An impulse changes an object's momentum. But in a collision every push comes in a pair, by Newton's third law, and that pairing makes the total momentum of the two objects stay fixed.
Why it works
During a collision, A pushes B with force $\vec{F}$ and B pushes A with $-\vec{F}$, for exactly the same time $\Delta t$ (they touch for as long as they touch). So $$\Delta\vec{p}_B = \vec{F}\Delta t, \qquad \Delta\vec{p}_A = -\vec{F}\Delta t.$$ Add them: $\Delta\vec{p}_A + \Delta\vec{p}_B = 0$. Whatever momentum A loses, B gains. Nothing in that argument cares whether the objects bounce, stick or crumple.
Key idea
If the net outside force on a system is zero (or its impulse is negligible during a short collision), the total momentum stays the same: $$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'.$$ Forces between parts of the system are internal and cancel in pairs.
- Choose the system. Include everything that pushes on everything else during the event.
- Check outside forces. On a level frictionless track, gravity and the normal force cancel. Friction from the road is outside, but during a quick crash its impulse is small.
- Pick a positive direction and write every velocity with a sign.
- Write total before = total after and solve for the unknown.
Worked examples
Worked example: Find a velocity after
On an air track, glider 1 ($0.40\ \text{kg}$) moves right at $1.5\ \text{m/s}$ and hits glider 2 ($0.60\ \text{kg}$) at rest. Afterward glider 1 moves right at $0.30\ \text{m/s}$. Find glider 2's velocity.
- Before: $$(0.40)(1.5) + (0.60)(0) = 0.60\ \text{kg·m/s}.$$
- After: $(0.40)(0.30) + 0.60\,v_2' = 0.60$, so $0.60\,v_2' = 0.48$.
- $v_2' = 0.80\ \text{m/s}$ to the right. Check: it is ahead of glider 1 and moving faster, as it must be.
Answer: $0.80\ \text{m/s}$ right
Worked example: Head-on, with signs
A $3.0\ \text{kg}$ cart moving right at $2.0\ \text{m/s}$ meets a $1.0\ \text{kg}$ cart moving left at $4.0\ \text{m/s}$. Afterward the $1.0\ \text{kg}$ cart moves right at $2.0\ \text{m/s}$. What does the $3.0\ \text{kg}$ cart do?
- Before: $$(3.0)(+2.0) + (1.0)(-4.0) = +2.0\ \text{kg·m/s}.$$
- After: $3.0\,v_1' + (1.0)(+2.0) = 2.0$, so $v_1' = 0$.
- The heavy cart stops dead. The total is still $+2.0\ \text{kg·m/s}$, now carried by the light cart.
Answer: It stops: $v_1' = 0$
Common mistake
Thinking momentum is conserved only in bouncy, elastic collisions. The third-law argument works for every collision, so momentum is conserved whether the objects bounce, stick or crumple. What can disappear is kinetic energy (into heat and sound). Check yourself: compute the total momentum before and after; if they differ, you used a wrong sign, or an outside force (such as friction over a long time) was not small.
Momentum conservation and Newton's third law are the same physics seen two ways. The Momentum and Collisions lab of the Science Lab lets you crash gliders and watch both bars.