To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.
By definition kinetic energy is defined as
[tex]KE = \frac{1}{2} mv^2[/tex]
Where,
m = Mass
v = Velocity
On the other hand we have the conservation of the moment, which for this case would be defined as
[tex]m*V_i = m_1V_1+m_2V_2[/tex]
Here,
m = Total mass (8Kg at this case)
[tex]m_1=m_2 =[/tex] Mass each part
[tex]V_i =[/tex] Initial velocity
[tex]V_2 =[/tex] Final velocity particle 2
[tex]V_1 =[/tex] Final velocity particle 1
The initial kinetic energy would be given by,
[tex]KE_i=\frac{1}{2}mv^2[/tex]
[tex]KE_i = \frac{1}{2}8*5^2[/tex]
[tex]KE_i = 100J[/tex]
In the end the energy increased 100J, that is,
[tex]KE_f = KE_i KE_{increased}[/tex]
[tex]KE_f = 100+100 = 200J[/tex]
By conservation of the moment then,
[tex]m*V_i = m_1V_1+m_2V_2[/tex]
Replacing we have,
[tex](8)*5 = 4*V_1+4*V_2[/tex]
[tex]40 = 4(V_1+V_2)[/tex]
[tex]V_1+V_2 = 10[/tex]
[tex]V_2 = 10-V_1[/tex](1)
In the final point the cinematic energy of EACH particle would be given by
[tex]KE_f = \frac{1}{2}mv^2[/tex]
[tex]KE_f = \frac{1}{2}4*(V_1^2+V_2^2)[/tex]
[tex]200J=\frac{1}{2}4*(V_1^2+V_2^2)[/tex](2)
So we have a system of 2x2 equations
[tex]V_2 = 10-V_1[/tex]
[tex]200J=\frac{1}{2}4*(V_1^2+V_2^2)[/tex]
Replacing (1) in (2) and solving we have to,
200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)
PART A: [tex]V_1 = 10m/s[/tex]
Then replacing in (1) we have that
PART B: [tex]V_2 = 0m/s[/tex]
