Momentum · AP/F=ma · 6 min
Momentum conservation is a system statement, not a universal magic spell.
01 The trap
Wrong path
Students write p_i = p_f before deciding the system or checking external impulse.
Why it feels right
Collision problems often conserve momentum, so it starts to feel automatic.
02 Correct model
Choose the system. During the time interval, compare external impulse to the internal impulses. Momentum is conserved only when external impulse is zero or negligible.
03 Mini-example
Prompt
Two carts collide and stick on a low-friction track. Why conserve momentum?
Common wrong answer
Because kinetic energy is conserved.
Correct reasoning
Because external impulse during the brief collision is negligible. Kinetic energy is not conserved.
Diagnostic cue
The phrase 'during the collision' usually invites an impulse-size argument, not a blind conservation equation.
04 Guided practice
Warm-up isolates the principle. Bridge changes the context. Contest-style requires a complete setup on less familiar geometry.
A 1 kg cart moving right at 6 m/s sticks to a 2 kg cart at rest. Find their final speed if external impulse is negligible.
Momentum survives the short collision; kinetic energy does not.
(1)(6) = (1+2)v, so v = 2 m/s right.
Trap: Conserving kinetic energy for a sticky collision.
During a 0.020 s collision, a 0.5 N external friction force acts on a cart system while internal collision forces are about 80 N. Why is momentum conservation reasonable?
Compare impulses over the same time interval.
External impulse is about 0.5(0.020) = 0.010 N·s, while internal impulses are of order 80(0.020) = 1.6 N·s and cancel inside the system. The external impulse is tiny.
Trap: Checking only whether any external force exists instead of whether its impulse matters.
A 3 kg cart moves horizontally at 4 m/s. A 1 kg bag of sand drops vertically into it and stays. Find the final horizontal speed.
Horizontal external impulse is negligible; vertical momentum is not the useful component.
Horizontal momentum: (3)(4) = (4)v, so v = 3 m/s. The vertical motion is handled by forces from the cart/ground, not horizontal momentum.
Trap: Treating the bag's vertical speed as part of horizontal momentum.
A 0.40 kg cart moving east at 3.0 m/s sticks to a 0.20 kg cart moving north at 4.0 m/s. With negligible external impulse, what is the magnitude of their final velocity?
The final momentum components are 1.2 kg·m/s east and 0.8 kg·m/s north.
The combined mass is 0.60 kg. The momentum magnitude is √(1.2² + 0.8²) = 1.442 kg·m/s, so v = 1.442/0.60 = 2.40 m/s.
Trap: Adding the two initial speeds as scalars.
05 Mock checks