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

`\vec{v}=(3)/(2)v_0\hat{i}-v_0\hat{j}+0\hat{k}`

Step-by-step 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=(2)/(3)mv_(x)\\v_x}=(3)/(2)v_0\\\\0=(1)/(3)m2v_0+(2)/(3)mv_(y)\\v_y=-(2)/(3)m(v_0)[(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}=(3)/(2)v_0\hat{i}-v_0\hat{j}+0\hat{k}`