Question

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

Answer 1

You have to add the speed of that ball plus the speed is running so 4.45 + 7.91= 12.36m/s

Answer 2

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.