What is the rocket's initial acceleration?

[JoeDGreat]

Homework Statement

A rocket is in outer space, far from any planet, when the rocket engine is turned on. In the first second of firing, the rocket ejects 1/120 of its mass with a relative speed of 2400m/s. What is the rocket's initial acceleration?

Relevant Equations

Vf-Vi = VeIn(Mi/Mf)

Help me solve... I'm getting errors here..

 

[Filip Larsen]

The rocket equation you have quoted is not very useful to find acceleration. For that you need to look at the fundamental conservation law that is behind the rocket equation, and specifically how the conserved quantities of the rocket and ejected propellant change the instant the rocket is turned on.

 

[haruspex]

Homework Statement: A rocket is in outer space, far from any planet, when the rocket engine is turned on. In the first second of firing, the rocket ejects 1/120 of its mass with a relative speed of 2400m/s. What is the rocket's initial acceleration?
Relevant Equations: Vf-Vi = VeIn(Mi/Mf)

Help me solve... I'm getting errors here..

Per forum rules, please post your attempt.

 

[JoeDGreat]

a = -Ve/Me × dM/dt
dM/dt = Mi/120 ÷ 1sec = -Mi/120sec
a =-2400/Mi ( -Mi/120) = 20m/s²

PS: This is the textbook solving but, I don't know how dM= Mi/120

 

[erobz]

a = -Ve/Me × dM/dt
dM/dt = Mi/120 ÷ 1sec = -Mi/120sec
a =-2400/Mi ( -Mi/120) = 20m/s²

PS: This is the textbook solving but, I don't know how dM= Mi/120

They are giving you the instantaneous rate of mass ejection at ##t=0##:

## \dot M(0) = -\frac{1}{120}M \frac{ \text{kg}}{ \text{s}} ##

Then apply "The Rocket Equation" at ##t = 0## (with no external forces).

 

[erobz]

In the first second of firing, the rocket ejects 1/120 of its mass with a relative speed of 2400m/s.

Is it this statement that is giving you interpretive issues? They should have just said something to the effect of " at the instant of firing", or we are just to assume the mass flow rate as constant over the first second for the sake of simplicity (i.e. being able to find a solution).

 

[haruspex]

A loss of 1/120th of total mass is sufficiently small that we don't need to worry about how it changes over the second, or, indeed, that it changes at all. Just use momentum conservation: m/120 * 2400m/s = m*v.
Then a=v/t.