Question

A rocket with mass m traveling with speed v0 along the x axis suddenly shoots out one-third its mass parallel to the y axis (as seen from an observer at rest) with speed 2v0. Find the x, y, and z components of the final velocity vector of the remaining 2/3 of the rocket's mass.

Answer 1

Answer:

$\vec{v}=\frac{3}{2}v_0\hat{i}-v_0\hat{j}+0\hat{k}$

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:

$mv_0=\frac{2}{3}mv_{x}\\v_x}=\frac{3}{2}v_0\\\\0=\frac{1}{3}m2v_0+\frac{2}{3}mv_{y}\\v_y=-\frac{2}{3}m(v_0)[\frac{3}{2m}]\\v_y=-v_0\\\\(0)_z=(0)_z\\v_z=0$

So, the velocity vector of the remaining rocket's mass is:

$\vec{v}=\frac{3}{2}v_0\hat{i}-v_0\hat{j}+0\hat{k}$