Elastic collision between two marbles

Problem

A marble of mass $m_1 = 0{,}20\;\mathrm{kg}$ travelling at $v_1 = 2{,}0\;\mathrm{m/s}$ strikes head-on and elastically a second, stationary marble of mass $m_2 = 0{,}10\;\mathrm{kg}$. Find the velocities of the two marbles after the collision.

Solution

In a 1D elastic collision with a stationary target ($v_2 = 0$), both momentum and kinetic energy are conserved. Solving the system gives the standard formulas: $v_1' = \frac{m_1 - m_2}{m_1 + m_2}\,v_1 \qquad v_2' = \frac{2\,m_1}{m_1 + m_2}\,v_1$ Substituting the data. The sum of the masses is $m_1 + m_2 = 0{,}30\;\mathrm{kg}$. $v_1' = \frac{0{,}20 - 0{,}10}{0{,}30}\cdot 2{,}0\;\mathrm{m/s} = \frac{0{,}10}{0{,}30}\cdot 2{,}0\;\mathrm{m/s}$ $v_1' \approx 0{,}67\;\mathrm{m/s}$ $v_2' = \frac{2\cdot 0{,}20}{0{,}30}\cdot 2{,}0\;\mathrm{m/s} = \frac{0{,}40}{0{,}30}\cdot 2{,}0\;\mathrm{m/s}$ $v_2' \approx 2{,}7\;\mathrm{m/s}$ Both marbles continue forwards (the first is heavier and does not bounce back). $\ev{v_1' \approx 0{,}67\;\mathrm{m/s};\quad v_2' \approx 2{,}7\;\mathrm{m/s}}$

Links

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