Posted by Hil on .
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.

physicS IMP PLS 
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/(1p))
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. 
physicS IMP PLS 
anonymus,
THE instantaenous acc is just the acceleration of the rocket

physicS IMP PLS 
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.

physicS IMP PLS 
Anonymous,
do you know how to solve the ruler question?

physicS IMP PLS 
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 
physicS IMP PLS 
Greco,
sorr that waw
1500*ln(1/0.9750) = 37.91 m/s 
physicS IMP PLS 
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.

physicS IMP PLS 
Greco,
always here Hill...if you can help with ruler problem it will pe magnificent

physicS IMP PLS 
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. 
physicS IMP PLS 
Greco,
i know this ,i already done it.thx for the efford!

physicS IMP PLS 
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 (1cos(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 
physicS IMP PLS 
socorro,
I=1/3*m*L*L?

physicS IMP PLS 
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!