A 50 -kg model rocket lifts off by expelling fuel at a rate of k=4.75 kg / s for 10 s . The fuel

A 50 -kg model rocket lifts off by expelling fuel at a rate...

A 50 -kg model rocket lifts off by expelling fuel at a rate of $k=4.75 \mathrm{kg} / \mathrm{s}$ for 10 $\mathrm{s}$ . The fuel leaves the end of the rocket with an exhaust velocity of $b=100 \mathrm{m} / \mathrm{s} .$ Let $m(t)$ be the mass of the rocket at time $t .$ From the law of conservation of momentum, we find the following differential equation for the rocket's velocity $v(t)($ in meters per second): $$m(t) v^{\prime}(t)=-9.8 m(t)+b \frac{d m}{d t}$$ (a) Show that $m(t)=50-4.75 t \mathrm{kg}$ . (b) Solve for $v(t)$ and compute the rocket's velocity at rocket burnout (after 10 $\mathrm{s}$ ).

A 50 -kg model rocket lifts off by expelling fuel at a rate of $k=4.75 \mathrm{kg} / \mathrm{s}$ for 10 $\mathrm{s}$ . The fuel leaves the end of the rocket with an exhaust velocity of $b=100 \mathrm{m} / \mathrm{s} .$ Let $m(t)$ be the mass of the rocket at time $t .$ From the law of conservation of momentum, we find the following differential equation for the rocket's velocity $v(t)($ in meters per second):
$$m(t) v^{\prime}(t)=-9.8 m(t)+b \frac{d m}{d t}$$
(a) Show that $m(t)=50-4.75 t \mathrm{kg}$ .
(b) Solve for $v(t)$ and compute the rocket's velocity at rocket burnout (after 10 $\mathrm{s}$ ).

Key Concepts

Conservation of Momentum in Variable Mass Systems

This principle is central to understanding how the momentum of a system changes when its mass is not constant. In problems like rocket motion, as the fuel is expelled, the mass decreases over time, and the conservation of momentum must account for both the changing mass and the velocity of the expelled fuel. This leads to differential equations that balance the momentum lost with the momentum gained through thrust.

Differential Equations and the Integrating Factor Method

Rocket motion with variable mass is modeled by differential equations that include time-varying coefficients, such as the mass of the rocket. Solving such first-order linear differential equations often involves using the integrating factor method. This technique transforms the original differential equation into an easily integrable form, allowing one to find an explicit solution for the velocity as a function of time.

Fuel Consumption and Its Effect on Mass

In variable mass problems, the fuel consumption rate determines how the mass of the rocket decreases with time. This rate, when applied to the initial mass, gives a linear relationship between the rocket’s mass and time. Understanding this relationship is essential to accurately model the system and solve the related differential equations.

Transcript

00:01 A 50 -kilogram model rocket lifts off by expelling fuel at a rate of 4 .75 kilograms per second for 10 seconds.

00:09 The fuel leaves the rocket at a rate of b equals minus 100 meters per second.

00:15 M is the mass, v is the velocity of the rocket, and then here's the differential equation.

00:21 We're going to solve that for v, but first we have to find out what m is.

00:25 So that's the mass of the rocket, but it's changing because the fuel is leaving.

00:29 All right so fuel is leaving at a rate of 4 .75 kilograms per second so that's the change in the mass dmdt minus 4 .75 so separate the variables dm equals minus 4 .75 d t so m equals minus 4 .75 d t plus c okay at times zero the mass of the rocket, including the fuel, was 50.

01:09 So 50 equals 0 plus c.

01:13 So c is 50.

01:14 So m is 50 minus 4 .75t.

01:22 Okay, next, take the differential equation and solve it.

01:26 So we have m, dv, dt equals minus 9 .8, m plus b which was minus 100 so minus 100 times dm d t well dmdt was this minus 4 .75 all right so first i'm getting rid of the red pen and now i'm going to divide everything by m so i have dvd t equals minus 9 .8 plus 475 over m but remember m was this which we have to change to because we need a function of t, not of m, so minus 9 .8 plus 475.

02:24 I'm just going to call that 50 minus 4 .75 t to the minus 1 power.

02:36 Okay, so now we got to separate the variables.

02:40 Oh, so i'm just going to have to move the dt over there.

02:52 Dt.

02:56 Okay, integrate.

02:58 So v is 9 .8.

03:00 T plus 475, 50 minus 4 .75.

03:08 Oh, i forgot.

03:09 If i make you be this, then i need a minus 4 .75 in here.

03:15 So a one over minus 4 .75 here...