# Conservation of Momentum

## Momentum of a body is defined as the product of its mass and velocity

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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 = m v (1)

where

M = momentum (kg m/s, lb ft/s)

m = mass (kg, lb)

v = velocity (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 moment 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_{T}= 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 = velocity before collision (m/s, ft/s)

u = velocity 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 _{T} = (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 - the Recoil Velocity after a Rifle Shoot

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

*M _{T} = 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 _{T} = (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) *

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