# Impulse & Momentum

## Practice

### practice problem 1

Now, the bears I live with average, the males, eight to twelve hundred pounds [360 to 540 kg]. They're the largest bears in the world…. They've been clocked at 41 [mph] and they've run a hundred meter dash in 5.85 seconds, which a human on steroids doesn't even approach.

- Compute the speed of a grizzly bear using Mr. Treadwell's hundred meter statement.
- Compute the momentum of a grizzly bear using the speed you calculated in part a and the average mass stated by Mr. Treadwell.
- How fast would a 250 lb man have to run to have the same momentum you calculated in part b? (Do not use a calculator to compute your answer.)
- How fast would a 4,000 lb car have to drive to have the same momentum you calculated in part b? (Do not use a calculator to compute your answer.)

#### solution

- Speed is distance over time.
*v*=∆ *s*∆ *t**v*=100 m 5.85 s *v*= 17 m/s - Momentum is mass times velocity. Let's use a mass in the middle of the range stated by Mr. Treadwell.
*p*=*mv**p*= (450 kg)(17 m/s)*p*= 7,700 kgm/s - Momentum is the product of mass and velocity, which makes the two quantities inversely proportional. Mass goes down when we replace the 1,000 pound grizzly bear with a 250 pound man. To keep the momentum constant, the man will have to run faster — faster by an amount that is inversely proportional to the decrease in weight.
Since our hypothetical man has ¼ the mass of a grizzly, he needs to run 4 times faster to have the same momentum. With numbers this simple, you should be able to compute the answers without a calculator.
*p*=*mv*⇒ *p*= (¼*m*)(4*v*)*v*_{man}= 4 *v*_{bear}*v*_{man}= 4(17 m/s) = 68 m/s *v*_{man}= 4(41 mph) = 164 mph - Use reasoning similar to part c. Mass goes up when we replace the 1,000 pound grizzly bear with a 4,000 pound car.
Four times the mass needs ¼ the speed to have the same momentum. Once again, the numbers are simple.
*p*=*mv*⇒ *p*= (4*m*)(¼*v*)*v*_{car}= ¼ *v*_{bear}*v*_{car}= ¼(17 m/s) = 4.25 m/s *v*_{car}= ¼(41 mph) = 10.25 mph

### practice problem 2

*Maṅgalayāna*or

*Mnglyan*), which is Sanskrit for "Mars craft". English speaking news agencies have been writing this as "Mangalyaan".

For the first month of its journey, the Mars Orbiter Spacecraft actually orbited the Earth. After a series of orbit raising maneuvers, the tiny main engine gave the spacecraft enough speed in the right direction to escape. For the next nine months the main engine was only used twice and only very briefly (less than a minute of total burn time) to correct the spacecraft's trajectory.

Nine months after leaving earth orbit, the spacecraft arrived at Mars. To enter orbit around Mars, the spacecraft needed to slow down — a maneuver called orbit insertion. Since the main engine hadn't been used much in nine months, a quick little test was needed.

Press Release: September 22, 2014

Mars Orbiter Spacecraft's Main Liquid Engine Successfully Test FiredThe 440 newton Liquid Apogee Motor (LAM) of India's Mars Orbiter Spacecraft, last fired on December 01, 2013, was successfully fired for a duration of 3.968 seconds at 1430 hrs IST today (September 22, 2014). This operation of the spacecraft's main liquid engine was also used for the spacecraft's trajectory correction and changed its velocity by 2.18 metre/second. With this successful test firing, Mars Orbiter Insertion (MOI) operation of the spacecraft is scheduled to be performed on the morning of September 24, 2014 at 07:17:32 hrs IST by firing the LAM along with eight smaller liquid engines for a duration of about 24 minutes.

- Using the press release, determine the mass of the Mars Orbiter Spacecraft at the time of this test firing.
- Why was the qualifying phrase "at the time of this test firing" added to the end of the previous question? Why does the mass of the spacecraft vary with time?

#### solution

Let's GUESS…

*G*iven…*F*=440 N *t*=3.968 s ∆ *v*=2.18 m/s *U*nknown…*m*=?? *E*quation…*F*∆*t*=*m*∆*v**S*ubstitute…(440 N)(3.968 s) =

*m*(2.18 m/s)*S*olve…*m*= 801 kg- A spacecraft with a rocket engine carries fuel and fuel has mass. As fuel is used up, the mass of the spacecraft decreases. The mass calculated in part a was the mass of the spacecraft plus fuel on 22 September 2014.

A quick surf of the Mars Orbiter Mission website takes us to a page with some specifications. The lift-off mass was 1337 kg. (Is the ISRO a fan of leet?) The mass we computed is 536 kg short of the lift off mass. That's the amount of fuel that was burned leaving Earth orbit.

∆m_{spacecraft} |
= | m_{fuel burned} |

1337 kg − 801 kg | = | 536 kg |

The web page with the specifications also gives a propellant mass of 852 kg. This means there was 316 kg of fuel available at the time of the test. That's probably just enough to slow the spacecraft for orbit insertion and also to adjust the orbit to the right shape and size for the mission.

∆m_{fuel} |
= | m_{fuel remaining} |

852 kg − 536 kg | = | 316 kg |

The main engine test was successful and so was the orbit insertion two days later. The Mars Orbiter Spacecraft is now a satellite of Mars. The ISRO is the first *Asian* space agency to have a successful mission to Mars and the only national space agency to have successful *first* mission to Mars. For comparison, the United States wasn't successful until its second mission (the 1964 flyby mission Mariner 4), the Soviet Union wasn't successful until its eleventh (the 1971 orbiter/lander combo Mars 3), Japan tried once and failed (the 1998 Nozomi mission), and the European Union was partly successful (the 2003 Mars Express orbiter made it into orbit, but the Beagle 2 lander failed to make radio contact after landing).

### practice problem 3

The density of the interstellar medium is about one hydrogen atom per cubic centimeter. Imagine a 1,000 tonne, 4 by 6 meter, classroom-sized interstellar spacecraft traveling at 60,000 km/s on its way to Proxima Centauri (the nearest solar system to our own) 4.243 light years away.After all, the faster we go, the more difficult it is to avoid collisions with small objects and the more damage such a collision will wreak. Even if we are fortunate enough to miss all sizable objects, we can scarcely expect to miss the dust and individual atoms that are scattered throughout space. At two-tenths of the speed of light, dust and atoms might not do significant damage even in a voyage of 40 years, but the faster you go, the worse it is — space begins to become abrasive. When you begin to approach the speed of light, each hydrogen atom becomes a cosmic ray particle and will fry the crew. (A hydrogen atom or its nucleus striking the ship at nearly the speed of light is a cosmic ray particle, and there is no difference if the ship strikes the hydrogen atom or nucleus at nearly the speed of light. As Sancho Panza: "Whether the stone strikes the pitcher, or the pitcher strikes the stone, it is bad for the pitcher.") So 60,000 kilometers per second may be the practical speed limit for space travel.

Kinematics:

- How long would it take our hypothetical spacecraft to complete its hypothetical journey?

- Determine the momentum of our spacecraft.
- What mass of interstellar medium is swept up during the journey?
- What impulse does the interstellar medium deliver to the spacecraft?
- How does this impulse compare to the momentum of the spacecraft?

- Determine the kinetic energy of our spacecraft.
- What is the effective drag force of the interstellar medium during the journey?
- How much work does the interstellar medium do on the spacecraft?
- How does this work compare to the kinetic energy of the spacecraft?

#### solution

Let's not bother doing the conversions ourselves. Let's let an online calculator do that work for us. The relationships of the quantities to one another through equations is always what's most important when solving a problem. Units are often a distraction.

- From the definition of speed…
*v̅*=∆ *s*∆ *t*∆ *t*=∆ *s**v̅*∆ *t*=4.243 light years 60,000 km/s ∆ *t*= 21.2 years - From the definition of momentum…
*p*=*mv**p*= (1,000 tonnes)(60,000 km/s)*p*= 6.00 × 10^{13}kgm/s - From the definition of density…
ρ = *m**V*

The volume swept up by the spacecraft on its journey is its area times the distance traveled.*m*= ρ*V*

Combine these two equations. Remember that a hydrogen atom is basically a proton.*V*=*A*∆*s*= ℓ*w*∆*s**m*= ρℓ*w*∆*s**m*= (1 proton mass/1 cm^{3})(6 m)(4 m)(4.243 light years) *m*= 1.61 grams - When our spacecraft strikes the interstellar medium, the medium changes its speed from zero to 60,000 km/s.
A change in momentum is caused by an impulse. The impulse on the interstellar medium is equal and opposite to the impulse on the spacecraft. We only care about the magnitudes in this problem, so we won't bother with a negative sign.
*J*=*m*∆*v**J*= (1.61 g)(60,000 km/s)*J*= 96,700 kgm/s - Taste and compare. It's only a few parts per billion, which hardly seems significant.
*J*_{medium}= 96,700 kgm/s *p*_{spacecraft}6.00 × 10 ^{13}kgm/s*J*_{medium}= 1.6 × 10 ^{−9}*p*_{spacecraft} - Use the definition of kinetic energy to start.
*K*= ½*mv*^{2}*K*= ½(1,000 tonnes)(60,000 km/s)^{2}*K*= 1.80 × 10^{21}J - Use the definition of impulse.

Solve for force and substitute in values.*J*=*F̅*∆*t**F̅*=*J*∆ *t**F̅*=96,700 kgm/s 21.2 years *F̅*= 145 µN - Use the definition of work.
*W*=*F̅*∆*s**W*= (145 µN)(2.243 light years)^{}*W*= 5.80 × 10^{12}J - Another incomparable comparison.
*W*_{medium}= 5.80 × 10 ^{12}J*K*_{spacecraft}1.80 × 10 ^{21}J*W*_{medium}= 3.2 × 10 ^{−9}*K*_{spacecraft}

### practice problem 4

#### solution

Answer it.