Question

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.) view available hint(s)
○ 1/3
○ 3/2
○ 1/2
○ 2/3
○ 1
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Explanation:

Step1: Recall Conservation of Momentum

The total momentum before and after the fluid is expelled must be equal (since initial momentum is zero, final momentum of rocket and fluid must be equal in magnitude and opposite in direction). Let the mass of the rocket in the video be \( m_r \), mass of water be \( m_w \), speed of water (and new fluid) be \( v_f \), speed of rocket in video be \( v_{r1} \), and speed of new rocket be \( v_{r2} \).

In the video: \( m_r v_{r1} = m_w v_f \) (momentum conservation, initial momentum = 0).

For the new experiment: Mass of rocket \( m_{r2} = 3m_r \), mass of fluid \( m_{f2} = 2m_w \) (since density is twice, mass is twice for same volume, assuming volume of fluid is same as water in video). So momentum conservation: \( m_{r2} v_{r2} = m_{f2} v_f \).

Step2: Solve for Ratios

From video: \( v_{r1} = \frac{m_w v_f}{m_r} \).

From new experiment: \( v_{r2} = \frac{m_{f2} v_f}{m_{r2}} = \frac{2m_w v_f}{3m_r} \).

Now find the ratio \( \frac{v_{r2}}{v_{r1}} \):

\( \frac{v_{r2}}{v_{r1}} = \frac{\frac{2m_w v_f}{3m_r}}{\frac{m_w v_f}{m_r}} \)

Simplify: The \( m_w \), \( v_f \), and \( m_r \) terms cancel, leaving \( \frac{2}{3} \).

Answer:

\(\frac{2}{3}\) (corresponding to the option "2/3")