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? (Ignore air resistance.)

2/3

Explanation:

In the case shown above, the result 2/3 is directly related to the fact that the speed of the rocket is proportional to the ratio between the mass of the fluid and the mass of the rocket.

In the case shown in the question above, the momentum will happen due to the influence of the fluid that is in the rocket, which is proportional to the mass and speed of the same rocket. If we consider the constant speed, this will result in an increase in the momentum of the fluid. Based on this and considering that rocket and fluid has momentum in opposite directions we can make the following calculation:

Rocket speed = rocket momentum / rocket mass.

As we saw in the question above, the mass of the rocket is three times greater than that of the rocket in the video. For this reason, we can conclude that the calculation should be done with the rocket in its initial state and another calculation with its final state:

Initial state: Speed = rocket momentum / rocket mass.

Final state: Speed = 2 rocket momentum / 3 rocket mass. > 2/3

6 cos 34, 6 sin 34 trig