# A ball with a mass of 2 kg is rolling at 9 m/s and elastically collides with a resting ball with a mass of 1 kg. What are the post-collision velocities of the balls?

Feb 19, 2016

No $\cancel{{v}_{1} = 3 \frac{m}{s}}$
No $\cancel{{v}_{2} = 12 \frac{m}{s}}$

the speed after collision of the two objects are see below fro explanation:
$\textcolor{red}{v {'}_{1} = 2.64 \frac{m}{s} , v {'}_{2} = 12.72 \frac{m}{s}}$

#### Explanation:

$\text{use the conversation of momentum}$
$2 \cdot 9 + 0 = 2 \cdot {v}_{1} + 1 \cdot {v}_{2}$
$18 = 2 \cdot {v}_{1} + {v}_{2}$
$9 + {v}_{1} = 0 + {v}_{2}$
${v}_{2} = 9 + {v}_{1}$
$18 = 2 \cdot {v}_{1} + 9 + {v}_{1}$
$18 - 9 = 3 \cdot {v}_{1}$
$9 = 3 \cdot {v}_{1}$
${v}_{1} = 3 \frac{m}{s}$
${v}_{2} = 9 + 3$
${v}_{2} = 12 \frac{m}{s}$

Because there are two unknown I am not sure how you able to solve the above without using, conservation of momentum and conservation of energy (elastic collision). The combination of the two yields 2 equation and 2 unknown which you then solve:

Conservation of "Momentum":
${m}_{1} {v}_{1} + {m}_{2} {v}_{2} = {m}_{1} v {'}_{1} + {m}_{2} v {'}_{2}$ =======> (1)

Let, m_1 = 2kg; m_2 = 1 kg; v_1=9m/s; v_2=0m/s

Conservation of energy (elastic collision):
$\frac{1}{2} {m}_{1} {v}_{1}^{2} + \frac{1}{2} {m}_{2} {v}_{2}^{2} = \frac{1}{2} {m}_{1} v {'}_{1}^{2} + \frac{1}{2} {m}_{2} v {'}_{2}^{2}$ =======> (2)

We have 2 equations and 2 unknowns:
From (1) ==> 2*9 = 2v'_1 + v'_2; color(blue)(v'_2 = 2(9-v'_1)) ==>(3)
From (2) ==> ${9}^{2} = v {'}_{1}^{2} + \frac{1}{2} v {'}_{2}^{2}$ ===================> (4)

Insert $\left(3\right) \implies \left(4\right)$:

${9}^{2} = v {'}_{1}^{2} + \frac{1}{2} \cdot {\left[\textcolor{b l u e}{2 \left(9 - v {'}_{1}\right)}\right]}^{2}$ expand
${9}^{2} = v {'}_{1}^{2} + 2 \left({9}^{2} - 18 v {'}_{1} + v {'}_{1}^{2}\right)$
$2 v {'}_{1}^{2} - 36 v {'}_{1} + {9}^{2} = 0$ solve the quadratic equation for $v {'}_{1}$
v'_1 = (b +-sqrt(b^2 - 4ac)/2a); v'_1 => (2.64, 15.36)
The solution that make sense is 2.64 (explain why?)
Insert in (3) and solve color(blue)(v'_2 = 2(9-color(red)2.64) = 12.72
So the speed after collision of the two objects are:
$v {'}_{1} = 2.64 \frac{m}{s} , v {'}_{2} = 12.72$

Feb 20, 2016

${v}_{1} = 3 \frac{m}{s}$
${v}_{2} = 12 \frac{m}{2}$

#### Explanation:

${m}_{1} \cdot {v}_{1} + {m}_{2} \cdot {v}_{2} = {m}_{1} \cdot {v}_{1} ' + {m}_{2} \cdot {v}_{2}^{'} \text{ (1)}$
cancel(1/2)*m_1*v_1^2+cancel(1/2)*m_2*v_2^2=cancel(1/2)*m_1*v_1^('2)+cancel(1/2)*m_2*v_2^('2) "
${m}_{1} \cdot {v}_{1}^{2} + {m}_{2} \cdot {v}_{2}^{2} = {m}_{1} \cdot {v}_{1}^{' 2} + {m}_{2} \cdot {v}_{2}^{' 2} \text{ (2)}$
${m}_{1} \cdot {v}_{1} - {m}_{1} \cdot {v}_{1}^{'} = {m}_{2} \cdot {v}_{2}^{'} - {m}_{2} \cdot {v}_{2} \text{ redeployment of (1)}$
${m}_{1} \left({v}_{1} - {v}_{1}^{'}\right) = {m}_{2} \left({v}_{2}^{'} - {v}_{2}\right) \text{ (3)}$
${m}_{1} \cdot {v}_{1}^{2} - {m}_{1} \cdot {v}_{1}^{' 2} = {m}_{2} \cdot {v}_{2}^{' 2} - {m}_{2} \cdot {v}_{2}^{2} \text{ redeployment of (2)}$
${m}_{1} \left({v}_{1}^{2} - {v}_{1}^{' 2}\right) = {m}_{2} \left({v}_{2}^{' 2} - {v}_{2}^{2}\right) \text{ (4)}$
$\text{divide :(3)/(4)}$
$\frac{{m}_{1} \left({v}_{1} - {v}_{1}^{'}\right)}{{m}_{1} \left({v}_{1}^{2} - {v}_{1}^{' 2}\right)} = \frac{{m}_{2} \left({v}_{2}^{'} - {v}_{2}\right)}{{m}_{2} \left({v}_{2}^{' 2} - {v}_{2}^{2}\right)}$
$\frac{{v}_{1} - {v}_{1}^{'}}{\left({v}_{1}^{2} - {v}_{1}^{' 2}\right)} = \frac{\left({v}_{2}^{'} - {v}_{2}\right)}{\left({v}_{2}^{' 2} - {v}_{2}^{2}\right)}$
v_1^2-v_1^('2)=(v_1+v_1^')*(v_1-v_1^') ; v_2^('2)=(v_2^'+v_2)*(v_2^'-v_2)
${v}_{1} + {v}_{1}^{'} = {v}_{2} + {v}_{2}^{'}$