Proof of the 1D elastic collision

Problem

Prove that, in a one-dimensional elastic collision between a mass $m_1$ with initial velocity $v_1$ and a mass $m_2$ initially at rest, the velocity of the first mass after the collision is $v_1' = \frac{m_1 - m_2}{m_1 + m_2}\,v_1$. Comment on the limiting cases $m_1 \gg m_2$ and $m_1 \ll m_2$.

Solution

With $m_2$ at rest ($v_2 = 0$), I write the two conservation laws of the elastic collision. Momentum: $m_1 v_1 = m_1 v_1' + m_2 v_2'$ Kinetic energy: $\tfrac{1}{2} m_1 v_1^2 = \tfrac{1}{2} m_1 v_1'^2 + \tfrac{1}{2} m_2 v_2'^2$ I rewrite both, bringing the $m_1$ terms to the left: $m_1 (v_1 - v_1') = m_2 v_2'$ $m_1 (v_1^2 - v_1'^2) = m_2 v_2'^2$ In (2’) I factor the difference of squares: $v_1^2 - v_1'^2 = (v_1 - v_1')(v_1 + v_1')$. I divide (2’) by (1’) term by term (legitimate because the collision occurs, so $v_1 \neq v_1'$ and $v_2' \neq 0$): $v_1 + v_1' = v_2'$ This is the relation on the relative velocities. I substitute (3) into (1’): $m_1 (v_1 - v_1') = m_2 (v_1 + v_1')$ $m_1 v_1 - m_1 v_1' = m_2 v_1 + m_2 v_1'$ I collect $v_1'$ on the right and $v_1$ on the left: $v_1 (m_1 - m_2) = v_1' (m_1 + m_2)$ $\ev{v_1' = \frac{m_1 - m_2}{m_1 + m_2}\,v_1}$ Limiting cases.

• $m_1 \gg m_2$: the fraction tends to $+1$, so $v_1' \approx v_1$. The heavy mass carries on almost undisturbed (like a lorry hitting a small ball).
• $m_1 \ll m_2$: the fraction tends to $-1$, so $v_1' \approx -v_1$. The light mass bounces back almost at the same speed (like a ball against a wall).

Links

Topics: Quantità di moto e urti Concepts: Urto elastico · Conservazione della quantità di moto · Energia cinetica Skills: Impostazione simbolica