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April 12th, 2015, 10:29 PM   #1
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Newton's Laws: Question 1

Again just looking for feedback. Thanks in advance.
I've included diagrams for assessment too.

A rocket of mass $\displaystyle 2.75\times 10^6kg$ exerts a vertical force of $\displaystyle 3.55\times 10^7N$ on the gases it expels.

Determine:
a) the rocket's acceleration
b)it's velocity after $\displaystyle 8s$
c) time taken to reach $\displaystyle 9500m$. Assuming $\displaystyle g$ is constant and mass of gas expelled is ignored.


$\displaystyle F_g =$ rocket's weight
$\displaystyle F_{RG} =$ force exerted by rocket on gas
$\displaystyle F_{GR} =$ force exerted by gas on rocket
$\displaystyle F_R = $resultant force

a)

$\displaystyle F = m\cdot{a}$, $\displaystyle F\propto{a}$

$\displaystyle a = \frac{F}{m}$

$\displaystyle a_R = a_{up} + a_{down}$

$\displaystyle a_{up} = \frac{3.55\times 10^7}{2.75\times 10^6}$

$\displaystyle a_{up} = 12.9m/s^2$

$\displaystyle a_{down} = -9.8m/s^2$

$\displaystyle a_R = 12.9 + [-9.8]$

$\displaystyle a_R = 3.1m/s^2$

b)

$\displaystyle x(t) = x_0 + v_{x_0}t + \frac{at^2}{2}$

$\displaystyle x(8.0) = 0 + 0(8.0) + \frac{3.1\cdot 8.0^2}{2}$

$\displaystyle x(8.0) = \frac{3.1\cdot 8.0^2}{2}$

$\displaystyle x(8.0) = 99.2m$

$\displaystyle v^2 = v_0^2 + 2a(y-y_0)$

$\displaystyle v^2 = 0 + 2\cdot 3.1(99.2-0)$

$\displaystyle v = \sqrt{2\cdot 3.1\cdot 99.2}$

$\displaystyle v = 25m/s$

c)
Not too sure how to go about this problem as I'm asked to ignore the mass of the expelled gas. This means $\displaystyle a_{rocket}$ of $\displaystyle 3.1m/s^2$ is ignored.

By Newton's Laws, in order for the rocket to move up, $\displaystyle a_{up} > a_{down}$ |$\displaystyle a_{rocket} > a_{gravity}$ | $\displaystyle a_{rocket} > 9.8m/s^2$ | $\displaystyle a_{rocket} \neq 9.8m/s^2$?
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Last edited by hyperbola; April 12th, 2015 at 10:34 PM.
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April 13th, 2015, 01:12 AM   #2
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Agree with your solutions for a) and b) ... could have calculated these much easier, however.

a) $F_{net}=ma$

$F_r-mg=ma$

$\displaystyle a=\frac{F_r-mg}{m}$

b) $v_f=v_0+at$

Since $v_0=0$, $v_f=at$

c) $\displaystyle \Delta x=v_0t+\frac{1}{2}at^2$

$\displaystyle v_0=0 \implies t=\sqrt{\frac{2 \Delta x}{a}} =\sqrt{\frac{2 \cdot 9500}{3.1}}$
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April 13th, 2015, 01:40 AM   #3
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Quote:
Originally Posted by skeeter View Post
Agree with your solutions for a) and b) ... could have calculated these much easier, however.
Thanks skeeter.

Haha yeah I'm taking the long route with the problems, I think it'll help me in the long run.

Just in regards to question (c), we're asked to ignore the mass of the expelled gas.

Would the solution therefore change if the mass of the expelled gas is taken into account?

Thanks again

Last edited by hyperbola; April 13th, 2015 at 01:49 AM.
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April 13th, 2015, 02:12 AM   #4
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simplerockets 1d
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