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))
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)
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
F = E / s
= (450 lb ft) / (0.125 ft)
= 3600 lb
velocity projectile (m/s, ft/s)
weight projectile (kg, lb)
weight pistol, gun, cannon .. (kg, lb)
recoil slow down distance (m, ft)
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)
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
Forces, acceleration, displacement, vectors, motion, momentum, energy of objects and more.
Change in velocity vs. time used.
Angular velocity and acceleration vs. power and torque.
Forces due to circular motion and centripetal / centrifugal acceleration.
Linear and angular (rotation) acceleration, velocity, speed and distance.
Impact forces acting on falling objects hitting the ground, cars crashing and similar cases.
Forces acting a very short time are called impulse forces.
Energy possessed by an object's motion is kinetic energy.
Calculate the range of a projectile - a motion in two dimensions.