Prelecture video impulse and momentum relationship C (Because the video only follows the free fall portion of the jump, the correct of the prelecture video, we outlined some of the key properties of momentum. . Which equation(s) correctly states the principle of conservation of momentum?. ANSWER: Correct Prelecture Reading Question Part A The quantity that is called What is the impulse–momentum relationship? ANSWER: Correct Video: Decreasing Momentum Over a Short Time Watch the video and then answer the. Learning Goal: To apply the law of conservation of energy to an .. Learning Goal: To learn about the impulse-momentum theorem and its.

My initial velocity is not This is 10 meters per second to the left, and momentum is a vector, it has direction, so you have to be careful with negative signs here. This is the most common mistake. People just plug in positive 10, then get the wrong answer.

But this ball changed directions, so the two velocities here have to have two different sides, so this has to be a negative 10 meters per second, if I'm assuming rightward is positive. This leftward velocity, and this leftward initial velocity, has to be negative And, if you didn't plug that in, you'd get a different answer, so you gotta be careful.

Physics Lecture 12 Today’s Concepts: a) Elastic Collisions - ppt video online download

So, what do I get here if I multiply this all out? I'm gonna get zero, no, sorry, I'm gonna get one kilogram meters per second, minus a negative two kilogram meters per second, and that's gonna give me positive three kilogram meters per second is the impulse, and that should make sense.

Impulse and Momentum

The impulse was positive. The direction of the impulse, which is a vector, is the same direction as the direction of the force. So, which way did our face exert a force on the ball? Our face exerted a force on the ball to the right. That's why the impulse on the ball is to the right. The impulse on this person's face is to the left, but the impulse on the ball is to the right, because the ball was initially going left and it had a force on it to the right that made it recoil and bounce back to the right.

That's why this impulse has a positive direction to it. Now, if you've been paying attention, you might be like, wait a minute, hold on. What we really did was we found the change in momentum of the ball, and when we do that, what we're finding is the net impulse on the ball. In other words, the impulse from all forces on the ball. But what this question was asking for was the impulse from a single force. The impulse from just the person's face. Now, aren't there other forces on this ball? Isn't there a force of gravity? And if there is, doesn't that mean what we really found here wasn't the impulse from just our face, but the impulse from the person's face and the force of gravity during this time period?

And the answer is no, not really, for a few reasons. Most important reason being that, what I gave you up here was the initial horizontal velocity. This 10 meters per second was in the X direction, and this five meters per second, I'm assuming is also in the X direction. When I do that, I'm finding the net impulse in the X direction, and there was only one X directed force during this time and that was our face on the ball, pushing it to the right.

There was a force of gravity. That force of gravity was downward. But what that force of gravity does, it doesn't add or subtract any impulse in the X direction. It tries to add impulse in the downward direction, in the Y direction, so it tries to add vertical component of velocity downward, and so we're not even considering that over here. We're just gonna consider that we're lookin' at the horizontal components of velocity. How much velocity does it add vertically, gravity? The particle is said to be in a state of equilibrium. This implies that the x component of the force on a particle at this location is negative, or that the force is directed to the left, just like at A.

What can you say now about the acceleration? This implies that the x component of the force on a particle at this location is positive, or that the force is directed to the right, just like at C.

What can you now say about the acceleration? Part G Name all labeled points on the graph corresponding to unstable equilibrium.

List your choices alphabetically, with no commas or spaces; for instance, if you choose points B, D, and E, type your answer as BDE. BF Part H Name all labeled points on the graph corresponding to stable equilibrium. Think of a ball rolling between two hills.

Impulse and momentum dodgeball example

DH Part I Name all labeled points on the graph where the acceleration of the particle is zero. BDFH Your answer, of course, includes the locations of both stable and unstable equilibrium. Part J Name all labeled points such that when a particle is released from rest there, it would accelerate to the left. If the acceleration is to the left, so is the force.

If the x component of the force at a point is negative, then the derivative of at that point is positive. Of these three points, which one corresponds to the greatest magnitude of acceleration of the particle?

As a reminder, on this graph, the total energy is shown by the horizontal line. Part L What point on the graph corresponds to the maximum kinetic energy of the moving particle? D It makes sense that the kinetic energy of the particle is maximum at one of the force equilibrium points. For example, think of a pendulum which has only one force equilibrium point--at the very bottom. Part M At what point on the graph does the particle have the lowest speed?

B As you can see, many different conclusions can be made about the particle's motion merely by looking at the graph. This problem should prove useful in improving such an understanding. Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length 20 m that makes an angle of with the vertical, steps off his tree limb, and swings down and then up to Jane's open arms. When he arrives, his vine makes an angle of with the vertical.

Part A Calculate Tarzan's speed just before he reaches Jane. You can ignore air resistance and the mass of the vine.

The platform and the pulleys have negligible mass. Assume that there are no friction losses. Part A Find the force he must exert. Take the free fall acceleration to be. Part B Find the increase in the energy of the system when he raises himself a distance. Answer by calculating the increase in potential energy. Part C Find the increase in the energy of the system when he raises himself. Answer by computing the product of the force on the rope and the length of the rope passing through his hands.

When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope. Part A What is the speed of the block as it slides along the horizontal surface after having left the spring? A car in an amusement park ride rolls without friction around the track shown in the figure.

It starts from rest at point at a height above the bottom of the loop. Treat the car as a particle. Part A What is the minimum value of in terms of such that the car moves around the loop without falling off at the top point? Part A If frictional forces do of work on her as she descends, how fast is she going at the bottom of the slope? Take free fall acceleration to be. Part B Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is.

If the patch is of width and the average force of air resistance on the skier ishow fast is she going after crossing the patch? Part C After crossing the patch of soft snow, the skier hits a snowdrift and penetrates a distance into it before coming to a stop. What is the average force exerted on her by the snowdrift as it stops her? Initially the truck is moving downhill at speed. After careening downhill a distance with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle. The truck ramp has a soft sand surface for which the coefficient of rolling friction is. Part A What is the distance that the truck moves up the ramp before coming to a halt? By slowly varying the force, a block with weight is moved through the anglaand the spring to which it is attached is stretched from position 1 to position 2.

The spring has negligible mass and force constant. The end of the spring moves in an arc of radius. Calculate the work done by the force. The coefficients of friction between the package and the incline are and. The mass of the spring is negligible. Part A What is the speed of the package just before it reaches the spring? And your sudden decision. Suppose you had waited twice as long to make up your mind.

Would you still have been acting on an impulse?

Chapter 6 Momentum and Impulse - PDF

The idea of momentum conveyed by the examples above does imply the motion of something and also has a quantitative quality. The idea of impulse conveys the feeling of quickness, of doing something in a short amount of time. How are such ideas related to the use of the definitions of momentum and impulse in physics? Are they similar or contradictory? We hope you will be able to answer that question when you finish this chapter. Block A will eventually come in contact with block B.

In accord with Newton's third law of motion, these two forces are equal in magnitude and oppositely directed. These forces are not constant. Both before and after contact these forces are zero.

The magnitude of the net forces varies with time by starting at zero at the instant of contact, t 1increasing to some maximum value, and then decreasing again to zero at the instant the blocks lose contact. A force of this type which is a function of time is called an impulsive force. Can you name or give examples of impulsive forces?

One example is kicking a football see Figure 6. The product of the net force and the time interval over which it acts is called the impulse of the force. This is represented in Figure 6.

Because F AB and F BA are equal in magnitude and opposite in direction, and the time of contact is the same for each of the bodies, the impulse of F AB is equal in magnitude and opposite in direction to the impulse of F BA. If the force is constant, the area under the curve is given by the product of the magnitude of the force times the change in time, F t 2 - t 1 where t 1 and t 2 are the times that contact begins and ends, respectively.

The impulse of a constant force is given by F t 2 - t 1. The units of impulse are newton-second N-sec in the SI system. Impulse is a vector quantity. So impulse has the same dimensions as the product of mass times velocity. The direction of impulse is determined by the direction of the net force, and the direction of momentum is given by the direction of the velocity.

An impulse may be applied to a body to change its momentum. In fact the impulse applied to a body is equal to its change in momentum: A body of mass 1. What is the impulse of the force? What is the change in momentum of the body and the change in velocity? The impulse of the force is represented by the triangular area. The current state of knowledge of human tolerances for impacts is not very complete.

In fatal collisions a head injury is usually the cause of death. If the head is subjected to an impulse of about N-sec in an abrupt collision, death may ensue. Consider a motorcyclist riding without a helmet to cushion his fall. Assume the mass of his head is 6 kg. If he has an abrupt collision, how slow must he be going to survive? Given the force-time graph for a sprinter's start as shown in Figure 6. After impact the ball is traveling with a velocity of What is the impulse imparted to the ball by the bat?

Let us take the direction of the velocity after impact v f as positive. The change in momentum is mv f - mv i where v i is the initial velocity. The above Equation 6. The relationship is called the principle of conservation of linear momentum. Remember that momentum and impulse are vector quantities. Glancing collisions occur more frequently than head-on collisions, and the situation is complex because of the vector properties of momentum. The total momentum of a system can only be changed by the action of external forces on the system.

The internal forces produce opposite and equal changes in momentum that cancel. Thus, we conclude that the total momentum of an isolated system is constant in magnitude and direction. Suppose body A is moving in the xdirection with velocity v Ai and collides with body B which is at rest.

The xcomponents of momentum are equal before and after the collision. A body A of mass 2. After collision, A is moving with a velocity of 2. What is the velocity of B after the collision? However, the same is not always true of kinetic energy. If the kinetic energy of the bodies involved in a collision is the same before and after impact, the collision is said to be a perfectly elastic collision. If the bodies stick together after impact, the collision is said to be completely inelastic.

These represent the two extreme cases. All intermediate cases are possible and are called partially elastic collisions. We have seen many examples of a partially elastic collision. Consider a collision between two automobiles, see Figure 6. Momentum is conserved, but the kinetic energy before the collision is greater than the kinetic energy after the collision.

Evidence of this is the change in shape of the vehicles as a result of the collision. These "transformations" require an expenditure of energy which comes from the kinetic energy of the system.