Hockey pucks

Problem

Two hockey pucks of masses $m_1 = 0{,}30$ kg and $m_2 = 0{,}10$ kg move along the same line with velocities $v_1 = +2{,}0$ m/s and $v_2 = -3{,}0$ m/s and collide elastically. Calculate their final velocities.

Solution

1D elastic collision: the general formulas hold (from the balances of $\vv{p}$ and $E_c$) $\begin{aligned} v_1' &= \frac{m_1-m_2}{m_1+m_2}\,v_1 + \frac{2 m_2}{m_1+m_2}\,v_2 \\ v_2' &= \frac{2 m_1}{m_1+m_2}\,v_1 + \frac{m_2-m_1}{m_1+m_2}\,v_2 \end{aligned}$ With $m_1+m_2 = 0{,}40$ kg: $v_1' = \frac{0{,}30-0{,}10}{0{,}40}\,(2{,}0) + \frac{2\cdot 0{,}10}{0{,}40}\,(-3{,}0) = 0{,}5\cdot 2{,}0 + 0{,}5\cdot(-3{,}0) = -0{,}5\ \mathrm{m/s}$ $v_2' = \frac{2\cdot 0{,}30}{0{,}40}\,(2{,}0) + \frac{0{,}10-0{,}30}{0{,}40}\,(-3{,}0) = 1{,}5\cdot 2{,}0 + (-0{,}5)\cdot(-3{,}0) = 4{,}5\ \mathrm{m/s}$

Momentum check: $p_i = 0{,}30\cdot 2{,}0 + 0{,}10\cdot(-3{,}0) = 0{,}30$; $p_f = 0{,}30\cdot(-0{,}5) + 0{,}10\cdot 4{,}5 = 0{,}30$. ✓ Relative velocity check (must reverse): before $v_1 - v_2 = 5{,}0$; after $v_1' - v_2' = -5{,}0$. ✓

Final result: $\ev{v_1' = -0{,}5\ \mathrm{m/s};\quad v_2' = +4{,}5\ \mathrm{m/s}}$

Note (source erratum)

The textbook gives $v_1' = -1{,}0$ m/s, $v_2' = +4{,}0$ m/s as the answer: incorrect values, which violate both conservation of momentum ($p_f = 0{,}1 \neq 0{,}3$) and conservation of kinetic energy. The correct values are $v_1' = -0{,}5$ m/s, $v_2' = +4{,}5$ m/s.

Links

Topics: Quantità di moto e urti Concepts: Urto elastico · Conservazione della quantità di moto Skills: Conservazione della quantità di moto Methods: Metodo delle due tabelle parallele