Rocket losing mass gaining speed
I'm not sure how to tackle this question as I can't use momentum or energy conservation
Attachment not found
I’m pretty sure it is a momentum question, you know the overall momentum of the rocket, and you know the final speed of both the rocket and the fuel capsule,
Surely you could just use the fact that 4000*7425 = (7500*(4000-Y)) + (1425 * Y)
Rearrange to find Y
Surely you could just use the fact that 4000*7425 = (7500*(4000-Y)) + (1425 * Y)
Rearrange to find Y
Original post by Daniel100499
I’m pretty sure it is a momentum question, you know the overall momentum of the rocket, and you know the final speed of both the rocket and the fuel capsule,
Surely you could just use the fact that 4000*7425 = (7500*(4000-Y)) + (1425 * Y)
Rearrange to find Y
Surely you could just use the fact that 4000*7425 = (7500*(4000-Y)) + (1425 * Y)
Rearrange to find Y
Yep, just did the question and I also used this method. My final answer I got was 49.4 which means the answer is 50kg?
(edited 6 years ago)
Original post by S.H.Rahman
Yep, just did the question and I also used this method. My final answer I got was 49.4 which I means the answer is 50kg?
Phew! Haven’t done physics since AS so I was hoping I was right, and yeah just put it in a calculator and got the same answer
Original post by Daniel100499
Phew! Haven’t done physics since AS so I was hoping I was right, and yeah just put it in a calculator and got the same answer
Haha, looks like you still got it!
Original post by Daniel100499
I’m pretty sure it is a momentum question, you know the overall momentum of the rocket, and you know the final speed of both the rocket and the fuel capsule,
Surely you could just use the fact that 4000*7425 = (7500*(4000-Y)) + (1425 * Y)
Rearrange to find Y
Surely you could just use the fact that 4000*7425 = (7500*(4000-Y)) + (1425 * Y)
Rearrange to find Y
Original post by S.H.Rahman
Yep, just did the question and I also used this method. My final answer I got was 49.4 which I means the answer is 50kg?
Unfortunately the answer is actually e , 200kg, I'm not sure about momentum since the the fuel stored chemical energy, the system is not closed
Original post by alvan15
Unfortunately the answer is actually e , 200kg, I'm not sure about momentum since the the fuel stored chemical energy, the system is not closed
Tried different methods and got 222kg once and 40kg using another
Let me know if you find the solution as I would be interested to see this! 
Original post by TheMindGarage
I got 200kg using a momentum method:
Let fuel ejected be x. Mass of the rest of the rocket is 4000-x.
Change in speed of the rocket is 75 m/s.
1425x = 75(4000-x)
1500x = 300000
x = 200
Let fuel ejected be x. Mass of the rest of the rocket is 4000-x.
Change in speed of the rocket is 75 m/s.
1425x = 75(4000-x)
1500x = 300000
x = 200
Ahh is that because once the mass is ejected , the system can be treated as isolated , therefore momentum is conserved? And using speeds relative toto 7425 .
(edited 6 years ago)
Original post by alvan15
Ahh is that because once the mass is ejected , the system can be treated as isolated , therefore momentum is conserved? And using speeds relative toto 7425 .
Yep. It's ejected in an instant, so you can just treat it as a single impulse.
Original post by alvan15
I'm not sure how to tackle this question as I can't use momentum or energy conservation
Attachment not found
This is a classic momentum question that just needs some algebraic manipulation
Initial momentum = 4000*7425 = 29700000 kgms^-1
Momentum is conserved:
so Initial momentum of system = Final momentum of system
Let m = mass of fuel ejected
-----> 29700000 = (4000-m)*7500 - 1425m
8925m = 300000
m = 33.61... = 34 kg (2.s.f)
So answer is B
Hope that was helpful
Original post by Daniel100499
I’m pretty sure it is a momentum question, you know the overall momentum of the rocket, and you know the final speed of both the rocket and the fuel capsule,
Surely you could just use the fact that 4000*7425 = (7500*(4000-Y)) + (1425 * Y)
Rearrange to find Y
Surely you could just use the fact that 4000*7425 = (7500*(4000-Y)) + (1425 * Y)
Rearrange to find Y
Be careful here, momentum is a vector quantity and the mass is ejected backwards to the direction of motion of the rocket so the equation would be
4000*7425 = (7500*(4000-Y)) - (1425 * Y)
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