Answer
The amount of energy that is released is $~~3.2\times 10^6~J$
Work Step by Step
We can find the initial kinetic energy:
$K_i = \frac{1}{2}mv^2$
$K_i = \frac{1}{2}(20.0~kg)(200~m/s)^2$
$K_i = 4.0\times 10^5~J$
In part (a), we found that the velocity of the third part is $(1000~\hat{i}-166.7~\hat{j})~m/s$
We can find the speed of the third part:
$v_3 = \sqrt{(1000~m/s)^2+(-166.7~m/s)^2}$
$v_3 = 1013.8~m/s$
We can find the total kinetic energy after the explosion:
$K_f = \frac{1}{2}m_1~v_1^2+\frac{1}{2}m_2~v_2^2+\frac{1}{2}m_3~v_3^2$
$K_f = \frac{1}{2}(10.0~kg)(100~m/s)^2+\frac{1}{2}(4.0~kg)(-500~m/s)^2+\frac{1}{2}(6.0~kg)(1013.8~m/s)^2$
$K_f = 3.6\times 10^6~J$
We can find the amount of energy that was released:
$\Delta K = K_f-K_i$
$\Delta K = (3.6\times 10^6~J)-(4.0\times 10^5~J)$
$\Delta K = 3.2\times 10^6~J$
The amount of energy that is released is $~~3.2\times 10^6~J$
