A firework rocket is launched vertically from rest and loses 60% of its mass during a burn time of 4 seconds. What is the final speed of the rocket relative to the ground (in m/s) when the gas relative to the rocket has a speed v= 80 m/s?
a. 15 m/s; b. 34 m/s; c. 68 m/s; d. 180 m/s
Thanks for your help.
Why did the firework lose weight? Because it went on a diet! But don't worry, it's still explosive with all that firework excitement!
Now, let's get to your question. We need to determine the final speed of the rocket relative to the ground.
Since the rocket is launched vertically from rest, and we know that the gas relative to the rocket has a speed of 80 m/s, it means that the gas is expelled with an initial velocity of -80 m/s (negative because it is opposite to the rocket's direction).
Now, since the rocket loses 60% of its mass during burn time, we can assume that the remaining 40% of the mass has a constant acceleration. By using the law of conservation of momentum, we can equate the momentum of the rocket to the momentum of the gas expelled.
But hold on, we're missing some crucial information here: the mass ratio of the gas to the rocket. Without that, I cannot provide you with a specific answer.
However, I must say that option d. 180 m/s seems a bit extreme for a firework rocket. We want a fun and celebratory explosion, not a rocket to the moon! So, let's not go with that option.
In the meantime, let's hope the rocket finds its lost weight and can enjoy some post-firework snacks!
To find the final speed of the rocket relative to the ground, we can use the principle of conservation of momentum.
First, let's denote the initial mass of the rocket as m0, the final mass as m (after losing 60% of its mass), the initial velocity as u (which is 0 since it starts from rest), the time taken as t (which is 4 seconds), and the relative speed of the gas as v (which is given as 80 m/s).
According to the principle of conservation of momentum, the momentum before the rocket burns its mass will be equal to the momentum after the rocket burns its mass. Mathematically, this can be written as:
m0 * u = m * v
Since the rocket loses 60% of its mass, we can write this as:
(1 - 0.6) * m0 * u = m * v
Simplifying this equation, we get:
(0.4) * m0 * u = m * v
The initial mass, m0, cancels out, and since u is zero, we are left with:
0.4 * 0 = m * v
Therefore, the final speed of the rocket, relative to the ground, is zero.
So, the correct answer is:
a. 15 m/s
To find the final speed of the rocket relative to the ground, we can use the principle of conservation of momentum.
The key idea is that the change in momentum of the rocket is equal to the impulse provided by the expelled gas.
First, let's find the initial mass of the rocket. We know that it loses 60% of its mass during its burn time of 4 seconds. So, the remaining mass after the burn would be 40% or 0.4 times the original mass.
Now, let's analyze the momentum of the rocket and the expelled gas.
The initial momentum of the rocket can be calculated as the product of its initial mass (m_initial) and initial velocity (u), which is zero since it is launched from rest:
Initial momentum of the rocket = m_initial * u = 0
The final mass of the rocket is 0.4 times the initial mass, and the final velocity (v) is what we need to find.
The final momentum of the rocket can be calculated as the product of the final mass (m_final) and the final velocity (v):
Final momentum of the rocket = m_final * v
The change in momentum of the rocket is given by the difference between the final and initial momentum:
Change in momentum = Final momentum of the rocket - Initial momentum of the rocket
Since the rocket is the only object involved, the change in momentum is equal to the impulse provided by the expelled gas.
Impulse = Change in momentum
Now, let's calculate the impulse provided by the gas. The impulse is equal to the product of the mass of the expelled gas (m_gas) and its speed (v_gas):
Impulse = m_gas * v_gas
Since the rocket loses 60% of its mass during the burn time, the mass of the gas would be the initial mass minus the final mass:
m_gas = m_initial - m_final
Substituting this into the impulse equation, we get:
Impulse = (m_initial - m_final) * v_gas
Since impulse is equal to the change in momentum, we can equate the two equations:
(m_initial - m_final) * v_gas = Final momentum of the rocket - Initial momentum of the rocket
Simplifying this equation, we get:
(m_initial - m_final) * v_gas = m_final * v
Now, we can solve for the final velocity (v) of the rocket:
v = [(m_initial - m_final) * v_gas] / m_final
Substituting the given values, with v_gas = 80 m/s, and assuming the initial mass of the rocket (m_initial) is 100 units:
m_final = 0.4 * 100 = 40 units
v = [(100 - 40) * 80] / 40 = 40 m/s
So, the final speed of the rocket relative to the ground is 40 m/s.
Therefore, the correct answer is not among the options provided.