Answer
After the explosion, the speed of the heavier fragment is 20 m/s and it is moving straight down.
Work Step by Step
We first find the velocity of the rocket after it accelerates upward for 2.00 seconds.
$v = a~t$
$v = (10.0~m/s^2)(2.00~s)$
$v = 20.0~m/s$
We then find the momentum of the rocket just before it explodes. We can call this momentum $p_0$.
$p_0 = m~v$
$p_0 = (1500~kg)(20.0~m/s)$
$p_0 = 30,000~N~s$
Next, we find the height $h$ of the rocket just before it explodes:
$h = \frac{1}{2}at^2$
$h = \frac{1}{2}(10.0~m/s^2)(2.00~s)^2$
$h = 20~m$
After the explosion, the lighter fragment (with mass 500 kg) reaches a height that is 510 meters above the point of the explosion. We can find the lighter piece's velocity $v_L$ just after the explosion.
$v_f^2 = v_L^2+2gy$
$v_L^2 = 0-2gy$
$v_L = \sqrt{-2gy}$
$v_L = \sqrt{-(2)(-9.80~m/s^2)(510~m)}$
$v_L = 100~m/s$
Just after the rocket explodes, the sum of the momenta of the two fragments will be equal to $p_0$ (by conservation of momentum). We can find the velocity $v_H$ of the heavier piece (with mass 1000 kg) just after the explosion.
$p_f=p_0$
$(1000~kg)~v_H+(500~kg)(100~m/s) = 30,000~N~s$
$(1000~kg)~v_H = -20,000~N~s$
$v_H = -20~m/s$
After the explosion, the speed of the heavier fragment is 20 m/s and it is moving straight down.
