sideways force: your 30 N. acm = 30 N / 6.0 kg = 5.0 m/s² right.
Why: the contact force between the boxes is internal, so it does not appear.
(b) Now choose the system "4.0 kg box only". The only sideways force on it is the push from the 2.0 kg box.
F = ma = 4.0 kg × 5.0 m/s² = 20 N right.
Why: changing the system turns the contact force into an external force we can solve for.
(c) You touch only the 2.0 kg box. Your 30 N must speed up all 6.0 kg; the 2.0 kg box keeps 2.0 × 5.0 = 10 N of it
(net force 30 − 20 = 10 N) and passes on 20 N.
Check: 10 N + 20 N = 30 N.
Example 4: a meter stick with a weight AP
A uniform 0.20 kg meter stick has a 0.30 kg clamp fixed at the 90 cm mark. Where is the center of mass of the
stick-plus-clamp system?
Show solutionRead the steps as text
A uniform stick acts like all its mass is at its middle: 0.20 kg at 50 cm.
Why: uniform and symmetric means the center of mass is at the geometric center.
So the stick balances on a finger at the 74 cm mark. Try it with the center of mass finder above (set two masses).
Practice (AP style)
1. Two carts are held together at rest on a level frictionless track with a compressed spring between
them. They are released and fly apart. The system is both carts and the spring. What happens to the velocity of the
system's center of mass?
Show answer
It stays zero. The spring forces are internal. The track and gravity are vertical and cancel. No net external
force means the center of mass keeps its starting velocity, which was zero.
2. A ball falls through the air. The system is the ball and Earth. Which force is external to this system?
Show answer
The air is not part of the system, so its push on the ball is external. Both gravitational forces act between two
objects that are inside the system (ball and Earth), so they are internal.
3. A 3.0 kg object is at x = 0 and a 1.0 kg object is at x = 8.0 m. Where is the center of mass?
Show answer
xcm = (3.0 × 0 + 1.0 × 8.0) / 4.0 = 2.0 m. It is closer to the heavier object, so 4.0 m (the middle) is wrong.
4. An astronaut floats at rest in space, holding a wrench. She throws the wrench away from her.
Which statement is correct about the astronaut-wrench system after the throw?
Show answer
The throw is an internal push. There is no external force, so the center of mass stays at rest. The wrench goes
one way and the astronaut drifts the other way, but the weighted average position does not change. A system still has
a center of mass when its parts are apart.
5. A 24 N horizontal push acts on a 1.0 kg block that is in contact with a 2.0 kg block on a frictionless
floor. What is the acceleration of the center of mass of the two blocks?
Show answer
acm = Fext / M = 24 N / 3.0 kg = 8.0 m/s². The contact forces between the blocks are internal and cancel.
6. A uniform thin ring (like a bracelet) lies flat on a table. Where is its center of mass?
Show answer
By symmetry each bit of the ring is balanced by the bit opposite it, so the center of mass is at the geometric center,
even though no material is there.
7. A person sits in a stopped car and pushes hard on the dashboard. Why does the car not move?
Show answer
The person's hands push the dashboard forward, but the person's back pushes the seat backward. Both are inside the
car-plus-person system. With no new external force, the center of mass cannot start moving.
8. (Short free response: qualitative reasoning) Two skaters at rest on frictionless ice pull on a rope
between them. Skater A has twice the mass of skater B. A student says: "They meet in the middle, because the rope pulls
each of them equally hard." Explain what is right and what is wrong in the student's reasoning, and say where they meet.
Show answer
Right: the rope does pull each skater with the same size force.
Wrong: the same force gives the lighter skater twice the acceleration (a = F/m), so B moves twice as far
as A in the same time. They do not meet in the middle.
Where: no external sideways forces act on the two-skater system, so its center of mass stays fixed.
They meet at the center of mass, which is one third of the way from A to B (closer to the heavier skater A).
Check with d = distance: xcm = (2m × 0 + m × d) / 3m = d/3.
9. (Short free response: experimental design) You are given a flat piece of cardboard cut into an odd
shape, a string, a pin and a small weight. Describe a procedure to find its center of mass, and explain why it works.
Show answer
1. Push the pin through a point near the edge and let the cardboard hang freely from it. 2. Hang the string with the weight
from the same pin so it makes a vertical line, and mark that line on the cardboard. 3. Repeat from a second, different point.
4. The center of mass is where the two lines cross.
Why: a freely hanging object comes to rest with its center of mass directly below the support point, so each vertical
line passes through the center of mass. Two lines meet at one point. A third point can be used to check.
10. (Short free response: mathematical routines) Three small blocks are glued to a very light rod that lies along the x axis on frictionless ice: 2.0 kg at x = 0, 3.0 kg at x = 2.0 m and 5.0 kg at x = 6.0 m. (a) Calculate the position of the center of mass. (b) A friend pushes the 5.0 kg block with a 12 N horizontal force along +x. Calculate the acceleration of the center of mass. (c) The 2.0 kg block is moved along the rod (the others stay put) so that the center of mass is at x = 4.0 m. Calculate the new position of the 2.0 kg block.
(b) System: all three blocks and the rod. The only horizontal external force is the 12 N push. acm = 12 N / 10 kg = 1.2 m/s² along +x. The glue and rod forces are internal.
(c) 4.0 = (2.0x + 6.0 + 30) / 10, so 2.0x + 36 = 40 and x = 2.0 m.
1 point: correct center of mass formula with all three masses (masses times positions over total mass).
1 point: xcm = 3.6 m.
1 point: uses only the external force and the total mass, acm = 1.2 m/s².
1 point: x = 2.0 m with correct algebra.
11. (Short free response: translation between representations) Cart A (1.0 kg) is at x = 0 and cart B (3.0 kg) is at x = 4.0 m on a level frictionless track, both at rest, with a compressed light spring between them held by a thread. The thread is cut and the carts fly apart. (a) Calculate the position of the center of mass of the two carts. (b) Which description matches the graph of xcm against time after the thread is cut? (A) a horizontal line at 3.0 m; (B) a straight line sloping up; (C) a straight line sloping down; (D) a curve that bends upward. Explain. (c) At one instant cart A is at x = −1.5 m. Where is cart B? (d) Describe the graph of vcm against time.
Show answer
(a) xcm = (1.0 × 0 + 3.0 × 4.0) / 4.0 = 3.0 m.
(b) (A). The spring forces are internal. Gravity and the track's normal force are vertical and cancel. With no net external force the center of mass keeps its starting velocity, zero, so xcm stays at 3.0 m.
(c) 3.0 = (1.0 × (−1.5) + 3.0 xB) / 4.0, so 3.0 xB = 12 + 1.5 = 13.5 and xB = 4.5 m. Cart A moved 1.5 m left and the heavier cart B moved only 0.5 m right.
(d) A horizontal line on the time axis: vcm = 0 at every moment.
Point guide (4 points):
1 point: xcm = 3.0 m.
1 point: picks (A) with the reason "internal forces only, so xcm does not change".
1 point: xB = 4.5 m.
1 point: vcm graph is a horizontal line at zero.
12. (Short free response: qualitative/quantitative translation) A person of mass m stands at one end of a raft of mass M and length L that floats at rest. Ignore water drag. The person walks to the other end. (a) Without equations, explain why the raft moves, and which way. (b) Derive an expression for the distance d the raft moves, in terms of m, M and L. (c) Show that your expression agrees with your reasoning in two cases: a very heavy raft, and m = M. (d) Calculate d for m = 60 kg, M = 240 kg, L = 5.0 m.
Show answer
(a) The feet push the raft backward while the raft pushes the person forward (internal pair). No horizontal external force acts on person + raft, so the center of mass stays put: as the person moves forward, the raft must move backward.
(b) Take the person's walking direction as +. Raft moves −d; the person moves L − d relative to the water. Center of mass fixed: m(L − d) − M d = 0, so d = mL / (m + M).
(c) Very heavy raft (M much bigger than m): d is close to 0, as expected, since the raft barely moves. m = M: d = L/2, so the person and raft each move half the length, which makes sense for equal masses.
(d) d = 60 × 5.0 / (60 + 240) = 300 / 300 = 1.0 m backward. The person moves 5.0 − 1.0 = 4.0 m forward. Check: 60 × 4.0 = 240 × 1.0 = 240.
Point guide (4 points):
1 point: internal forces only, center of mass stays fixed, so the raft moves opposite to the person.
1 point: correct equation using displacements relative to the water.
1 point: d = mL/(m + M) with a correct limit check.
1 point: d = 1.0 m backward.
Common mistakes
The mistake: "The center of mass is always in the middle between the objects."
Why it is wrong: the center of mass is a weighted average; heavier objects count more.
How to spot it: if the masses are different, your answer must be closer to the heavier one. Check that it is.
The mistake: believing internal forces can move a system ("pushing your own car from inside").
Why it is wrong: internal forces come in equal and opposite pairs inside the system and cancel.
How to spot it: ask "does this force come from something outside my chosen system?" If not, it cannot change acm.
The mistake: not saying what the system is, then mixing forces from different choices.
Why it is wrong: a force can be internal for one system and external for another (the contact force in Example 3).
How to spot it: every free body diagram and every F = ma line should start with "System: ...".
The mistake: thinking the center of mass must be on the object.
Why it is wrong: rings, boomerangs, horseshoes and two separated skaters all have their center of mass in empty space.
How to spot it: trust the formula and symmetry, not where the material is.
The mistake: using the total mass with only one part's force (or one part's mass with the total force).
Why it is wrong: Fnet and m in F = ma must belong to the same system.
How to spot it: circle the system, then check that both the force and the mass refer to exactly that system.