The final velocity (
) of the first astronaut will be greater than the <em>final velocity</em> of the second astronaut (
) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts <em>after throwing the ball</em>.
The given parameters;
- Mass of the first astronaut, = m₁
- Mass of the second astronaut, = m₂
- Initial velocity of the first astronaut, = v₁
- Initial velocity of the second astronaut, = v₂ > v₁
- Mass of the ball, = m
- Speed of the ball, = u
- Final velocity of the first astronaut, =
- Final velocity of the second astronaut, =
The final velocity of the first astronaut relative to the second astronaut after throwing the ball is determined by applying the principle of conservation of linear momentum.
if v₂ > v₁, then 
, to conserve the linear momentum.
Thus, the final velocity () of the first astronaut will be greater than the <em>final velocity</em> of the second astronaut () to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts after throwing the ball.
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Answer:
Boyle's Law
Explanation:
Given that:
<u><em>initially:</em></u>
pressure of gas, 
volume of gas, 
<em><u>finally:</u></em>
pressure of gas, 
volume of gas, 
<u>To solve for final volume</u> 
<em>According to Avogadro’s law the volume of an ideal gas is directly proportional to the no. of moles of the gas under a constant temperature and pressure.</em>
<em>According to the Charles' law, at constant pressure the volume of a given mass of an ideal gas is directly proportional to its temperature.</em>
But here we have a change in the pressure of the Gas so we cannot apply Avogadro’s law and Charles' law.
Here nothing is said about the temperature, so we consider the Boyle's Law which states that <em>at constant temperature the volume of a given mass of an ideal gas is inversely proportional to its pressure.</em>
Mathematically:
<span>Answer:
Assuming that I understand the geometry correctly, the combine package-rocket will move off the cliff with only a horizontal velocity component. The package will then fall under gravity traversing the height of the cliff (h) in a time T given by
h = 0.5*g*T^2
However, the speed of the package-rocket system must be sufficient to cross the river in that time
v2 = L/T
Conservation of momentum says that
m1*v1 = (m1 + m2)*v2
where m1 is the mass of the rocket, v1 is the speed of the rocket, m2 is the mass of the package, and v2 is the speed of the package-rocket system.
Expressing v2 in terms of v1
v2 = m1*v1/(m1 + m2)
and then expressing the time in terms of v1
T = (m1 + m2)*L/(m1*v1)
substituting T in the first expression
h = 0.5*g*(m1 + m2)^2*L^2/(m1*v1)^2
solving for v1, the speed before impact is given by
v1 = sqrt(0.5*g/h)*(m1 + m2)*L/m1</span>
