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Did you know that a swarm of jellyfish once managed to shut down a nuclear power plant, in Japan, after getting stuck in the cooling system? No, probably not, and now you're wondering what jellyfish have to do with physics, right? Well, what if I told you that jellyfish apply the principle of conservation of momentum every time they move? When a jellyfish wants to move, it fills its umbrella-like section with water and then pushes the water out. This motion creates a backward momentum that in turn creates an equal and opposite forward momentum that allows the jellyfish to push itself forward. Therefore, let us use this example as a starting point in understanding momentum.
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Jetzt kostenlos anmeldenDid you know that a swarm of jellyfish once managed to shut down a nuclear power plant, in Japan, after getting stuck in the cooling system? No, probably not, and now you're wondering what jellyfish have to do with physics, right? Well, what if I told you that jellyfish apply the principle of conservation of momentum every time they move? When a jellyfish wants to move, it fills its umbrella-like section with water and then pushes the water out. This motion creates a backward momentum that in turn creates an equal and opposite forward momentum that allows the jellyfish to push itself forward. Therefore, let us use this example as a starting point in understanding momentum.
Figure 1: Jellyfish use momentum to move.
Momentum is a vector quantity related to the motion of objects. It can be linear or angular depending on the motion of a system. Linear motion, one-dimensional motion along a straight path, corresponds to linear momentum which is the topic of this article.
Linear momentum is the product of an object's mass and velocity.
Linear momentum is a vector; it has magnitude and direction.
The mathematical formula corresponding to the definition of linear momentum is $$p=mv$$ where \( m \) is mass measured in \( \mathrm{kg} \) , and \( v \) is velocity measured in \( \mathrm{\frac{m}{s}} \). Linear momentum has SI units of \( \mathrm{kg\,\frac{m}{s}} \). Let's check our understanding with a quick example.
A \( 3.5\,\mathrm{kg} \) soccerball is kicked with a speed of \( 5.5\,\mathrm{\frac{m}{s}} \). What is the linear momentum of the ball?
Figure 2: Kicking a soccer ball to demonstrate linear momentum.
Using the linear momentum equation, our calculations are $$\begin{align}p&=mv\\p&= (3.5\,\mathrm{kg})\left(5.5\,\mathrm{\frac{m}{s}}\right)\\p&=19.25\,\mathrm{{kg\,\frac{m}{s}}}\\\end{align}.$$
When discussing momentum, the term impulse will arise. Linear impulse is a term used to describe how force affects a system with respect to time.
Linear impulse is defined as the integral of a force exerted on an object over a time interval.
The mathematical formula corresponding to this definition is
$$\Delta \vec{J}= \int_{t_o}^{t}\vec{F}(t)dt,$$
which can be simplified to
$$J=F\Delta{t}$$, when \( F \) doesn't vary with time, i.e. a constant force.
Note \( F \) is force, \( t \) is time, and the corresponding SI unit is \( \mathrm{Ns}. \)
Impulse is a vector quantity, and its direction is the same as that of the net force acting on an object.
Impulse and momentum are related by the impulse-momentum theorem. This theorem states that the impulse applied to an object is equal to the object's change in momentum. For linear motion, this relationship is described by the equation \( J=\Delta{p}. \) Newton's second law of motion can be derived from this relationship. To complete this derivation, we must use the equations corresponding to the impulse-momentum theorem in conjunction with the individual formulas of linear momentum and linear impulse. Now, let us derive Newton's second law for linear motion starting with the equation \( J=\Delta{p} \) and rewriting it as \( F\Delta{t}=m\Delta{v}. \)
$$\begin{align}J&=\Delta{p}\\F\Delta{t}&=\Delta{p}\\F\Delta{t}&=m\Delta{v}\\F&=\frac{m\Delta{v}}{\Delta{t}}\\\end{align}$$
Be sure to recognize that \( \frac{\Delta_v}{\Delta_t} \) is the definition of acceleration so the equation can be written as $$\begin{align}F&= ma\\\end{align},$$ which we know to be Newton's second law for linear motion. As a result of this relationship, we can define force in terms of momentum. Force is the rate at which the momentum of an object changes with respect to time.
To distinguish linear momentum from angular momentum, let us first define angular momentum. Angular momentum corresponds to rotational motion, circular motion about an axis.
Angular momentum is the product of angular velocity and rotational inertia.
The mathematical formula corresponding to this definition is $$L=I\omega$$ where \( \omega \) is angular velocity measures in \( \mathrm{\frac{rad}{s}} \) and \( I \) is inertia measured in \( \mathrm{kg\,m^2}. \) Angular momentum has SI units of \( \mathrm{kg\,\frac{m^2}{s}} \).
This formula can only be used when the moment of inertia is constant.
Again, let's check our understanding with a quick example.
A student vertically swings a conker, attached to a string, above their head. The conker rotates with an angular velocity of \( 5\,\mathrm{\frac{rad}{s}}. \) If its moment of inertia, which is defined in terms of the distance from the center of rotation, is \( 6\,\mathrm{kg\,m^2} \), calculate the angular momentum of the conker,
Using the equation for angular momentum, our calculations are $$\begin{align}L&=I\omega\\L&=(5\,\mathrm{kg\,m^2})\left(6\,\mathrm{\frac{rad}{s}}\right)\\L&= 30\,\mathrm{kg\,\frac{m^2}{s}}\\\end{align}$$
Linear momentum and angular momentum are related because their mathematical formulas are of the same form as angular momentum is the rotational equivalent of linear momentum. However, the main difference between each is the type of motion they are associated with. Linear momentum is a property associated with objects traveling a straight-line path. Angular momentum is a property associated with objects traveling in a circular motion.
Collisions are divided into two categories, inelastic and elastic, in which each type produces different results.
Inelastic collisions are characterized by two factors:
Elastic collisions are characterized by two factors:
Note that the equations associated with elastic collisions can be used in conjunction with one another to calculate an unknown variable if needed such as final velocity or final angular velocity.
Two important principles related to these collisions are the conservation of momentum and the conservation of energy.
The conservation of momentum is a law in physics that states momentum is conserved as it is neither created nor destroyed as stated in Newton's third law of motion. In simple terms, the momentum before the collision will be equal to the momentum after the collision. This concept is applied to elastic and inelastic collisions. However, it is important to note that conservation of momentum only applies when no external forces are present. When no external forces are present, we refer to this as a closed system. Closed systems are characterized by conserved quantities, meaning that no mass or energy is lost. If a system is open, external forces are present and quantities are no longer conserved. To check our understanding, let's do an example.
A \( 2\,\mathrm{kg} \) billiard ball moving with a speed of \( 4\,\mathrm{\frac{m}{s}} \) collides with a stationary \( 4\,\mathrm{kg} \) billiard ball, causing the stationary ball to now move with a velocity of \( -6\,\mathrm{\frac{m}{s}}. \) What is the final velocity of the \( 2\,\mathrm{kg} \) billiard ball after the collision?
Figure 4: A game of billiards demonstrates the concept of collisions.
Using the equation for conservation of momentum corresponding to an elastic collision and linear motion, our calculations are $$\begin{align}m_1v_{1i} + m_2v_{2i}&= m_1v_{1f}+m_2v_{2f}\\(2\,\mathrm{kg})\left(4\,\mathrm{\frac{m}{s}}\right) + 0 &= ( 2\,\mathrm{kg})(v_{1f}) + (4\,\mathrm{kg})\left(-6\,\mathrm{\frac{m}{s}}\right)\\8\,\mathrm{kg\,\frac{m}{s}}+ 0&=(2\,\mathrm{kg})(v_{1f}) - 24\,\mathrm{kg\,\frac{m}{s}}\\8 +24 &=(2\,\mathrm{kg})(v_{1f})\\\frac{32}{2}&=(v_{1f})=16\,\mathrm{\frac{m}{s}}\\\end{align}.$$
To better understand the conservation of momentum works, let us perform a quick thought experiment involving the collision of two objects. When two objects collide, we know that according to Newton's third law, the forces acting on each object will be equal in magnitude but opposite in direction, \( F_1 = -F_2 \), and logically, we know that the time it takes for \( F_1 \) and \( F_2 \) to act on the objects will be the same, \( t_1 = t_2 \). Therefore, we can further conclude that the impulse experienced by each object will also be equal in magnitude and opposite in direction, \( F_1{t_1}= -F_2{t_2} \). Now, if we apply the impulse-momentum theorem, we can logically conclude that changes in momentum are equal and opposite in direction as well. \( m_1v_1=-m_2v_2 \). However, although momentum is conserved in all interactions, the momentum of individual objects which make up a system can change when they are imparted with an impulse, or in other words, an
object's momentum can change when it experiences a non-zero force. As a result, momentum can change or be constant.
Note that the impulse exerted by one object on a second object is equal and opposite to the impulse exerted by the second object on the first. This is a direct result of Newton's third law.
Therefore, if asked to calculate the total momentum of a system, we must consider these factors. As a result, some important takeaways to understand are:
An example of an application that uses the law of conservation of momentum is rocket propulsion. Before launching, a rocket will be at rest indicating that its total momentum relative to the ground equals zero. However, once the rocket is fired, chemicals within the rocket are burnt in the combustion chamber producing hot gases. These gases are then expelled through the rocket's exhaust system at extremely high speeds. This produces a backward momentum which in turn produces an equal and opposite forward momentum that thrusts the rocket upwards. In this case, the change in the momentum of the rocket consists in part due to a change in mass in addition to a change in velocity. Rember, it is the change in the momentum which is associated with a force, and momentum is the product of mass and velocity; a change in either one of these quantities will contribute terms to Newton's second law: $$\frac{\mathrm{d}p}{\mathrm{d}t}=\frac{\mathrm{d}(mv)}{\mathrm{d}t}=m\frac{\mathrm{d}v}{\mathrm{d}t}+\frac{\mathrm{d}m}{\mathrm{d}t}v.$$
Momentum is important because it can be used to analyze collisions and explosions as well as describe the relationship between speed, mass, and direction. Because much of the matter we deal with has mass, and because it is often moving with some velocity relative to us, momentum is a ubiquitous physical quantity. The fact that momentum is conserved is a convenient fact that allows us to deduce velocities and masses of particles in collisions and interactions given the total momentum. We can always compare systems before and after a collision or interaction involving forces, because the total momentum of the system before will always be equal to the momentum of the system after.
The conservation of energy is a principle within physics that states that energy cannot be created or destroyed.
Conservation of energy: The total mechanical energy, which is the sum of all potential and kinetic energy, of a system remains constant when excluding dissipative forces.
Dissipative forces are nonconservative forces, such as friction or drag forces, in which work is dependent on the path an object travels.
The mathematical formula corresponding to this definition is
$$K_i + U_i = K_f + U_f$$
where \( K \) is kinetic energy and \( U \) is potential energy.
However, when discussing collisions, we focus only on the conservation of kinetic energy. Thus, the corresponding formula is
$$\begin{align}\frac{1}{2}m_1{v_{1i}}^2 + \frac{1}{2}m_2{v_{2i}}^2 =\frac{1}{2}m_1{v_{1f}}^2+ \frac{1}{2}m_1{v_{2f}}^2\\\end{align}$$
This formula will not apply to inelastic collisions.
The total energy of a system is always conserved, however, energy can be transformed in collisions. Consequently, these transformations affect the behavior and motion of objects. For example, let us look at collisions where one object is at rest. The object at rest initially has potential energy because it is stationary, thus meaning its velocity is zero indicating no kinetic energy. However, once a collision occurs, potential energy transforms into kinetic energy as the object now has motion. In elastic collisions, energy is conserved, however, for inelastic collisions energy is lost to the environment as some is transformed to heat or sound energy.
An application of the law of conservation of linear momentum is rocket propulsion.
Momentum is important because it can be used to analyze collisions and explosions as well as describe the relationship between speed, mass, and direction.
For momentum to be constant, the mass of a system must be constant throughout an interaction and the net forces exerted on the system must equal zero.
Linear momentum is defined as the product of an object's mass times velocity.
Impulse is defined as the integral of a force exerted on an object over a time interval.
Total linear momentum is the sum of the linear momentum before and after an interaction.
As momentum is a vector quantity, it has which of the following.
Magnitude and Direction.
Momentum is conserved in all interactions.
True
Momentum is constant within a system, if which two factors occur?
the mass of a system is constant and the net forces exerted on the system equal zero.
If the momentum of a system changes, what has to occur?
A net force is exerted on the system.
Impulse is the force applied to an object during a specific time period which causes the momentum of an object to change.
True.
Impulse and momentum are related by the impulse-momentum theorem.
True
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