# physicS IMP PLS

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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 -

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.

• physicS IMP PLS -

THE instantaenous acc is just the acceleration of the rocket

• physicS IMP PLS -

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 -

do you know how to solve the ruler question?

• physicS IMP PLS -

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 -

sorr that waw

1500*ln(1/0.9750) = 37.91 m/s

• physicS IMP PLS -

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 -

always here Hill...if you can help with ruler problem it will pe magnificent

• physicS IMP PLS -

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 -

i know this ,i already done it.thx for the efford!

• physicS IMP PLS -

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

• physicS IMP PLS -

I=1/3*m*L*L?

• physicS IMP PLS -

@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!