A ball with a mass of 2 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
May 9, 2016

vec v_1^'=-7.64 m/s
vec v_2^'=4.36 m/s

Explanation:

"momentums before collision"
"................................................................................................"
m_1=2kg
vec v_1=12m/s
vec P_1=m_1*v_1" (momentum before collision for the first object)"
vec P-1=2*12=24 kg*m/s

m_2=9kg
vec v_2=0
vec P_2=m_2*v_2"(momentum before collision for the second object)"
vec P_2=9.0
vec P_2=0

Sigma vec P_b=vec P_1+vec P_2"( total momentum before collision)"
Sigma vec P_b=24+0
Sigma vec P_b=24 kg*m/s

"momentums after collision"
"................................................................................................"

P_1^'=m_1*vec v_1^'" momentum after collision for the first object"
P_1^'=2*vec v_1^'
P_2^'=m_2*vec v_2^'" momentum after collision for the second object"

P_2^'=9*vec v_2^'

Sigma vec P_a=vec P_1^'+vec P_2^'

Sigma vec P_a=2*vec v_1^'+9*vec v_2^'" (total momentum after collision)"

"conservation of momentum"
Sigma vec vec P_b=Sigma vec P_a

24=2*vec v_1^'+9*vec v_2^' " (1)"

Solution 1:
vec v_1+vec v_1^'=vec v_2+vec v_2^'
12+vec v_1^'=0+v_2^'
v_2^'=12+vec v_1^'

"using (1)"

24=2*vec v_1^'+9(12+vec v_1^')
24=2 vec v_1^'+108+9 vec v_1^'
24-108=11 vec v_1^'
-84=11v_1^'

vec v_1^'=-7.64 m/s

vec v_2^'=12+vec v_1^'

vec v_2^'=12-7.64

vec v_2^'=4.36 m/s

"solution -2:"

v_1^'=(2* vec P_b)/(m_1+m_2)- vec v_1

v_1^'=(2*24)/(2+9)-12" "vec v_1^'=48/11-12

v_1^'=-7.64 m/s

vec v_2^'=(2*vec P_b)/(m_1+m_2) - vec v_2

vec v_2^'=(2*24)/(2+9)-0

vec v_2^'=48/11

vec v_2^'=4.36 m/s