# Conservation of Momentum

## Momentum of a body is defined as 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

*velocity projectile (m/s, ft/s)*

* weight projectile (kg, lb)*

* weight pistol, gun, cannon .. (kg, lb)*

* recoil slow down distance (m, ft)*

### 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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