Worked exercise — partially inelastic collision

Problem

Two balls ($m_1 = 3\;\text{kg}$, $v_{1A} = 4\;\text{m/s}$; $m_2 = 5\;\text{kg}$, $v_{2A} = -7\;\text{m/s}$) collide. 30% of the internal kinetic energy becomes heat. Find the final velocities.

Solution

Table of $\vv{p}$ (1D, $x$ component; $p$ in $\text{kg m/s}$):

	$p_1$	$p_2$	$p_\text{tot}$
A	$12$	$-35$	$-23$
B	$3v_{1B}$	$5v_{2B}$	$-23$

$3v_{1B} + 5v_{2B} = -23 \qquad\text{(I)}$

Velocity of the centre of mass (conserved): $v_\text{CM} = \frac{p_\text{tot}}{M_\text{tot}} = \frac{-23}{8} = -2{,}875\;\text{m/s}$

Internal kinetic energy (the only part that can become heat, i.e. that of the motion relative to the centre of mass). Total kinetic energy and that of the CM: $E_\text{kin,tot} = \tfrac{1}{2}\cdot 3\cdot 4^2 + \tfrac{1}{2}\cdot 5\cdot 7^2 = 24 + 122{,}5 = 146{,}5\;\text{J}$ $E_\text{kin,CM} = \tfrac{1}{2}\,M_\text{tot}\,v_\text{CM}^2 = \tfrac{1}{2}\cdot 8\cdot 2{,}875^2 = 33{,}1\;\text{J}$ $E_\text{kin,int} = E_\text{kin,tot} - E_\text{kin,CM} = 146{,}5 - 33{,}1 = 113{,}4\;\text{J}$

30% becomes heat: $E_\text{heat} = 0{,}3\cdot 113{,}4 = 34{,}0\;\text{J}$

Table of $E$ ($E$ in $\text{J}$):

	$E_{\text{kin},1}$	$E_{\text{kin},2}$	$E_\text{heat}$	$E_\text{tot}$
A	$24$	$122{,}5$	$0$	$146{,}5$
B	$\tfrac{3}{2}v_{1B}^2$	$\tfrac{5}{2}v_{2B}^2$	$34{,}0$	$146{,}5$

$\tfrac{3}{2}v_{1B}^2 + \tfrac{5}{2}v_{2B}^2 = 146{,}5 - 34{,}0 = 112{,}5 \qquad\text{(II)}$

Solving the system (I)–(II). From (I): $v_{1B} = \dfrac{-23 - 5v_{2B}}{3}$. Substituting into (II) gives a second-degree equation in $v_{2B}$: $\tfrac{3}{2}\!\left(\frac{-23 - 5v_{2B}}{3}\right)^{\!2} + \tfrac{5}{2}v_{2B}^2 = 112{,}5$ $20v_{2B}^2 + 115\,v_{2B} - 73 = 0 \quad\Rightarrow\quad v_{2B} = \frac{-115 \pm \sqrt{115^2 + 4\cdot 20\cdot 73}}{2\cdot 20}$ $v_{2B} = \frac{-115 \pm 138{,}1}{40}$ The two roots correspond to “before” and “after” the collision; the physical one (the bodies separate, with the relative velocity reversing) is $\ev{v_{2B} \approx 0{,}58\;\mathrm{m/s},\quad v_{1B} \approx -8{,}63\;\mathrm{m/s}}$ The other root ($v_{2B} \approx -6{,}33$, $v_{1B} \approx 2{,}88$), with the bodies still approaching, is the “before-collision” solution: it is discarded. The two-table method provided the two equations (I) and (II) needed to close the problem.

Links

Topics: Momentum and collisions Concepts: Inelastic collision · Conservation of momentum · Kinetic energy · Centre of mass Skills: Two-table method Methods: Parallel two-table method