Answer
(a) The speed of the 0.28-kg fragment is 71.5 m/s and the speed of the 1.40-kg fragment is 14.3 m/s.
(b) The two fragments are a distance of 347 meters apart when they land on the ground.
Work Step by Step
(a) Let $m_1$ be the lighter piece and let $m_2$ be the heavier piece. We can use conservation of momentum to express $v_1$ in terms of $v_2$.
$m_1v_1 + m_2v_2 = 0$
$m_1v_1 = -m_2v_2$
$v_1 = \frac{-m_2v_2}{m_1}$
We can use an energy equation to find $v_2$, the speed of the larger piece after the explosion.
$K = 860~J$
$\frac{1}{2}m_1~v_1^2+\frac{1}{2}m_2~v_2^2 = 860~J$
$m_1~(\frac{-m_2v_2}{m_1})^2+m_2~v_2^2 = 1720~J$
$(\frac{m_2^2}{m_1})v_2^2+m_2~v_2^2 = 1720~J$
$v_2^2 = \frac{(1720~J)(m_1)}{(m_2^2)+m_1~m_2}$
$v_2 = \sqrt{\frac{(1720~J)(m_1)}{(m_2^2)+m_1~m_2}}$
$v_2 = \sqrt{\frac{(1720~J)(0.28~kg)}{(1.40~kg)^2+(0.28~kg)(1.40~kg)}}$
$v_2 = 14.3~m/s$
We can use $v_2$ to find $v_1$.
$v_1 = \frac{-m_2v_2}{m_1}$
$v_1 = \frac{-(1.40~kg)(14.3~m/s)}{0.28~kg}$
$v_1 = -71.5~m/s$
The speed of the 0.28-kg fragment is 71.5 m/s and the speed of the 1.40-kg fragment is 14.3 m/s.
(b) Since the two fragments hit the ground at the same time, the fragments must have been moving horizontally in opposite directions after the explosion. We can find the time it takes for the fragments to fall to the ground.
$y = \frac{1}{2}gt^2$
$t = \sqrt{\frac{2y}{g}}$
$t = \sqrt{\frac{(2)(80.0~m)}{9.80~m/s^2}}$
$t = 4.04~s$
As the two fragments move away from each other, the relative speed is 85.8 m/s. We can find the distance $x$ between the fragments when they land on the ground.
$x = (85.8~m/s)(4.04~s)$
$x = 347~m$
The two fragments are a distance of 347 meters apart when they land on the ground.
