Newton’s second law $\vec{F}=m\vec{a}$ applies to objects whose mass does not change. That’s an excellent assumption for balls and bicycles, but what about something like a rocket that loses a significant amount of mass as its fuel is burned? Problems of varying mass are solved with momentum rather than acceleration. We’ll look at one important example.

FIGURE 11.29 shows a rocket being propelled by the thrust of burning fuel but not influenced by gravity or drag. Perhaps it is a rocket in deep space where gravity is very weak in comparison to the rocket’s thrust. This may not be highly realistic, but ignoring gravity allows us to understand the essentials of rocket propulsion without making the mathematics too complicated. Rocket propulsion with gravity is a Challenge Problem in the end-of-chapter problems.

FIGURE 11.29 A before-and-after pictorial representation of a rocket burning a small amount of fuel.

The system rocket $+$ exhaust gases is an isolated system, so its total momentum is conserved. The basic idea is simple: As exhaust gases are shot out the back, the rocket “recoils” in the opposite direction. Putting this idea on a mathematical footing is fairly straightforward—it’s basically the same as analyzing an explosion—but we have to be extremely careful with signs.

We’ll use a before-and-after approach, as we do with all momentum problems. The Before state is a rocket of mass m (including all onboard fuel) moving with velocity $v_x$ and having initial momentum $P_{\text{i}x}=m{v}_x$. During a small interval of time $dt$, the rocket burns a small mass of fuel $m_{\text{fuel}}$ and expels the resulting gases from the back of the rocket at an exhaust speed $v_{\text{ex}}$ relative to the rocket. That is, a space cadet on the rocket sees the gases leaving the rocket at speed $v_{\text{ex}}$ regardless of how fast the rocket is traveling through space.

After this little packet of burned fuel has been ejected, the rocket has new velocity $v_x+d{v}_x$ and new mass $m+dm$. Now you’re probably thinking that this can’t be right; the rocket loses mass rather than gaining mass. But that’s our understanding of the physical situation. The mathematical analysis knows only that the mass changes, not whether it increases or decreases. Saying that the mass is $m+dm$ at time $t+dt$ is a formal statement that the mass has changed, and that’s how analysis of change is done in calculus. The fact that the rocket’s mass is decreasing means that $dm$ has a negative value. That is, the minus goes with the value of $dm$, not with the statement that the mass has changed.

After the gas has been ejected, both the rocket and the gas have momentum. Conservation of momentum tells us that

The mass of this little packet of burned fuel is the mass lost by the rocket: $m_{\text{fuel}}=-dm$. Mathematically, the minus sign tells us that the mass of the burned fuel (the gases) and the rocket mass are changing in opposite directions. Physically, we know that $dm<0$, so the exhaust gases have a positive mass.

The gases are ejected toward the left at speed $v_{\text{ex}}$ relative to the rocket. If the rocket’s velocity is $v_x$, then the gas’s velocity through space is $v_x-v_{\text{ex}}$. When we assemble all these pieces of information, the momentum conservation equation is

Multiplying this out gives

You can see that several terms cancel, leading to $md{v}_x+v_{\text{ex}}dm+dmd{v}_x=0$. We can drop the third term; it is the product of two infinitesimal terms and thus is negligible compared to the first two terms. With one final algebraic rearrangement, we’re left with

Remember that $dm$ is negative—it’s the mass lost by the rocket when a small amount of fuel is burned—and so $d{v}_x$ is positive. Physically, Equation 11.40 is telling us the amount by which the rocket’s velocity increases when it burns a small amount of fuel. Not surprisingly, a lighter rocket (smaller m) gains more velocity than a heavier rocket (larger m).

There are a couple of ways to use Equation 11.40. First, divide both sides by the small interval of time $dt$ in which the fuel is burned, which will make this a rate equation:

where $R=|dm/dt|$ is the rate—in kg/s—at which fuel is burned. The fuel burn rate is reasonably constant for most rocket engines.

The left side of Equation 11.41 is the rocket’s acceleration: $a_x=d{v}_x/dt$. Thus from Newton’s second law, $a_x=F_x/m$, the numerator on the right side of Equation 11.41 must be a force. This is the thrust of the rocket engine:

So Equation 11.41 is just Newton’s second law, $a=F_{\text{thrust}}/m$, for the instantaneous acceleration, which will change as the rocket’s mass m changes. But now we know how the thrust force is related to physical properties of the rocket engine.

Returning to Equation 11.40, we can find out how the rocket’s velocity changes as fuel is burned by integrating. Suppose the rocket starts from rest $(v_x=0)$ with mass $m_0=m_{\text{R}}+m_{\text{F}0}$, where $m_{\text{R}}$ is the mass of the empty rocket and $m_{\text{F}0}$ is the initial mass of the fuel. At a later time, when the mass has been reduced to m, the velocity is v.

Integrating between this Before and After, we find

where $\text{ln }m$ is the natural logarithm (logarithm with base e) of m. Evaluating this between the limits, and using the properties of logarithms, gives

Thus the rocket’s velocity when its mass has decreased to m is

Initially, when $m=m_0\text{,}v=0$ because $\text{ln}1=0$. The maximum speed occurs when the fuel is completely gone and $m=m_{\text{R}}$. This is

Notice that the rocket speed can exceed $v_{\text{ex}}$ if the fuel-mass-to-rocket-mass ratio is large enough. Also, because $v_{\text{max}}$ is greatly improved by reducing $m_{\text{R}}$, we can see why rockets that need enough speed to go into orbit are usually multistage rockets, dropping off the mass of the lower stages when their fuel is depleted.

CHAPTER 11 CHALLENGE EXAMPLE A rebounding pendulum

A 200 g steel ball hangs on a 1.0-m-long string. The ball is pulled sideways so that the string is at a $45\text{°}$ angle, then released. At the very bottom of its swing the ball strikes a 500 g steel paperweight that is resting on a frictionless table. To what angle does the ball rebound?

MODEL We can divide this problem into three parts. First the ball swings down as a pendulum. Second, the ball and paperweight have a collision. Steel balls bounce off each other very well, so we will model the collision as perfectly elastic. Third, the ball, after it bounces off the paperweight, swings back up as a pendulum. For Parts 1 and 3, we define the system as the ball and the earth. This is an isolated, nondissipative system, so its mechanical energy is conserved. In Part 2, let the system consist of the ball and the paperweight, which have a perfectly elastic collision.

VISUALIZE FIGURE 11.30 shows four distinct moments of time: as the ball is released, an instant before the collision, an instant after the collision but before the ball and paperweight have had time to move, and as the ball reaches its highest point on the rebound. Call the ball A and the paperweight B, so $m_{\text{A}}=0.20\text{kg}$ and $m_{\text{B}}=0.50\text{kg}$.

FIGURE 11.30 Four moments in the collision of a pendulum with a paperweight.

FIGURE 11.30 Full Alternative Text

SOLVE Part 1: The first part involves the ball only. Its initial height is

$${\left(y_0\right)}_{\text{A}}=L-L\text{cos}{\theta }_0=L(1-\text{cos}{\theta }_0)=0.293\text{m}$$

We can use conservation of mechanical energy to find the ball’s velocity at the bottom, just before impact on the paperweight:

$$\frac{1}{2}{m}_{\text{A}}{\left(v_1\right)}_{\text{A}}{}^2+m_{\text{A}}g{\left(y_1\right)}_{\text{A}}=\frac{1}{2}{m}_{\text{A}}{\left(v_0\right)}_{\text{A}}{}^2+m_{\text{A}}g{\left(y_0\right)}_{\text{A}}$$

We know ${\left(v_0\right)}_{\text{A}}=0$. Solving for the velocity at the bottom, where ${\left(y_1\right)}_{\text{A}}=0$, gives

$${\left(v_1\right)}_{\text{A}}=\sqrt{2g{\left(y_0\right)}_{\text{A}}}=2.40\text{m}\text{/}\text{s}$$

Part 2: The ball and paperweight undergo a perfectly elastic collision in which the paperweight is initially at rest. These are the conditions for which Equations 11.31 were derived. The velocities immediately after the collision, prior to any further motion, are

$$\begin{array}{l}{\left(v_{2x}\right)}_{\text{A}}={\displaystyle \frac{m_{\text{A}}-m_{\text{B}}}{m_{\text{A}}+m_{\text{B}}}}{\left(v_{1x}\right)}_{\text{A}}=-1.03\text{ m}\text{/}\text{s }\\ {\left(v_{2x}\right)}_{\text{B}}={\displaystyle \frac{2m_{\text{A}}}{m_{\text{A}}+m_{\text{B}}}}{\left(v_{1x}\right)}_{\text{A}}=+1.37\text{ m}\text{/}\text{s}\end{array}$$

The ball rebounds toward the left with a speed of $1.03\text{m}\text{/}\text{s}$ while the paperweight moves to the right at $1.37\text{m}\text{/}\text{s}$. Kinetic energy has been conserved (you might want to check this), but it is now shared between the ball and the paperweight.

Part 3: Now the ball is a pendulum with an initial speed of $1.03\text{m}\text{/}\text{s}$. Mechanical energy is again conserved, so we can find its maximum height at the point where ${\left(v_3\right)}_{\text{A}}=0:$

$${\displaystyle \frac{1}{2}}{m}_{\text{A}}{\left(v_3\right)}_{\text{A}}{}^2+m_{\text{A}}g{\left(y_3\right)}_{\text{A}}={\displaystyle \frac{1}{2}}{m}_{\text{A}}{\left(v_2\right)}_{\text{A}}{}^2+m_{\text{A}}g{\left(y_2\right)}_A$$

Solving for the maximum height gives

$${\left(y_3\right)}_{\text{A}}=\frac{{\left(v_2\right)}_{\text{A}}{}^2}{2g}=0.0541\text{m}$$

The height ${\left(y_3\right)}_{\text{A}}$ is related to angle ${\theta }_3$ by ${\left(y_3\right)}_{\text{A}}=L(1-\text{cos}{\theta }_3)$. This can be solved to find the angle of rebound:

$${\theta }_3={\text{cos}}^{-1}(1-{\displaystyle \frac{{\left(y_3\right)}_{\text{A}}}{L}})=19\text{°}$$

The paperweight speeds away at $1.37\text{m}\text{/}\text{s}$ and the ball rebounds to an angle of $19\text{°}$.

REVIEW The ball and the paperweight aren’t hugely different in mass, so we expect the ball to transfer a significant fraction of its energy to the paperweight when they collide. Thus a rebound to roughly half the initial angle seems reasonable.