A conservation law applies to a system, not to a single object. Draw the boundary first, list the forces that cross it, and only then choose momentum or energy.
This lesson is part of selecting and checking a physical model in SPM Physics. It builds on explaining momentum and impulse.
What is the boundary-first routine?
Use these four steps each time.
- Name the objects inside the boundary.
- List every force that crosses the boundary from outside.
- If the resultant outside force is zero, momentum is conserved. If no outside force does work that removes energy, mechanical energy is conserved.
- Write the conclusion in one sentence, then use the equation.
Step 2 is the easiest step to skip, and it explains why a conservation equation fails.
Worked example 1: momentum with the right boundary
Two skaters stand still on smooth ice. A 60 kg skater pushes a 40 kg friend, and the friend moves away at 1.5 m s⁻¹. Find the velocity of the first skater.
Boundary: both skaters. Outside forces: weight and the ice’s normal reaction cancel, and friction is negligible, so the resultant outside force is zero.
Total momentum before = 0. After: 60v + 40(1.5) = 0, so 60v = −60 and v = −1.0 m s⁻¹. The first skater moves at 1.0 m s⁻¹ in the opposite direction.
If you had drawn the boundary around the friend alone, the push from the first skater would be an outside force, and momentum of the friend would not be conserved.
Worked example 2: energy with the wrong boundary
A 5.0 kg box slides down a rough ramp from a height of 2.0 m and reaches the bottom at 4.0 m s⁻¹. Take g = 10 m s⁻².
Boundary: the box only. Initial potential energy = 5.0 × 10 × 2.0 = 100 J. Final kinetic energy = ½ × 5.0 × 16 = 40 J. The 60 J difference shows that mechanical energy is not conserved, because friction from the ramp acts from outside the boundary.
Boundary: box and ramp. The 60 J becomes thermal energy in the box and the ramp, so the total energy is conserved. The boundary decides whether the statement “energy is conserved” is true.
The mistake to avoid
The common mistake is to apply “kinetic energy + potential energy = constant” to any moving object. The table shows when each statement holds.
| Situation | Mechanical energy conserved? |
|---|---|
| Ball in free fall, air resistance ignored | Yes |
| Box on a rough ramp | No, friction removes energy |
| Two trolleys sticking together | No, kinetic energy is lost in the collision |
State your assumption in the answer. The units and significant figure checker can confirm the units of the quantities you compare.
Check yourself
A 3.0 kg trolley moving at 2.0 m s⁻¹ hits a stationary 1.0 kg trolley, and they stick together. Explain the boundary you would draw and find the common velocity.
Answer
Boundary: both trolleys. Friction on the track is negligible, so the resultant outside force is zero and momentum is conserved.
Momentum before = 3.0 × 2.0 = 6.0 kg m s⁻¹. After: 4.0v = 6.0, so v = 1.5 m s⁻¹. Kinetic energy is not conserved, because the trolleys stick together.
What to study next
Move on to explaining why constant speed does not always mean constant velocity. Record the assumptions you forget to state with the mistake log and paper-error review tool.
If you would like a teacher to question your boundaries with you, see online one-to-one Physics tuition.