A charging bull elephant with a mass of 5230 kg comes directly toward you with a speed of 4.45 m/s. You toss a 0.150-kg rubber ball at the elephant with a speed of 7.91 m/s. (a) When the ball bounces back toward you, what is its speed? (b) How do you account for the fact that the ball’s kinetic energy has increased
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-
\(m_{1} \ and\ m_{2} \ respectively\)
Let the initial velocity of the elephant and rubber ball is denoted as -
\(u_{1} \ and\ u_{2}\ respectively\)
Let the final velocities of the elephant and rubber ball is denoted as-
\(v_{1} \ and\ v_{2} \ respectively\)
As per the question-
\(m_{1} = 5230 kg\) \(u_{1} =4.45 m/s\)
\(m_{2} =0.150 kg\) \(u_{2} =7.91 m/s\)
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.
\(v_{1} =\frac{m_{1}- m_{2}} {m_{1}+ m_{2}}*u_{1} +\frac{2m_{2}} {m_{1}+ m_{2}}*u_{2}\)
\(v_{1} =\frac{5230-0.150}{5230+0.150}*[4.45] +\frac{2*0.150}{5230+0.150} *[-7.91]\\\) [ Here the velocity of rubber ball is taken as negative as it is opposite to the direction of motion of elephant.]
\(v_{1} =\frac{5229.85}{5230.15} *[4.45]-\frac{0.3}{5230.15} *[7.91]\)
\(v_{1} =4.449291034m/s\)
Now we have to calculate velocity of rubber ball.
\(v_{2} =\frac{m_{2}- m_{1}} {m_{1}+ m_{2}}*u_{2} +\frac{2m_{1}} {m_{1}+ m_{2}} * u_{1}\)
\(v_{2} =\frac{0.150-5230}{5230+0.150} *[-7.91]+\frac{2*5230}{5230+.150} *[4.45]\)
\(v_{2} =\frac{-5229.85}{5230.15} *[-7.91]+\frac{10460}{5230.15} *[4.45]\)
\(v_{2} =16.80929103 m/s\)
Here we see that velocity of rubber ball is increased.
The kinetic energy of the rubber ball is given as -
\(kinetic energy [K.E]=\frac{1}{2} mv^2\)
As the velocity of the ball is increased,hence, its kinetic energy is increased.
