A rocket of mass M when empty carries a mass M of fuel. The rocket and fuel travel at speed v.
The engine of the rocket is fired and all of the fuel is expelled. The speed of the rocket increases
to 2v.
What happens to the kinetic energy of the rocket?
A It doubles.
B It halves.
C It increases by a factor of four.
D It stays the same.
#kinetic energy
56 messages · Page 1 of 1 (latest)
!status
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
4
!show
Show your work, and if possible, explain where you are stuck.
So as M+M js 2M
And kinetic energy is mv^2 /2
So we get mv^2
But when fuel is exhausted and velocity doubles
So
M(2v)^2 /2
We get 2mv^2
And it doubles
But the marking scheme says otherwise
it looks good to me
but the marking scheme said it's 4 times?
Yes
the flaw is that
i mean the tricky part
is that
"what happens to the KE of the rocket
without saying anything about the fuel
I did this but teacher said u need to take fuel in equation
Marking scheme js certified like it js never wrong
K
So
What i thought
I initially was
That fuel mass is just for confusion
As a trick
We only take rocket mass as M
And velocity as V
So ee get 1/2 Mv^2
Then we do mass as M/2 and velocity as 2v
So we get 2mv^2
And 1/2 : 2 is 4
So this is the approach?
thinking
Initial
1/2 M v² (KE of rocket) + 1/2 M v² (KE of fuel)
Final
1/2 M (2v²) (KE of rocket) + ?(KE of fuel)
oh good, i think this approach looks nice
Fuel kinetic energy would be 0
As 0 mass
to be fair, this is math server, im not that good at physics