Answer
(a) The other fragment strikes the ground a distance of 848 meters from the point of firing.
(b) The amount of energy released in the explosion is 16000 J.
Work Step by Step
(a) We can find the initial horizontal speed $v_{0x}$.
$v_{0x} = v_0~\cos(\theta)$
$v_{0x} = (80.0~m/s)~\cos(60.0^{\circ})$
$v_{0x} = 40.0~m/s$
We can use conservation of momentum to find the speed of the fragment that does not drop straight down after the explosion.
$(\frac{m}{2})(v_{x2}) = m~v_{0x}$
$v_{x2} = 2v_{0x}$
$v_{x2} = (2)(40.0~m/s)$
$v_{x2} = 80.0~m/s$
We can find the time for the projectile to go up to maximum height.
$t = \frac{v_0~\sin(\theta)}{g}$
$t = \frac{(80.0~m/s)~\sin(60.0^{\circ})}{9.80~m/s^2}$
$t = 7.07~s$
Note that the time to fall back down is also 7.07 seconds.
We can find the total horizontal distance $x$.
$x = (40.0~m/s)(7.07~s)+(80.0~m/s)(7.07~s)$
$x = 848~m$
The other fragment strikes the ground a distance of 848 meters from the point of firing.
(b) We can find the kinetic energy just before the explosion.
$K_1 = \frac{1}{2}mv_{0x}^2$
$K_1 = \frac{1}{2}(20.0~kg)(40.0~m/s)^2$
$K_1 = 16000~J$
We can find the kinetic energy just after the explosion.
$K_2 = \frac{1}{2}\frac{m}{2}~v_{x2}^2$
$K_2 = \frac{1}{2}(10.0~kg)(80.0~m/s)^2$
$K_2 = 32000~J$
The amount of energy released in the explosion is 16000 J.
