Question

Could someone please help me with a momentum problem? A rocket mass m travels w speed Vo along x axis. The rocket shoots out fuel equal to 1/3 its mass perpindicular to x axis wth speed 2Vo. Express final velocity of rocket in i,j,k notation.

Answer 1

The conservation of momentum in the y (j) direction is such that -(2Vo)(m/3) fuel momentum will be balanced by the momentum of the rocket in the positive y direction of (Vo)(2m/3), giving the rocket a y component of velocity of Vo. I think that it continues with a speed ov Vo in the x (I) direction, as well, but I am not sure, because it has become 1/3 lighter than before. Conservation of momentum in the x direction would suggest that Vo m becomes = (3/2)Vo (2/3)m, speeding up because it is lighter. My best guess is that the velocity components become Vo in the y (j) direction and (3/2) Vo in the x (i) direction, but this latter is in question.

Answer 2

I agree with @douglaswinslowcooper 's answers: however, since the problem doesn't give any specifications about whether the thrust is in the y or z or both directions, it might be a good idea to make it as general as possible, so defining theta to be the angle between the negative z coordinate and the thrust direction in the yz plane



Cons. of Momentum in x (i=initial, f=final)
\[p_{ix}=p_{fx}\]
\[ m_iv_{ix} = m_fv_{fx} \]
\[v_{fx}=\frac{m_i}{m_f}v_{ix}\]
\[\quad m_i=m \quad ; \quad m_f=2m/3 \quad ; \quad v_{ix}=v_0\]

Cons. of Momentum in y (r=rocket, f=fuel)
\[p_{initial \ y}=p_{final \ y}\]
\[0=m_fv_{fy}+m_rv_{ry}\]
\[ v_{ry}=-\frac{m_f}{m_r}v_{fy}\]
\[ \quad m_f = m/3 \quad ; \quad m_r = 2m/3 \quad ; \quad v_{fy} = -v_0 \sin \theta\]

Similarly, Cons. of Momentum in z (r=rocket, f=fuel)
\[p_{initial \ z}=p_{final \ z}\]
\[0=m_fv_{fz}+m_rv_{rz}\]
\[ v_{rz}=-\frac{m_f}{m_r}v_{fz}\]
\[\quad v_{fz}=-v_0 \cos \theta\]

Then ultimately
\[\textbf v = v_{final \ x}\hat{\textbf i}+v_{ry}\hat{\textbf j}+v_{rz}\hat{\textbf k}\]

Answer 3

derp derp
\[v_{fy} = -2v_0 \sin \theta\]
\[v_{fz} = -2v_0 \cos \theta\]

Answer 4

Thanks again!