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23 December, 18:03

Suppose we repeat the experiment from the video, but this time we use a rocket three times as massive as the one in the video, and in place of water we use a fluid that is twice as massive (dense) as water. If the new fluid leaves the rocket at the same speed as the water in the video, what will be the ratio of the horizontal speed of our rocket to the horizontal speed of the rocket in the video after all the fluid has left the rocket?

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  1. 23 December, 19:18
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    2/3

    Explanation:

    It is important to bear in mind that momentum is generated by the fluid. Momentum is proportional to the mass and velocity, therefore increasing the mass (density) of the liquid by a factor of 2, while the velocity is constant, increases the momentum of fluid (and rocket) by 2. For any closed system, the momentum is conserved. With that in mind, the momentum of the fluid is equal to but opposite in direction to the momentum of the rocket. Velocity of the rocket given by v=p/m (Where p=momentum of the rocket initially and m is its initial mass). The mass of the rocket increases 3 times. So ...

    Initially:

    vi=p/m

    Now:

    vf=2p/3m

    The ratio is given by:

    vf:vi = 2p/3m:p/m
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