Three balls against the wall

Problem

Three balls hit a wall perpendicularly with the same velocity $v$:

• A: mass $m$, bounces back with velocity $v$;
• B: mass $m$, stops;
• C: mass $2m$, bounces back with velocity $v$.

Order the change in momentum $|\Delta p|$ undergone by the ball from smallest to largest.

Solution

I take as positive the direction of approach to the wall. Each ball arrives with momentum $+m_i v$; I calculate $\Delta p = p_\text{dopo} - p_\text{prima}$.

A (mass $m$, bounces back at $v$): $p_\text{prima} = +mv$, $p_\text{dopo} = -mv$. $|\Delta p_A| = |-mv - mv| = 2mv$

B (mass $m$, stops): $p_\text{prima} = +mv$, $p_\text{dopo} = 0$. $|\Delta p_B| = |0 - mv| = mv$

C (mass $2m$, bounces back at $v$): $p_\text{prima} = +2mv$, $p_\text{dopo} = -2mv$. $|\Delta p_C| = |-2mv - 2mv| = 4mv$

Bouncing back “costs” more momentum than stopping (it reverses the direction), and more mass amplifies the effect. Increasing order: $mv < 2mv < 4mv \;\Longrightarrow\; |\Delta p_B| < |\Delta p_A| < |\Delta p_C|$

Final result: $\ev{\Delta p_C > \Delta p_A > \Delta p_B\quad (4mv,\ 2mv,\ mv)}$

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Topics: Quantità di moto e urti Concepts: Quantità di moto · Impulso Skills: Ragionamento di ranking