Modeling · F=ma · 6 min
The fastest solution often comes from choosing the right system boundary.
01 The trap
Wrong path
Students isolate every block immediately and get buried in internal forces.
Why it feels right
Free-body diagrams train you to isolate objects, which is correct, but sometimes the combined system reveals the acceleration first.
02 Correct model
Use the combined system when internal forces cancel and the question asks about shared motion. Then isolate a piece only when you need an internal force.
03 Mini-example
Prompt
Blocks of 2 kg and 3 kg touch on a frictionless floor. A 10 N force pushes the pair. Find the contact force.
Common wrong answer
Guess the contact force is 10 N because that is the applied force.
Correct reasoning
Combined system gives a = 2 m/s². Isolate the 3 kg block: contact force = 6 N.
Diagnostic cue
If a force acts between two objects inside your chosen system, it should not appear in the system equation.
04 Guided practice
Warm-up isolates the principle. Bridge changes the context. Contest-style requires a complete setup on less familiar geometry.
A 12 N force pushes a 2 kg block against a 4 kg block on a frictionless floor. Find the acceleration of the pair.
Use both blocks as the system.
The total mass is 6 kg, so a = 12/6 = 2 m/s². The contact force is internal for the combined system.
Trap: Including the contact force in the combined-system equation.
For the same 2 kg and 4 kg blocks pushed by 12 N on the 2 kg block, find the contact force on the 4 kg block.
After finding acceleration, isolate the 4 kg block.
The pair accelerates at 2 m/s². The only horizontal force on the 4 kg block is contact, so F_contact = (4)(2) = 8 N.
Trap: Saying the contact force equals the applied 12 N.
Three carts of masses 1 kg, 2 kg, and 3 kg are pulled by a 12 N force on the 1 kg cart. Find the tension between the 2 kg and 3 kg carts on a frictionless track.
First find the acceleration of all three carts. Then isolate the 3 kg cart.
Total mass is 6 kg, so a = 2 m/s². The tension pulling the 3 kg cart is T = (3)(2) = 6 N.
Trap: Trying to solve all internal tensions before finding the shared acceleration.
A 24 N horizontal force pushes a 2.0 kg block against a 6.0 kg block on a frictionless floor. The force acts on the 2.0 kg block. What is the contact force on the 6.0 kg block?
The pair accelerates together at 24/(2.0 + 6.0) m/s².
The acceleration is 3.0 m/s². The contact force is the only horizontal force on the 6.0 kg block, so F = (6.0)(3.0) = 18 N.
Trap: Assigning the full 24 N applied force to the second block.
A pendulum of length ℓ and bob mass m hangs in a cart accelerating horizontally to the right with constant acceleration A. After transients die out, the bob is stationary relative to the cart. (a) Derive the string angle from vertical. (b) Derive the tension. (c) Reproduce the result in the inertial ground frame and explain why no physical horizontal force labeled 'pseudo-force' appears there.
In the cart frame, add a backward inertial force mA; in the ground frame, the bob accelerates rightward with the cart.
In the cart frame, T sinθ = mA and T cosθ = mg, so tanθ = A/g and T = m√(g² + A²). In the ground frame, horizontal acceleration A requires T sinθ = mA while vertical acceleration is zero, giving the same equations without a pseudo-force.
Trap: Mixing an accelerating-frame pseudo-force with an inertial-frame ma term in one equation.
A small block of mass m starts from rest at height h on a smooth wedge of mass M. The wedge rests on a frictionless horizontal floor. At the bottom, the block's velocity relative to the wedge is horizontal. (a) State the system assumptions and conserved quantities. (b) Derive the ground-frame speeds of the block and wedge at the bottom. (c) Check the limits M → ∞ and m → 0.
Horizontal momentum gives mv = MV; total kinetic energy includes both bodies.
With rightward block speed v and leftward wedge speed V, horizontal momentum gives mv = MV. Energy gives mgh = ½mv² + ½MV². Therefore v = √[2ghM/(M + m)] and V = (m/M)v = √[2ghm²/(M(M + m))]. As M → ∞, v → √(2gh) and V → 0. As m → 0, the wedge speed also tends to zero while the block approaches √(2gh).
Trap: Using v = √(2gh) for a finite wedge and ignoring the wedge's kinetic energy.
05 Mock checks
A compact check for the force-direction, normal-force, friction, and circular-motion habits that decide early contest mechanics.
6 questions · 15 pts
A bridge set for pulleys, system boundaries, friction direction, and accelerating frames.
6 questions · 19 pts