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Conservation of Momentum

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Linear Momentum

The momentum of a body is defined as the product of its mass and velocity. Since velocity is a vector quantity - momentum is a vector quantity.

The momentum of a body can be expressed as

ML = m v                                      (1)


ML = linear momentum (kg m/s, lb ft/s)

m = mass of body (kg, lb)

v = velocity of body (m/s, ft/s)

The momentum of a body remains the same as long as there is no external forces acting on it. The principle of conservation of momentum can be stated as

the total linear momentum of a system is a constant

The total momentum of two or more bodies before collision in a given direction is equal to the total momentum of the bodies after collision in the same direction and can be expressed as

ML = m1 v1 + m2 v2 + ..  + mn vn

        =  m1 u1 + m2 u2 + ..  + mn un                                       (2)


v = velocities of bodies before collision (m/s, ft/s)

u = velocities of bodies after collision (m/s, ft/s)


Example - Linear Momentum

A body with mass 30 kg and velocity 30 m/s collides with a body with mass 20 kg and velocity 20 m/s. The velocities of both bodies are in the same direction. Assuming the both bodies have same velocity after impact - the resulting velocity can be calculated as:

   ML = (30 kg) (30 m/s)  + (20 kg)  (20 m/s) 

        =  u ((30 kg) + (20 kg))

        = constant


    u = (30 kg) (30 m/s)  + (20 kg) (20 m/s)  /  ((30 kg) + (20 kg))

        = 26 (m/s)

Example - Recoil Velocity after Rifle Shoot

A rifle weighing 6 lb fires a bullet weighing 0.035 lb with muzzle speed 2100 ft/s. The total momentum of the rifle and bullet can be expressed as

ML = mr vr + mb vb

    = mr ur + mb ub

The rifle and the bullet are initially at rest (vr = vb = 0):

ML = (6 lb) (0 ft/s) + (0.035 lb) (0 ft/s)

     = (6 lb) ur (ft/s) + (0.035 lb) (2100 ft/s)

     = 0

The rifle recoil speed:

ur = - (0.035 lb) (2100 ft/s) / (6 lb)

    = - 12.25 (ft/s)

The recoil kinetic energy in the rifle can be calculated as

E = 1/2 m ur2

   = 1/2 (6 lb) (-12.25)2

   = 450 lb ft

The  force required to slow down a recoil depends on the slow down distance. With a slow down distance s = 1.5 in (0.125 ft) - the acting force can be calculated as

F = E / s

  = (450 lb ft) / (0.125 ft)

  = 3600 lb


Recoil Calculator

Angular Momentum

Angular momentum for a rotating body can be expressed as

MR = ω I                          (3)


MR = angular momentum of rotating body (kg m2 / s, lbf ft s)

ω = angular velocity (rad/s)

I = moment of inertia - an object's resistance to changes in rotation direction (kg m2, slug ft2)

The conservation of angular momentum can be expressed as

ω1 I1 = ω2 I2                          (4a)


n1 I1 = n2 I2     (4b)


n = rotation velocity in revolutions per minute (rpm)

For a small point mass I = m r2 and ω = v / r and 4a can be modified to

m v1 r1 = m v2 r2                  (4c)


v1 r1 = v2 r2                       (4d)


ω1 r1 = ω2 r2                          (4e)


n1 r1 = n2 r2                          (4f)


v = tangential velocity (m/s, ft/s)

r = radius (m, ft)

Example - Small Sphere

A small iron sphere rotates around an axis with velocity 100 rpm when connected to a rope with length 1.5 m. The rope is extended to 2.5 m and due to conservation of angular momentum the new velocity can be calculated by modifying 4f to

n2 = n1 r1 / r2

    = (100 rpm) (1.5 m) / (2.5 m)

    = 60 rpm

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