Rocket on a Spring: Solving for Speed and Compression

[wulfsdad]

Homework Statement

This is generalized from problem # 11.52 from "Physics: For Scientists and Engineers: A Strategic Approach" By Randall D. Knight 2nd Ed.

An (M) kg weather rocket generates a thrust of (F) N. The rocket, pointing upward, is clamped to the top of a vertical spring. The bottom of the spring, whose spring constant is (K) N/m, is anchored to the ground.
There are 3 parts:

A) finding the spring's initial compression (y) with the rocket resting on it,
B) find rocket's speed (V) when spring is stretched (d) m, and
C) "For comparison, what would be the rocket's speed after traveling this distance if it weren't attached to the spring? "

Homework Equations

K$$_{f}$$ + U$$_{f}$$ + $$\Delta$$E$$_{th}$$ = K$$_{i}$$ + U$$_{i}$$ + W$$_{ext}$$

$$\Delta$$K = W$$_{net}$$

W = $$\vec{F}$$ $$\bullet$$ $$\Delta$$$$\vec{s}$$

The Attempt at a Solution

I got part A: finding the spring's initial compression (y).

Mgy = Ky$$^{2}$$ $$\rightarrow$$ Mg = Ky $$\rightarrow$$ y = $$\frac{Mg}{K}$$

Then part B: finding rockets speed (V) when the spring is stretched (d) m.

$$\frac{1}{2}$$MV$$^{2}$$ + Mg( y + d ) + $$\frac{1}{2}$$Kd$$^{2}$$ = $$\frac{1}{2}$$Ky$$^{2}$$ + F( y + d )

So for part C: find the rocket's speed (v) after traveling this distance if it weren't attached to the spring?
I used the same equation as B but with the spring removed:

$$\frac{1}{2}$$MV$$^{2}$$ + Mg( y + d ) = F( y + d )

This was incorrect. I also tried it with just ( d ), instead of ( y + d ).
Then I tried reasoning that:

F = MA --> A = F/M

Then using kinematics:

2$$\frac{F}{M}$$( y + d ) = v$$^{2}$$

This was also incorrect.

 

[issacnewton]

did you try solving the equations after plugging the value of y ?

 

[wulfsdad]

Yes, I solved for V after plugging in y.
I believe that my error may be in assuming that the spring was removed, when the question ambiguously stated that the rocket was no longer attached to it. However, when I calculated
(1/2)(M)(V)^2 + (M)(g)( y + d ) = (1/2)(K)(y)^{2} + F( y + d )
I still got the wrong answer.

 

[issacnewton]

which book is this problem from ?

 

[wulfsdad]

This is generalized from problem # 11.52 from "Physics: For Scientists and Engineers: A Strategic Approach" By Randall D. Knight 2nd Ed.

If you'd like some numbers:
M = 11.2 kg
F = 200 N
K = 490 N/m
y = 22.4 cm
d = 38.0 cm
V = 2.37 m/s <--attached to spring
v = 3.45 m/s <--unattached to spring, this is the answer I'm trying to get

 

[issacnewton]

Hi, for the numbers given by you, I got V=3.12 m/s for the unattached case. I have downloaded the book mentioned by you in djvu format and the book uses different numbers. Did you just come up with your own numbers ?

 

[wulfsdad]

For my class, we use <masteringphysics.com> to do the homework, which provides each student with unique numbers. The numbers I gave you are from that.
3.12 was one of the results I came up with too, that was marked as incorrect. 3.45 is the answer the program was expecting.

 

[issacnewton]

oh strange... let me check again...

 

[issacnewton]

Hi I found the mistake, I was assuming that in the unattached case, spring will bounce back
beyond its original equilibrium, so I didn't include the elastic potential energy term.
I also found the solution manual for the Knight's book. The manual agrees with my solution.
send me pm. I can send you the solution manual.