by Gary Gende and Owen "O-Dog" Zahorcak, January 1998
Here's the quantities you can know:
These quantities are defined and explained on other pages
execept Impulse
and momentum. They are defined below.
Momentum is defined as the product of an object's mass and
velocity
This means p(momentum)=m(mass)*V(velocity) by the definition of
momentum
Momentum is measured in N(Newton)m(meters)/s(per second)
Impulse is the product of force and time.
As this suggests the unit for measuring impulse is
N(Newton)s(seconds)
A formula for finding impulse is shown and explained below.
From other chapters we know that
We can use these formulas to show that /\p(Impulse)=Ft
*Several of the formula in the work above become important*
F=/\p/t is Newton's Second Law of Motion.
So now we have all the formulas we need for solving impulse
and momentum
problems:
Go back to: Table of Contents
Go back to: Table of Contents
Joe the alien fires a 20 Kg photon torpedo at 200 m/s from his stationary UFO in deep space. The torpedo has special no fuel engines that can give 650 N of force. What is the momentum of the torpedo? How long must Joe fire the engines to get 3000 Ns of impulse. What speed is Joe attempting to get the torpedo to go? If Joe is firing at you and your UFO can accelerate at a rate of 30 m/s/s, will you be able to out accellerate the torpedo?
a. Momentum is just p=mv so plug in the known m (20) and V (200) 20 * 200 = 4000 Kgm/s
b. The needed formula is /\p = Ft. F (650) and /\p (1500) are given. divide both sides of the equation by F to get /\p/F = t then plug in the known values to get 3000/650 = t = 4.62 Ns
c. The needed formula now is F = m(/\V/t). F is still 650 N, the torpedo still has a mass of 20 Kg and time can be found from part b. When you plug them in you get 650 = 20(/\V/2.31). Multiplying both sides by 2.31/20 gives you (4.62 * 650)/20 = /\V = 150 m/s. Since this is the change in speed (hence the /\), 150 needs to be added to the original 200. The final speed is 350 m/s.
d. F = ma is the importanr equation now. You know F (650 N)
and m (20 Kg) so just divide both sides of the equation by m to
solve it for a and plug in the numbers to get 650/20 = a = 32.5
m/s/s. So you better hope you have a big enough head start.
The answers to each problem follow it in parentheses. They
also link to a solution to the problem.
Try the problem, check your answer, and go to the solution if you
do not understand.
What is the momentum of a 23 Kg cannon shell going 530 m/s? (12190 Kgm/s)
What speed must a 5 Kg object go to have 24 Kgm/s of momentum? (4.8 m/s)
A bullet going 640 m/s has 42 Kgm/s of momentum. What is its mass? (66 g or .066 Kg)
What is the impulse imparted by a rocket that exerts 4.8 N for 1.63 seconds? (7.8 Ns)
For what time must you exert a force of 45 N to get an impulse of 16 Ns? (.36 s) Go back to: Table of Contents
What force exerted over 6 seconds gives you an impulse of 64 Ns? (10.7 N)
What is the change in velocity of a .35 Kg air track cart if you exert a force of 1.2 N on it for 3 seconds? (10.3 m/s)
A rocket engine exerts a force of 500 N on a space probe (in outer space!) for 5 seconds. The probe speeds up from rest to a speed of 21 m/s. What is its mass? (119 Kg)
What force exerted for .12 seconds will make a .54 Kg baseball change its velocity 80 m/s. (360 N)
How long must the space probe in question 8 fire its engine to change its velocity by 3 m/s? (.71 s) Go back to: Table of Contents
A rocket engine burns 5 Kg of fuel per second. The exhaust gas velocity is 608 m/s. What is the thrust of the engine? What time must it burn to impart an impulse of 12,000 Ns? How much fuel will it burn to do this? (3040 N, 3.95 s, 19.7 Kg)
An 11 Ns rocket engine has 12.5 g of fuel. What is the exhaust velocity? (880 m/s)
A rocket generates 25 N of thrust, and the exhaust gas velocity is 1250 m/s. At what rate does it consume fuel in Kg/s? How much fuel has it burned in 5 minutes? (.02 Kg/s, 6 Kg)
A small rocket probe in deep space has a mass of 68.5 Kg, 45.2 Kg of which is fuel. Its engine consumes .250 Kg of fuel per second, and it has an exhaust velocity of 720 m/s. For how much time will the engine burn? What is the initial acceleration of the rocket engine? What is the acceleration just before it runs out of fuel? (180.8 s, 2.63 m/s/s, 7.73 m/s/s)
A rocket takes off from the surface of the Earth straight up. The total mass of the rocket is 5000 Kg, 3500 Kg of which is fuel. The exhaust gas velocity is 3000 m/s, and the rocket consumes 25 Kg of fuel per second. For how long do the engines burn? What is the thrust of the engine? What is the initial and final accelerations of the rocket? (Don't forget gravity) (140 s, 75000 N, 5.2 m/s/s, 40.2 m/s/s) Go back to: Table of Contents
What is the momentum of a 23 Kg cannon shell going 530 m/s?
23 * 530 = 12190 kgm/s....p = mV
Go to: Problem Formulas Table of Contents
What speed must a 5 Kg object go to have 24 Kgm/s of momentum?
24 = 5 * V....p = mV V = 24/5 or 4.8 m/s
Go to: Problem Formulas Table of Contents
A bullet going 640 m/s has 42 Kgm/s of momentum. What is its mass?
42 = 640 * m....p = mV m = 42/640 or .066 kg
Go to: Problem Formulas Table of Contents
What is the impulse imparted by a rocket that exerts 4.8 N for 1.63 seconds?
4.8 * 1.63 = 7.824 or 7.8 Ns..../\p = Ft
Go to: Problem Formulas Table of Contents
For what time must you exert a force of 45 N to get an impulse of 16 Ns?
16 = 45 * t..../\p = Ft t = 16/45 or .36 s
Go to: Problem Formulas Table of Contents
What force exerted over 6 seconds gives you an impulse of 64 Ns?
64 = 6 * N..../\p = Ft N = 64/6 or 10.7 N
Go to: Problem Formulas Table of Contents
What is the change in velocity of a .35 Kg air track cart if you exert a force of 1.2 N on it for 3 seconds?
.35 * /\V = 1.2 * 3....F = m(/\V/t) /\V = 3.6/.35 or 10.3 m/s
Go to: Problem Formulas Table of Contents
A rocket engine exerts a force of 500 N on a space probe (in outer space!) for 5 seconds. The probe speeds up from rest to a speed of 21 m/s. What is its mass?
21 * m = 500 * 5....F = m(/\V/t) m = 2500/21 or 119 kg
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What force exerted for .12 seconds will make a .54 Kg baseball change its velocity 80 m/s.
N * .12 = .54 * 80....F = m(/\V/t) N = 43.2/.12 or 360 N
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How long must the space probe in question 8 fire its engine to change its velocity by 3 m/s?
500 * t = 119 * 3....F = m(/\V/t) t = 357/500 or .71 s
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A rocket engine burns 5 Kg of fuel per second. The exhaust gas velocity is 608 m/s. What is the thrust of the engine? What time must it burn to impart an impulse of 12,000 Ns? How much fuel will it burn to do this?
a. 5 * 608 = 3040 N....F = m(/\V/t)
b. 12000 = 3040 * t..../\p = Ft
t = 12000/3040 or 3.95 s
c. 3.95 * 5 = 19.7 s....fuel burned per second multiplied by the number of seconds it is
burning
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An 11 Ns rocket engine has 12.5 g of fuel. What is the exhaust velocity?
11 = .0125 * V....F = m(/\V/t) V = 11/.0125 or 880 m/s
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A rocket generates 25 N of thrust, and the exhaust gas velocity is 1250 m/s. At what rate does it consume fuel in Kg/s? How much fuel has it burned in 5 minutes?
a. 25 = 1250 * V....F = m(/\V/t)
kg/s = 25/1250 or .02 Kg/s
b. .02 * 5 min * 60 sec/min = 6 kg....fuel burned per second multiplied by the number of
seconds it is burning
Go to: Problem Formulas Table of Contents
A small rocket probe in deep space has a mass of 68.5 Kg, 45.2 Kg of which is fuel. Its engine consumes .250 Kg of fuel per second, and it has an exhaust velocity of 720 m/s. For how much time will the engine burn? What is the initial acceleration of the rocket engine? What is the acceleration just before it runs out of fuel?
a. 45.2/.25 = t....Amount of fuel divided by the burn rate
b. F = (720 * .25) = 180....F = m(/\V/t)
180 = 68.5 * a....F = ma
a = 180/68.5
c. F = (720 * .25) = 180....F = m(/\V/t)
180 = (68.5 - 45.2) * a....F = ma
a = 180/23.3
Go to: Problem Formulas Table of Contents
A rocket takes off from the surface of the Earth straight up. The total mass of the rocket is 5000 Kg, 3500 Kg of which is fuel. The exhaust gas velocity is 3000 m/s , and the rocket consumes 25 Kg of fuel per second. For how long do the engines burn? What is the thrust of the engine? What is the initial and final accelerations of the rocket? (don't forget gravity)
a. 3500/25 = t....Amount of fuel divided by the burn rate
b. F = (25 * 3000)....F = m(/\V/t)
c1. 75000(from b) - (9.8 * 5000)(gravity) = 5000 * a....F = ma
26000/5000 = a
c2. 75000(from b) - (9.8 * (5000 - 3500))(gravity) = (5000 - 3500) * a....F = ma
60300/1500 = a
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