Conservation of Momentum
The momentum of a body is the product of its mass and velocity  recoil calculator.
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
M_{L} = m v (1)
where
M_{L} = 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
M_{L} = m_{1} v_{1} + m_{2} v_{2} + .. + m_{n} v_{n}
= m_{1} u_{1} + m_{2} u_{2} + .. + m_{n} u_{n} (2)
where
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:
M_{L} = (30 kg) (30 m/s) + (20 kg) (20 m/s)
= u ((30 kg) + (20 kg))
= constant
transformed:
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
M_{L} = m_{r} v_{r} + m_{b} v_{b}
= m_{r} u_{r} + m_{b} u_{b}
The rifle and the bullet are initially at rest (v_{r} = v_{b} = 0):
M_{L} = (6 lb) (0 ft/s) + (0.035 lb) (0 ft/s)
= (6 lb) u_{r} (ft/s) + (0.035 lb) (2100 ft/s)
= 0
The rifle recoil speed:
u_{r} =  (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 u_{r}^{2}
= 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
M_{R} = ω I (3)
where
M_{R} = angular momentum of rotating body (kg m^{2} / s, lb_{f} ft s)
ω = angular velocity (rad/s)
I = moment of inertia  an object's resistance to changes in rotation direction (kg m^{2}, slug ft^{2})
The conservation of angular momentum can be expressed as
ω_{1} I_{1 }= ω_{2} I_{2} (4a)
or
n_{1} I_{1 }= n_{2} I_{2} (4b)
where
n = rotation velocity in revolutions per minute (rpm)
For a small point mass I = m r^{2} and ω = v / r and 4a can be modified to
m v_{1} r_{1} = m v_{2} r_{2} (4c)
or
v_{1} r_{1} = v_{2} r_{2} (4d)
or
ω_{1} r_{1 }= ω_{2} r_{2} (4e)
or
n_{1} r_{1 }= n_{2} r_{2} (4f)
where
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
n_{2} = n_{1} r_{1 }/ r_{2}
= (100 rpm) (1.5 m) / (2.5 m)
= 60 rpm
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