User Name Remember Me? Password

 Physics Physics Forum

April 12th, 2015, 10:29 PM   #1
Senior Member

Joined: Dec 2014
From: The Asymptote

Posts: 142
Thanks: 6

Math Focus: Certainty
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, $\displaystyle a_{up} > a_{down}$ |$\displaystyle a_{rocket} > a_{gravity}$ | $\displaystyle a_{rocket} > 9.8m/s^2$ | $\displaystyle a_{rocket} \neq 9.8m/s^2$?
Attached Images Rocket.png (4.6 KB, 17 views)

Last edited by hyperbola; April 12th, 2015 at 10:34 PM. April 13th, 2015, 01:12 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1588 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) $\displaystyle \Delta x=v_0t+\frac{1}{2}at^2$ $\displaystyle v_0=0 \implies t=\sqrt{\frac{2 \Delta x}{a}} =\sqrt{\frac{2 \cdot 9500}{3.1}}$ Thanks from hyperbola April 13th, 2015, 01:40 AM   #3
Senior Member

Joined: Dec 2014
From: The Asymptote

Posts: 142
Thanks: 6

Math Focus: Certainty
Quote:
 Originally Posted by skeeter 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. April 13th, 2015, 02:12 AM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1588 Thanks from hyperbola Tags laws, newton, question Search tags for this page

### www.scienceforums.net applied math newton laws

Click on a term to search for related topics.
 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post hyperbola Physics 5 April 10th, 2015 08:46 AM SamFe Calculus 3 October 18th, 2013 11:12 PM OriaG Probability and Statistics 1 February 13th, 2013 01:10 AM aaron-math Calculus 1 February 23rd, 2012 07:12 PM rooster Applied Math 1 October 27th, 2009 08:30 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      