Rocket Problem - Law Of Conservation Of Momentum

[adabistanesoophia]

Hi,

I am unable to solve this problem. :zzz:

I applied Law of Conservation of Momentum but unable to get the related answer.

The problem is as. :!)

"A rocket is moving vertically upward. % Kg fuel gas is released out of it in downward direction with a velocity of 5000 m/s. If the mass of rocket is 1000 Kg, what will be its velocity?"

The answer is –25 m/s.

Regards,

Adabistan-e-Soophai

 

[Galileo]

Since the momentum of the fuel that the rocket spits out is (velocity times mass): 5000*5=25000 kgm/s. (I take it the % is a 5).

Then since total momentum is conserved, the rocket gains a momentum of 25000 kgm/s in the opposite direction. Use the momentum equation for the rocket to get it's velocity.

This question does assume the rocket has zero velocity to start with (so you're viewing from the rocket's initial frame of reference).
Also, the ejection assumed to be instantanuous. That is, a big clumb of fuel is spit out at an instant. If the ejection was gradual (as in reality) the velocity would be different, because the mass of the fuel had to be incorperated into the mass of the rocket.

 

[wais]

Hi Adabistan-e-Soophai,

I understand that this problem may seem difficult, but let's break it down together. The key to solving this problem is to apply the Law of Conservation of Momentum, which states that the total momentum of a closed system remains constant.

In this problem, we have a closed system consisting of the rocket and the fuel gas. Initially, the total momentum of the system is 0, as the rocket is at rest and the fuel gas has not been released yet. However, when the fuel gas is released, it exerts a downward force on the rocket and causes it to accelerate upwards.

To solve for the final velocity of the rocket, we can use the equation:

m1v1 + m2v2 = (m1 + m2)v

Where m1 and v1 represent the mass and initial velocity of the rocket, m2 and v2 represent the mass and velocity of the fuel gas, and v represents the final velocity of the rocket.

Plugging in the values from the problem, we get:

(1000 kg)(0 m/s) + (-50 kg)(5000 m/s) = (1000 kg + (-50 kg))v

-250,000 kg m/s = 950 kg v

v = -250,000 kg m/s / 950 kg

v = -26.316 m/s

However, we need to take note of the direction of the velocity. Since the rocket is moving upwards, we consider the positive direction as upwards. Therefore, the final velocity of the rocket is actually 26.316 m/s upwards.

I believe the answer given in the problem may be incorrect. I hope this helps you understand the concept better. Keep practicing and you'll get the hang of it!