Linear Momentum

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Previously Asked Questions

Q:     A box with mass 5 kg and moving at a speed v = 10 m/s slides across a frictionless floor.  It suddenly explodes into two pieces.  One piece moves at a velocity of 9 m/s in the positive x direction and has a mass of  3 kg.  What is the velocity of the second piece?

A:     
If the first piece had a mass of 3 kg (m2 = 3 kg), the second piece must have a mass of 2 kg.
Momentum is conserved so Pi = Pf

Pi = mv                                  Pf = m1v1 + m2v2
mv = m1v1 + m2v2  
v2  =  (mv - m1v1) / m2  =   ( 5 kg * 10 m/s - 3 kg * 9 m/s ) / (2kg) =  11.5 m/s

The second piece moves with a velocity of 11.5 m/s, since it is positive it moves in the positive x direction.

Q:     There is a scene in the movie "Eraser" in which Arnold Schwarzenegger wields 2 electromagnetic railguns, one on each arm.  A railgun works by accelerating aluminum pellets to near the speed of light.  Doesn't conservation of linear momentum mean that Arnold would be shot back with the same momentum that the pellets were shot forward?  Would this type of weapon be realistic?

Eraser3.jpg (7174 bytes)A:     Yes, conservation of linear momentum would give Arnold a recoil backwards with the same magnitude of momentum as that of the forward going pellet.  The force applied on him depends on his mass.

   This type of gun is not realistic since accelerating material particles in a straight line gun barrel would require a barrel of tremendous length.  Even small particles like electrons, to be accelerated to speeds close to that of light require particle accelerators of many miles in length.

In reality the US Military has designed electromagnetic (EM) guns that fire projectiles of about 1 kg mass at a velocities between 2 and 6 kilometers per second (about 1/50,000 the speed of light).  These guns are fairly large and mounted on battleships or tanks.  A hypothetical handheld gun would give the person holding it a recoil of  6,000 (Ns) at a typical velocity 88 meters/second (or about 197 miles/hour)  for a person of average weight of 150 pounds.

Example:

A rail gun fires a 100 gram (.1 kg) pellet at 5 kilometers/second.   What mass would the gun and its supports have to have in order to move only 0.5 meters/second after from the firing of the gun

ppellet = pgun

ppellet = mpelletvpellet = 0.1kg * 5000 m/s = 500 kg m /s = 500 N s.

pgun = mgunvgun =  0.5 m/s * mgun =  ppellet

500 N s = 0.5 m/s * mgun

mgun= 1000 kg  or  2204.5 pounds or about 1.10 tons

 

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References

The Law of Conservation of Linear Momentum: In the absence of any
external forces, the momentum of a system remains unchanged.

 

Equations

Center of Mass

 9-11a.gif (229 bytes)9-11b.gif (233 bytes)  , 9-11c.gif (223 bytes)

9-8.gif (238 bytes)

Center of Mass for a mass that is continuously distributed 9-9a.gif (230 bytes)9-9b.gif (244 bytes)9-9c.gif (227 bytes)
If density is uniform CM can be written as 9-11a.gif (229 bytes)9-11b.gif (233 bytes), 9-11c.gif (223 bytes)
Newton's Second Law for a System of Particles sigma.gif (836 bytes) Fext = M acm
Linear Momentum and Newton's Second Law
p = mv 9-28.gif (215 bytes)
Linear Momentum and Newton's Second Law for a system of particles
P = Mvcm 9-28.gif (215 bytes)
Relativistic Momentum 9-24.gif (264 bytes)
Conservation of Linear Momentum P = constant or  Pi = Pf
Variable Mass System (rockets)
Ru = Ma 9-47.gif (212 bytes)
External Forces and Internal Energy Change delta.gif (839 bytes)Eint = - Fext dcm cos  

delta.gif (839 bytes)Kcm = Fext dcm cos

delta.gif (839 bytes)Eint = delta.gif (839 bytes)Kcmdelta.gif (839 bytes)Ucm = Fext dcm cos

 

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Mechanics List of Topics

Measurements Newton's Laws Potential Energy and Conservation of Energy Rotation of
Rigid Bodies
Elasticity
Vectors Forces and Fields Linear Momentum Angular Momentum Mechanical
Oscillations
Motion of Point-Mass Objects in One Dimension The Gravitational Field Collisions Torque Mechanical Waves
Motion of Point-Mass Objects in Two and Three Dimensions Kinetic Energy
and Work
Circular Motion of Point-Mass Objects Equilibrium Sound

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