Conservation and change of reference frame

Problem

Conservation and change of reference frame. A person standing still on a $40\ \text{kg}$ cart jumps off onto the ground. The cart recoils at $0{,}5\ \text{m/s}$. Explain why momentum is conserved in the ground reference frame and what changes if you observe the scene from the cart itself after the jump.

Solution

Ground reference frame. Before the jump everything is at rest: total momentum is zero. It is an internal “explosion” (the person pushes on the cart, the cart pushes on the person: internal forces, equal and opposite). No horizontal external force, so $\vv{p}$ is conserved: $\vv{p}_\text{prima} = 0 = m_p \vv{v}_p + m_c \vv{v}_c$ The cart recoils: $m_c v_c = 40 \cdot 0{,}5 = 20\ \mathrm{kg\,m/s}$ in one direction, and the person carries exactly $-20\ \mathrm{kg\,m/s}$ in the other. The sum stays zero: conserved.

Change of reference frame. If I observe the scene from an inertial frame in motion (for example the one moving with the cart after the jump), all velocities change by a constant $\vv{u}$: $\vv{v}\,' = \vv{v} - \vv{u}$ The total momentum now has a value different from zero: $\vv{p}\,' = \sum_i m_i (\vv{v}_i - \vv{u}) = \vv{p} - M\vv{u} = -M\vv{u}$ but it still remains constant over time (before and after the jump it is the same), because $M$ and $\vv{u}$ do not change. So the momentum balance holds in every inertial frame: the numerical value changes, not the fact that it is conserved.

Final result: $\ev{\text{ground: } \vv{p} = 0 \to m_p v_p + m_c v_c = 0;\ \text{other frame: } \vv{p} \neq 0 \text{ but always constant}}$

Links

Topics: Quantità di moto e urti Concepts: Conservazione della quantità di moto · Esplosione Skills: Cambio di sistema di riferimento