Answer:
[tex]\vec{v}=\frac{3}{2}v_0\hat{i}-v_0\hat{j}+0\hat{k}[/tex]
Explanation:
Initially the rocket velocity is only on the x-axis. After the rocket suddenly shoots out one-third its mass parallel to the y axis, its velocity is in the x and y axes. Thus, According to momentum conservation principle, we have:
[tex]mv_0=\frac{2}{3}mv_{x}\\v_x}=\frac{3}{2}v_0\\\\0=\frac{1}{3}m2v_0+\frac{2}{3}mv_{y}\\v_y=-\frac{2}{3}m(v_0)[\frac{3}{2m}]\\v_y=-v_0\\\\(0)_z=(0)_z\\v_z=0[/tex]
So, the velocity vector of the remaining rocket's mass is:
[tex]\vec{v}=\frac{3}{2}v_0\hat{i}-v_0\hat{j}+0\hat{k}[/tex]
