Momentum momentum Momentum and Its Conservation - Lesson 1 - The Impulse-Momentum Change Theorem Momentum Real-World Applications The sports announcer says, "Going into the all-star break, the Chicago White Sox have the momentum." The headlines declare "Chicago Bulls Gaining Momentum." The coach pumps up his team at half-time, saying "You have the momentum; the critical need is that you use that momentum and bury them in this third quarter." Momentum is a commonly used term in sports. A team that has the momentum is on the move and is going to take some effort to stop. A team that has a lot of momentum is really on the move and is going to be hard to stop. Momentum is a physics term; it refers to the quantity of motion that an object has. A sports team that is on the move has the momentum. If an object is in motion (on the move) then it has momentum. Momentum can be defined as "mass in motion." All objects have mass; so if an object is moving, then it has momentum - it has its mass in motion. The amount of momentum that an object has is dependent upon two variables: how much stuff is moving and how fast the stuff is moving.

Momentum depends upon the variables mass and velocity . In terms of an equation, the momentum of an object is equal to the mass of the object times the velocity of the object. Momentum = mass • velocity In physics, the symbol for the quantity momentum is the lower case p. Thus, the above equation can be rewritten as p = m • v where m is the mass and v is the velocity. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity. The units for momentum would be mass units times velocity units. The standard metric unit of momentum is the kg•m/s. While the kg•m/s is the standard metric unit of momentum, there are a variety of other units that are acceptable (though not conventional) units of momentum. Examples include kg•mi/hr, kg•km/hr, and g•cm/s. In each of these examples, a mass unit is multiplied by a velocity unit to provide a momentum unit. This is consistent with the equation for momentum. Momentum as a Vector Quantity Momentum is a vector quantity. As discussed in an earlier unit, a vector quantity is a quantity that is fully described by both magnitude and direction.

To fully describe the momentum of a 5-kg bowling ball moving westward at 2 m/s, you must include information about both the magnitude and the direction of the bowling ball. It is not enough to say that the ball has 10 kg•m/s of momentum; the momentum of the ball is not fully described until information about its direction is given. The direction of the momentum vector is the same as the direction of the velocity of the ball. In a previous unit, it was said that the direction of the velocity vector is the same as the direction that an object is moving. If the bowling ball is moving westward, then its momentum can be fully described by saying that it is 10 kg•m/s, westward. As a vector quantity, the momentum of an object is fully described by both magnitude and direction. The Momentum Equation as a Guide to Thinking From the definition of momentum, it becomes obvious that an object has a large momentum if both its mass and its velocity are large. Both variables are of equal importance in determining the momentum of an object. Consider a Mack truck and a roller skate moving down the street at the same speed. The considerably greater mass of the Mack truck gives it a considerably greater momentum.

Yet if the Mack truck were at rest, then the momentum of the least massive roller skate would be the greatest. The momentum of any object that is at rest is 0. Objects at rest do not have momentum - they do not have any " mass in motion ." Both variables - mass and velocity - are important in comparing the momentum of two objects. The momentum equation can help us to think about how a change in one of the two variables might affect the momentum of an object. Consider a 0.5-kg physics cart loaded with one 0.5-kg brick and moving with a speed of 2.0 m/s. The total mass of loaded cart is 1.0 kg and its momentum is 2.0 kg•m/s. If the cart was instead loaded with three 0.5-kg bricks, then the total mass of the loaded cart would be 2.0 kg and its momentum would be 4.0 kg•m/s. A doubling of the mass results in a doubling of the momentum.

Similarly, if the 2.0-kg cart had a velocity of 8.0 m/s (instead of 2.0 m/s), then the cart would have a momentum of 16.0 kg•m/s (instead of 4.0 kg•m/s). A quadrupling in velocity results in a quadrupling of the momentum. These two examples illustrate how the equation p = m•v serves as a "guide to thinking" and not merely a "plug-and-chug recipe for algebraic problem-solving." Check Your Understanding Express your understanding of the concept and mathematics of momentum by answering the following questions. Click the button to view the answers. 1. Determine the momentum of a ... a. 60-kg halfback moving eastward at 9 m/s. b. 1000-kg car moving northward at 20 m/s. c. 40-kg freshman moving southward at 2 m/s. A. p = 40 000 units (doubling the velocity will double the momentum) B. p = 60 000 units (tripling the velocity will triple the momentum) C.

p = 40 000 units (doubling the mass will double the momentum) D. p = 80 000 units (doubling the velocity will double the momentum and doubling the mass will also double the momentum; the combined result is that the momentum is doubled twice -quadrupled) 3. A halfback (m = 60 kg), a tight end (m = 90 kg), and a lineman (m = 120 kg) are running down the football field. Consider their ticker tape patterns below. Compare the velocities of these three players. How many times greater are the velocity of the halfback and the velocity of the tight end than the velocity of the lineman? See Answer A. The tight end travels twice the distance of the lineman in the same amount of time. Thus, the tight end is twice as fast (vtight end = 6 m/s). The halfback travels three times the distance of the lineman in the same amount of time. Thus, the halfback is three times as fast (vhalfback = 9 m/s). B. Both the halfback and the tight end have the greatest momentum.

The each have the same amount of momentum - 540 kg*m/s. The lineman only has 360 kg*m/s. How is force related to momentum? How is force related to momentum? Asked by: Melissa Thomas Answer Momentum measures the 'motion content' of an object, and is based on the product of an object's mass and velocity. Momentum doubles, for example, when velocity doubles. Similarly, if two objects are moving with the same velocity, one with twice the mass of the other also has twice the momentum. Force, on the other hand, is the push or pull that is applied to an object to CHANGE its momentum. Newton's second law of motion defines force as the product of mass times ACCELERATION (vs. velocity). Since acceleration is the change in velocity divided by time, you can connect the two concepts with the following relationship: force = mass x (velocity / time) = (mass x velocity) / time = momentum / time Multiplying both sides of this equation by time: force x time = momentum To answer your original question, then, the difference between force and momentum is time.

Knowing the amount of force and the length of time that force is applied to an object will tell you the resulting change in its momentum. Answered by: Paul Walorski, B.A., Part-time Physics Instructor Answer They are related by the fact that force is the rate at which momentum changes with respect to time (F = dp/dt). Note that if p = mv and m is constant, then F = dp/dt = m*dv/dt = ma. On the other hand, you can also say that the change in momentum is equal to the force multiplied by the time in which it was applied (or the integral of force with respect to time, if the force is not constant over the time period). Interestingly enough, this, along with Newton's Third law, gives us conservation of momentum. Newton's Third law says that for a force exerted by object 1 on object 2, object 2 exerts a force on object 1 that is equal in magnitude and opposite in direction to the force object 1 exerts. Or, more succinctly, F[1->2] = -F[2->1].

Now the total change in momentum for any interaction is the integral of F[1->2] over time plus the integral of F[2->1] over time, which equals the integral of F[1->2] minus the integral of F[1->2], which equals zero - always! A similar argument for conservation of energy can be made using the fact that energy is the integral of force with respect to position. Answered by: Gregory Ogin, Physics Undergraduate Student, UST, St. Paul, MN Answer Newton's 2nd Law tells us that force = mass x acceleration ( F = ma ). Since acceleration is just how velocity changes over time, we can write this as F = m * v/t Multiply both sides by time to arrive at F t = m v Since mv is momentum, we can see that the momentum conferred to an object by a force equals the force times the time the force is applied. Thus if a 15 Newton force to the right is applied to an initially stationary object for 3 seconds, it will have a momentum of 45 kg m/s to the right. Most students who ask this question are usually trying to figure out the reverse situation, however.

If an object hits me with a certain amount of momentum, how much force does it hit me with? Note that due to Newton's 3rd Law, this can be calculated the same way. If a thrown egg hits your hand with a momentum of 5 kg m/s, the force it applies to your hand depends on the time it takes for your hand to absorb the momentum. If you hold your hand very stiffly (and try to make the egg stop in a very short period of time) the ball exerts a high force on your hand, e.g. 100 N for 1/20th of a second. However as anyone who has ever played in an egg toss knows, if you let your hand 'give' and extend the amount of time it takes to absorb the momentum, the egg exerts a smaller force on your hand, e.g. 10 N for 1/2 a second. Answered by: Rob Landolfi, Science Teacher, Washington, DC If light has no mass, and mass is energy, how does light carry energy? If light has no mass, and momentum is mass multiplied by velocity, how does light carry momentum?

- Quora Quora Classically the energy of a moving object (Kinetic energy) is: [math]KE = \frac{1}{2}mv^2 =\frac{p^2}{2m}[/math] Naively one would think that you could plug in [math]m=0[/math] and [math]v=c[/math] for a photon and find that it has no energy. We could do the same for momentum, [math]p=mv[/math], and find that photons can't push anything either. According to classical mechanics, photons pretty much don't exist! But... we see. Our eyes interact with energy and momentum. How can this be? A photon is a relativistic particle, so classical mechanics is incapable of telling us much. We must examine things from a relativistic perspective. I'll use an old method of teaching this(relativistic mass increase), but we'll get all the right answers!

The method is simple: anytime you see a formula with a mass, [math]m[/math], we replace it with a [math]\gamma m[/math], where [math]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/math] is a relativistic factor. It is equal to one when the velocity is zero, and is very close to one for most velocities we encounter day to day, which is why it wasn't noticed for such a long time in physics experiments. It grows to be infinite as we approach the speed of light. Here's a plot of the relativistic factor as a function of the fraction of the speed of light: For example, a particle at rest has energy [math]E=mc^2[/math].

If the particle is moving then: [math]E = \gamma mc^2[/math] [math]= (1-\frac{v^2}{c^2})^{-\frac{1}{2}}mc^2[/math] [math]\approx (1+\frac{1}{2}\frac{v^2}{c^2}+\frac{3}{8}\frac{v^4}{c^4}+\cdots)mc^2[/math] [math]= mc^2+\frac{1}{2}mv^2+\frac{3}{8}m\frac{v^4}{c^2}+\cdots[/math] (where I used a Taylor expansion in the second line). Look at that! The classical kinetic energy is the smallest correction to the total energy of a moving body! Now one might think that they could plug in [math]m=0[/math] into the energy formula, but hold your horses. For a photon [math]v=c[/math], and the relativistic factor is infinite. Multiplying [math]0\cdot \infty[/math] requires a bit of caution.

To get to the heart of the matter we have to also use the concept of relativistic momentum, [math]p=\gamma mv[/math]. Together, energy, mass, and momentum form a triad of physical quantities, intertwined with one another. I will let the interested reader use the formulas for relativistic energy and momentum to derive the famous Einstein Energy-Momentum relation: [math](\frac{E}{c})^2 = (mc)^2+p^2[/math] If anything, this formula should look a little familiar. It looks just like the Pythagorean theorem! Energy (divided by c), plays the role of the hypoteneuse, while mass (times c) and momentum play the role of the legs of the right triangle. One can interpret this as a particle that has some total energy must divy it up amongst an energy that flows through time (mass) and an energy that flows through space(momentum) (I'm going to disregard factors of c from now on). Since energy MUST be divied up this way, we begin to see what photons are like (no pun intended). A massive object at rest only has energy flowing through time, and we call that energy mass.

A moving massive object has both mass and momentum, energy flowing through both space and time. A massless object only has energy flowing through space, it solely has momentum! Photons are weird like that: they are timeless. A photon with a pocketwatch would never see the hands on the clock move. To a photon, everything happens at once because there is no before or after. This is the world a photon, and all other massless particles, experience according to special relativity. Rick A. Baartman , Physicist, designs particle accelerators. The energy and momentum of particles have very little to do with mass. All particles are waves, and therefore have wavefunctions. The energy of a particle is given by how quickly its wavefunction oscillates in time, and the momentum of a particle is given by how quickly its wavefunction oscillates in space. Note that this gives a photon (a particle of light) energy and momentum for the exact same reason that an electron (or any other particle) has energy and momentum.

In quantum mechanics, the energy operator (or Hamiltonian) is the derivative of the wavefunction [math]\psi[/math] with respect to time: [math]\hat{E}=i\hbar \frac{\partial}{\partial t}[/math], which is simply mathematics for saying that the energy of a particle is how quickly its wavefunction changes in time ([math]t[/math]). So to find the energy of a photon, we take the temporal derivative of its wavefunction. That is, we act on it with the energy operator and find the eigenvalue: [math]\hat{E}\psi = i\hbar \frac{\partial}{\partial t}\psi[/math] The wavefunction of a photon is a plane-wave: [math]\psi = e^{ikx-i\omega t}[/math] where [math]k=2\pi/\lambda[/math] is the wavevector ([math]\lambda[/math] is the wavelength) and [math]\omega=2\pi f[/math] is the angular frequency ([math]f[/math] is the frequency).

So: [math]\hat{E}\psi=i\hbar \frac{\partial}{\partial t}e^{ikx-i\omega t}[/math] [math]\hat{E}\psi = \hbar \omega\psi[/math] which gives the energy of the photon as [math]\hat{E}=\hbar\omega =hf[/math]. In a similar way, the momentum operator is given by the derivative of the wavefunction [math]\psi[/math] with respect to space: [math]\hat{p}=-i\hbar \frac{\partial}{\partial x}[/math]. So, to find the momentum of a photon, we take the spatial derivative of its wavefunction. That is, we act on it with the momentum operator and find the eigenvalue: [math]\hat{p}\psi = -i\hbar \frac{\partial}{\partial x}\psi[/math], which is simply mathematics for saying that the momentum of a particle is how quickly its wavefunction changes in space ([math]x[/math]). The wavefunction of a photon is still a plane-wave: [math]\psi = e^{ikx-i\omega t}[/math].

So: [math]\hat{p}\psi=-i\hbar \frac{\partial}{\partial x}e^{ikx-i\omega t}[/math] [math]\hat{p}\psi = \hbar k\psi[/math] This gives the momentum of a photon as [math]\hat{p}=\hbar k = h/\lambda[/math]. Written Jan 13, 2015 Neither of your statements are correct, though they do represent the way things work in our everyday lives. Mass and energy are two aspects of a common thing, mass/energy. You can view mass as crystallized energy. Photons carry energy in its uncrystallized form, massive particles in the crystallized. The energy content of mass is m*c^2, the energy content of a photon is h*nu, where nu is the frequencey of the photon. Momentum can be carried in both forms. In the mass form, p = mv. In a photon, p = h/lambda, where lambda is the wavelength. Dynamics | Physics For Idiots Physics For Idiots 7 Projectile Motion Dynamics is the name give to the rules of motion.

It’s something that you would think would be one of the first things to be figured out, but wasn’t fully locked down until fairly recently. That being said the rules have’t changed much and are fairly predictable, at least on large scales. Someone once told me that all you needed to know for a Dynamics exam was: and everything else could be derived from that. I never found out if they was right, I learnt these ones too, just in case: If you’re already familiar with the equations you might want to skip ahead to the next section, otherwise I’ll explain where they came from and how to use them. When dealing with measurements you can use scalar or vector quantities. Scalar quantities: Energy, Length, Mass, Speed, Temperature and Time are all scalar quantities. Vector quantities: Have both a magnitude and a direction Displacement, Force, Velocity, Acceleration and Momentum are all vector quantities. It might seem sometimes that speed and velocity are the same thing (often they’re equal to each other), but they are actually slightly different. Speed is just how fast something is going, it doesn’t matter if it’s going up, down, left or right, all that matters is how far it travels in a set time.

Probably the best way to consider velocity is if you think or an ordinary x,y-axis. If a body is travelling horizontally in a straight line with a speed of 10 then stops and goes in the exact opposite direction, with a speed of 10 there has obviously been a change, however the speed does not reflect this. The speed before it turned around is the same as after. The velocity however is not the same. If we said that the velocity to begin with was the same as the speed: 10 , then when the body is travelling in exactly the opposite direction, with the same speed, the velocity would be –10 . Newtons Laws Isaac Newton was a clever bloke. We have him to thank for gravity (I should probably add he discovered, not invented it, otherwise people will start blaming him every time they fall over). The things Newton is most famous for (aside from the incident with the apple) are his laws of motion: A particle will remain at rest or continue with its motion, unless acted upon by an external force. The force on an object is equal to its mass multiplied by its acceleration ( ). Every action has an equal and opposite reaction. All of this is fine, but what do these laws really mean?

1. A particle will remain at rest or continue with its motion, unless acted upon by an external force. This just means that provided no external force acts upon a particle it won’t change it’s motion in any way. If there was no friction or air resistance, then a particle moving at 5 would carry on indefinitely. Obviously in real life this does not happen as there are air resistance and friction, so it’s almost impossible to have no external force on a moving particle. However, if you think about a stationary particle, this makes far more sense. Unless a force is applied to a stationary particle, it doesn’t start moving. 2. The force on an object is equal to its mass multiplied by its acceleration. More simply known as , this is probably one of the most fundamental formulae in Dynamics. It’s one of those which pops up all over the place in Dynamics and is a really good idea to learn. It’s not too difficult to understand either. It makes sense that if something has a greater mass, it would take a larger force to give it the same acceleration as something with less mass. 3.

Every action has an equal and opposite reaction This law basically means that if you push against a wall, it pushes you back, which is a good job really because otherwise you’d go straight through! Constant Acceleration Formulae These go by so many different names it’s hard to keep up sometimes. You may have heard them referred to as kinematic equations, equations of motion, SUVAT equations, or maybe you’ve not heard of them at all. First of all lets take a look at them: (1) (4) (5) There may seem a lot to remember there, but believe me, it’s not as difficult as it seems. Like these equations are incredibly important in Dynamics. SUVAT Equation 1 As you probably already know, velocity divided by time is equal to acceleration and velocity multiplied by time is equal to displacement. This means that on a speed vs. time graph, the gradient of the line is equal to acceleration and the area under the line is equal to displacement. If you have an initial velocity and a final velocity the graph would look something like this: Plot showing u against t As I previously said, the gradient of the line is equal to acceleration .

So the subject, gives us our first constant acceleration formula: SUVAT Equation 2 Okay, so one down, just four to go! We know that the area under the graph is equal to the displacement. So we know that multiplied by gives us the bottom rectangle of the area and divided by 2, gives us the top triangle. This gives us: Now we already know that so we can rearrange that to give and then substitute this into our equation for displacement. From this we have . If we just multiply out the bracket that provides us with our second formula: For those of you who like to find maths where they can, you might be interested to know that is the integral of with respect to . If that makes no sense to you, why not go take a look at the wonderful Integration section, where all will become clear! SUVAT Equation 3 Now those of you who are keen on spotting patterns may have noticed that this equation looks a lot like the last one. This is because it’s a lot like the last one. Those of you who decided not to go to the Integration page might regret that now.

If you rearrange the subject you get: Now you just need to integrate this result with respect to time to give you our 3rd equation: We’ve already established that the area under the graph (equal to displacement ) is equal to: If we multiply out the bracket we get: which is the same as: Finally we just factorise this to give: this eventually give us the final form of And that’s that! These equations are definitely worth learning since they’re useful time and time again. There are a couple of rules, for instance they can only be used in cases where there is constant acceleration. This means that if the acceleration is something like 12 ms-2 they’re fine, but if the acceleration is along the lines of 12 ms-2 then they won’t work since the acceleration varies with . What a Drag A lot of dynamics is done by neglecting air resistance, and while this makes things a lot easier to deal with it’s always worth knowing what affect it would have.

For any object moving in a fluid the drag force on it can be calculated using: is the density of the fluid (998.2071 kg m for water at 30 degrees and 1.204 kg m for air), is the velocity of the object, is the objects cross sectional area and is the drag coefficient. The drag coefficient it a number that relates to how aerodynamic the object is, with a cube having and a sphere having . An object falling to the Earth it will eventually (if it’s falling for long enough) reach a speed at which the drag force equals the force of gravity pulling it down. This is called Terminal Velocity and you can get an expression for it by equating the drag force to and then rearranging for For a human falling through the air ( from above) we have 70kg, 0.5m area and a drag coefficient of about 0.8 (rough guess somewhere around a angle cube or cylinder) we get a terminal velocity of around 53 m s (which turns out to be a pretty good rough estimate). Motion on a flat surface This is the most simple instance in dynamics. A body moves along a flat surface in a straight line.

For example: 1. The Rev is driving his car, when suddenly the engine stops working! If he is travelling at 10 ms-1 and his deceleration is 2 ms-2 how long will it take for the car to come to rest? Okay, it’s always a good idea with these kind of problems to list what you know. We’re given initial velocity, , and acceleration, . We also know that if the car is going to finish at rest, that final velocity, , must be 0ms-1. We want to find out time, . Personally I find it’s best to lay this information out like this: u = 10ms-1 a = -2 ms-2 t = ? s From this we can see which equation we need. In this case we can see that the equation we want is . We rearrange this to make the subject, giving us Finally, we put the numbers into the equation: . 2. Michael steps into the road, 30 metres from where the engine stops working. The Rev's glasses have fallen off and he does not see Michael. Will The car stop in time to miss hitting Michael?

Once again, it’s best to lay out all the information we have: u = 10 ms-1 t = 5 s s = ? m This time we want to find displacement, s, so we need to pick an equation with that in. I’m going to use . I could have used or , however, since we’re not given time, but have instead worked it out ourselves, any error made in previous calculations would be carried forward into this one. Again, I’m going to rearrange the equation, this time to give as the subject. This is a good habit to get into, it might not make much difference now whether you rearrange the equation before or after putting numbers in, but with more complex formulae it can get really messy if you don’t rearrange it first. Also in exam situations, if you make a mistake, you may still get method marks if the examiner can see what you’ve done. Anyway, this gives us Putting the numbers into the equation gives us: , so Michael will not be hit! (Phew!) In the above example, friction has been completely ignored.

In the real world we can’t do this (quite fortunate really because we’d fall over all the time and people would think we were drunk). So now we’d better have a look at a situation with friction. The coefficient of friction is given the symbol μ. The resultant (normal) force of the weight, balances the weight of the car (so it doesn’t go through the road). The force due to friction is μ (or μN). 3. The Rev's car has broken down on the M1. He needs to push it to the hard shoulder. The car weighs 5000N. The Rev can push about 1800N. The coefficient of friction between the car and the road is 0.6. Will The Rev be able to push the car to the hard shoulder? Okay, so firstly in a situation like this it’s good to draw a little sketch of what’s going on. A force diagram to show what is going on in example 3. From this we know that in order for the car to move, The Rev must be pushing with a force of at least μR.

By simply multiplying the coefficient of friction by the resultant force, we find that the force due to friction is 3000N, so The Rev won’t be able to push the car to the side of the road. 4. A body-builder happens to be passing by and in an attempt to ease congestion on the ever busy M1, he decides to help. He can push with a force of 3200N. With both the body-builder and The Rev pushing, what will be the acceleration of the car? NB – Take the mass of the car to be 510kg So, same situation as before really, only this time the forces don’t balance and so there will be an acceleration. We get this from the oh-so-clever Isaac Newton’s, . Remember that to find the overall force you need to take away the frictional force. So that’s (3200 + 1800) – 3000. That leaves an overall force of 2000N. Again we need to rearrange the formula to give a as the subject this time. This gives us .

Putting the numbers in we get: a = 3.9 ms-2 (2 s.f.) Motion on a Inclined Plane This is pretty similar to motion on a flat surface, just one or two more variables… oh and we won’t be talking about The Rev’s car any more since I’m not sure that would make it up a hill! Anyway, I fear I may have wandered off track a little there. Introducing the “inclined plane”, or “slope” as it’s known to most of us, means you’re going to have to brush up on your trigonometry. On the plus side, you’ll find out why people have been trying to drill it into you for years! Provided you’re familiar with the good old order of operations you should be fine. So lets start with a nice easy example. Inclined plane example The picture above shows us a block sitting on a slope. A good place to start with this (probably the only place to start if you want a chance in hell of getting anywhere with a question) is to resolve forces.

Assuming that the block is at rest, we know that it is in equilibrium, so the horizontal forces must be equal and so must the vertical forces (unless it’s one of those lovely levitating blocks). Projectile Motion Projectiles aren’t completely dissimilar to Motion in a straight line, just instead of a body just moving from left to right, it’s going up or down aswell. First lets have a look at a typical example of projectile motion: A ball is thrown at an an angle of 30°. It has an initial velocity of 20ms-1. Find the maximum height the ball would reach. Okay so as per usual, we draw a diagram: Projectile Motion example Now lets list what we know: u = 20 sin30 ms-1 a = -9.81 ms-2 s = ? m Now we pick one of the kinematic formulae, the one which is going to get us the result in the most direct way is: , and rearrange it to make the subject: Linear Momentum Cut to the Chase Momentum Momentum is defined as the mass multiplied by the velocity. p= mv. Since the momentum depends on the velocity, it is also a vector quantity.

The units for momentum are [kg.m.s-1] Momentum is the mass multiplied by the velocity. p=mv. The rate of change in momentum, we explained in the section on Newton's laws of motion is another definition of the force. If we measure the momentum of a object over a small period of time, F = (mv - mu)/δt. Impulse Taking the time to the left-hand-side of the equation, F δt = mv - mu. The force multiplied by the time has a special name, the impulse. The impulse is the change in the momentum. Impulse is a vector with units of [kg.m.s-1] or since [kg.ms-2] is equivalent to a Newton [N], the units of impulse are [N.s]. Impulse is the force multiplied by time the force acts on an object. I=Ft Consider a ball that is thrown into the air. We can make the ball go higher in two ways, firstly, we can apply a larger force, F or we can make the same force act on the ball for a longer time.

If the force acting on the object is not constant over the time for the time in which the force acts, the impulse is given by the integral of the force with respect to time. Conservation of Momentum The conservation is an important principle in physics that allows us to determine what happens in collisions or explosions. If a body A of mass mA and velocity uA collides with another body B of mass mB and velocity vB moving in the same direction. If A exerts a force F to the right on B for a time t then by Newton's third law, body B will exert an equal and opposite force to the left on A also for a time t but to the left. Thus the bodies receive equal but opposite impulse Ft. The changes of momentum must be equal and opposite. Therefore, the total momentum change is zero. In other words, the total momentum of A and B together remains constant. In mathematical terms: mAuA+ mBuB= mAvA+ mBvB This is the general case. For motion in 1-dimension, we can consider the momenta in one direction only which allows us to drop the vector notation.

It is important to release that the momentum is conserved only when there are no external agent acting on the system that could add momentum to the system. They must be included in the system before the conservation of momentum can be applied. A good example of this is a ball that is dropped from a height. If we are considering the ball alone, then the momentum is mbvb before the ball collides and -mbvb when the ball hits the Earth. The change in momentum is 2mbvb not zero. It appears to gain momentum from nowhere. However, we need to consider the change in momentum of the Earth. It moves with a velocity vE after the collision. The conservation of momentum, mbvb + mE.0 = mb(-v)b+mEvE. The Earth has to gain the same momentum as the ball but because the mass of the Earth is so large, the movement is unmeasurable. Elastic and Inelastic Collisions Momentum is always conserved in any closed system, however kinetic energy is not always conserved and this brings us neatly to the subject of elastic and inelastic collisions. An elastic collision is a collision where the both the momentum and the kinetic energy are conserved.

In an inelastic collision the momentum is conserved but the kinetic energy is not. In fact the kinetic energy will be lower after the collision because some energy has been used in creating sound, deformation and heat. The reality is that most collisions are inelastic to some extent. However, collisions between snooker balls come quite close to being elastic collisions. Collisions in 1-d. Flash. Change the masses mA, mB of the trolleys and the initial velocity uA press play to watch the collision. In the flash animation above, the conservation of momentum is illustrated by the collision between two trolleys A and B. Initially, trolley B is stationary and trolley has a velocity vA also the masses mA and mB are the same. Applying the conservation of momentum gives, mA uA+ 0 = mAvA + mBvB. In the event that the two masses are the same, then mass mB will move off at the same speed and mass mA will stop. To calculate what happens when the masses are different we need to use a second equation. uA = vA+vB. This states that the incoming velocity is equal to the sum of the final velocities.

Using this equation we obtain a general solutions for vA and vB. When both masses are allowed to move, their final velocities are given by: Limiting Cases MA>>MB In the limiting cases where a large mass collides with a much smaller mass, the final velocity of the small mass has a limiting value of twice the velocity of the large mass. This can be seen by looking at equation (3) since mA+mB &approx; mA then vb=2 (mA/mA) uA = 2uA. MA<<MB In the case where a small mass collides with a much larger mass, the large mass hardly moves and the small mass rebounds with a velocity similar to its entry velocity. Looking at equation (4) this can be seen mA-mB~-mB and mA+mB~mB therefore, vA~-1 uA. Totally inelastic Collisions So far we have considered the elastic collisions where both the kinetic energy and momentum are conserved. Most collisions are inelastic. How do we deal with those? We can measure the how much of the energy remains as kinetic energy after the collision. We take the ratio of the outgoing kinetic energy to the incoming kinetic energy. This is known as the coefficient of restitution.

e=kinetic energy after collision/kinetic energy before collision. Explosions In an explosion, the inital momentum is zero, after the explosion the outgoing pieces move out in opposite directions so the momentum is conserved. The idea of an explosion is replicated somewhat less dramatical with two trolleys A and B with masses mA and mB respectively, which are initially held together. Momentum is given to the trolleys by a compressed spring. When the spring is released the momentum is given to the trolleys which move apart in opposite directions. Applying the conservation of momentum, the momentum before the explosion is 0 since neither trolley is moving and after the explosion the momentum is mAvA-mBvB. Therefore, the change in momentum is mAvA-mB-0 or mAvA=mBvB. Flash Animation 2. Change the masses and potential energy stored in the spring. Play releases trolleys. Stop button resets the trolleys. As the mass of trolley A is less than trolley B it moves away at a faster velocity.

Experimental Confirmation of the Conservation of Momentum The predictions of the conservation of momentum can be verified experimentally using gliders on an air track which essentially removes the effects of friction from the experiment. The gliders can be arranged to collide with one another. The glider's mass can be changed by adding additional mass. The velocity of the gliders is determined using a photogate which measures the time for the glider to pass through a beam of light. Two photogates are required to determine both of the glider's velocities. what Is equal to mass multiplied by velocity? what Is equal to mass multiplied by velocity? mass multiplied by velocity is momentum Get an answer what Is equal to mass multiplied by velocity? Original conversation User: what Is equal to mass multiplied by velocity? Weegy: mass multiplied by velocity is momentum Expert answered| kakakamal |Points 28| User: will Fluid speed up as it passes into constricted pipe because of the pressure difference? There are no new answers. Comments Weegy: A person has 20 coins, [ all nickels and dimes. x= number of nicle y= number of ...

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Question: Mass multiplied by velocity gives what?
Answer: Momentum