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?

1 Answer
Aug 19, 2016

vec v_1^'=-6" "m/s
vec v_2^'=6" "m/s
"The kinetic energy and momentum are reserved"

Explanation:

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"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"