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.

Momentum - body mass and velocity

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 = m₁ v₁ + m₂ v₂ + .. + m_n v_n

= m₁ u₁ + m₂ u₂ + .. + 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_r²

= 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 m² / s, lb_f ft s)

ω = angular velocity (rad/s)

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

Angular moment

The conservation of angular momentum can be expressed as

ω₁ I₁= ω₂ I₂ (4a)

n₁ I₁= n₂ I₂ (4b)

where

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)

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

n₂ = n₁ r₁/ r₂

= (100 rpm) (1.5 m) / (2.5 m)

= 60 rpm