# A ball with a mass of  5 kg is rolling at 5 m/s and elastically collides with a resting ball with a mass of 2 kg. What are the post-collision velocities of the balls?

Mar 7, 2017

$\text{The answer " vec v_("2_after")=50/7" "m/s" , "vec v_("1_after")=15/7" } \frac{m}{s}$

#### Explanation:

$\text{The definition of the momentum is given as } \vec{P} = m \cdot \vec{v}$
$\text{Where ;}$
$\text{m is mass of object and v is its velocity}$

$\text{if object has not a velocity,it will not have any momentum}$
$\left(\vec{P} = m \cdot 0 = 0\right)$

$\text{The momentum of any object is a vectorial quantity,}$
$\text{it has a magnitude and direction.}$

$\text{The momentum problems is solved by conservation of momentum}$

$\text{let us find momentums of object for before impact}$
$\text{.........................................................................................}$
${\vec{P}}_{\text{1_before")=m_1* vec v_("1_before}}$
$\text{(momentum before impact for a mass of 5 kg)}$
$\text{plug m=5 kg and v=5 } \frac{m}{s}$

vec P_("1_before")=5*5=25 " "kg*m/s

${\vec{P}}_{\text{2_before")=m_2* vec v_("2_before}}$
$\text{(momentum before impact for a mass of 2 kg)}$
$\text{plug m=2 kg and v=0 } \frac{m}{s}$

vec P_("2_before")=2*0=0 " "kg*m/s

$\text{The vectorial sum of the momentums before impact is}$

 color(red)(Sigma vec P_("before"))=vec P_("1_before")+vec P_("2_before")

$\text{if collision is centered, the total momentum}$
$\text{will be "color(red)(Sigma vec P_("before"))=25+0=25 " } k g \cdot \frac{m}{s}$

$\text{now let us find the momentums of objects for after impact}$
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .$
${\vec{P}}_{\text{1_after")=m_1* vec v_("1_after}}$
$\text{(momentum after impact for a mass of 5 kg)}$

${\vec{P}}_{\text{1_after")=5*vec v_("1_after}}$

${\vec{P}}_{\text{2_after")=m_2* vec v_("2_after}}$
$\text{(momentum after impact for a mass of 2 kg)}$

${\vec{P}}_{\text{2_after")=2*vec v_("2_after}}$

$\text{The vectorial sum of the momentums after impact is}$

color(blue)(Sigma vec P_("after"))=vec P_("1_after")+vec P_("2_after")

color(blue)(Sigma vec P_("after"))=5*vec v_("1_after") +2*vec v_("2_after")

$\text{Let us write the conservation of momentum}$
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . .$

$\textcolor{red}{\Sigma {\vec{P}}_{\text{before"))=color(blue)(Sigma vec P_("after}}}$

25=5*vec v_("1_after") +2*vec v_("2_after")" " "(1)"

vec v_("1_before")+ vec v_("1_after")=vec v_("2_before")+vec v_("2_after")" "(2)

$\text{note: you can proof the equation (2) using together momentum }$$\text{and kinetic energy conservations equations. }$

5+vec v_("1_after")=0+vec v_("2_after")" "(3)

${\vec{v}}_{\text{2_after")=5+vec v_("1_after}}$

$\text{now let us use (1)}$

25=5*vec v_("1_after") +2(5+vec v_("1_after"))

$25 = 5 \cdot {\vec{v}}_{\text{1_after") +10+2*vec v_("1_after}}$

$25 - 10 = 7 \cdot {\vec{v}}_{\text{1_after}}$

$15 = 7 \cdot {\vec{v}}_{\text{1_after}}$

vec v_("1_after")=15/7" "m/s

$\text{use (3)}$

$5 + \frac{15}{7} = {\vec{v}}_{\text{2_after}}$

vec v_("2_after")=50/7" "m/s