A 2 kg cart rolls right at 1 m/s. A 6 kg cart charges at it leftward at 3 m/s. They couple. Which way does the wreck go? The full method answers in five steps, and the sign of the answer does the last one for you.
Everything the topic built now runs in one sequence: check that momentum is conserved, classify the interaction, set up the equation, solve, and read the sign.
Step one, conserved? The track is nearly frictionless, so the net external force on the two-cart system is roughly zero. Momentum is conserved.
Step two, classify. The carts couple: the collision is perfectly inelastic, so the after side is one combined mass.
Step three, set up, taking rightward as positive: 2 × 1 + 6 × (−3) = 8 v'.
Step four, solve: 2 − 18 = 8 v', so −16 = 8 v' and v' = −2.
Step five, read the sign. Rightward is positive, so v' = −2 m/s means the coupled pair moves leftward at 2 m/s. The sign is the direction; the size is the speed.
The same five steps run a separation: two objects push apart from rest, the total starts at zero, and the signs sort out who goes which way.
Take rightward as positive. A 1 kg cart at +2 m/s meets a 4 kg cart at −2 m/s on a low-friction track, and they couple. Run the full analysis.
Conserved? The track is low-friction, so the net external force is roughly zero: yes.
Classify: the carts couple, so the collision is perfectly inelastic.
Write down the values in the question:
m_1 = 1 kg, v_1 = +2 m/s; m_2 = 4 kg, v_2 = −2 m/s; v' = ?
Write down the equation:
m_1 v_1 + m_2 v_2 = (m_1 + m_2) v'
Substitute in the values:
1 × 2 + 4 × (−2) = 5 v'
2 − 8 = 5 v'
Simplify:
v' = −6 ÷ 5 = −1.2
v' = −1.2 m/s
Read the sign: the pair moves leftward at 1.2 m/s.
The coupled pair moves at 1.2 m/s leftward (v' = −1.2 m/s).
Take rightward as positive. A 5 kg cart and a 2 kg cart push apart from rest on a low-friction track, and the 2 kg cart leaves at +7 m/s. Run the full analysis for the 5 kg cart.
Conserved? Low friction, net external force roughly zero: yes.
Classify: a separation from rest, so the total momentum is zero throughout.
Write down the values in the question:
m_1 = 5 kg, v_1' = ?; m_2 = 2 kg, v_2' = +7 m/s
Write down the equation:
0 = m_1 v_1' + m_2 v_2'
Substitute in the values:
0 = 5 v_1' + 2 × 7
0 = 5 v_1' + 14
Simplify:
v_1' = −14 ÷ 5 = −2.8
v_1' = −2.8 m/s
Read the sign: the 5 kg cart moves leftward at 2.8 m/s.
The 5 kg cart moves at 2.8 m/s leftward (v' = −2.8 m/s).
Take rightward as positive. On a low-friction track, a 3 kg cart at +2 m/s meets a 5 kg cart at −2 m/s, and they couple. Enter the pair's velocity after the collision in m/s (include the sign).
Piet solves a coupling collision, taking eastward as positive, and gets v' = −1.5 m/s for the pair. What does the negative sign say?
Take rightward as positive. On a low-friction track, a 6 kg cart and a 1.5 kg cart push apart from rest, and the 6 kg cart moves at −0.9 m/s. Enter the 1.5 kg cart's velocity in m/s (include the sign).
Take rightward as positive. On a low-friction track, a 2 kg cart at +2.5 m/s meets an 8 kg cart at −1.5 m/s, and they couple. Enter the pair's velocity after the collision in m/s (include the sign).
Nico begins a collision problem. Before writing the conservation equation, which check does the method ask for first?
Take rightward as positive. On a low-friction track, a 4 kg cart at +2 m/s meets a 2 kg cart at −5.5 m/s, and they bounce apart. Afterward the 4 kg cart moves at −2.5 m/s. Enter the 2 kg cart's velocity after the collision in m/s (include the sign).