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Making connections: force and momentum

Force and momentum are intimately related. Force acting over time can change momentum, and Newton’s second law of motion, can be stated in its most broadly applicable form in terms of momentum. Momentum continues to be a key concept in the study of atomic and subatomic particles in quantum mechanics.

This statement of Newton’s second law of motion includes the more familiar F net = m a as a special case. We can derive this form as follows. First, note that the change in momentum Δ p size 12{Δp} {} is given by

Δ p = Δ ( m v ) . size 12{Δp=Δ left (mv right )} {}

If the mass of the system is constant, then

Δ ( m v ) = m Δ v . size 12{Δ left (mv right )=mΔv} {}

So that for constant mass, Newton’s second law of motion becomes

F net = Δ p Δ t = m Δ v Δ t . size 12{ F rSub { size 8{"net"} } = { {Δp} over {Δt} } = { {mΔv} over {Δt} } "." } {}

Because Δ v Δ t = a size 12{ { {Δv} over {Δt} } =a} {} , we get the familiar equation

F net = m a

when the mass of the system is constant .

Newton’s second law of motion stated in terms of momentum is more generally applicable because it can be applied to systems where the mass is changing, such as rockets, as well as to systems of constant mass. We will consider systems with varying mass in some detail ; however, the relationship between momentum and force remains useful when mass is constant, such as in the following example.

Calculating force: venus williams’ racquet

During the 2007 French Open, Venus Williams hit the fastest recorded serve in a premier women’s match, reaching a speed of 58 m/s (209 km/h). What is the average force exerted on the 0.057-kg tennis ball by Venus Williams’ racquet, assuming that the ball’s speed just after impact is 58 m/s, that the initial horizontal component of the velocity before impact is negligible, and that the ball remained in contact with the racquet for 5.0 ms (milliseconds)?

Strategy

This problem involves only one dimension because the ball starts from having no horizontal velocity component before impact. Newton’s second law stated in terms of momentum is then written as

F net = Δ p Δ t . size 12{ F rSub { size 8{"net"} } = { {Δp} over {Δt} } = { {mΔv} over {Δt} } "." } {}

As noted above, when mass is constant, the change in momentum is given by

Δ p = m Δ v = m v f v i . size 12{Δp=mΔv=m left (v rSub { size 8{f} } - v rSub { size 8{i} } right )} {}

In this example, the velocity just after impact and the change in time are given; thus, once Δ p size 12{Δp} {} is calculated, F net = Δ p Δ t size 12{ F rSub { size 8{"net"} } = { {Δp} over {Δt} } } {} can be used to find the force.

Solution

To determine the change in momentum, substitute the values for the initial and final velocities into the equation above.

Δ p = m v f v i = 0.057 kg 58 m/s 0 m/s = 3 .306 kg · m/s 3.3 kg · m/s

Now the magnitude of the net external force can determined by using F net = Δ p Δ t size 12{ F rSub { size 8{"net"} } = { {Δp} over {Δt} } } {} :

F net = Δ p Δ t = 3.306 kg m/s 5 . 0 × 10 3 s = 661 N 660 N, alignl { stack { size 12{ F rSub { size 8{"net"} } = { {Δp} over {Δt} } = { {3 "." "306"`"kg" cdot "m/s"} over {5 "." 0 times "10" rSup { size 8{ - 3} } `s} } } {} #" "="661 N" approx "660"`"N," {} } } {}

where we have retained only two significant figures in the final step.

Discussion

This quantity was the average force exerted by Venus Williams’ racquet on the tennis ball during its brief impact (note that the ball also experienced the 0.56-N force of gravity, but that force was not due to the racquet). This problem could also be solved by first finding the acceleration and then using F net = ma size 12{F rSub { size 8{"net"} } " = " ital "ma"} {} , but one additional step would be required compared with the strategy used in this example.

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Section summary

  • Linear momentum ( momentum for brevity) is defined as the product of a system’s mass multiplied by its velocity.
  • In symbols, linear momentum p is defined to be
    p = m v , size 12{p=mv} {}
    where m size 12{m} {} is the mass of the system and v size 12{v} {} is its velocity.
  • The SI unit for momentum is kg · m/s size 12{"kg" cdot "m/s"} {} .
  • Newton’s second law of motion in terms of momentum states that the net external force equals the change in momentum of a system divided by the time over which it changes.
  • In symbols, Newton’s second law of motion is defined to be
    F net = Δ p Δ t , size 12{ F rSub { size 8{"net"} } = { {Δp} over {Δt} } = { {mΔv} over {Δt} } "." } {}
    F net is the net external force, Δ p size 12{Δp} {} is the change in momentum, and Δ t size 12{Δt} {} is the change time.

Conceptual questions

An object that has a small mass and an object that has a large mass have the same momentum. Which object has the largest kinetic energy?

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An object that has a small mass and an object that has a large mass have the same kinetic energy. Which mass has the largest momentum?

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Professional Application

Football coaches advise players to block, hit, and tackle with their feet on the ground rather than by leaping through the air. Using the concepts of momentum, work, and energy, explain how a football player can be more effective with his feet on the ground.

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How can a small force impart the same momentum to an object as a large force?

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Problems&Exercises

(a) Calculate the momentum of a 2000-kg elephant charging a hunter at a speed of 7 . 50 m/s size 12{7 "." "50"``"m/s"} {} . (b) Compare the elephant’s momentum with the momentum of a 0.0400-kg tranquilizer dart fired at a speed of 600 m/s size 12{"600"``"m/s"} {} . (c) What is the momentum of the 90.0-kg hunter running at 7 . 40 m/s size 12{7 "." "40"``"m/s"} {} after missing the elephant?

(a) 1.50 × 10 4 kg m/s size 12{1 "." "50" times "10" rSup { size 8{4} } `"kg" cdot "m/s"} {}

(b) 625 to 1

(c) 6 . 66 × 10 2 kg m/s size 12{6 "." "66" times "10" rSup { size 8{2} } `"kg" cdot "m/s"} {}

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(a) What is the mass of a large ship that has a momentum of 1 . 60 × 10 9 kg · m/s , when the ship is moving at a speed of 48.0 km/h? size 12{"48" "." 0``"km/h?"} {} (b) Compare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s size 12{"1200"``"m/s"} {} .

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(a) At what speed would a 2 . 00 × 10 4 -kg size 12{2 "." "00" times "10" rSup { size 8{4} } "-kg"} {} airplane have to fly to have a momentum of 1 . 60 × 10 9 kg · m/s size 12{1 "." "60" times "10" rSup { size 8{9} } "kg" cdot "m/s"} {} (the same as the ship’s momentum in the problem above)? (b) What is the plane’s momentum when it is taking off at a speed of 60.0 m/s size 12{"60" "." 0``"m/s"} {} ? (c) If the ship is an aircraft carrier that launches these airplanes with a catapult, discuss the implications of your answer to (b) as it relates to recoil effects of the catapult on the ship.

(a) 8 . 00 × 10 4 m/s size 12{8 "." "00" times "10" rSup { size 8{4} } " m/s"} {}

(b) 1 . 20 × 10 6 kg · m/s size 12{1 "." "20" times "10" rSup { size 8{6} } " kg" cdot "m/s"} {}

(c) Because the momentum of the airplane is 3 orders of magnitude smaller than of the ship, the ship will not recoil very much. The recoil would be 0 . 0100 m/s size 12{ - 0 "." "0100"`"m/s"} {} , which is probably not noticeable.

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(a) What is the momentum of a garbage truck that is 1 . 20 × 10 4 kg size 12{1 "." "20" times "10" rSup { size 8{4} } " kg"} {} and is moving at 10 . 0 m/s size 12{10 "." "0 m/s"} {} ? (b) At what speed would an 8.00-kg trash can have the same momentum as the truck?

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A runaway train car that has a mass of 15,000 kg travels at a speed of 5 .4 m/s size 12{5 "." 4`"m/s"} {} down a track. Compute the time required for a force of 1500 N to bring the car to rest.

54 s

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The mass of Earth is 5 . 972 × 10 24 kg size 12{5 "." "972" times 10 rSup { size 8{"24"} } " kg"} {} and its orbital radius is an average of 1 . 496 × 10 11 m size 12{1 "." "496" times 10 rSup { size 8{"11"} } " m"} {} . Calculate its linear momentum.

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Questions & Answers

what is the fumula for calculating specific heat capacity, fusion,fission and vaporization?
Dohn Reply
Q=cm(∆t)
Emmanuel
Q=cm∆T
Muhammad
what is difference b/w vaporization and evaporation
Muhammad
evaporation is the process of extracting moisture while vaporization is process of becoming a vapor or gas
Emmanuel
From a molecular standpoint they are both cooling processes. Also, you may want to explore states of matter😊 #myTwoCents ~Shi~
Shii
cooling is a similarlity in both process I am confused in difference
Muhammad
1- Evaporation is a process where a liquid change to gas without reaching its boiling point. 2- Vaporization is a process where a liquid change to gas after reaching its boiling point. 3- Sublimation is a process where a solid changes into vapour without passing through a liquid state
Victor
I see. Evaporation is a type of vaporization, that occurs on the surface of a liquid as it changes into the gaseous phase before reaching its boiling point. hope that aids
Shii
vaporisation is cooling process while vaporization is heating process
Emmanuel
I mean to write evaporation is an heating process while vaporization is cooling process
Emmanuel
Yea here are two applications. 1- your wet washed clothes dry under the sun, the water EVAPORATES 2- when u are cooking, it reaches a point where u need to add more water because the water you added previously is getting dried. this is VAPORIZATION. Am not sure which is a cooling or heating process
Victor
vaporization occur only when the evaporation get to level where the above cloud is been (saturated) so cooling take place and started to change to liquid (eg rain fall)
Emmanuel
They are both properties of the same process so they're both cooling
Shii
what about sublimation? cooling or heating process?
Victor
exact
Muhammad
evaporation is the increase in kinetic energy of the liquid which can be gone by adding heat
Emmanuel
so its an heating process
Emmanuel
sublimation is when a solid change to gas
Emmanuel
evaporation is very definitely a cooling process. respectfully@Emmanuel when liquid turns to gas it requires more energy from its surroundings, this energy is in the form of heat, and when heat energy leaves the evaporating liquid it leaves it cooler. Thus, cooling process.
Shii
.
Shii
evaporation is very definitely a cooling process. respectfully@Emmanuel
Shii
kk
Emmanuel
You're right @Shi. I get your point
Victor
eascape velocity on the surface of Earth is 11.2 kms-1 the escape velocity on the surface of another planet of same mass as that of Earth but of 1/4 times of radius of Earth is a5.6kms-1 b11.2 kms-1 c22.4kms-1 d5.6ms-1
Muhammad
Emm.. is that a question? or..
Victor
it is McQ
Muhammad
a)5.6km/s
Alvis
c= Q/cm◇T
A.d
it's answer is 22.4
Muhammad
units...
Shii
vital
Shii
the time period of the artificial satellite is given by ?
raza
Why is there no 2nd harmonic in the classical electron orbit?
Shree Reply
how to reform magnet after been demagneted
Inuwa Reply
A petrol engine has a output of 20 kilowatts and uses 4.5 kg of fuel for each hour of running. The energy given out when 1 kg of petrol is burnt is 4.8 × 10 to the power of 7 Joules. a) What is the energy output of the engine every hour? b) What is the energy input of the engine every hour?
Morris Reply
what is the error during taking work done of a body..
Aliyu Reply
what kind of error do you think? and work is held by which force?
Daniela
I am now in this group
smart
theory,laws,principles and what-a-view are not defined. why? you
Douglas Reply
A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces it 10 o from the equilibrium position, and releases it. Using a s
Emmanuel Reply
so what question are you passing across... sir?
Olalekan
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed?
Emmanuel Reply
54 joule
babar
how?
rakesh
Reduce that two body problem into one body problem. Apply potential and k. E formula to get total energy of the system
rakesh
i dont think dere is any potential energy... by d virtue of no height present
Olalekan
there is compressed energy,dats only potential energy na?
rakesh
yes.. but... how will u approach that question without The Height in the question?
Olalekan
Can you explain how you get 54J?
Emmanuel
Because mine is 36J
Emmanuel
got 36J too
Douglas
OK the answer is 54J Babar is correct
Emmanuel
Conservation of Momentum
Emmanuel
woow i see.. can you give the formula for this
joshua
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed? Asume there is no external force.
Emmanuel Reply
Please help!
Emmanuel
please help find dy/dx 2x-y/x+y
Inuwa
By using the Quotient Rule dy/dx = 3y/(x +y)²
Emmanuel
3y/(x+y)²
Emmanuel
may be by using MC^2=MC^2 and Total energy=kinetic energy +potential energy so 1st find kinetic energy and den find potential energy which is stored energy
rakesh
i think i m correct
rakesh
But how?
Emmanuel
3y/(x+y)²
Douglas
what's the big bang?
kwame Reply
yes what is it?
LamaBbake
it is the explanation of how the universe began
Zainab
yes
Ana
explain
Chinagorom
in
Chinagorom
it is a theory on how the universe began. to understand more I would suggest researching the topic online.
david
thanks guys
kwame
if a force of 12N is applied to load of 200g what us the work done
Joshua Reply
We can seek accelation first
Nancy
we are given f=12 m=200g which is 0.2kg now from 2nd law of newton a= f/m=60m/s*2 work done=force applied x displacement cos (theta) w= 12x60 =720nm/s*2
Mudang
this very interesting question very complicated for me, í need urgent help. 1,two buses A and B travel along the same road in the same direction from Harper city (asume They both started from the same point) to Monrovia. if bus A maintains a Speedy of 60km/h and bus B a Speedy of 75km/h, how many
mohammed
hours Will it take bus B to overtake bus A assuming bus B starts One hour after bus A started. what is the distance travelled by the buses when They meet?.
mohammed
pls í need help
mohammed
4000 work is done
Ana
speed=distance /time distance=speed/time
Ana
now use this formula
Ana
what's the answer then
Julius
great Mudang
Kossi
please Ana explain 4000 ?
babar
hey mudang there is a product of force and acceleration not force and displacement
babar
@Mohammed answer is 0.8hours or 48mins
Douglas
nice
A.d
its not possible
Olalekan
í want the working procedure
mohammed
the answer is given but how Will One arrive at it. the answers are 4hours and 300m.
mohammed
physics is the science that studies the non living nature
isidor Reply
ancient greek language physis = nature
isidor
what is phyacs
technical Reply
if i am going to start studying physics where should i start?
BRIAN Reply
I think from kinematics
Nancy
You can find physics books at the library or online. That's how I started.
Chelsea
And yes, kinematics is usually where you can begin.
Chelsea
study basic algebra and calculus and can start from classical mechanics
Mudang
yes think so but dimension is the best starting point
Obed
3 formula's of equations of motion
benjamin Reply
vf=vi+at........1 s=vit+1/2(at)2 vf2=vi2+2as
Ana
solve the formula's please
benjamin
those are the three .. what you wanna solve ?
Nihrantz
For first equation simply integrate formula of acceleration in the limit v and u
Tripti
For second itegrate velocity formula by ising first equation
Tripti
similarly for 3 one integrate acceleration again by multiplying and dividing term ds
Tripti
any methods can take to solve this eqtions
a=vf-vi/t vf-vi=at vf=vi+at......1
Ana
suppose a body starts with an initial velocity vi and travels with uniform acceleration a for a period of time t.the distance covered by a body in this time is "s" and its final velocity becomes vf
Ana
what is the question dear
Zeeshan
average velocity=(vi+vf)/2 distance travelled=average velocity ×time therefore s=vi+vf/2×t from the first equation of motion ,we have vf =vi+at s=[vi+(vi+at)]/2×t s=(2vi+at)/2×t s=bit+1/2at2
Ana
find the distance
Ana
how
Zeeshan
v=u+at,v^2=(u^2) +2as,s=2u +(1/2)gt^2
Emmanuel
Practice Key Terms 2

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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