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?

#"Before"#
#"................................................................"#
#m_1:3" "kg" ""'mass of the first object'"#
#vec v_1:12" "m/s" ""'velocity of the first object'"#
#vec P_1:" 'the first object's momentum before collision'"#
#vec P_1=m_1*vec v_1#
#vec P_1=3*12=36 " "kg*m/s#

#m_2:9""kg" ""'mass of the second object'"#
#vec v_2=0#
#vec P_2" 'the second object's momentum before collision'"#
#vec P_2=m_2*vec v_2#
#vecP_2=9*0=0#

#Sigma P_b:" '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" "kg*m/s #

#"After"#
#......................................................................"#
#vec v_1^':" 'the first object's velocity after collision'"#
#vec P_1^':" 'the first object's momentum after collision'"#

#vec P_1^'=m_1*v_1^'=3*v_1^'#

#vec v_2^':" 'the second object's velocity after collision"#
#vec P_2^':"' the second object's momentum after collision'"#

#vec P_2^'=m_2*v_2^'=9*v_2^'#

#Sigma P_b:" 'The vectorial sum of the momentums after collision'"#

#vec Sigma P_a=P_1^'+P_2^'#
#vec Sigma P_a⁼3v_1^'+9v_2^'#

#"momentum is reserved in elastic collisions."#

#vec Sigma P_b=vec Sigma P_a#

#36=3v_1^'+9v_2^'" (1)"#

#m_1*v_1+m_2*v_2=m_1*v_1^'+m_2*v_2^'" (2)"#

#1/2*m_1*v_1^2+1/2*m_2*v_2^2=1/2*m_1*v_1^('2)+1/2*m_2*v_2^('2)" (3)"#

#"you can obtain the equation " v_1+v_1^'=v_2+v_2^'" using (2) and (3)"#

#"Thus;"#

#12+v_1^'=0+v_2^'" (4)"#

#color(red)(v_2^'=12+v_1^')#

#"plug into (1)"#

#36=3v_1^'+9(color(red)(12+v_1^'))#

#36=3v_1^'+108+9v_1^'#

#12*vec v_1^'=36-108#
#12*vec v_1^'=-72#

#vec v_1^'=-72/12#

#vec v_1^'=-6" "m/s#

#"now use (4)"#

#vec v_2^'=12-6#

#vec v_2^'=6" "m/s#

#"Testing momentum..."#

#"Total momentum before:"36 " "kg*m/s#

#"Total momentum after:"3*(-6)+9*6=-18+54=36" "kg*m/s#

#Sigma vec P_b=Sigma vec P_a#
#"Momentum has conserved"#

#"Testing the kinetic energy..."#

#E_b=1/2*3*12^2+0=72*3=216J" 'Before'"#
#E_a=1/2*3*(-6)^2+1/2*9*6^2#

#E_a=54+162#
#E_a=216J#

#E_b=E_a#
#"The Kinetic energy has conserved"#