## 9 Linear Momentum and Collisions

9.7 rocket propulsion, learning objectives.

By the end of this section, you will be able to:

- Describe the application of conservation of momentum when the mass changes with time, as well as the velocity
- Calculate the speed of a rocket in empty space, at some time, given initial conditions
- Calculate the speed of a rocket in Earth’s gravity field, at some time, given initial conditions

Now we deal with the case where the mass of an object is changing. We analyze the motion of a rocket, which changes its velocity (and hence its momentum) by ejecting burned fuel gases, thus causing it to accelerate in the opposite direction of the velocity of the ejected fuel (see (Figure) ). Specifically: A fully fueled rocket ship in deep space has a total mass [latex] {m}_{0} [/latex] (this mass includes the initial mass of the fuel). At some moment in time, the rocket has a velocity [latex] \overset{\to }{v} [/latex] and mass m ; this mass is a combination of the mass of the empty rocket and the mass of the remaining unburned fuel it contains. (We refer to m as the “instantaneous mass” and [latex] \overset{\to }{v} [/latex] as the “instantaneous velocity.”) The rocket accelerates by burning the fuel it carries and ejecting the burned exhaust gases. If the burn rate of the fuel is constant, and the velocity at which the exhaust is ejected is also constant, what is the change of velocity of the rocket as a result of burning all of its fuel?

Figure 9.32 The space shuttle had a number of reusable parts. Solid fuel boosters on either side were recovered and refueled after each flight, and the entire orbiter returned to Earth for use in subsequent flights. The large liquid fuel tank was expended. The space shuttle was a complex assemblage of technologies, employing both solid and liquid fuel, and pioneering ceramic tiles as reentry heat shields. As a result, it permitted multiple launches as opposed to single-use rockets. (credit: modification of work by NASA)

## Physical Analysis

Here’s a description of what happens, so that you get a feel for the physics involved.

- As the rocket engines operate, they are continuously ejecting burned fuel gases, which have both mass and velocity, and therefore some momentum. By conservation of momentum, the rocket’s momentum changes by this same amount (with the opposite sign). We will assume the burned fuel is being ejected at a constant rate, which means the rate of change of the rocket’s momentum is also constant. By (Figure) , this represents a constant force on the rocket.
- However, as time goes on, the mass of the rocket (which includes the mass of the remaining fuel) continuously decreases. Thus, even though the force on the rocket is constant, the resulting acceleration is not; it is continuously increasing.
- So, the total change of the rocket’s velocity will depend on the amount of mass of fuel that is burned, and that dependence is not linear.

The problem has the mass and velocity of the rocket changing; also, the total mass of ejected gases is changing. If we define our system to be the rocket + fuel, then this is a closed system (since the rocket is in deep space, there are no external forces acting on this system); as a result, momentum is conserved for this system. Thus, we can apply conservation of momentum to answer the question ( (Figure) ).

Figure 9.33 The rocket accelerates to the right due to the expulsion of some of its fuel mass to the left. Conservation of momentum enables us to determine the resulting change of velocity. The mass m is the instantaneous total mass of the rocket (i.e., mass of rocket body plus mass of fuel at that point in time). (credit: modification of work by NASA/Bill Ingalls)

At the same moment that the total instantaneous rocket mass is m (i.e., m is the mass of the rocket body plus the mass of the fuel at that point in time), we define the rocket’s instantaneous velocity to be [latex] \overset{\to }{v}=v\hat{i} [/latex] (in the + x -direction); this velocity is measured relative to an inertial reference system (the Earth, for example). Thus, the initial momentum of the system is

[latex] {\overset{\to }{p}}_{\text{i}}=mv\hat{i}. [/latex]

The rocket’s engines are burning fuel at a constant rate and ejecting the exhaust gases in the − x -direction. During an infinitesimal time interval dt , the engines eject a (positive) infinitesimal mass of gas [latex] d{m}_{g} [/latex] at velocity [latex] \overset{\to }{u}=\text{−}u\hat{i} [/latex]; note that although the rocket velocity [latex] v\hat{i} [/latex] is measured with respect to Earth, the exhaust gas velocity is measured with respect to the (moving) rocket. Measured with respect to the Earth, therefore, the exhaust gas has velocity [latex] (v-u)\hat{i} [/latex].

As a consequence of the ejection of the fuel gas, the rocket’s mass decreases by [latex] d{m}_{g} [/latex], and its velocity increases by [latex] dv\hat{i} [/latex]. Therefore, including both the change for the rocket and the change for the exhaust gas, the final momentum of the system is

Since all vectors are in the x -direction, we drop the vector notation. Applying conservation of momentum, we obtain

Now, [latex] d{m}_{g} [/latex] and dv are each very small; thus, their product [latex] d{m}_{g}dv [/latex] is very, very small, much smaller than the other two terms in this expression. We neglect this term, therefore, and obtain:

Our next step is to remember that, since [latex] d{m}_{g} [/latex] represents an increase in the mass of ejected gases, it must also represent a decrease of mass of the rocket:

Replacing this, we have

Integrating from the initial mass m i to the final mass m of the rocket gives us the result we are after:

and thus our final answer is

This result is called the rocket equation . It was originally derived by the Soviet physicist Konstantin Tsiolkovsky in 1897. It gives us the change of velocity that the rocket obtains from burning a mass of fuel that decreases the total rocket mass from [latex] {m}_{0} [/latex] down to m . As expected, the relationship between [latex] \text{Δ}v [/latex] and the change of mass of the rocket is nonlinear.

## Problem-Solving Strategy: Rocket Propulsion

In rocket problems, the most common questions are finding the change of velocity due to burning some amount of fuel for some amount of time; or to determine the acceleration that results from burning fuel.

- To determine the change of velocity, use the rocket equation (Figure) .
- To determine the acceleration, determine the force by using the impulse-momentum theorem, using the rocket equation to determine the change of velocity.

## Thrust on a Spacecraft

A spacecraft is moving in gravity-free space along a straight path when its pilot decides to accelerate forward. He turns on the thrusters, and burned fuel is ejected at a constant rate of [latex] 2.0\,×\,{10}^{2}\,\text{kg/s} [/latex], at a speed (relative to the rocket) of [latex] 2.5\,×\,{10}^{2}\,\text{m/s} [/latex]. The initial mass of the spacecraft and its unburned fuel is [latex] 2.0\,×\,{10}^{4}\,\text{kg} [/latex], and the thrusters are on for 30 s.

- What is the thrust (the force applied to the rocket by the ejected fuel) on the spacecraft?
- What is the spacecraft’s acceleration as a function of time?
- What are the spacecraft’s accelerations at t = 0, 15, 30, and 35 s?
- The force on the spacecraft is equal to the rate of change of the momentum of the fuel.
- Knowing the force from part (a), we can use Newton’s second law to calculate the consequent acceleration. The key here is that, although the force applied to the spacecraft is constant (the fuel is being ejected at a constant rate), the mass of the spacecraft isn’t; thus, the acceleration caused by the force won’t be constant. We expect to get a function a ( t ), therefore.
- We’ll use the function we obtain in part (b), and just substitute the numbers given. Important: We expect that the acceleration will get larger as time goes on, since the mass being accelerated is continuously decreasing (fuel is being ejected from the rocket).

The ejection velocity [latex] v=2.5\,×\,{10}^{2}\text{m/s} [/latex] is constant, and therefore the force is

Now, [latex] \frac{d{m}_{g}}{dt} [/latex] is the rate of change of the mass of the fuel; the problem states that this is [latex] 2.0\,×\,{10}^{2}\text{kg/s} [/latex]. Substituting, we get

The force is constant and the empty rocket mass [latex] {m}_{R} [/latex] is constant, but the fuel mass [latex] {m}_{g} [/latex] is decreasing at a uniform rate; specifically:

This gives us

Notice that, as expected, the acceleration is a function of time. Substituting the given numbers:

At [latex] t=15\,\text{s},\,a(15\,\text{s})={2.9\,\text{m/s}}^{2} [/latex].

At [latex] t=30\,\text{s},\,a(30\,\text{s})=3.6\,{\text{m/s}}^{2} [/latex].

## Significance

Notice that the acceleration is not constant; as a result, any dynamical quantities must be calculated either using integrals, or (more easily) conservation of total energy.

## Check Your Understanding

What is the physical difference (or relationship) between [latex] \frac{dm}{dt} [/latex] and [latex] \frac{d{m}_{g}}{dt} [/latex] in this example?

The notation [latex] {m}_{g} [/latex] stands for the mass of the fuel and m stands for the mass of the rocket plus the initial mass of the fuel. Note that [latex] {m}_{g} [/latex] changes with time, so we write it as [latex] {m}_{g}(t) [/latex]. Using [latex] {m}_{R} [/latex] as the mass of the rocket with no fuel, the total mass of the rocket plus fuel is [latex] m={m}_{R}+{m}_{g}(t) [/latex]. Differentiation with respect to time gives

[latex] \frac{dm}{dt}=\frac{d{m}_{R}}{dt}+\frac{d{m}_{g}(t)}{dt}=\frac{d{m}_{g}(t)}{dt} [/latex]

where we used [latex] \frac{d{m}_{R}}{dt}=0 [/latex] because the mass of the rocket does not change. Thus, time rate of change of the mass of the rocket is the same as that of the fuel.

## Rocket in a Gravitational Field

Let’s now analyze the velocity change of the rocket during the launch phase, from the surface of Earth. To keep the math manageable, we’ll restrict our attention to distances for which the acceleration caused by gravity can be treated as a constant g .

The analysis is similar, except that now there is an external force of [latex] \overset{\to }{F}=\text{−}mg\hat{j} [/latex] acting on our system. This force applies an impulse [latex] d\overset{\to }{J}=\overset{\to }{F}dt=\text{−}mgdt\hat{j} [/latex], which is equal to the change of momentum. This gives us

where we have again neglected the term [latex] d{m}_{g}dv [/latex] and dropped the vector notation. Next we replace [latex] d{m}_{g} [/latex] with [latex] \text{−}dm [/latex]:

Dividing through by m gives

and integrating, we have

Unsurprisingly, the rocket’s velocity is affected by the (constant) acceleration of gravity.

Remember that [latex] \text{Δ}t [/latex] is the burn time of the fuel. Now, in the absence of gravity, (Figure) implies that it makes no difference how much time it takes to burn the entire mass of fuel; the change of velocity does not depend on [latex] \text{Δ}t [/latex]. However, in the presence of gravity, it matters a lot. The − g [latex] \text{Δ}t [/latex] term in (Figure) tells us that the longer the burn time is, the smaller the rocket’s change of velocity will be. This is the reason that the launch of a rocket is so spectacular at the first moment of liftoff: It’s essential to burn the fuel as quickly as possible, to get as large a [latex] \text{Δ}v [/latex] as possible.

- A rocket is an example of conservation of momentum where the mass of the system is not constant, since the rocket ejects fuel to provide thrust.
- The rocket equation gives us the change of velocity that the rocket obtains from burning a mass of fuel that decreases the total rocket mass.

## Key Equations

Conceptual questions.

It is possible for the velocity of a rocket to be greater than the exhaust velocity of the gases it ejects. When that is the case, the gas velocity and gas momentum are in the same direction as that of the rocket. How is the rocket still able to obtain thrust by ejecting the gases?

Yes, the rocket speed can exceed the exhaust speed of the gases it ejects. The thrust of the rocket does not depend on the relative speeds of the gases and rocket, it simply depends on conservation of momentum.

(a) A 5.00-kg squid initially at rest ejects 0.250 kg of fluid with a velocity of 10.0 m/s. What is the recoil velocity of the squid if the ejection is done in 0.100 s and there is a 5.00- N frictional force opposing the squid’s movement?

(b) How much energy is lost to work done against friction?

(a) 0.413 m/s, (b) about 0.2 J

A rocket takes off from Earth and reaches a speed of 100 m/s in 10.0 s. If the exhaust speed is 1500 m/s and the mass of fuel burned is 100 kg, what was the initial mass of the rocket?

Repeat the preceding problem but for a rocket that takes off from a space station, where there is no gravity other than the negligible gravity due to the space station.

How much fuel would be needed for a 1000-kg rocket (this is its mass with no fuel) to take off from Earth and reach 1000 m/s in 30 s? The exhaust speed is 1000 m/s.

What exhaust speed is required to accelerate a rocket in deep space from 800 m/s to 1000 m/s in 5.0 s if the total rocket mass is 1200 kg and the rocket only has 50 kg of fuel left?

Unreasonable Results Squids have been reported to jump from the ocean and travel 30.0 m (measured horizontally) before re-entering the water.

(a) Calculate the initial speed of the squid if it leaves the water at an angle of 20.0°, assuming negligible lift from the air and negligible air resistance.

(b) The squid propels itself by squirting water. What fraction of its mass would it have to eject in order to achieve the speed found in the previous part? The water is ejected at 12.0 m/s; gravitational force and friction are neglected.

(c) What is unreasonable about the results?

(d) Which premise is unreasonable, or which premises are inconsistent?

## Additional Problems

Two 70-kg canoers paddle in a single, 50-kg canoe. Their paddling moves the canoe at 1.2 m/s with respect to the water, and the river they’re in flows at 4 m/s with respect to the land. What is their momentum with respect to the land?

Which has a larger magnitude of momentum: a 3000-kg elephant moving at 40 km/h or a 60-kg cheetah moving at 112 km/h?

the elephant has a higher momentum

A driver applies the brakes and reduces the speed of her car by 20%, without changing the direction in which the car is moving. By how much does the car’s momentum change?

You friend claims that momentum is mass multiplied by velocity, so things with more mass have more momentum. Do you agree? Explain.

Answers may vary. The first clause is true, but the second clause is not true in general because the velocity of an object with small mass may be large enough so that the momentum of the object is greater than that of a larger-mass object with a smaller velocity.

Dropping a glass on a cement floor is more likely to break the glass than if it is dropped from the same height on a grass lawn. Explain in terms of the impulse.

Your 1500-kg sports car accelerates from 0 to 30 m/s in 10 s. What average force is exerted on it during this acceleration?

[latex] 4.5\,×\,{10}^{3}\,\text{N} [/latex]

A ball of mass [latex] m [/latex] is dropped. What is the formula for the impulse exerted on the ball from the instant it is dropped to an arbitrary time [latex] \tau [/latex] later? Ignore air resistance.

Repeat the preceding problem, but including a drag force due to air of [latex] {f}_{\text{drag}}=\text{−}b\overset{\to }{v}. [/latex]

[latex] \overset{\to }{J}={\int }_{0}^{\tau }[m\overset{\to }{g}-m\overset{\to }{g}(1-{e}^{\text{−}bt\text{/}m})]dt=\frac{{m}^{2}}{b}\overset{\to }{g}({e}^{\text{−}b\tau \text{/}m}-1) [/latex]

A 5.0-g egg falls from a 90-cm-high counter onto the floor and breaks. What impulse is exerted by the floor on the egg?

A car crashes into a large tree that does not move. The car goes from 30 m/s to 0 in 1.3 m. (a) What impulse is applied to the driver by the seatbelt, assuming he follows the same motion as the car? (b) What is the average force applied to the driver by the seatbelt?

a. [latex] \text{−}(2.1\,×\,{10}^{3}\,\text{kg}·\text{m/s})\hat{i} [/latex], b. [latex] \text{−}(24\,×\,{10}^{3}\,\text{N})\hat{i} [/latex]

Two hockey players approach each other head on, each traveling at the same speed [latex] {v}_{\text{i}} [/latex]. They collide and get tangled together, falling down and moving off at a speed [latex] {v}_{\text{i}}\text{/}5 [/latex]. What is the ratio of their masses?

You are coasting on your 10-kg bicycle at 15 m/s and a 5.0-g bug splatters on your helmet. The bug was initially moving at 2.0 m/s in the same direction as you. If your mass is 60 kg, (a) what is the initial momentum of you plus your bicycle? (b) What is the initial momentum of the bug? (c) What is your change in velocity due to the collision with the bug? (d) What would the change in velocity have been if the bug were traveling in the opposite direction?

a. [latex] (1.1\,×\,{10}^{3}\,\text{kg}·\text{m/s})\hat{i} [/latex], b. [latex] (0.010\,\text{kg}·\text{m/s})\hat{i} [/latex], c. [latex] \text{−}(0.00093\,\text{m/s})\hat{i} [/latex], d. [latex] \text{−}(0.0012\,\text{m/s})\hat{i} [/latex]

A load of gravel is dumped straight down into a 30 000-kg freight car coasting at 2.2 m/s on a straight section of a railroad. If the freight car’s speed after receiving the gravel is 1.5 m/s, what mass of gravel did it receive?

Two carts on a straight track collide head on. The first cart was moving at 3.6 m/s in the positive x direction and the second was moving at 2.4 m/s in the opposite direction. After the collision, the second car continues moving in its initial direction of motion at 0.24 m/s. If the mass of the second car is 5.0 times that of the first, what is the mass and final velocity of the first car?

0.10 kg, [latex] \text{−}(130\,\text{m/s})\hat{i} [/latex]

A 100-kg astronaut finds himself separated from his spaceship by 10 m and moving away from the spaceship at 0.1 m/s. To get back to the spaceship, he throws a 10-kg tool bag away from the spaceship at 5.0 m/s. How long will he take to return to the spaceship?

Derive the equations giving the final speeds for two objects that collide elastically, with the mass of the objects being [latex] {m}_{1} [/latex] and [latex] {m}_{2} [/latex] and the initial speeds being [latex] {v}_{\text{1,i}} [/latex] and [latex] {v}_{\text{2,i}}=0 [/latex] (i.e., second object is initially stationary).

[latex] {v}_{\text{1,f}}={v}_{\text{1,i}}\frac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}},\,{v}_{\text{2,f}}={v}_{\text{1,i}}\frac{2{m}_{1}}{{m}_{1}+{m}_{2}} [/latex]

Repeat the preceding problem for the case when the initial speed of the second object is nonzero.

A child sleds down a hill and collides at 5.6 m/s into a stationary sled that is identical to his. The child is launched forward at the same speed, leaving behind the two sleds that lock together and slide forward more slowly. What is the speed of the two sleds after this collision?

For the preceding problem, find the final speed of each sled for the case of an elastic collision.

A 90-kg football player jumps vertically into the air to catch a 0.50-kg football that is thrown essentially horizontally at him at 17 m/s. What is his horizontal speed after catching the ball?

Three skydivers are plummeting earthward. They are initially holding onto each other, but then push apart. Two skydivers of mass 70 and 80 kg gain horizontal velocities of 1.2 m/s north and 1.4 m/s southeast, respectively. What is the horizontal velocity of the third skydiver, whose mass is 55 kg?

Two billiard balls are at rest and touching each other on a pool table. The cue ball travels at 3.8 m/s along the line of symmetry between these balls and strikes them simultaneously. If the collision is elastic, what is the velocity of the three balls after the collision?

final velocity of cue ball is [latex] \text{−}(0.76\,\text{m/s})\hat{i} [/latex], final velocities of the other two balls are 2.6 m/s at ±30° with respect to the initial velocity of the cue ball

A billiard ball traveling at [latex] (2.2\,\text{m/s})\hat{i}-(0.4\,\text{m/s})\hat{j} [/latex] collides with a wall that is aligned in the [latex] \hat{j} [/latex] direction. Assuming the collision is elastic, what is the final velocity of the ball?

Two identical billiard balls collide. The first one is initially traveling at [latex] (2.2\,\text{m/s})\hat{i}-(0.4\,\text{m/s})\hat{j} [/latex] and the second one at [latex] \text{−}(1.4\,\text{m/s})\hat{i}+(2.4\,\text{m/s})\hat{j} [/latex]. Suppose they collide when the center of ball 1 is at the origin and the center of ball 2 is at the point [latex] (2R,0) [/latex] where R is the radius of the balls. What is the final velocity of each ball?

ball 1: [latex] \text{−}(1.4\,\text{m/s})\hat{i}-(0.4\,\text{m/s})\hat{j} [/latex], ball 2: [latex] (2.2\,\text{m/s})\hat{i}+(2.4\,\text{m/s})\hat{j} [/latex]

Repeat the preceding problem if the balls collide when the center of ball 1 is at the origin and the center of ball 2 is at the point [latex] (0,2R) [/latex].

Repeat the preceding problem if the balls collide when the center of ball 1 is at the origin and the center of ball 2 is at the point [latex] (\sqrt{3}R\text{/}2,R\text{/}2) [/latex]

ball 1: [latex] (1.4\,\text{m/s})\hat{i}-(1.7\,\text{m/s})\hat{j} [/latex], ball 2: [latex] \text{−}(2.8\,\text{m/s})\hat{i}+(0.012\,\text{m/s})\hat{j} [/latex]

Where is the center of mass of a semicircular wire of radius R that is centered on the origin, begins and ends on the x axis, and lies in the x , y plane?

Where is the center of mass of a slice of pizza that was cut into eight equal slices? Assume the origin is at the apex of the slice and measure angles with respect to an edge of the slice. The radius of the pizza is R .

[latex] (r,\theta )=(2R\text{/}3,\pi \text{/}8) [/latex]

If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon-population system move? Assume the population is 7 billion, the average human has a mass of 65 kg, and that the population is evenly distributed over both the Earth and the Moon. The mass of the Earth is [latex] 5.97\,×\,{10}^{24}\text{kg} [/latex] and that of the Moon is [latex] 7.34\,×\,{10}^{22}\text{kg} [/latex]. The radius of the Moon’s orbit is about [latex] 3.84\,×\,{10}^{5}\text{m} [/latex].

You friend wonders how a rocket continues to climb into the sky once it is sufficiently high above the surface of Earth so that its expelled gasses no longer push on the surface. How do you respond?

Answers may vary. The rocket is propelled forward not by the gasses pushing against the surface of Earth, but by conservation of momentum. The momentum of the gas being expelled out the back of the rocket must be compensated by an increase in the forward momentum of the rocket.

To increase the acceleration of a rocket, should you throw rocks out of the front window of the rocket or out of the back window?

A 65-kg person jumps from the first floor window of a burning building and lands almost vertically on the ground with a horizontal velocity of 3 m/s and vertical velocity of [latex] -9\,\text{m/s} [/latex]. Upon impact with the ground he is brought to rest in a short time. The force experienced by his feet depends on whether he keeps his knees stiff or bends them. Find the force on his feet in each case.

- First find the impulse on the person from the impact on the ground. Calculate both its magnitude and direction.
- Find the average force on the feet if the person keeps his leg stiff and straight and his center of mass drops by only 1 cm vertically and 1 cm horizontally during the impact.
- Find the average force on the feet if the person bends his legs throughout the impact so that his center of mass drops by 50 cm vertically and 5 cm horizontally during the impact.
- Compare the results of part (b) and (c), and draw conclusions about which way is better.

You will need to find the time the impact lasts by making reasonable assumptions about the deceleration. Although the force is not constant during the impact, working with constant average force for this problem is acceptable.

a. [latex] 617\,\text{N}·\text{s} [/latex], 108°; b. [latex] {F}_{x}=2.91\,×\,{10}^{4}\,\text{N} [/latex], [latex] {F}_{y}=2.6\,×\,{10}^{5}\,\text{N} [/latex]; c. [latex] {F}_{x}=5265\,\text{N} [/latex], [latex] {F}_{y}=5850\,\text{N} [/latex]

Two projectiles of mass [latex] {m}_{1} [/latex] and [latex] {m}_{2} [/latex] are ﬁred at the same speed but in opposite directions from two launch sites separated by a distance D . They both reach the same spot in their highest point and strike there. As a result of the impact they stick together and move as a single body afterwards. Find the place they will land.

Two identical objects (such as billiard balls) have a one-dimensional collision in which one is initially motionless. After the collision, the moving object is stationary and the other moves with the same speed as the other originally had. Show that both momentum and kinetic energy are conserved.

Conservation of momentum demands [latex] {m}_{1}{v}_{\text{1,i}}+{m}_{2}{v}_{\text{2,i}}={m}_{1}{v}_{\text{1,f}}+{m}_{2}{v}_{\text{2,f}} [/latex]. We are given that [latex] {m}_{1}={m}_{2} [/latex], [latex] {v}_{\text{1,i}}={v}_{\text{2,f}} [/latex], and [latex] {v}_{\text{2,i}}={v}_{\text{1,f}}=0 [/latex]. Combining these equations with the equation given by conservation of momentum gives [latex] {v}_{\text{1,i}}={v}_{\text{1,i}} [/latex], which is true, so conservation of momentum is satisfied. Conservation of energy demands [latex] \frac{1}{2}{m}_{1}{v}_{\text{1,i}}^{2}+\frac{1}{2}{m}_{2}{v}_{\text{2,i}}^{2}=\frac{1}{2}{m}_{1}{v}_{\text{1,f}}^{2}+\frac{1}{2}{m}_{2}{v}_{\text{2,f}}^{2} [/latex]. Again combining this equation with the conditions given above give [latex] {v}_{\text{1,i}}={v}_{\text{1,i}} [/latex], so conservation of energy is satisfied.

A ramp of mass M is at rest on a horizontal surface. A small cart of mass m is placed at the top of the ramp and released.

What are the velocities of the ramp and the cart relative to the ground at the instant the cart leaves the ramp?

Find the center of mass of the structure given in the figure below. Assume a uniform thickness of 20 cm, and a uniform density of [latex] 1{\,\text{g/cm}}^{3}. [/latex]

Assume origin on centerline and at floor, then [latex] ({x}_{\text{CM}},{y}_{\text{CM}})=(0,86\,\text{cm}) [/latex]

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## Collection of Solved Problems in Physics

Task number: 4318.

A rocket of a mass \(M_r=2\,\mathrm{t}\) loaded with a propellant of a mass \(M_p=12\,\mathrm{t}\) is powered by a rocket engine whose exhaust velocity is \(v_e=5000\,\mathrm{km\cdot h^{-1}}\). What is the maximal fuel consumption per second possible, how long can the rocket engine under such a consumption operate and what is the final speed of the rocket?

The maximal acceleration of the crew allowed is \(7 \, g\), where \(g\) stands for the gravitational acceleration.

\(M_r=2\,\mathrm{t}\) … mass of the rocket without the propellant

\(M_p=12\,\mathrm{t}\) … mass of the propellant

\(v_e = 5000\,\mathrm{km\cdot h^{-1}}\) … exhaust velocity of the rocket

\(a_{max}=7g\) … maximal acceleration

\(\left|\frac{dm}{dt}\right|_{max} =? \mathrm{kg\cdot s^{-1}}\) … maximal fuel consumption per second

\(\tau= ? \mathrm{s}\) … time of operation of the rocket engine

\(v_f= ? \mathrm{km\cdot h^{-1}}\) … final speed of the rocket

We will derive the movement of the rocket in a few steps.

We first consider the rocket to be in an isolated system, i.e., we consider only the rocket to be located in our system with no external forces acting on it and not being in any force field. The rocket is moving with velocity \(\vec v\) with respect to an inertial frame of reference. It must expel propellant (exhaust gases) to accelerate. Let’s assume that it expels a mass \(\mathrm{d}m\) of the exhaust gases during the time interval \(\mathrm{d}t\) and that the velocity of those gases with respect to the inertial frame of reference is \(\vec {v_p}\). The velocity of the rocket is thus increased by \(\mathrm{d}\vec v\).

We can obtain the velocity of the expelled exhaust gases with respect to the inertial frame of reference as a superposition of the velocity of the rocket \(\vec v\) and the velocity \(\vec {v_e}\) of the exhaust gases with respect to the rocket. We write \(\vec {v_p}=\vec v + \vec{v_e}\). The velocity \(\vec {v_e}\) is called the exhaust velocity.

The law of conservation of momentum holds in our isolated system. We can denote it as follows:

The motion of the rocket is depicted in the following figure; we can see the situation in the initial time \(t\) and then after passing the time interval \(\mathrm{d}t\) when the exhaust gases of the mass \(\mathrm{d}m\) are expelled.

We write the law of conservation of momentum for the rocket in an isolated system as follows:

\[m\vec v = \left(m-|\mathrm{d}m|\right)\left(\vec v + \mathrm{d}\vec v\right) + |\mathrm{d}m|\vec{v_p}.\]

Let us now consider the motion of the rocket in a gravitational field so that the gravitational force acts against the direction in which the rocket was launched. We will use Meshchersky’s equation in this case.

To derive the equation, we begin with change in momentum of a body being equal to the impulse of the force acting on it:

Simplifying the equation and neglecting the term \(\mathrm{d}m \mathrm{d}\vec v\) we get:

We divide the equation by \(\mathrm{d}t\), remove the absolute value of \(\mathrm{d}m\) and substitute the exhaust velocity \(\vec {v_e}\) for the difference of velocities \(\vec{v_p} - \vec v\). We obtain the Meshchersky’s equation :

\[m\frac{\mathrm{d}\vec v}{\mathrm{d}t}=\vec F + \vec {v_e}\frac{\mathrm{d}m}{\mathrm{d}t},\]

where \(\vec F\) is the force acting on the body in the given field. In our case, it is the gravitational force acting on the rocket. We can thus substitute the expression \(m\vec g\) for the gravitational force into the Meshchersky’s equation :

\[m\frac{\mathrm{d}\vec v}{\mathrm{d}t}=m\vec g + \vec {v_e}\frac{\mathrm{d}m}{\mathrm{d}t}.\]

We know that the gravitational force is acting in a direction opposite to the direction of the motion of the rocket. The direction of the velocity of the exhaust gases is also opposite to the direction of the motion of the rocket. Choosing the direction of the motion of the rocket to have the positive sign, we write:

\[m\frac{\mathrm{d}v}{\mathrm{d}t}=-mg - {v_e}\frac{\mathrm{d}m}{\mathrm{d}t}.\tag{1}\]

We now aim to determine the final speed of the rocket \(v_{f}\).

Multiplying the equation (1) by \(\mathrm{d}t\) we get:

Dividing the equation by \(m\) gives us:

\[\mathrm{d}v= -v_e\frac{\mathrm{d}m}{m} - g\mathrm{d}t,\tag{2}\]

where \(\mathrm{d}v\) is a very small change in the rocket’s speed.

To determine the final speed of the rocket \(v_{f}\) we must integrate the equation over the whole time of operation of the rocket engine. The engine starts operating at time 0 and stops its activity at the time we denote as \(\tau\). The speed of the rocket changes during that time interval from the initial speed \(v_0\) to the rocket’s final and maximal speed \(v_{f}\). The mass changes during that time interval from the initial value \(m_0\) (which is the mass of the rocket and the propellant together) to the final value \(m_f\) (which is the mass of the rocket alone).

From the equation (2) we thus get:

\[\int_{v_0}^{v_{f}}1\,\mathrm{d}v=-v_e\,\int_{m_0}^{m_f}{\frac{1}{m}}\,\mathrm{d}m -g\int_0^{\tau}1\,\mathrm{d}t,\]

\[[v]_{v_0}^{v_{f}}=-v_e [\ln m]_{m_0}^{m_f}-g[t]_0^{\tau},\]

\[v_{f}-v_0=v_e\ln{\frac{m_0}{m_f}}-g\tau.\tag{3}\]

The equation we have just obtained is termed the Tsiolkovsky rocket equation and it gives the relationship for the maximal speed of the rocket which can be obtained after a given time of the operation of the rocket engine. The maximal speed is a function of the mass of the propellant and the mass of the rocket alone.

Start with Meshchersky’s equation and express the acceleration of the rocket from it. For which mass of the rocket will the acceleration be maximal?

## Hint I solution

Meshchersky’s equation for the movement in the gravitational field is (see theory ):

\[m\frac{\mathrm{d}v}{\mathrm{d}t}=-mg - {v_e}\frac{\mathrm{d}m}{\mathrm{d}t}.\]

We can express acceleration as the change in speed with respect to a change in time. Thus dividing the equation by \(m\) we get the expression of acceleration \(a\):

\[a=\frac{\mathrm{d}v}{\mathrm{d}t}=-g - \frac{v_e}{m}\frac{\mathrm{d}m}{\mathrm{d}t}.\]

We achieve the maximal acceleration when moving the body of a minimal mass. The rocket is the lightest when it burns all the propellant and carries only its own mass. We know the maximal acceleration allowed (\(7\,\mathrm{g}\)) and we are able to express the maximal fuel consumption per second possible \(\left(\frac{\mathrm{d}m}{\mathrm{d}t}\right)_{max}\).

We are able to express the maximal fuel consumption and we know the mass of the propellant. Use it to express the time of operation of the rocket engine.

## Hint II solution

We get the time of operation of the rocket engine by dividing the mass of the propellant by the maximal fuel consumption:

\[\tau = \frac{M_p}{\left(\frac{\mathrm{d}m}{\mathrm{d}t}\right)_{max}}.\]

Use the Tsiolkovsky rocket equation derived in the theory part (see theory ) to find the final speed of the rocket.

## Hint III solution

We get the final speed of the rocket by substituting the time of operation of the rocket engine (which we found in the previous hint) into the Tsiolkovsky rocket equation (3) .

## Complete solution

We start with equation (2) for a very small change in the rocket’s speed which we derived from Meshchersky’s equation in the theory (see theory ) section.

\[dv=-v_e\frac{\mathrm{d}m}{m}-g\mathrm{d}t.\]

The task assignment gives us a condition for maximal acceleration. Acceleration is the change in speed with respect to a change in time and thus adjusting the above equation we get:

\[a=\frac{\mathrm{d}v}{\mathrm{d}t}=-\frac{v_e}{m}\frac{\mathrm{d}m}{\mathrm{d}t}-g.\tag{4}\]

The term \(\frac{\mathrm{d}m}{\mathrm{d}t}\) stands for fuel consumption per second which can be considered constant. Following steps lead us to determine the maximal fuel consumption per second possible:

The acceleration \(a\) is maximal when the mass of the rocket as a whole (i.e. with the propellant) is minimal. The rocket is lightest just before the end of operation of the rocket engine when it burns all the propellant and carries only its own mass, i.e. when \(m=M_r\).

We know that the maximal acceleration allowed is a multiple of the gravitational acceleration \(g\):

\[a_{max}=7g.\]

Let’s denote the acceleration generally as an n -multiple of the gravitational acceleration ( n stands for a natural number). Substituting this and the minimal mass (which leads to the maximal acceleration of the rocket) to the equation (4) , we get:

\[a=ng=-\frac{v_e}{M_r}\frac{\mathrm{d}m}{\mathrm{d}t}-g.\tag{5}\]

We want to determine the maximal fuel consumption per second, i.e. the value of \(\left(\frac{\mathrm{d}m}{\mathrm{d}t}\right)_{max}\). From the equation (5) , we have:

\[\left(\frac{\mathrm{d}m}{\mathrm{d}t}\right)_{max}= -\left(n+1\right)g\frac{M_r}{v_e}.\]

The mass of the fuel decreases, i.e. \(\left(\frac{\mathrm{d}m}{\mathrm{d}t}\right)_{max}\) is negative. When substituting numbers, we take the absolute value:

\[\left|\left(\frac{\mathrm{d}m}{\mathrm{d}t}\right)_{max}\right|=\left(7+1\right)\cdot9.81\cdot\frac{2000}{1388.89}\,\mathrm{kg\cdot s^{-1}}=113.0\,\mathrm{kg\cdot s^{-1}}.\]

Knowing the mass of the fuel consumed per \(1\,\mathrm{s}\) of the rocket engine operation, we want to determine the total time of operation of the engine. To do so, we divide the total mass of the propellant by its consumption per second. Thus:

\[\tau = \frac{M_p}{\left|\left(\frac{\mathrm{d}m}{\mathrm{d}t}\right)_{max}\right|}=\frac{M_pv_e}{(n+1)gM_r}.\]

After substitution, we have:

\[\tau =\frac{12000{\cdot} 1388.89}{(7+1)\cdot 9.81{\cdot} 2000}\,\mathrm {s}=106.2\,\mathrm{s}.\]

It remains to determine the final speed of the rocket \(v_f\). We use the Tsiolkovsky rocket equation (3) , whose derivation is described in detail in the section theory . Substituting the mass from the task assignment and the expression of the total time of operation \(\tau\), we get:

\[v_f=v_e\ln{\frac{M_r +M_p}{M_r}}-g\frac{M_pv_e}{(n+1)gM_r}=v_e\ln{\frac{M_r +M_p}{M_r}}-\frac{M_pv_e}{(n+1)M_r}.\]

We substitute all the values:

\[v_f= (5000\cdot\ln{\frac{2000 +12000}{2000}}-\frac{12000{\cdot}5000}{(7+1)\cdot 2000})\,\mathrm{km\cdot h^{-1}}\doteq 6000\,\mathrm{km\cdot h^{-1}},\]

which gave us the answer to the last question from the task assignment.

The maximal fuel consumption per second is generally given by the following equation:

\[\left|\left(\frac{\mathrm{d}m}{\mathrm{d}t}\right)_{max}\right|=\left(n+1\right)g\frac{M_r}{v_e}\]

and for our rocket, it equals \(113\,\mathrm{kg\cdot s^{-1}}\).

The time of operation of the rocket engine with such power is:

\[\tau=\frac{M_pv_e}{(n+1)gM_r}=106\,\mathrm{s}.\]

The final speed of the rocket is \(6000\,\mathrm{km\cdot h^{-1}}\) and generally, it is described as follows:

\[v_f=v_e\ln{\frac{M_r +M_p}{M_r}}-\frac{M_pv_e}{(n+1)M_r}.\]

9 Linear Momentum and Collisions

## 9.7 Rocket Propulsion

Learning objectives.

By the end of this section, you will be able to:

- Describe the application of conservation of momentum when the mass changes with time, as well as the velocity
- Calculate the speed of a rocket in empty space, at some time, given initial conditions
- Calculate the speed of a rocket in Earth’s gravity field, at some time, given initial conditions

Now we deal with the case where the mass of an object is changing. We analyze the motion of a rocket, which changes its velocity (and hence its momentum) by ejecting burned fuel gases, thus causing it to accelerate in the opposite direction of the velocity of the ejected fuel (see Figure ). Specifically: A fully fueled rocket ship in deep space has a total mass [latex]{m}_{0}[/latex] (this mass includes the initial mass of the fuel). At some moment in time, the rocket has a velocity [latex]\mathbf{\overset{\to }{v}}[/latex] and mass m ; this mass is a combination of the mass of the empty rocket and the mass of the remaining unburned fuel it contains. (We refer to m as the “instantaneous mass” and [latex]\mathbf{\overset{\to }{v}}[/latex] as the “instantaneous velocity.”) The rocket accelerates by burning the fuel it carries and ejecting the burned exhaust gases. If the burn rate of the fuel is constant, and the velocity at which the exhaust is ejected is also constant, what is the change of velocity of the rocket as a result of burning all of its fuel?

## Physical Analysis

Here’s a description of what happens, so that you get a feel for the physics involved.

- As the rocket engines operate, they are continuously ejecting burned fuel gases, which have both mass and velocity, and therefore some momentum. By conservation of momentum, the rocket’s momentum changes by this same amount (with the opposite sign). We will assume the burned fuel is being ejected at a constant rate, which means the rate of change of the rocket’s momentum is also constant. By Figure , this represents a constant force on the rocket.
- However, as time goes on, the mass of the rocket (which includes the mass of the remaining fuel) continuously decreases. Thus, even though the force on the rocket is constant, the resulting acceleration is not; it is continuously increasing.
- So, the total change of the rocket’s velocity will depend on the amount of mass of fuel that is burned, and that dependence is not linear.

The problem has the mass and velocity of the rocket changing; also, the total mass of ejected gases is changing. If we define our system to be the rocket + fuel, then this is a closed system (since the rocket is in deep space, there are no external forces acting on this system); as a result, momentum is conserved for this system. Thus, we can apply conservation of momentum to answer the question ( Figure ).

At the same moment that the total instantaneous rocket mass is m (i.e., m is the mass of the rocket body plus the mass of the fuel at that point in time), we define the rocket’s instantaneous velocity to be [latex]\mathbf{\overset{\to }{v}}=v\mathbf{\hat{i}}[/latex] (in the + x -direction); this velocity is measured relative to an inertial reference system (the Earth, for example). Thus, the initial momentum of the system is

[latex]{\mathbf{\overset{\to }{p}}}_{\text{i}}=mv\mathbf{\hat{i}}.[/latex]

The rocket’s engines are burning fuel at a constant rate and ejecting the exhaust gases in the − x -direction. During an infinitesimal time interval dt , the engines eject a (positive) infinitesimal mass of gas [latex]d{m}_{g}[/latex] at velocity [latex]\mathbf{\overset{\to }{u}}=\text{−}u\mathbf{\hat{i}}[/latex]; note that although the rocket velocity [latex]v\mathbf{\hat{i}}[/latex] is measured with respect to Earth, the exhaust gas velocity is measured with respect to the (moving) rocket. Measured with respect to the Earth, therefore, the exhaust gas has velocity [latex](v-u)\mathbf{\hat{i}}[/latex].

As a consequence of the ejection of the fuel gas, the rocket’s mass decreases by [latex]d{m}_{g}[/latex], and its velocity increases by [latex]dv\mathbf{\hat{i}}[/latex]. Therefore, including both the change for the rocket and the change for the exhaust gas, the final momentum of the system is

Since all vectors are in the x -direction, we drop the vector notation. Applying conservation of momentum, we obtain

Now, [latex]d{m}_{g}[/latex] and dv are each very small; thus, their product [latex]d{m}_{g}dv[/latex] is very, very small, much smaller than the other two terms in this expression. We neglect this term, therefore, and obtain:

Our next step is to remember that, since [latex]d{m}_{g}[/latex] represents an increase in the mass of ejected gases, it must also represent a decrease of mass of the rocket:

Replacing this, we have

Integrating from the initial mass m i to the final mass m of the rocket gives us the result we are after:

and thus our final answer is

This result is called the rocket equation . It was originally derived by the Soviet physicist Konstantin Tsiolkovsky in 1897. It gives us the change of velocity that the rocket obtains from burning a mass of fuel that decreases the total rocket mass from [latex]{m}_{0}[/latex] down to m . As expected, the relationship between [latex]\Delta v[/latex] and the change of mass of the rocket is nonlinear.

## Problem-Solving Strategy: Rocket Propulsion

In rocket problems, the most common questions are finding the change of velocity due to burning some amount of fuel for some amount of time; or to determine the acceleration that results from burning fuel.

- To determine the change of velocity, use the rocket equation Figure .
- To determine the acceleration, determine the force by using the impulse-momentum theorem, using the rocket equation to determine the change of velocity.

## Thrust on a Spacecraft

A spacecraft is moving in gravity-free space along a straight path when its pilot decides to accelerate forward. He turns on the thrusters, and burned fuel is ejected at a constant rate of [latex]2.0\times {10}^{2}\,\text{kg/s}[/latex], at a speed (relative to the rocket) of [latex]2.5\times {10}^{2}\,\text{m/s}[/latex]. The initial mass of the spacecraft and its unburned fuel is [latex]2.0\times {10}^{4}\,\text{kg}[/latex], and the thrusters are on for 30 s.

- What is the thrust (the force applied to the rocket by the ejected fuel) on the spacecraft?
- What is the spacecraft’s acceleration as a function of time?
- What are the spacecraft’s accelerations at t = 0, 15, 30, and 35 s?
- The force on the spacecraft is equal to the rate of change of the momentum of the fuel.
- Knowing the force from part (a), we can use Newton’s second law to calculate the consequent acceleration. The key here is that, although the force applied to the spacecraft is constant (the fuel is being ejected at a constant rate), the mass of the spacecraft isn’t; thus, the acceleration caused by the force won’t be constant. We expect to get a function a ( t ), therefore.
- We’ll use the function we obtain in part (b), and just substitute the numbers given. Important: We expect that the acceleration will get larger as time goes on, since the mass being accelerated is continuously decreasing (fuel is being ejected from the rocket).

The ejection velocity [latex]v=2.5\times {10}^{2}\text{m/s}[/latex] is constant, and therefore the force is

Now, [latex]\frac{d{m}_{g}}{dt}[/latex] is the rate of change of the mass of the fuel; the problem states that this is [latex]2.0\times {10}^{2}\text{kg/s}[/latex]. Substituting, we get

The force is constant and the empty rocket mass [latex]{m}_{R}[/latex] is constant, but the fuel mass [latex]{m}_{g}[/latex] is decreasing at a uniform rate; specifically:

This gives us

Notice that, as expected, the acceleration is a function of time. Substituting the given numbers:

At [latex]t=15\,\text{s},\,a(15\,\text{s})={2.9\,\text{m/s}}^{2}[/latex].

At [latex]t=30\,\text{s},\,a(30\,\text{s})=3.6\,{\text{m/s}}^{2}[/latex].

## Significance

Notice that the acceleration is not constant; as a result, any dynamical quantities must be calculated either using integrals, or (more easily) conservation of total energy.

## Check Your Understanding

What is the physical difference (or relationship) between [latex]\frac{dm}{dt}[/latex] and [latex]\frac{d{m}_{g}}{dt}[/latex] in this example?

The notation [latex]{m}_{g}[/latex] stands for the mass of the fuel and m stands for the mass of the rocket plus the initial mass of the fuel. Note that [latex]{m}_{g}[/latex] changes with time, so we write it as [latex]{m}_{g}(t)[/latex]. Using [latex]{m}_{R}[/latex] as the mass of the rocket with no fuel, the total mass of the rocket plus fuel is [latex]m={m}_{R}+{m}_{g}(t)[/latex]. Differentiation with respect to time gives

[latex]\frac{dm}{dt}=\frac{d{m}_{R}}{dt}+\frac{d{m}_{g}(t)}{dt}=\frac{d{m}_{g}(t)}{dt}[/latex]

where we used [latex]\frac{d{m}_{R}}{dt}=0[/latex] because the mass of the rocket does not change. Thus, time rate of change of the mass of the rocket is the same as that of the fuel.

## Rocket in a Gravitational Field

Let’s now analyze the velocity change of the rocket during the launch phase, from the surface of Earth. To keep the math manageable, we’ll restrict our attention to distances for which the acceleration caused by gravity can be treated as a constant g .

The analysis is similar, except that now there is an external force of [latex]\mathbf{\overset{\to }{F}}=\text{−}mg\mathbf{\hat{j}}[/latex] acting on our system. This force applies an impulse [latex]d\mathbf{\overset{\to }{J}}=\mathbf{\overset{\to }{F}}dt=\text{−}mgdt\mathbf{\hat{j}}[/latex], which is equal to the change of momentum. This gives us

where we have again neglected the term [latex]d{m}_{g}dv[/latex] and dropped the vector notation. Next we replace [latex]d{m}_{g}[/latex] with [latex]\text{−}dm[/latex]:

Dividing through by m gives

and integrating, we have

Unsurprisingly, the rocket’s velocity is affected by the (constant) acceleration of gravity.

Remember that [latex]\Delta t[/latex] is the burn time of the fuel. Now, in the absence of gravity, Figure implies that it makes no difference how much time it takes to burn the entire mass of fuel; the change of velocity does not depend on [latex]\Delta t[/latex]. However, in the presence of gravity, it matters a lot. The − g [latex]\Delta t[/latex] term in Figure tells us that the longer the burn time is, the smaller the rocket’s change of velocity will be. This is the reason that the launch of a rocket is so spectacular at the first moment of liftoff: It’s essential to burn the fuel as quickly as possible, to get as large a [latex]\Delta v[/latex] as possible.

- A rocket is an example of conservation of momentum where the mass of the system is not constant, since the rocket ejects fuel to provide thrust.
- The rocket equation gives us the change of velocity that the rocket obtains from burning a mass of fuel that decreases the total rocket mass.

## Key Equations

Conceptual questions.

It is possible for the velocity of a rocket to be greater than the exhaust velocity of the gases it ejects. When that is the case, the gas velocity and gas momentum are in the same direction as that of the rocket. How is the rocket still able to obtain thrust by ejecting the gases?

Yes, the rocket speed can exceed the exhaust speed of the gases it ejects. The thrust of the rocket does not depend on the relative speeds of the gases and rocket, it simply depends on conservation of momentum.

(a) A 5.00-kg squid initially at rest ejects 0.250 kg of fluid with a velocity of 10.0 m/s. What is the recoil velocity of the squid if the ejection is done in 0.100 s and there is a 5.00- N frictional force opposing the squid’s movement?

(b) How much energy is lost to work done against friction?

(a) 0.413 m/s, (b) about 0.2 J

A rocket takes off from Earth and reaches a speed of 100 m/s in 10.0 s. If the exhaust speed is 1500 m/s and the mass of fuel burned is 100 kg, what was the initial mass of the rocket?

Repeat the preceding problem but for a rocket that takes off from a space station, where there is no gravity other than the negligible gravity due to the space station.

How much fuel would be needed for a 1000-kg rocket (this is its mass with no fuel) to take off from Earth and reach 1000 m/s in 30 s? The exhaust speed is 1000 m/s.

What exhaust speed is required to accelerate a rocket in deep space from 800 m/s to 1000 m/s in 5.0 s if the total rocket mass is 1200 kg and the rocket only has 50 kg of fuel left?

Unreasonable Results Squids have been reported to jump from the ocean and travel 30.0 m (measured horizontally) before re-entering the water.

(a) Calculate the initial speed of the squid if it leaves the water at an angle of 20.0°, assuming negligible lift from the air and negligible air resistance.

(b) The squid propels itself by squirting water. What fraction of its mass would it have to eject in order to achieve the speed found in the previous part? The water is ejected at 12.0 m/s; gravitational force and friction are neglected.

(c) What is unreasonable about the results?

(d) Which premise is unreasonable, or which premises are inconsistent?

## Additional Problems

Two 70-kg canoers paddle in a single, 50-kg canoe. Their paddling moves the canoe at 1.2 m/s with respect to the water, and the river they’re in flows at 4 m/s with respect to the land. What is their momentum with respect to the land?

Which has a larger magnitude of momentum: a 3000-kg elephant moving at 40 km/h or a 60-kg cheetah moving at 112 km/h?

the elephant has a higher momentum

A driver applies the brakes and reduces the speed of her car by 20%, without changing the direction in which the car is moving. By how much does the car’s momentum change?

You friend claims that momentum is mass multiplied by velocity, so things with more mass have more momentum. Do you agree? Explain.

Answers may vary. The first clause is true, but the second clause is not true in general because the velocity of an object with small mass may be large enough so that the momentum of the object is greater than that of a larger-mass object with a smaller velocity.

Dropping a glass on a cement floor is more likely to break the glass than if it is dropped from the same height on a grass lawn. Explain in terms of the impulse.

Your 1500-kg sports car accelerates from 0 to 30 m/s in 10 s. What average force is exerted on it during this acceleration?

[latex]4.5\times {10}^{3}\,\text{N}[/latex]

A ball of mass [latex]m[/latex] is dropped. What is the formula for the impulse exerted on the ball from the instant it is dropped to an arbitrary time [latex]\tau[/latex] later? Ignore air resistance.

Repeat the preceding problem, but including a drag force due to air of [latex]{f}_{\text{drag}}=\text{−}b\mathbf{\overset{\to }{v}}.[/latex]

[latex]\mathbf{\overset{\to }{J}}={\int }_{0}^{\tau }[m\mathbf{\overset{\to }{g}}-m\mathbf{\overset{\to }{g}}(1-{e}^{\text{−}bt\text{/}m})]dt=\frac{{m}^{2}}{b}\mathbf{\overset{\to }{g}}({e}^{\text{−}b\tau \text{/}m}-1)[/latex]

A 5.0-g egg falls from a 90-cm-high counter onto the floor and breaks. What impulse is exerted by the floor on the egg?

A car crashes into a large tree that does not move. The car goes from 30 m/s to 0 in 1.3 m. (a) What impulse is applied to the driver by the seatbelt, assuming he follows the same motion as the car? (b) What is the average force applied to the driver by the seatbelt?

a. [latex]\text{−}(2.1\times {10}^{3}\,\text{kg}\cdot \text{m/s})\mathbf{\hat{i}}[/latex], b. [latex]\text{−}(24\times {10}^{3}\,\text{N})\mathbf{\hat{i}}[/latex]

Two hockey players approach each other head on, each traveling at the same speed [latex]{v}_{\text{i}}[/latex]. They collide and get tangled together, falling down and moving off at a speed [latex]{v}_{\text{i}}\text{/}5[/latex]. What is the ratio of their masses?

You are coasting on your 10-kg bicycle at 15 m/s and a 5.0-g bug splatters on your helmet. The bug was initially moving at 2.0 m/s in the same direction as you. If your mass is 60 kg, (a) what is the initial momentum of you plus your bicycle? (b) What is the initial momentum of the bug? (c) What is your change in velocity due to the collision with the bug? (d) What would the change in velocity have been if the bug were traveling in the opposite direction?

a. [latex](1.1\times {10}^{3}\,\text{kg}\cdot \text{m/s})\mathbf{\hat{i}}[/latex], b. [latex](0.010\,\text{kg}\cdot \text{m/s})\mathbf{\hat{i}}[/latex], c. [latex]\text{−}(0.00093\,\text{m/s})\mathbf{\hat{i}}[/latex], d. [latex]\text{−}(0.0012\,\text{m/s})\mathbf{\hat{i}}[/latex]

A load of gravel is dumped straight down into a 30 000-kg freight car coasting at 2.2 m/s on a straight section of a railroad. If the freight car’s speed after receiving the gravel is 1.5 m/s, what mass of gravel did it receive?

Two carts on a straight track collide head on. The first cart was moving at 3.6 m/s in the positive x direction and the second was moving at 2.4 m/s in the opposite direction. After the collision, the second car continues moving in its initial direction of motion at 0.24 m/s. If the mass of the second car is 5.0 times that of the first, what is the mass and final velocity of the first car?

0.10 kg, [latex]\text{−}(130\,\text{m/s})\mathbf{\hat{i}}[/latex]

A 100-kg astronaut finds himself separated from his spaceship by 10 m and moving away from the spaceship at 0.1 m/s. To get back to the spaceship, he throws a 10-kg tool bag away from the spaceship at 5.0 m/s. How long will he take to return to the spaceship?

Derive the equations giving the final speeds for two objects that collide elastically, with the mass of the objects being [latex]{m}_{1}[/latex] and [latex]{m}_{2}[/latex] and the initial speeds being [latex]{v}_{\text{1,i}}[/latex] and [latex]{v}_{\text{2,i}}=0[/latex] (i.e., second object is initially stationary).

[latex]{v}_{\text{1,f}}={v}_{\text{1,i}}\frac{{m}_{1}-{m}_{2}}{{m}_{1}+{m}_{2}},\,{v}_{\text{2,f}}={v}_{\text{1,i}}\frac{2{m}_{1}}{{m}_{1}+{m}_{2}}[/latex]

Repeat the preceding problem for the case when the initial speed of the second object is nonzero.

A child sleds down a hill and collides at 5.6 m/s into a stationary sled that is identical to his. The child is launched forward at the same speed, leaving behind the two sleds that lock together and slide forward more slowly. What is the speed of the two sleds after this collision?

For the preceding problem, find the final speed of each sled for the case of an elastic collision.

A 90-kg football player jumps vertically into the air to catch a 0.50-kg football that is thrown essentially horizontally at him at 17 m/s. What is his horizontal speed after catching the ball?

Three skydivers are plummeting earthward. They are initially holding onto each other, but then push apart. Two skydivers of mass 70 and 80 kg gain horizontal velocities of 1.2 m/s north and 1.4 m/s southeast, respectively. What is the horizontal velocity of the third skydiver, whose mass is 55 kg?

Two billiard balls are at rest and touching each other on a pool table. The cue ball travels at 3.8 m/s along the line of symmetry between these balls and strikes them simultaneously. If the collision is elastic, what is the velocity of the three balls after the collision?

final velocity of cue ball is [latex]\text{−}(0.76\,\text{m/s})\mathbf{\hat{i}}[/latex], final velocities of the other two balls are 2.6 m/s at ±30° with respect to the initial velocity of the cue ball

A billiard ball traveling at [latex](2.2\,\text{m/s})\mathbf{\hat{i}}-(0.4\,\text{m/s})\mathbf{\hat{j}}[/latex] collides with a wall that is aligned in the [latex]\mathbf{\hat{j}}[/latex] direction. Assuming the collision is elastic, what is the final velocity of the ball?

Two identical billiard balls collide. The first one is initially traveling at [latex](2.2\,\text{m/s})\mathbf{\hat{i}}-(0.4\,\text{m/s})\mathbf{\hat{j}}[/latex] and the second one at [latex]\text{−}(1.4\,\text{m/s})\mathbf{\hat{i}}+(2.4\,\text{m/s})\mathbf{\hat{j}}[/latex]. Suppose they collide when the center of ball 1 is at the origin and the center of ball 2 is at the point [latex](2R,0)[/latex] where R is the radius of the balls. What is the final velocity of each ball?

ball 1: [latex]\text{−}(1.4\,\text{m/s})\mathbf{\hat{i}}-(0.4\,\text{m/s})\mathbf{\hat{j}}[/latex], ball 2: [latex](2.2\,\text{m/s})\mathbf{\hat{i}}+(2.4\,\text{m/s})\mathbf{\hat{j}}[/latex]

Repeat the preceding problem if the balls collide when the center of ball 1 is at the origin and the center of ball 2 is at the point [latex](0,2R)[/latex].

Repeat the preceding problem if the balls collide when the center of ball 1 is at the origin and the center of ball 2 is at the point [latex](\sqrt{3}R\text{/}2,R\text{/}2)[/latex]

ball 1: [latex](1.4\,\text{m/s})\mathbf{\hat{i}}-(1.7\,\text{m/s})\mathbf{\hat{j}}[/latex], ball 2: [latex]\text{−}(2.8\,\text{m/s})\mathbf{\hat{i}}+(0.012\,\text{m/s})\mathbf{\hat{j}}[/latex]

Where is the center of mass of a semicircular wire of radius R that is centered on the origin, begins and ends on the x axis, and lies in the x , y plane?

Where is the center of mass of a slice of pizza that was cut into eight equal slices? Assume the origin is at the apex of the slice and measure angles with respect to an edge of the slice. The radius of the pizza is R .

[latex](r,\theta )=(2R\text{/}3,\pi \text{/}8)[/latex]

If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon-population system move? Assume the population is 7 billion, the average human has a mass of 65 kg, and that the population is evenly distributed over both the Earth and the Moon. The mass of the Earth is [latex]5.97\times {10}^{24}\text{kg}[/latex] and that of the Moon is [latex]7.34\times {10}^{22}\text{kg}[/latex]. The radius of the Moon’s orbit is about [latex]3.84\times {10}^{5}\text{m}[/latex].

You friend wonders how a rocket continues to climb into the sky once it is sufficiently high above the surface of Earth so that its expelled gasses no longer push on the surface. How do you respond?

Answers may vary. The rocket is propelled forward not by the gasses pushing against the surface of Earth, but by conservation of momentum. The momentum of the gas being expelled out the back of the rocket must be compensated by an increase in the forward momentum of the rocket.

To increase the acceleration of a rocket, should you throw rocks out of the front window of the rocket or out of the back window?

A 65-kg person jumps from the first floor window of a burning building and lands almost vertically on the ground with a horizontal velocity of 3 m/s and vertical velocity of [latex]-9\,\text{m/s}[/latex]. Upon impact with the ground he is brought to rest in a short time. The force experienced by his feet depends on whether he keeps his knees stiff or bends them. Find the force on his feet in each case.

- First find the impulse on the person from the impact on the ground. Calculate both its magnitude and direction.
- Find the average force on the feet if the person keeps his leg stiff and straight and his center of mass drops by only 1 cm vertically and 1 cm horizontally during the impact.
- Find the average force on the feet if the person bends his legs throughout the impact so that his center of mass drops by 50 cm vertically and 5 cm horizontally during the impact.
- Compare the results of part (b) and (c), and draw conclusions about which way is better.

You will need to find the time the impact lasts by making reasonable assumptions about the deceleration. Although the force is not constant during the impact, working with constant average force for this problem is acceptable.

a. [latex]617\,\text{N}\cdot \text{s}[/latex], 108°; b. [latex]{F}_{x}=2.91\times {10}^{4}\,\text{N}[/latex], [latex]{F}_{y}=2.6\times {10}^{5}\,\text{N}[/latex]; c. [latex]{F}_{x}=5265\,\text{N}[/latex], [latex]{F}_{y}=5850\,\text{N}[/latex]

Two projectiles of mass [latex]{m}_{1}[/latex] and [latex]{m}_{2}[/latex] are ﬁred at the same speed but in opposite directions from two launch sites separated by a distance D . They both reach the same spot in their highest point and strike there. As a result of the impact they stick together and move as a single body afterwards. Find the place they will land.

Two identical objects (such as billiard balls) have a one-dimensional collision in which one is initially motionless. After the collision, the moving object is stationary and the other moves with the same speed as the other originally had. Show that both momentum and kinetic energy are conserved.

Conservation of momentum demands [latex]{m}_{1}{v}_{\text{1,i}}+{m}_{2}{v}_{\text{2,i}}={m}_{1}{v}_{\text{1,f}}+{m}_{2}{v}_{\text{2,f}}[/latex]. We are given that [latex]{m}_{1}={m}_{2}[/latex], [latex]{v}_{\text{1,i}}={v}_{\text{2,f}}[/latex], and [latex]{v}_{\text{2,i}}={v}_{\text{1,f}}=0[/latex]. Combining these equations with the equation given by conservation of momentum gives [latex]{v}_{\text{1,i}}={v}_{\text{1,i}}[/latex], which is true, so conservation of momentum is satisfied. Conservation of energy demands [latex]\frac{1}{2}{m}_{1}{v}_{\text{1,i}}^{2}+\frac{1}{2}{m}_{2}{v}_{\text{2,i}}^{2}=\frac{1}{2}{m}_{1}{v}_{\text{1,f}}^{2}+\frac{1}{2}{m}_{2}{v}_{\text{2,f}}^{2}[/latex]. Again combining this equation with the conditions given above give [latex]{v}_{\text{1,i}}={v}_{\text{1,i}}[/latex], so conservation of energy is satisfied.

A ramp of mass M is at rest on a horizontal surface. A small cart of mass m is placed at the top of the ramp and released.

What are the velocities of the ramp and the cart relative to the ground at the instant the cart leaves the ramp?

Find the center of mass of the structure given in the figure below. Assume a uniform thickness of 20 cm, and a uniform density of [latex]1{\,\text{g/cm}}^{3}.[/latex]

Assume origin on centerline and at floor, then [latex]({x}_{\text{CM}},{y}_{\text{CM}})=(0,86\,\text{cm})[/latex]

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