Topic 2.5: Newton's Second Law, Part 2 — Multi-Object Systems

Name: _______________________________ Date: _______________

Before We Start: Yesterday's Recap

Yesterday every problem had just one object. Today we connect two objects with a string, sharing one acceleration. Quick guess: is it easier to solve for that shared acceleration by looking at each object separately, or by treating the whole connected system as one bigger object?

Today's New Concepts

AP CED Alignment: Unit 2, Topic 2.5 (Part 2 of 2)

Two Ways to Solve a Connected System

Method 1 — Whole system: combine both objects into one total mass, use ONLY external forces: $F_{net,ext} = m_{total},a$. Internal forces (like the string's tension) cancel out completely and never appear.
Method 2 — Isolate one object: use only the forces on THAT object (including internal forces like tension), and its own mass: $F_{net,1} = m_1,a$
Use Method 1 to find the shared acceleration quickly; use Method 2 (after finding $a$) to solve for an internal force like tension

Worked Example: Two blocks connected by a string, pulled by force $F$: whole system gives $a = \dfrac{F}{m_1+m_2}$; isolating the trailing block (only tension acts on it) gives $F_T = m_2 a = \dfrac{m_2 F}{m_1+m_2}$.

Side-by-side comparison: the same two connected blocks analyzed as a whole system (only F_ext counts, tension cancels) versus an isolated single block (tension appears explicitly)

Worked Example: A block on a table connects over a pulley to a hanging mass. Isolating each object gives one equation apiece; adding them together eliminates tension and solves directly for the shared acceleration.

FBDs for a block-and-pulley (Atwood-style) system: F_T and F_f on the table block, F_T and F_g on the hanging block, same tension and same acceleration magnitude on both sides

Center of Mass Acceleration Depends Only on External Force

$a_{cm} = \dfrac{F_{net,ext}}{m_{total}}$ — depends ONLY on the net force from OUTSIDE the system, never on internal interactions between the system's own parts
Internal forces, no matter how complicated, can never change a system's own center-of-mass motion

Worked Example: Two stars orbit their shared center of mass, far from anything else. Each star individually moves in a circle (accelerating), but since there's no external force on the two-star system, the system's center of mass has zero acceleration.

Two orbiting stars of different mass wobbling around their shared center of mass from mutual gravity (internal, cancels for the system), with an external force shown accelerating the CM itself

The "String Breaks" Setup

Before breaking: the connected objects share ONE acceleration — solve as a whole system
After breaking: each object now has its OWN acceleration, from whatever forces still act on it alone — solve each object completely separately
The center-of-mass equation still applies to the whole system throughout, using whatever total external force remains

Worked Example: Two crates connected by a string, one pulled by force $F$, both experiencing friction. After the string breaks, the trailing crate's acceleration is just $a_1 = f_1/m_1$ (friction alone), while the system's center-of-mass acceleration is still $a_{cm} = (F-f_1-f_2)/(m_1+m_2)$ — generally a different number.

Keep This Sheet!

These sheets build into your semester study guide. Keep them in order in a binder or folder — you'll want to flip back through them before quizzes, unit tests, and when AP exam review starts in the spring.