Question

First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question at right. You can watch the video again at any point. 

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.)

Answer 1

Concepts and reason

The main concept used is momentum conservation.

Initially, assume the rocket and fluid as the system and the momentum in vertical direction will no be conserved due to gravity, Later, the momentum is conserved in the horizontal direction. Finally use the conservation of momentum to calculate the ratio of the horizontal speed of our rocket to the horizontal speed of the rocket.

Fundamentals

The conservation of momentum states that the momentum of a system is constant when there is no external force acting on the system. It is expressed as follows:

${m_e}{v_{e.f}} = {m_r}{v_{r.f}}$

Here, ${m_e}$ is the mass of exhaust, ${m_r}$ is the mass of rocket, ${v_{e.f}}$ is the velocity of the exhaust, and ${v_{r.f}}$ is the velocity of the rocket.

Calculate the ratio of the horizontal speed of our rocket to the horizontal speed of the rocket.

The conservation of momentum is expressed as follows:

${m_e}{v_{e.f}} = {m_r}{v_{r.f}}$

Here, ${m_e}$ is the mass of exhaust, ${m_r}$ is the mass of rocket, ${v_{e.f}}$ is the velocity of the exhaust, and ${v_{r.f}}$ is the velocity of the rocket.

Rewrite the equation for the rocket’s velocity ${v_{r.f}}$ .

$\frac{{{v_{r.f}}}}{{{v_{e.f}}}} = \frac{{{m_e}}}{{{m_r}}}$

The equation for the rocket’s velocity ${v_{r.f}}$ is expressed as follows:

$\frac{{{v_{r.f}}}}{{{v_{e.f}}}} = \frac{{{m_e}}}{{{m_r}}}$

Substitute $2{m_e}$ for ${m_e}$ and $3{m_r}$ for ${m_r}$ in expression $\frac{{{v_{r.f}}}}{{{v_{e.f}}}} = \frac{{{m_e}}}{{{m_r}}}$ .

$\frac{{{v_{r.f}}}}{{{v_{e.f}}}} = \frac{{2{m_e}}}{{3{m_r}}}$

Ans:

The ratio of the horizontal speed of our rocket to the horizontal speed of the rocket is $\frac{2}{3}$ .