When two objects stick together, momentum conservation alone tells you everything: after the collision there is only one velocity to find.
Key idea
For objects that stick: $$m_1 v_1 + m_2 v_2 = (m_1 + m_2)\,v.$$ This is a perfectly inelastic collision. Momentum is conserved, but kinetic energy is not: the most kinetic energy that momentum allows turns into thermal energy, sound and bent material.
Why it works
Why must kinetic energy be lost? Take a cart hitting an identical cart at rest. Momentum forces the pair to move at $v/2$. Kinetic energy before: $\tfrac{1}{2}mv^2$. After: $\tfrac{1}{2}(2m)(v/2)^2 = \tfrac{1}{4}mv^2$. Half of it is gone. In general, when a moving $m_1$ sticks to a resting $m_2$, the fraction lost is $\dfrac{m_2}{m_1 + m_2}$. The energy hasn't vanished; it went into the Velcro, the putty or the crumpled bumper as heat and sound.
Worked examples
Worked example: Railroad cars couple
A $30{,}000\ \text{kg}$ freight car rolling at $2.0\ \text{m/s}$ couples with a $20{,}000\ \text{kg}$ car at rest. How fast do they roll together?
- Before: $(30{,}000)(2.0) + 0 = 60{,}000\ \text{kg·m/s}$.
- After: $(50{,}000\ \text{kg})\,v = 60{,}000\ \text{kg·m/s}$, so $v = 1.2\ \text{m/s}$.
Answer: $1.2\ \text{m/s}$
Worked example: A head-on tackle
A $100\ \text{kg}$ player running east at $5.0\ \text{m/s}$ tackles an $80\ \text{kg}$ player running west at $6.0\ \text{m/s}$, and they hold on. Find their velocity just after (east positive).
- Before: $$(100)(+5.0) + (80)(-6.0) = 500 - 480 = +20\ \text{kg·m/s}.$$
- $v = \dfrac{20\ \text{kg·m/s}}{180\ \text{kg}} = +0.11\ \text{m/s}$: the pair drifts slowly east.
Answer: $0.11\ \text{m/s}$ east
Worked example: Where the energy goes
For the railroad cars, how much kinetic energy is lost?
- Before: $\tfrac{1}{2}(30{,}000)(2.0)^2 = 60{,}000\ \text{J}$.
- After: $\tfrac{1}{2}(50{,}000)(1.2)^2 = 36{,}000\ \text{J}$.
- Lost: $24{,}000\ \text{J}$, which is $\dfrac{20{,}000}{50{,}000} = 40\%$, just as the rule $\dfrac{m_2}{m_1 + m_2}$ predicts.
Answer: $24{,}000\ \text{J}$ (40%)
Sticking collisions are useful for measuring speeds that are hard to time. Fire a dart into a block on ice, measure how fast the block slides off, and momentum gives the dart's speed: $v = \dfrac{(m + M)V}{m}$. The heavier the block, the slower and easier to measure its motion.
Common mistake
Conserving kinetic energy in a sticking collision. Setting $\tfrac{1}{2}m_1v_1^2 = \tfrac{1}{2}(m_1 + m_2)v^2$ gives a speed that is too big and breaks momentum conservation. In any collision, start with momentum. Check: after a sticking collision, the kinetic energy must come out smaller than before.