# A ball with a mass of  3 kg is rolling at 12 m/s and elastically collides with a resting ball with a mass of 9 kg. What are the post-collision velocities of the balls?

Aug 19, 2016

${\vec{v}}_{1}^{'} = - 6 \text{ } \frac{m}{s}$
${\vec{v}}_{2}^{'} = 6 \text{ } \frac{m}{s}$
$\text{The kinetic energy and momentum are reserved}$

#### Explanation:

$\text{Before}$
$\text{................................................................}$
${m}_{1} : 3 \text{ "kg" ""'mass of the first object'}$
${\vec{v}}_{1} : 12 \text{ "m/s" ""'velocity of the first object'}$
${\vec{P}}_{1} : \text{ 'the first object's momentum before collision'}$
${\vec{P}}_{1} = {m}_{1} \cdot {\vec{v}}_{1}$
${\vec{P}}_{1} = 3 \cdot 12 = 36 \text{ } k g \cdot \frac{m}{s}$

${m}_{2} : 9 \text{kg" ""'mass of the second object'}$
${\vec{v}}_{2} = 0$
${\vec{P}}_{2} \text{ 'the second object's momentum before collision'}$
${\vec{P}}_{2} = {m}_{2} \cdot {\vec{v}}_{2}$
${\vec{P}}_{2} = 9 \cdot 0 = 0$

$\Sigma {P}_{b} : \text{ 'The vectorial sum of the momentums before collision'}$

$\vec{\Sigma} {P}_{b} = {\vec{P}}_{1} + {\vec{P}}_{2}$
$\vec{\Sigma} {P}_{b} = 36 + 0 = 36 \text{ } k g \cdot \frac{m}{s}$

$\text{After}$
......................................................................"
${\vec{v}}_{1}^{'} : \text{ 'the first object's velocity after collision'}$
${\vec{P}}_{1}^{'} : \text{ 'the first object's momentum after collision'}$

${\vec{P}}_{1}^{'} = {m}_{1} \cdot {v}_{1}^{'} = 3 \cdot {v}_{1}^{'}$

${\vec{v}}_{2}^{'} : \text{ 'the second object's velocity after collision}$
${\vec{P}}_{2}^{'} : \text{' the second object's momentum after collision'}$

${\vec{P}}_{2}^{'} = {m}_{2} \cdot {v}_{2}^{'} = 9 \cdot {v}_{2}^{'}$

$\Sigma {P}_{b} : \text{ 'The vectorial sum of the momentums after collision'}$

$\vec{\Sigma} {P}_{a} = {P}_{1}^{'} + {P}_{2}^{'}$
vec Sigma P_a⁼3v_1^'+9v_2^'

$\text{momentum is reserved in elastic collisions.}$

$\vec{\Sigma} {P}_{b} = \vec{\Sigma} {P}_{a}$

$36 = 3 {v}_{1}^{'} + 9 {v}_{2}^{'} \text{ (1)}$

${m}_{1} \cdot {v}_{1} + {m}_{2} \cdot {v}_{2} = {m}_{1} \cdot {v}_{1}^{'} + {m}_{2} \cdot {v}_{2}^{'} \text{ (2)}$

$\frac{1}{2} \cdot {m}_{1} \cdot {v}_{1}^{2} + \frac{1}{2} \cdot {m}_{2} \cdot {v}_{2}^{2} = \frac{1}{2} \cdot {m}_{1} \cdot {v}_{1}^{' 2} + \frac{1}{2} \cdot {m}_{2} \cdot {v}_{2}^{' 2} \text{ (3)}$

$\text{you can obtain the equation " v_1+v_1^'=v_2+v_2^'" using (2) and (3)}$

$\text{Thus;}$

$12 + {v}_{1}^{'} = 0 + {v}_{2}^{'} \text{ (4)}$

$\textcolor{red}{{v}_{2}^{'} = 12 + {v}_{1}^{'}}$

$\text{plug into (1)}$

$36 = 3 {v}_{1}^{'} + 9 \left(\textcolor{red}{12 + {v}_{1}^{'}}\right)$

$36 = 3 {v}_{1}^{'} + 108 + 9 {v}_{1}^{'}$

$12 \cdot {\vec{v}}_{1}^{'} = 36 - 108$
$12 \cdot {\vec{v}}_{1}^{'} = - 72$

${\vec{v}}_{1}^{'} = - \frac{72}{12}$

${\vec{v}}_{1}^{'} = - 6 \text{ } \frac{m}{s}$

$\text{now use (4)}$

${\vec{v}}_{2}^{'} = 12 - 6$

${\vec{v}}_{2}^{'} = 6 \text{ } \frac{m}{s}$

$\text{Testing momentum...}$

$\text{Total momentum before:"36 " } k g \cdot \frac{m}{s}$

$\text{Total momentum after:"3*(-6)+9*6=-18+54=36" } k g \cdot \frac{m}{s}$

$\Sigma {\vec{P}}_{b} = \Sigma {\vec{P}}_{a}$
$\text{Momentum has conserved}$

$\text{Testing the kinetic energy...}$

${E}_{b} = \frac{1}{2} \cdot 3 \cdot {12}^{2} + 0 = 72 \cdot 3 = 216 J \text{ 'Before'}$
${E}_{a} = \frac{1}{2} \cdot 3 \cdot {\left(- 6\right)}^{2} + \frac{1}{2} \cdot 9 \cdot {6}^{2}$

${E}_{a} = 54 + 162$
${E}_{a} = 216 J$

${E}_{b} = {E}_{a}$
$\text{The Kinetic energy has conserved}$