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Conservation of Energy and Momentum

- Astrophysics
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Imagine a ball pushed down two different hills of the same height. The first hill runs smoothly down to the level ground at the bottom, whereas the second hill has lots of bumps and divots before ending up on the same level. Ignoring friction, will the ball have a greater speed when it rolls down the first hill or the second? Maybe climbing out of those divots will cause the final speed to decrease? Or will falling from the extra height of these bumps increase it? In fact, because of the conservation of energy, both balls will have the same speed at the bottom of each hill!

But what would happen if one ball was in the middle of the trajectory and gets hit by another one? For this case, we need to consider the conservation of energy and momentum. Let's explore these two concepts in modern detail and learn about them by looking at some day-to-day examples.

In physics, we focus our attention on a system. There are different types of systems, but particularly especially one is an isolated system.

An **isolated system** is a system where no energy or matter can enter or leave.

In an isolated system, the total amount of energy is always the same. This is known as the Law of Conservation of Energy.

The **L****aw of Conservation of Energy** states that energy is always conserved in an isolated system.

Therefore, energy is only converted between different forms. A gain in one type of energy must mean an equal energy loss in another type of energy, such that the total is always the same.

The different forms of energy come under one of two categories: **kinetic **and **potential energy**. Thus, the total energy of a system or an object is the sum of its kinetic energy and its potential energy.

Kinetic energy is associated with motion.

The **Kinetic energy** of an object is the energy that it has in virtue of its motion.

We have to transfer energy to an object to get it moving. For instance, when you are riding a bike, your leg muscles are doing work to move forward, and you and the bike gain kinetic energy as you move faster.

There are different types of kinetic energy, such as thermal, sound, and electrical energy.

Potential energy is associated with a system's arrangement.

The **potential energy** of a system is the energy stored within it due to its configuration.

An object in a gravitational field has **gravitational potential energy** according to its relative position with respect to the source of the gravitational field. For example, as we lift an object, it gains gravitational potential energy because its distance with respect to the source of the gravitational field (the Earth) increases. Similarly, an object in an electric field has electric potential energy.

Other forms of potential energy include chemical energy and elastic potential energy.

Consider dropping a ball to the ground. When holding the ball, it has a maximum amount of gravitational potential energy but no kinetic energy because it isn't moving. After you release it, the ball starts to fall losing potential energy but gaining speed, and thus, kinetic energy. The potential energy becomes zero as the ball reaches the ground but the kinetic energy reaches its maximum value. The original amount of potential energy became kinetic energy, but the total amount of energy remained the same!

But what about when it reaches the ground and just sits there? It has no kinetic energy anymore, but its potential energy is also gone! It seems that energy was lost. However, the Earth and the ball form an isolated system so this can't be true. Where did the energy go?

There are a number of possibilities. You probably heard a thud when the ball hit the ground, so some energy went into sound waves. The impact should have caused vibrations in the ground, which also used some energy. Moreover, there is non-negligible friction with the surface of contact. Some of our kinetic energy surely became heat, or maybe the ball got deformed using part of the energy. We end up with what seems to be an energy loss, but our energy was simply transformed into other forms!

The **linear momentum **of an object is a quantity that depends on its mass and velocity.

$$\vec p=m\vec v$$

In the above equation, \( m \) is the mass of the object in \( \mathrm{kg} \) and \( \vec v \) is its velocity in \( \mathrm m/\mathrm s \). The units for momentum, \( \mathrm{\vec p}\) are \( \mathrm{kg\,m}/\mathrm s \). Momentum and velocity are both vectors so their direction must be taken into account. The momentum vector of a moving object points in the same direction as its velocity.

A vector quantity has a magnitude *and* a direction.

A particle with a mass of \( 2\;\mathrm{kg} \) is moving at a speed of \( 20\;\mathrm m/\mathrm s \) along the \( x \) axis in the negative direction. What is the momentum of the particle?

We can use the equation for linear momentum presented above.

$$\vec p=m\vec v.$$

As we need to find the momentum along the \( x \) axis and the particle is moving in the negative \( x \) direction, we can add a minus sign to the speed to represent the direction of the velocity:

$$\vec v=-20\;\mathrm m/\mathrm s.$$

The momentum can then be calculated as

$$2\times-20=-40\;\mathrm{kgm}/\mathrm s.$$

Therefore, the momentum of the particle is \( 40\;\mathrm{kg\,m}/\mathrm s \) in the negative direction.

Similarly as with energy, there is even a conservation law for momentum!

The L**aw of Conservation of Momentum **states that momentum is always conserved when no external forces act on the system.

**External forces** are forces that act on an object or system from its surroundings.

Consider a system of two objects that collide. This law implies that the total momentum of the two objects before they collide equals the total momentum after the collision since there are no external forces acting on the system.

Considering the system of the bullet and gun. Before the bullet is fired, the momentum is zero. Therefore, the momentum of the gun must be equal but opposite to the momentum of the bullet, sot hat the total momentum is also zero. This is why the gun recoils.

The only condition for the conservation momentum of a system is that no external forces act on it. However, this does not mean that all forces are forbidden. Momentum is still conserved if there are *internal *forces acting.

**Internal forces** are forces that originate within a system.

Internal forces cannot change the momentum of a system as they are always canceled out by each other.

We can represent the Law of Conservation of Energy with the following equation:

$$E_{\mathrm i}=E_{\mathrm f},$$

where \( E_{\mathrm i} \) is the initial energy of the system and \( E_{\mathrm f} \) is the final energy.

Remember that the total energy is the sum of the kinetic \( KE \) and potential energy \( PE \) of the system. Let's see how to calculate each of them.

$$\begn{aligne}KE &= \frac12mv^2\\ PE_{\text{gravitaitonal}}&=mgh$$

where \( m \) is the mass of the object in \( \mathrm{kg} \) and \( v \) is its speed in \( \mathrm m/\mathrm s \), \( g \) is the acceleration due to Earth's gravity, \(9.8 \mathrm{m/s^2}\), and \( h \) is the height of the object in meters (\( \mathrm{m} \)).

Let us consider our ball again. What is the speed of the ball when it hits the ground if we drop it from a height \( \mathrm H \)? This can be worked out using conservation of energy. All the gravitational potential energy is converted into kinetic energy at the moment it reaches the floor. This means that we can set the two above equations equal to each other:

$$\frac12mv^2=mg\mathrm H.$$

The masses on either side of the expression cancel out and we can then solve for the speed

$$v=\sqrt{2g\mathrm H}.$$

Similarly, the law of conservation of momentum can be represented by the formula:

$$\vec{p_{\mathrm i}}=\vec{p_{\mathrm f}},$$

As discussed above, momentum is a vector quantity so you must pay attention to the direction of the object's motion when calculating its momentum.

Both conservation principles are very similar but they have an important difference. Energy is a scalar quantity; it has a numerical value and no direction. On the other hand, momentum is a vector quantity and so it has a direction. Usually, we use the coordinate system with the \( x \), \( y \) and \( z \) axes. The momentum of an object can be calculated along each axis and it will be conserved in each direction, independently.

The laws of conservation of energy and momentum are useful when dealing with collisions. While momentum is always conserved in collisions (as long as no external forces exist!) energy can be wasted in some types of collisions.

No energy is lost in elastic collisions. All of the energy is kinetic.

**Elastic collisions **are those in which kinetic energy is conserved.

The collisions of billiard balls can be considered to be elastic collisions to a good approximation - just a small proportion of kinetic energy is lost.

The general equations for the final velocities of particles colliding in an elastic collision can be found from the equations for conservation of momentum and kinetic energy. Consider an elastic collision between particle \( 1 \) with a mass \( m_1 \) and particle \( 2 \) with a mass \( m_2 \). They collide along a straight line - they do not change direction. Particle \( 1 \) has an initial speed of \( u_1 \) and a final speed of \( v_1 \). Particle \( 2 \) has an initial speed of \( u_2 \) and a final speed of \( v_2 \). This means that we can write the conservation of momentum equation as

$$m_1u_1+m_2u_2=m_1v_1+m_2v_2.$$

The collision is elastic so the kinetic energy before and after is the same, so we can also write

$$m_1u_1^2+m_2u_2^2=m_1v_1^2+m_2v_2^2.$$

The equation for conservation of momentum for two particles is

$$m_1u_1+m_2u_2=m_1v_1+m_2v_2,$$

which can be rearranged to

$$m_1(u_1-v_1)=m_2(v_2-u_2).$$

Let this be equation A. The equation for conservation of kinetic energy is

$$m_1u_1^2+m_2u_2^2=m_1v_1^2+m_2v_2^2,$$

which can be rearranged to$$m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2).$$This expression can be factorized to get$$m_1(u_1+v_1)(u_1-v_1)=m_2(v_2+u_2)(v_2-u_2).$$Let this be equation B. We can now divide equation B by equation A, which gives$$u_1+v_1=v_2+u_2$$and hence$$v_2=u_1+v_1-u_2.$$The previous expression for \( v_2 \) can be substituted into the conservation of momentum equation to find \( v_1 \):$$\begin{aligned}m_1u_1^2+m_2u_2^2 &=m_1v_1^2+m_2v_2^2\\[6pt] m_1u_1+m_2u_2 &=m_1v_1+m_2v_1+m_2u_1-m_2u_2\\[6pt] (m_1+m_2)v_1 &=(m_1-m_2)u_1+2m_2u_2\\[6pt]v_1 &=\frac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2} \end{aligned}$$Notice that we could have just as well named particle \( 1 \) as particle \( 2 \). This means that we can swap all of the subscripts around to find \( v_2 \)!$$v_2=\frac{2m_1u_1+(m_2-m_1)u_2}{m_1+m_2}$$A particle of mass \( 2\;\mathrm kg \) moving at a speed of \( 2\;\mathrm m/\mathrm s \) collides head-on with another particle of mass \( 1\;\mathrm kg \) moving in the same direction at a speed of \( 1\;\mathrm m/\mathrm s \). What is the final speed of each particle?

We can use the equations derived above for this question. Let the \( 1\;\mathrm{kg} \) particle be particle \( 1 \) and the \( 2\;\mathrm{kg} \) one be particle \( 2 \). The final velocity of particle \( 1 \) is given by

$$v_1=\frac{(m_1-m_2)u_1+2m_2u_2}{m_1+m_2}.$$

The question gives the masses and the initial velocity so they can be substituted into the equation for \( v_1 \) to find its value as

$$v_1=\frac{((1\;\mathrm{kg}\;-\;2\;\mathrm{kg})\times1\;\mathrm m/\mathrm s)+(2\times2\;\mathrm{kg}\times2\;\mathrm m/\mathrm s)}{1\;\mathrm{kg}+2\;\mathrm{kg}}=\frac{-1\;\mathrm{kg\,m}/\mathrm s+8\mathrm{kg\,m}/\mathrm s}{3\;\mathrm{kg}}=\frac{7\;\mathrm{kg\,m}/\mathrm s}{3\;\mathrm{kg}}=2.3\;\mathrm m/\mathrm s.$$

The equation for the final velocity of particle \( 2 \) is

$$v_2=\frac{2m_1u_1+(m_2-m_1)u_2}{m_1+m_2}.$$

We again have all of the quantities required to compute the value of \( v_2 \), which is

$$v_2=\frac{(2\times1\;\mathrm{kg}\times1\;\mathrm m/\mathrm s)+((2\;\mathrm{kg}-1\;\mathrm{kg})\times2\;\mathrm m/\mathrm s)}{1\;\mathrm{kg}+2\;\mathrm{kg}}=\frac{2\;\mathrm{kgm}/\mathrm s+2\;\mathrm{kgm}/\mathrm s}{3\;\mathrm{kg}}=\frac43\mathrm m/\mathrm s=1.3\;\mathrm m/\mathrm s.$$

Kinetic energy is lost in all inelastic collisions. However, kinetic energy is lost the most at completely inelastic collisions where the objects stick together after they collide.

A** completely inelastic collision **is a collision for which the maximum amount of lost kinetic energy happens.

A bullet that comes to a complete stop when fired into a block of wood is an example of a completely inelastic collision.

Consider a completely inelastic collision between two particles, with the same notation as before. However, this time the particles stick together after the collision so they have the same final velocity \( V \). The equation of conservation of momentum for this case becomes

$$m_1u_1+m_2u_2=(m_1+m_2)V.$$

And we can solve for the final velocity of the particles:

$$V=\frac{m_1u_1+m_2u_2}{m_1+m_2}.$$

A car of mass \( 1000.0\;\mathrm kg \) traveling at a speed of \( 20.0\;\mathrm m/\mathrm s \) collides head on with a truck of mass \( 10000.0\;\mathrm kg \) traveling at a speed of \( 10.0\;\mathrm m/\mathrm s \) in the same direction. The two vehicles stick together after the collision. What is their final speed?

The two vehicles stick together after the collision so they have the same final speed which can be calculated from the equation

$$V=\frac{m_1u_1+m_2u_2}{m_1+m_2}.$$

We are given all of the values needed in the question so we can find the speed as

$$\frac{(1000.0\;\mathrm{kg}\times20.0\;\mathrm m/\mathrm s)+(10000.0\;\mathrm{kg}\times10.0\;\mathrm m/\mathrm s)}{11000.0\;\mathrm{kg}}=10.9\;\mathrm m/\mathrm s.$$

A ballistic pendulum is a device used for measuring the speed of a projectile using the conservation of energy and momentum. The projectile is fired into a large block of wood suspended by two cords, which causes the block to swing with the motion of a pendulum. There are angle markings that follow the path of the block's swing so that the height of the swing can be calculated.

Suppose that a bullet of mass \( m \) reaches the block with a speed \( u \) (which is what we want to find) and the wooden block has a mass \( M \). The bullet comes to a complete stop *within *the block, so the collision is completely inelastic. We can use the equation derived above for the final speed of two objects after a completely inelastic collision:

$$V=\frac{m_1u_1+m_2u_2}{m_1+m_2}.$$

Let the speed of the bullet and block after the collision be \( v \) so that this equation gives

$$v=\frac{mu}{m+M}$$

After the impact, the block and bullet together have kinetic energy.

$$\begin{aligned}KE&=\frac12mv^2\\[8pt] KE&=\frac12(m+M){\left(\frac{mu}{m+M}\right)}^2\\[8pt]KE&=\frac12\frac{m^2u^2}{m+M}\end{aligned}$$

As they begin to rise, this kinetic energy is converted into gravitational potential energy. Once the block is at the end of its pendulum motion, at its highest point, \( h \), all of the initial kinetic energy will have been converted to gravitational potential energy.

$$PE_{\text{gravitational}}=(m+M)gh,$$

The potential energy can be equated to the initial kinetic energy of the block and bullet to find the bullet's speed!

$$\begin{aligned}KE&=PE\\[8pt]\frac12\frac{m^2u^2}{m+M} &=(m+M)gh\\[8pt]u^2&=\frac{2{(m+M)}^2gh} {m^2}\\[8pt] u&=\frac{(m+M)}m\sqrt{2gh}\end{aligned}$$

- The law of the conservation of energy states that energy is always conserved in an isolated system.
- Energy is a scalar quantity.
- An isolated system is a system where no energy or matter can enter or leave.

- The linear momentum of an object is a vector quantity defined by \( \vec p=m\vec v \).
- Since momentum is a vector quantity it has a direction as well as a magnitude.
- The law of conservation of linear momentum
- The different components of momentum are conserved independently.
- Kinetic energy is conserved in elastic collisions.
- The maximum amount of kinetic energy is lost in completely inelastic collisions and is characterized by the objects moving together as one after the collision.

- Fig. 1 - a ball will always have the same speed when rolled down a hill of a certain height (https://upload.wikimedia.org/wikipedia/commons/thumb/4/44/Zorb_02.jpg/1280px-Zorb_02.jpg) by Harry Malsch (https://www.flickr.com/people/12229484@N02) is licensed by CC BY 2.0 (https://creativecommons.org/licenses/by/2.0)
- Fig. 4 - the momentum of a fired gun due to recoil balances the momentum of the bullet OGL v1.0OGL v1.0 (https://upload.wikimedia.org/wikipedia/commons/thumb/b/b2/Firing_a_Sig_Sauer_P226_Handgun_MOD_45156988.jpg/1365px-Firing_a_Sig_Sauer_P226_Handgun_MOD_45156988.jpg) by SAC Daniel Herrick is licensed by (https://nationalarchives.gov.uk/doc/open-government-licence/version/1/)
- Fig. 6 - the velocity vector (and hence the momentum vector) can be split up into components along three perpendicular directions (https://upload.wikimedia.org/wikipedia/commons/thumb/e/e7/Velocity-vector-3D-and-components.svg/1831px-Velocity-vector-3D-and-components.svg.png) by MikeRun (https://commons.m.wikimedia.org/w/index.php?title=User:MikeRun&redlink=1) is licensed by CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/deed.en)
- Fig. 8 - "Ballistic pendulum" (https://upload.wikimedia.org/wikipedia/commons/thumb/8/81/Ballistic_pendulum.jpg/1280px-Ballistic_pendulum.jpg) by Steven Keys (https://commons.wikimedia.org/wiki/User:Steevven1) is licensed by CC BY 4.0 (https://creativecommons.org/licenses/by/4.0)

*E* would be *E*_{start}=*E*_{end}, and the formula for conservation of momentum *p* would be *p*_{start}=*p*_{end}.

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