Momentum

1. Introducing momentum

• Momentum is a measure of an object’s mass and velocity.
• It’s a vector quantity, meaning it has both magnitude (amount of movement) and direction.
• The momentum of an object is calculated as:
• [math] \text{Momentum} \, (\vec{p}) = \text{Mass} \, (m) \times \text{Velocity} \, (\vec{v}) \\ \vec{p} = m \vec{v} [/math]
• The unit of momentum is typically measured in kilogram-meters per second (kg · m/s).
• Figure 1 Momentum is transferred from one object to another
• Here are some key aspects of momentum in physics:
  1. Newton’s laws: Momentum is closely related to Newton’s laws of motion.
  2. The second law states that force ([math]\vec{F}[/math]) is equal to the rate of change of momentum ([math] \Delta \vec{p} / \Delta t [/math]).
  3. Momentum transfer: When objects interact, momentum is transferred from one object to another. This is why a baseball bat can transfer momentum to a ball, making it fly!
  4. Relativity: In special relativity, momentum is affected by an object’s speed and mass. As an object approaches the speed of light, its momentum increases exponentially.
• In the example of the exploding firework, chemical potential energy is transferred to thermal energy, light energy and kinetic energy of the exploding fragments.
• However, during the explosion the momentum remains the same.

Example

A ball of mass 0.1kg hits the ground with a velocity of 6m/s and sticks to the ground. Calculate its change of momentum.
Given data:
Mass of a ball = m = 0.1 kg
Change in velocity = [math]\Delta \vec{v}[/math] = 6m/s
Find data:
Change in momentum=[math]\Delta \vec{p}[/math]  = ?
Formula:

[math] \Delta \vec{p} = m \Delta \vec{v} [/math]

Solution:

[math] \Delta \vec{p} = m \Delta \vec{v} [/math]

Put values

[math] \Delta \vec{p} = 0.1(6) [/math]

[math] \Delta \vec{p} = 0.6 kg .\frac{m}{s} [/math]

2. Momentum and impulse

• Momentum and impulse are closely related concepts in physics.
• Impulse is a measure of the change in momentum of an object.
• In other words, impulse is the product of the force applied to an object and the time over which it’s applied.
• The unit of impulse is typically measured in newton-seconds (N·s).
• Here are some key points about the relationship between momentum and impulse:
  1. Impulse changes momentum: When an impulse is applied to an object, it changes the object’s momentum.
  2. Momentum conservation: If the total impulse is zero, the total momentum remains constant.
  3. Force and time tradeoff: A smaller force applied over a longer time can produce the same impulse as a larger force applied over a shorter time.
• Impulse- momentum theorem: The impulse applied to an object is equal to the change in momentum of the object.
• Newton’s second law can be used to link an applied force to a change of momentum:
• [math] \vec{F} = m \vec{a} [/math]
• Substituting
• [math] \vec{a} = \frac{\Delta \vec{v}}{\Delta t} [/math]
• Gives
•  [math] \vec{F} = \frac{ m\Delta \vec{v}}{\Delta t} [/math]

• Or

  [math] \vec{F} = \frac{\Delta (m\vec{v})}{\Delta t} [/math]

• So,

• Force equals the rate of change of momentum.
• This is a more general statement of Newton’s second law of motion.
• The last equation may also be written in the form:
• [math] \vec{F} \Delta t = \Delta (m \vec{v}) [/math]
• or
• [math] \vec{F} \Delta t = m \vec{v}_2 – m \vec{v}_1 [/math]
• where [math]\vec{v_2}[/math] is the velocity after a force has been applied and [math]\vec{v_1}[/math] the velocity before the force was applied.
• The quantity [math] \vec{F} \Delta t = I [/math] is called the impulse. Where [math] \vec{F} [/math]  is constant act.
• 
  Figure 2 Impulse and momentum

Figure 3 Force–time graphs for two passengers in a car crash.

3. Conservation of linear momentum

• Conservation of linear momentum states that in a closed system, the total linear momentum remains constant over time.
• This means that the total momentum before an event (like a collision) is equal to the total momentum after the event.
• Mathematically, this is expressed as:
  Before collusion (initial) total momentum ([math]P_i[/math]) = After collusion (final) total momentum ([math]P_f[/math])
• [math]P_i = P_f \\ m\vec{v-i} = m\vec{v-f}[/math]
• For two bodies
• [math] m_1 \vec{v}_1 + m_2 \vec{v}_2 = m_1 \vec{v}’_1 + m_2 \vec{v}’_2 [/math]
• The total momentum of two bodies in a collision (or explosion) is the same after the collision (or explosion) as it was before.
• 
  Figure 4 Law of conservation of momentum
• This law applies to all closed systems, where no external forces are acting. In other words, the total momentum is conserved if there are no external influences that can change the momentum.
• Momentum is always conserved in bodies involved in collisions and explosions provided no external forces act. This is the principle of conservation of momentum.
• Here are some key aspects of conservation of linear momentum:
  1. Closed system: The system must be closed, meaning no objects can enter or leave, and no external forces can act upon it.
  2. Total momentum: The total momentum is the vector sum of all individual momenta in the system.
  3. Conservation: The total momentum remains constant over time, unless acted upon by an external force.
  4. Collisions: In collisions, momentum is transferred between objects, but the total momentum remains conserved.
  5. Explosions: In explosions, momentum is conserved, but the total momentum is distributed among the products.
• Figure 5 shows a demonstration of a small one-dimensional explosion.
• The head of a match has been wrapped tightly in aluminium foil.
• A second match is used to heat the foil and the head of the first match, which ignites and explodes inside the foil.
• The gases produced cause the foil to fly rapidly one way, and the matchstick to fly in the opposite direction. (It is necessary to use ‘strike-anywhere’ matches, the heads of which contain phosphorous.
• The experiment is safe because the match blows out as it flies through the air, but common-sense safety precautions should be taken – do this in the laboratory, not in a carpeted room at home, and dispose of the matches afterwards.)
• 
  Figure 5 A simple experiment that demonstrates the conservation of linear momentum
• In all collisions and explosions, both total energy and momentum are conserved, but kinetic energy is not always conserved.
• In this case, the chemical potential energy in the match head is transferred to the kinetic energy of the foil and matchstick, and also into thermal, light and sound energy.
• As the match head explodes, Newton’s third law of motion tells us that both the matchstick and foil experience equal and opposite forces, [math] \vec{F} [/math].
• Since the forces act for the same interval of time, Δt, both the matchstick and foil experience equal and opposite impulses, [math] \vec{F} \Delta t[/math].
• Since [math] \vec{F} \Delta t = \Delta(m\vec{v})[/math], it follows that the foil gains exactly the same positive momentum as the matchstick gain’s negative momentum.
• We can now do a vector sum to find the total momentum after the explosion (see Figure 5c):
• [math] + \Delta (m \vec{v}) + \left[ – \Delta (m \vec{v}) \right] = 0[/math]
• So, there is conservation of momentum: the total momentum of the foil and matchstick was zero before the explosion, and the combined momentum of the foil and matchstick is zero after the explosion.

⇒ Collisions on an air track

• Collision experiments can be carried out using gliders on linear air tracks.
• These can demonstrate the conservation of linear momentum.
• The air blowing out of small holes in the track lifts the gliders so that frictional forces are very small.

Figure 6 A laboratory air track with gliders

4. Momentum and energy:

• Momentum (p): Measures an object’s mass and velocity ([math]\vec{p} =m\vec{v}[/math]).
• Energy (E): Measures an object’s capacity to do work (e.g., kinetic energy, potential energy, thermal energy).
• Here are some key connections between momentum and energy:
• Kinetic energy:
  The energy of motion, calculated as
• [math] \text{Energy (K.E)} = \frac{1}{2} mv^2[/math]
• The more massive and faster an object, the more kinetic energy it has.
• Work-energy theorem: The work done on an object is equal to its change in kinetic energy.
• [math] W = \Delta K [/math]
• Conservation of energy: In a closed system, the total energy remains constant over time.
• [math] E_initial = E_final[/math]
• Momentum-energy relation: For an object with mass m and speed v, the energy-momentum relation is
• [math] E^2 = (pc)^2 + (mc^2)^2 [/math]
• where c is the speed of light.
• In elastic collisions kinetic energy is conserved.
• In inelastic collisions kinetic energy is not conserved. Some or all of the kinetic energy is transferred to heat or other types of energy.

⇒Some useful equations:

• Although momentum and kinetic energy are very different quantities, they are linked by some useful equations.
• Momentum is given the symbol [math]\vec{p}[/math].
• [math] \vec{p} =m\vec{v} \qquad(1) [/math]
• Kinetic energy is given the symbol [math] E_k[/math].
• [math] E_k =\frac{1}{2} mv^2\qquad(2) [/math]
• multipling and dividing by m with equation 2 on right side then
• [math] E_k = \frac{1}{2} mv^2 \left(\frac{m}{m}\right) [/math]
• But [math]E_k [/math] can also be written in this form:
• [math] E_k = \frac{m^2 v^2}{2m} [/math]
• giving
• [math] E_k = \frac{p^2}{2m} [/math]
• It is important to remember that momentum is a vector quantity and kinetic energy is a scalar quantity.
• We give momentum a direction: for example, +p to the right and −p to the left.
• A vehicle travelling with momentum p has as much kinetic energy when travelling to the right as it does to the left.
• To the right
• [math] E_k = \frac{(+p)^2}{2m} = \frac{p^2}{2m}\\E_k = \frac{(-p)^2}{2m} = \frac{p^2}{2m} [/math]

• When a vector is squared, it becomes a scalar quantity.

5. Elastic and inelastic collisions

• Elastic and inelastic collisions are types of collisions that differ in how energy is transferred and conserved.
• Elastic Collisions:
  – Momentum is conserved.
  – Kinetic energy is conserved.
  – Objects bounce off each other.
  – No energy is lost or transferred.
• 
  Figure 7 Elastic collusion in one dimension
• Inelastic collisions are a type of collision where:
  1. Momentum is conserved
  2. Kinetic energy is NOT conserved
  3. Objects stick together or deform
  4. Energy is transferred or lost (e.g., heat, sound, vibrations)
• A helium nucleus of mass 4m collides head-on with a stationary proton of mass m. Use the information in Figure 8 to show that this is an elastic collision.

Figure 8 elastic collusion between a helium nucleus and a stationary proton

Solution

• The velocity of the proton can be calculated as follows. The momentum before the collision equals the momentum after the collision.
• [math]  4mu + 0 = 4m \times 0.6u + mv\\
  mv = 4mu – 2.4mu \\
  v = 1.6u [/math]
• The total kinetic energy before the collision was:
• [math] E_k = \frac{1}{2} \times 4mu^2\\
  E_k = 2mu^2 [/math]
• The total kinetic energy after the collision was:
• [math] E_k = \frac{1}{2} \times 4m (0.6u)^2 + \frac{1}{2} \times m (1.6u)^2\\
  E_k = 2m (0.36u^2) + \frac{1}{2} m (2.56u^2)\\
  E_k = 0.72mu^2 + 1.28mu^2\\
  E_k = 2mu^2 [/math]
• So, the total kinetic energy is conserved in this collision. It is a useful rule to know that in an elastic collision the relative speeds of the two particles is the same before and after the collision.

⇒More advanced problems in momentum:

• The generalized form of Newton’s second law of motion enables you to solve some more complicated problems.
• [math] force = rate\, of\, change \,of\, momentum\\
  F = \frac{\Delta (mv)}{\Delta t} [/math]
• Example
• Calculate the force exerted by water leaving a fire hose on the firefighter holding the hose. The water leaves with a velocity of 17m/s; the radius of the hose is 3cm; the density of water is 1000kgm^-3.
• Solution:
• Given data:
• Velocity = 17m/s
• Radius = 3cm = 0.03m
• Density of water = 1000 kg.m^-3
• Find data:
• Force exerted by water = F =?
• Formula:
• The mass of water leaving the hose each second is [math] \frac{\Delta m}{\Delta t} [/math].
• This is equal to the density of water (ρ) × the volume of water, V, flowing per second:
• [math] \frac{\Delta m}{\Delta t} = \rho \frac{\Delta V}{\Delta t} [/math]
• But the volume flow per second is the cross-sectional area of the hose (A) multiplied by the velocity of the water (v). So:
• [math] \frac{\Delta m}{\Delta t} = \rho Av [/math]
• Thus the force, or change of momentum per second, is:
• [math] F = \frac{\Delta \vec{p}}{\Delta t} = \rho A \vec{v} \times \vec{v}[/math]
• [math]\vec{F} = \rho A v^2 [/math]
• Solve:
• [math] \vec{F} = \rho A v^2 [/math]
• [math] \vec{F} = 1000 \, \text{kg/m}^-3 \times \pi \times (0.03 \, \text{m})^2 \times (17 \, \text{m/s})^2 \\ \vec{F} = 820 \, \text{N} [/math]
• This large force explains why a hose is sometimes held by two firefighters, and why large hoses have handles.

⇒ Collisions in two dimensions:

• Collisions in two dimensions involve objects moving in a plane, with both momentum and energy conserved. This adds a new level of complexity, as velocities and momenta have both magnitude and direction.
• Key aspects of collisions in two dimensions:
  1. Vector addition: Momenta and velocities are vectors, so addition and subtraction involve vector operations.
  2. Components: Break down velocities and momenta into x- and y-components to analyze collisions.
  3. Conservation of momentum: Total momentum is conserved in both x- and y-directions.
  4. Conservation of energy: Kinetic energy is conserved, but may be redistributed between objects.
  5. Scattering: Objects may change direction and/or speed after the collision.

Example

• If the two particles (protons for example) collide head-on, then the momentum and kinetic energy of the moving proton (A) is transferred completely to the stationary proton (B). See Figure 9.

Figure 11 In vector terms,

p = pcos θ + psinθ.

• This conserves kinetic energy as well as momentum.
• In all other situations, both particles will be travelling after the impact; this only occurs when the particles have the same mass.
• The momentum of particle A can be resolved into two components:
• In figure 10, pcosθ along the line of collision and psinθ  perpendicular to the line of the collision.
• As in Figure 9, all of the momentum of A along the line of the collision (here pcosθ) is transferred to particle B.
• This leaves particle A with the component psinθ, which is at right angles to the line of the collision.
• So, in a non-head-on elastic collision between two particles of the same mass, they always move at right angles to each other.
• Figure 11 shows how momentum is conserved as a vector quantity.
• Kinetic energy is also conserved.
• Before the collision the kinetic energy is:
• [math] E_k = \frac{p^2}{2m} [/math]
• After the collision the kinetic energy of the two particles is:
• [math] = \frac{p^2}{2m} \cos^2 \theta + \frac{p^2}{2m} \sin^2 \theta\\
   = \frac{p^2}{2m} (\cos^2 \theta + \sin^2 \theta)\\\cos^2 \theta + \sin^2 \theta = 1\\
  So,\\
   = \frac{p^2}{2m}\\
  [/math]
• The kinetic energy after the collision is the same as before [math] \frac{p^2}{2m} [/math]