Two stages, two laws. During a quick collision the outside forces are small, so momentum is conserved. After it, on a smooth ramp, energy is conserved. The trap is to use energy across the collision itself. Each problem here is one short scene and a chain of questions. Answer them in order: each later question uses what you found before, so a slip early on carries through. Keep full precision between steps and round only at the end.
Key idea
Sticking collision: $$m_1v = (m_1 + m_2)V.$$ Kinetic energy is lost: the fraction lost is $\dfrac{m_2}{m_1 + m_2}$ when $m_2$ starts at rest. After the collision: $\tfrac{1}{2}V^2 = gh$, so $h = \dfrac{V^2}{2g}$.
Worked example: Carts on a track
Cart A ($0.6\ \text{kg}$, $3.0\ \text{m/s}$) locks onto cart B ($0.4\ \text{kg}$, at rest). Find (1) their speed, (2) the kinetic energy lost and (3) how high they roll up a smooth ramp.
- (1) $V = \dfrac{(0.6)(3.0)}{1.0} = 1.80\ \text{m/s}$.
- (2) $K_0 = \tfrac{1}{2}(0.6)(3.0)^2 = 2.70\ \text{J}$ and $K_1 = \tfrac{1}{2}(1.0)(1.80)^2 = 1.62\ \text{J}$, so $1.08\ \text{J}$ is lost (40%, which is $m_2/(m_1+m_2)$).
- (3) $h = \dfrac{(1.80)^2}{2(9.8)} = 0.165\ \text{m}$.
Answer: $1.80\ \text{m/s}$, $1.08\ \text{J}$, $0.165\ \text{m}$
Common mistake
Using $\tfrac{1}{2}m_1v^2 = (m_1 + m_2)gh$ straight from the start. That ignores the energy lost when the carts stick, so the height comes out too large. Check: an answer above $v^2/2g$ is impossible.