Movement in outer space via Newton's law of every action has an equal and opposite reaction What is more effective for travel in outer space ignoring all other factors like air radiation etc. I have a 10 kg bag of rice would I travel faster throwing the whole bag at once or throwing a grain at a time compounding my tiny acceleration or would they end up being equal?
 A: You need a model for how you throw the rice.  The obvious one is that you can expel any mass at the same velocity $v$ relative to you.  Letting $M$ be your mass (without the rice), $V$ your velocity in the CM frame, if you throw it as one lump we have momentum conservation.  You start with no momentum in the CM frame, so $10v=MV, V=\frac {10v}M$.  If you throw it as bits, the later bits have less velocity in the CM frame because they start with some velocity in your direction.  Your velocity is therefore lower.  If you can throw small bits faster than the whole bag, we need to reassess, but you need to specify the model.
A: This has a simple closed-form solution. Denoting $m_0,m_1$ as the initial and final person's mass, $v_r$ as the rice speed and $\delta=m_0/m_1$, if the bag is thrown in one single parcel, we have
$$\Delta v_1=(\delta-1)v_r$$
By the rocket equation, if the rice is thrown continuously, we have
$$\Delta v_2=v_r\text{Log}(\delta).$$
But
$$\text{Log}(\delta)\leq \delta-1$$
for all $1\leq\delta\leq\infty$, so $\Delta v_1\geq\Delta v_2$.
A: Given a bag of rice of mass $m_b$ that you can throw with a maximum acceleration $\vec{a}_b$, by Newton's second law, the most force $\vec{F}_b$ you could exert on the rice is given by
$$\vec{F}_b = m_b \vec{a}_b$$
By Newton's third law, the reaction force (acting on you) $\vec{F}_{you}$ is given by
$$\vec{F}_{you} = - \vec{F}_b = - m_b \vec{a}_b$$
Again by Newton's second law, if you have mass $M$, your acceleration would be given by
$$\vec{a}_{you} = - \frac{m_b}{M} \vec{a}_b$$
Now imagine you throw the $N$ individual rice grains with masses $m_i$ one at a time, each with the same acceleration $\vec{a}_b$ that we imparted on the bag. Because the mass of the bag of rice is just the sum of the masses of all the little grains of rice, 
$$\sum_{i = 1}^N m_i = m_b$$
Newton's second law says that the force $\vec{F}_i$ on each grain of rice is given by
$$\vec{F}_i = m_i \vec{a}_b$$
By Newton's third law,
$$\vec{F}_{you, i} = - m_i \vec{a}_b$$
So your acceleration due to each grain of rice is 
$$\vec{a}_{you, i} = -\frac{m_i}{M} \vec{a}_b$$
Summing over $i$ to find your total acceleration gives
$$\vec{a}_{you} = \sum_{i = 1}^N \vec{a}_{you, i} = \sum_{i = 1}^N -\frac{m_i}{M} \vec{a}_b = -\frac{1}{M} \vec{a}_b \left(\sum_{i = 1}^N m_i \right) = -\frac{m_b}{M} \vec{a}_b$$
Therefore the acceleration caused by throwing the whole bag at once and that caused by throwing each grain of rice individually are equal (provided the rice grains are all thrown in the same direction, of course).
