Asked by Hil

Consider a rocket in space that ejects burned fuel at a speed of Vex = 1.5 km/s with respect to the rocket. The rocket burns 5% of its mass in 280 seconds.
(a) What is the speed V of the rocket after a burn time of 140 seconds (in m/s)?
(b) What is the instantaneous acceleration a of the rocket at time 140.0 seconds after start of the engines (in m/s^20)? I have tried to solve, but not been successful. Thanks for your help.
Answered by Hil
I have tried to solve this problem the following way:
5% mass is burned in 280s.
1.5 km/s = 1500 m/s
u=fuel speed m/s
Solution:
V= u*ln(1/(1-p))
1500*ln(1/0.950) = 76.9399416 m/s
a= V/t = 76.9399416/280 = 0.274785506 m/s^2
What is confusing to me what is the speed after 140s and instantaneous acceleration a 140 s after the start of the engines. Thanks for your help.
Answered by anonymus
THE instantaenous acc is just the acceleration of the rocket
Answered by Hil
Hi Anonymus, So my question is V= 76.939416 m/s and a=0.2747 m/s^2 correct. Is the other information just trying to through us off? Thanks.
Answered by Anonymous
do you know how to solve the ruler question?
Answered by Greco
no it is wrong!!! he ask you "What is the speed V of the rocket after a burn time of 140 seconds (in m/s)? " that is the half of time and half of % . so ..

1500*ln(1/0.750) = 37.91 m/s
a=v/t=37.91/140=0.27
Answered by Greco
sorr that waw

1500*ln(1/0.9750) = 37.91 m/s
Answered by Hil
Hi Anonymous, I tried that problem and was unsuccessful. My answer was 8 radians. I knew it was wrong. I put that off to the side. I am just trying to get a passing grade. Thanks for your help. If I learn any thing I will let you know. Also Thanks Greco and everyone who was kind enough to help me out.
Answered by Greco
always here Hill...if you can help with ruler problem it will pe magnificent
Answered by Hil
Hi, Greco. To be honest Questions 1 and the ruler are by far the most difficult for me. when I started the test I knew I was in trouble. I saw this approach:
Apply the conservation of energy
U= m*g*h cm
Ek = m*g* (L/2)+0
l=1/3*m*L
Einitial= m*g*(L/2)+0
Efinal= m*g*(2/L)cos30+1/2*L*w^2
Solve for Einitial = Efinal -> w =
Maybe you saw this. I hope it helps.
Answered by Greco
i know this ,i already done it.thx for the efford!
Answered by Greco
I=1/3*m*L^2
Eini= mg(L/2) + 0
Efin= mg(L/2)cos30 + 1/2*I*w^2
Eini=Efin ->
w=sqrt(3*g (1-cos(theta))/L)

b)
alpha=3*mg/2*sin(theta)/L
ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha
Fx=m*ax
Fy=m*g*cos(theta)-m*w^2*(L/2)

c)
cos(theta)=2/3
theta=48.19
Answered by socorro
I=1/3*m*L*L?
Answered by Anon
@Greco I tried it and it workrd for a and c and for Fy but Fx was wrong!why did you use alpha? please help with this question!
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