Answer :
You have to add the speed of that ball plus the speed is running so 4.45 + 7.91= 12.36m/s
Let us consider the masses of elephant and rubber ball was denoted as-
[tex]m_{1} \ and\ m_{2} \ respectively[/tex]
Let the initial velocity of the elephant and rubber ball is denoted as -
[tex]u_{1} \ and\ u_{2}\ respectively[/tex]
Let the final velocities of the elephant and rubber ball is denoted as-
[tex]v_{1} \ and\ v_{2} \ respectively[/tex]
As per the question-
[tex]m_{1} = 5230 kg[/tex] [tex]u_{1} =4.45 m/s[/tex]
[tex]m_{2} =0.150 kg[/tex] [tex]u_{2} =7.91 m/s[/tex]
We are asked to calculate the velocities of rubber ball and charged elephant.
From law of conservation of momentum and kinetic energy, we know that-
First we have to calculate the velocity of elephant.
[tex]v_{1} =\frac{m_{1}- m_{2}} {m_{1}+ m_{2}}*u_{1} +\frac{2m_{2}} {m_{1}+ m_{2}}*u_{2}[/tex]
[tex]v_{1} =\frac{5230-0.150}{5230+0.150}*[4.45] +\frac{2*0.150}{5230+0.150} *[-7.91]\\[/tex] [ Here the velocity of rubber ball is taken as negative as it is opposite to the direction of motion of elephant.]
[tex]v_{1} =\frac{5229.85}{5230.15} *[4.45]-\frac{0.3}{5230.15} *[7.91][/tex]
[tex]v_{1} =4.449291034m/s[/tex]
Now we have to calculate velocity of rubber ball.
[tex]v_{2} =\frac{m_{2}- m_{1}} {m_{1}+ m_{2}}*u_{2} +\frac{2m_{1}} {m_{1}+ m_{2}} * u_{1}[/tex]
[tex]v_{2} =\frac{0.150-5230}{5230+0.150} *[-7.91]+\frac{2*5230}{5230+.150} *[4.45][/tex]
[tex]v_{2} =\frac{-5229.85}{5230.15} *[-7.91]+\frac{10460}{5230.15} *[4.45][/tex]
[tex]v_{2} =16.80929103 m/s[/tex]
Here we see that velocity of rubber ball is increased.
The kinetic energy of the rubber ball is given as -
[tex]kinetic energy [K.E]=\frac{1}{2} mv^2[/tex]
As the velocity of the ball is increased,hence, its kinetic energy is increased.