A worked solution sits on the page, every line tidy, and the answer comes out faster than either cart went in. The arithmetic is clean. Something in the set-up is wrong.
A conservation solve can only lie in a few places, and each place has its own check. This lesson teaches the three checks that find the flaw.
Here is the base problem, taking rightward as positive: a 2 kg cart at +3 m/s meets a 1 kg cart moving leftward at 3 m/s, and they couple. Done right: 2 × 3 + 1 × (−3) = 3 v', so 6 − 3 = 3 v' and v' = +1 m/s.
Check one: the signs. Every velocity enters with the sign the convention gives it. The classic flaw writes the leftward cart as +3 m/s, turning the before total into 9 kg·m/s and the answer into a too-fast +3 m/s.
Check two: the masses. Each velocity carries its own object's mass, and a coupled pair's total divides by the combined mass. The classic flaw divides by one cart's mass, and a flaw one step earlier adds bare velocities with no masses at all.
Check three: the sides. Before-values on one side, after-values on the other. The classic flaw lets a measured after-velocity sneak into the before total.
A solution that passes all three checks and still looks odd is usually telling the truth: a slow pair, a negative answer, or a zero total can all be right.
An answer faster than every input, though, is a flag: a dropped sign or a missing mass usually made it.
Take rightward as positive. A 4 kg cart at +2 m/s meets a 2 kg cart at −1 m/s, and they couple. Nico writes: 4 × 2 + 2 × (+1) = 6 v', so v' = 10 ÷ 6. Which check catches the flaw?
Check one, the signs: the 2 kg cart moves leftward, so its velocity must enter as −1 m/s. Nico wrote +1.
The sign check catches it. Done right: 4 × 2 + 2 × (−1) = 6 v', so 8 − 2 = 6 v' and v' = +1 m/s.
Check one finds the flaw: the leftward velocity went in without its negative sign.
Take rightward as positive. A 3 kg cart at +4 m/s hits a parked 3 kg cart, and they couple. Margit writes: 3 × 4 = 12, so v' = 12 ÷ 3 = 4 m/s. Which check catches the flaw?
Check one, the signs: both entries are rightward or zero, so the signs pass.
Check two, the masses: the coupled pair's mass is 3 + 3 = 6 kg, and Margit divided by 3 kg.
The mass check catches it. Done right: v' = 12 ÷ 6 = 2 m/s.
Check two finds the flaw: the total momentum was divided by one cart's mass instead of the combined 6 kg.
Take rightward as positive. A 3 kg cart moves right at 2 m/s and a 2 kg cart moves left at 2.5 m/s; they couple. A student writes: 3 × 2 + 2 × 2.5 = 5 v'. What is the error?
Take rightward as positive. A 2 kg cart moving right at 4 m/s hits a parked 6 kg cart, and they couple. A student writes: total = 8 kg·m/s, so v' = 8 ÷ 2 = 4 m/s. What is the error?
Take rightward as positive. A 3 kg cart at +2 m/s meets a 2 kg cart at −4 m/s, and they couple. Oriana writes: 3 × 2 + 2 × (−4) = 5 v'. Is her set-up correct?
Take rightward as positive. A 5 kg cart at +4 m/s meets a 1 kg cart at −2 m/s, and they bounce apart. A student writes: 4 + (−2) = v_1' + v_2'. What is the error?
Take rightward as positive. A 2 kg cart at +5 m/s hits a parked 4 kg cart, and they bounce apart; afterward the 2 kg cart moves at −0.5 m/s. Piet computes the BEFORE total as 2 × (−0.5) + 4 × 0. What is the error?
Take rightward as positive. A 1.5 kg cart at +4 m/s meets a 2.5 kg cart at −2 m/s, and they couple. Four students set it up. Whose set-up has no error?