A $1500kg$weather rocket accelerates upward at$10m/s^2$. It explodes 2.0 s after liftoff and breaks into two fragments, one twice as massive as the other. Photos reveal that the lighter fragment traveled straight up and reached a maximum height of 530 m. What were the speed and direction of the heavier fragment just after the explosion?

The mass of the rocket, $m=1500kg$

Acceleration, $a=10m/s^2$

Time, $t=2s$

Maximum height,$h=530m$

The formula for the velocity of the lighter fragment is given:

$v_{\text{lighter}}^2=2g\left(h_{\text{max}}-h_{\text{rocket}}\right)$

Calculating the distance traveled by the rocket prior to its detonation.

$\begin{array}{l}{v}_{\text{rocket}}=a_{\text{rocket}}{t}_{\text{rocket}}=20m{d}_{\text{rocket}}\\ =\frac{v_{\text{rocket}}^2}{2a_{\text{rocket}}}=\frac{a_{\text{rocket}}^2{t}_{\text{rocket}}^2}{2a_{\text{rocket}}}\\ =\frac{100m^2/s^4\cdot 4s^2}{2\cdot 10m/s^2}=20m\end{array}$

So, the speed of the lighter fragment is:

$v_{\text{lighter}}=\sqrt{2\cdot 9.81m/s^2\cdot \left(530m-20m\right)}=100m/s$

We can now use the rule of conservation of momentum to calculate the velocity of the heavier fragment:

$\begin{array}{l}\left(m_{\text{heavier}}+m_{\text{lighter}}\right)v_{\text{rocket}}=m_{\text{heavier}}{v}_{\text{heavier}}+m_{\text{lighter}}{v}_{\text{lighter}}\\ 3m_{\text{lighter}}{v}_{\text{rocket}}=2m_{\text{lighter}}{v}_{\text{heavier}}+m_{\text{lighter}}{m}_{\text{lighter}}\\ 3\times 20m/s=2\times v_{\text{heavier}}+100m/s\\ \Rightarrow v_{\text{heavier}}=-20m/s\end{array}$

where the negative sign indicates that the heavier piece is moving downwards (opposite direction of rocket and lighter fragment).