Systems · F=ma · 7 min
Energy and momentum answer different questions and survive different interactions.
01 The trap
Wrong path
Students use whichever conservation law gives the easiest algebra, especially in collisions.
Why it feels right
Both laws are powerful, both use before/after states, and both often appear in the same unit.
02 Correct model
Momentum survives short collisions when external impulse is negligible. Mechanical energy survives only when nonconservative work is absent. A single problem may need both laws on different intervals.
03 Mini-example
Prompt
In a ballistic pendulum, what law applies during the sticky collision?
Common wrong answer
Conserve kinetic energy because the block rises afterward.
Correct reasoning
Conserve momentum during the collision, then use mechanical energy for the swing.
Diagnostic cue
Sticky collision means kinetic energy is not conserved, even when momentum is.
04 Guided practice
Warm-up isolates the principle. Bridge changes the context. Contest-style requires a complete setup on less familiar geometry.
Two carts stick together after colliding on a nearly frictionless track. Which quantity is conserved during the collision: momentum, kinetic energy, both, or neither?
Sticky is the keyword.
Momentum is conserved if external impulse is negligible. Kinetic energy is not conserved because the carts stick and deform.
Trap: Assuming every collision conserves kinetic energy.
A projectile embeds in a hanging block, and the block-projectile pair swings upward. Which law applies during the collision, and which applies during the swing?
Use separate time intervals.
During the short collision, conserve momentum if external impulse is negligible. During the swing after the collision, use mechanical energy if air resistance and pivot losses are negligible.
Trap: Using mechanical energy through the embedding collision.
A block slides down a frictionless ramp, reaches the bottom, and sticks to a cart. Which model should be used for each interval?
The ramp motion and the collision are different intervals.
Use energy on the frictionless ramp to find the block's speed at the bottom. Then use momentum during the sticky collision with the cart. Do not conserve mechanical energy across the sticking event.
Trap: Trying to use one conservation law for the entire story.
A satellite of mass m is in a circular orbit of radius r around mass M. (a) Derive v(r). (b) Derive total mechanical energy E(r). (c) If drag slowly removes mechanical energy, explain whether r increases or decreases.
For a circular orbit, GMm/r² = mv²/r.
From GMm/r² = mv²/r, v = √(GM/r). Then K = (1/2)mv² = GMm/(2r), U = -GMm/r, so E = -GMm/(2r). If drag removes energy, E becomes more negative, which corresponds to a smaller circular-orbit radius.
Trap: Thinking losing energy must mean moving to a larger orbit because the satellite slows down.
A pendulum bob is released, strikes a stationary lump of clay, sticks to it, and the pair rises. Which sequence of models is valid?
Mechanical energy does not survive the sticking collision.
Use mechanical energy for the first swing, momentum during the short collision, and mechanical energy for the rise after the collision.
Trap: Applying one conservation law continuously across a sticking event.
A projectile of mass m moving horizontally at speed u embeds in a stationary block of mass M suspended by a light string. The pair rises through height h. (a) Derive u in terms of m, M, g, and h. (b) Find the fraction of the projectile's initial kinetic energy lost during the collision. (c) State the assumptions that permit each conservation law.
Use momentum only during the collision and mechanical energy only during the subsequent swing.
The swing gives V = √(2gh). Collision momentum gives mu = (M + m)V, so u = (M + m)√(2gh)/m. The post-collision kinetic-energy fraction is m/(M + m), so the lost fraction is M/(M + m).
Trap: Conserving mechanical energy from before impact to the top of the swing.
A particle of mass m moves under gravity from a fixed mass M with nonzero angular momentum L. Its radial motion can be described by U_eff(r) = L²/(2mr²) - GMm/r. (a) Find the radius of a circular orbit. (b) Show that the orbit is stable against small radial displacements. (c) Find the circular-orbit energy and check its sign.
Set dU_eff/dr = 0, then inspect the second derivative at that radius.
The stationary point is r₀ = L²/(GMm²). At r₀, d²U_eff/dr² = GMm/r₀³ > 0, so the orbit is stable. Substitution gives E = U_eff(r₀) = -G²M²m³/(2L²) = -GMm/(2r₀).
Trap: Treating the gravitational potential alone as the radial potential when angular momentum is fixed.
05 Mock checks
A decision-making set for choosing the right conservation law on the right interval.
6 questions · 20 pts
A concentrated test of torque arms, rolling constraints, angular momentum, and rotational energy.
6 questions · 22 pts