Santa wants to deliver his presents but his reindeers are on strike. In order to still be able to get somewhere he decides to sacrifice some of his presents to gain speed. He decides to throw them straight off the back with velocity $v$. Would he be better off throwing all of the presents at once or should he he throw $\frac{1}{2}$ of the presents twice? (Ignore friction)

Hint: Mass of the sacrificed presents is $\frac{1}{2}M$ where $M$ is the total mass: $M=m_{santa}+m_{presents}+m_{sleigh}$

This is my solution to the problem:

  1. Throw everything at once:

$$m_sv_s+m_1v_1=m_s v_s'+m_1v_1' \\ \iff 0=m_sv_s'+m_1v_1' \\ \iff v_s'=\frac{m_1}{m_s}v_1'=\frac{\frac{1}{2}M}{\frac{1}{2}M} \\ \iff v'_s=-v_1' $$

Whre $m_s=\frac{1}{2}M \space $(mass of the sleigh+santa) and $m_1=\frac{1}{2}M \space $(mass of presents thrown off the back). This result makes perfect sense and is exactly what I expected. The sleigh and santa move with a velocity that is equal to the velocity of the sacrificed presents but in the other direction.

  1. Throw twice:

This was a pretty long calculation but I arrive at the exact same result (I can write it down in case it helps answering my question) $v_s'=-v_1'$ which I thought to be logical because we have conservation of momentum and it shouldn't matter if I split up the presents into $n$ throws.

However, this is not the correct answer. The solution we got says that the second scenario is worse and that the velocity of the sleigh+santa will be $$\boxed{v_{santa}=\frac{5}{6}v_{presents}}$$

How can that be?


1 Answer 1


The problem is that by throwing only half of the presents, Santa needs to accelerate himself, the sleigh, and the other half of the presents that are still in the sleigh. If he throws them all at once, he only needs to accelerate himself and the sleigh. Santa is losing out on velocity by accelerating his propellant!

  • $\begingroup$ That makes sense. I really didn't think of that. Thank you very much! $\endgroup$
    – Nullspace
    Dec 6, 2018 at 20:36
  • 1
    $\begingroup$ My first thought was that it wouldn't matter either way as well, I thought the solution manual might be wrong before I thought of the answer. $\endgroup$ Dec 7, 2018 at 1:02

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