Unit 4 Energy and Momentum

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## Unit 4 Energy and Momentum

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**Unit 4 Energy and Momentum**This unit is a continuation of the study of mechanics that trainees studied in Units 2 and 3. It involves looking at mechanics from a different point of view (using work end energy) that can simplify some otherwise complicated situations.**Module 4.1 – Momentum and Impulse**Momentum and Impulse is another way of looking at the motion of objects, and is a continuation of the last unit. There are some situations where it is difficult to determine all of the individual forces acting on the objects. Looking at these situations from the point of view of momentum and impulse can allow us to more easily analyze the interaction between objects and to predict their behaviour.**Module 4.1 Objectives**• To be able to calculate the momentum of an object. • To understand the relationship between change in momentum and impulse. • To be able to apply conservation of momentum to a variety of one-dimensional collisions and explosions.**Momentum**• Momentum depends on mass and velocity • Momentum is a vector and is defined as: • where momentum p is measured in kg·m/s when mass is measured in kg and velocity is measured in m/s**Example**Calculate the momentum of a 1200 kg car travelling at 90.0 km/h north. Solution:**Check Your Learning**• Which has more momentum, a falling raindrop or a transport truck parked along the side of the road? Since the truck is not moving (using the Earth as the frame of reference), it has zero momentum; the raindrop does have momentum, since it is actually moving. Even though the truck has a much larger mass, it does not have any momentum.**Check Your Learning**• Calculate the momentum of a 0.160 kg cricket ball that has a velocity of 110 km/h. The positive sign on the momentum indicates the direction of the momentum. The velocity could also be left in km/h, which would result in a momentum of**Change in Momentum**• Change in velocity causes change in momentum • This product is referred to as the impulse**Impulse**• Change in Momentum and Impulse are equivalent: • an impulse of 1 N·s will provide a change in momentum of 1 kg·m/s.**Collisions**• Collisions usually occur in a short time; force changes quickly • The area of the graph represents the impulse**Application**• Smaller time Larger force • Larger time Smaller force since impulse is usually fixed**Example 1**An egg when thrown against a brick wall will break; however, an egg thrown at the same speed into a sagging sheet will likely survive. Explain why using the concepts of impulse and momentum. Solution: The change in momentum of the egg is the same in each case, since the mass of the egg does not change and the egg goes from the same initial velocity to rest in each case. Since the change in momentum is the same, the impulse must also be the same; however, the brick wall applies the impulse in a very short time, which means the force must be large. The large force causes the egg to break. The sheet brings the egg to a stop more gradually, thereby lengthening the time that the impulse is applied over. This allows a smaller force to be applied to the egg and does not cause the egg to break.**Example 2**A tennis player hits a ball with a racket, giving it a speed of 43 m/s. If the ball has a mass of 0.085 kg and is in contact with the racket for 4.5×10-3 s, what is the average force exerted on the ball by the racket?**Check Your Learning**A 0.15 kg cricket ball is approaching a bat with a speed of 95 km/h. After it is hit by the bat, it has a speed in the opposite direction of 115 km/h. The interaction between the bat and the ball lasted 0.015 s. • What was the change in momentum of the ball? It is important to remember that since the ball changes direction, the final velocity must be made negative (if we take the ball’s original direction to be positive)**Check Your Learning**• What impulse was applied to the ball? Since impulse is equivalent to change in momentum, What was the average force exerted on the ball?**Conservation of Momentum**• Law of Conservation of Momentum - the total momentum of an isolated system remains constant.**Example 1**Someone throws a heavy ball to you when you are standing on a skateboard. You catch it and roll backward with the skateboard. If you were standing on the ground, however, you would be able to avoid moving. Explain both using momentum conservation. Solution: Consider first the case of the person standing on the skateboard when he catches the ball. In this case there is very little friction, so little that it can be ignored compared to the force exerted on the person by the heavy ball. If we define our system to be the ball and the person on the skateboard, conservation of momentum says that the total momentum must stay the same. Before the collision of the ball with the person, the ball has momentum since it is moving. After the person catches the ball, the ball and the person act as one object. The combined momentum of the ball and the person must still be equal to the momentum of the ball before the collision; therefore, the person and the ball must be moving.**Example 1**Now consider the case of the person standing on the ground. If we again define our system to be the ball and the person, there is now a significant outside force acting on the person since friction is now increased. Since there is a significant outside force, the system is no longer isolated and conservation of momentum does not apply. We can, however, define a different system where conservation of momentum will apply. If we now consider our system to be the ball, the person, and the Earth then the friction between the person and the Earth is now an internal force and our system is again isolated. Now the momentum of the ball, person, and the Earth (since they can be considered to be one object after the collision) will be the same as the momentum of the ball before the collision; however, because this combined mass is so much larger than the mass of the ball alone the velocity will be so small that it will be negligible.**Example 2**In the previous example of the person catching the ball while on the skateboard, assume that the ball has a mass of 6.0 kg and is thrown with a speed of 3.4 m/s. The combined mass of the person and the skateboard is 68.0 kg and the person is initially at rest. Find the velocity of the person on the skateboard after catching the ball. Solution: Before the collision, only the ball is moving. After the collision, the ball, person, and the skateboard are moving as one object. It is the total velocity of this combined mass that we are trying to find**Example 3**The person on the skateboard is now standing on the skateboard at rest and holding the heavy ball. The mass of the person and the skateboard is still 68.0 kg and the mass of the ball is still 6.0 kg. The person throws the ball, giving it a velocity of 4.2 m/s to the right, and is amazed to see that he begins rolling to the left. What is the person’s velocity after throwing the ball? Solution: This is an example of an “explosion”. The initial momentum of the system is zero, and must remain at zero since it is an isolated system. We will take right to be the positive direction.**Check Your Learning**A 455 g rubber ball is thrown at a 2.10 kg block with a speed of 5.0 m/s. Assume that the block is at rest on a frictionless surface. The rubber ball rebounds in the opposite direction with a speed of 3.2 m/s. What is the speed of the block after the collision?**Module Summary**In this module, you have learned that • Momentum is calculated using the formula • The change in momentum can be calculated from the change in velocity • Impulse is equal to the change in momentum and is given by the formula**Module Summary**• Momentum is conserved in all isolated systems. This is known as the law of conservation of momentum**Module 4.2 Work, Power and Energy**Until now, forces and acceleration have been used to describe and understand motion. We will now look at another model that is used to analyze motion – work and energy. This new model allows us to study situations where the forces are changing throughout the problem, and simplifies some otherwise complex situations. You will also learn in this module what machines can and cannot do to help us to complete tasks.**Mechanical Energy**Mechanical Energy Potential Energy Kinetic Energy Work Transfer of Mechanical Energy**Work Requirements**• Work requires 3 things: • There must be a force • There must be a displacement • The displacement must be a result of (or caused by) the force.**Work**General Equation**Special Cases**• Force and Displacement in the same direction (θ = 0°, cosθ = 1)**Example 1**A person uses their hand to lift a 5.3 kg box 1.2 m into the air (at a constant speed). How much work did the person do? Solution:**Special Cases**• Force and Displacement are perpendicular to each other (θ = 90°, cosθ = 0)**Example 2**A person walks 3.2 m across the room carrying a 5.3 kg box at a constant speed. How much work was done by the person? Solution:**Special Cases**• Force and Displacement are in opposite directions (θ = 180°, cosθ = -1)**Example 3**A person holding a 5.3 kg box in the air lowers the box 1.2 m. How much work is done by the person? Solution:**Check Your Learning**A 12 kg box is being pulled by Bill across the floor. Bill is exerting a force of 45 N, and the coefficient of friction is 0.22. He pulls the box a distance of 7.2 m. • How much work was done by Bill?**Check Your Learning**• How much work was done by friction?**Check Your Learning**• Calculate the net work done on the box. or Although Bill did 320 J of work, only 140 J of it was actually transferred to the box – the rest was lost to friction as heat energy**Example 1**Consider a 75 kg person who walks up a flight of stairs (shown below) compared to a similar person (same mass) who runs up the stairs. The person walking takes 6.3 s, and the person running takes 2.6 s. Assume both are moving at a constant speed. How much work does each person do?**Example 1 - Solution**Person 1 Since time was not used, person 2 does the same amount of work!!**Power**• Since Person 2 did the work more quickly, we say that they used more power. • Power is defined as the rate of doing work (or transferring energy): • 1 watt (W) = 1 J/s**Example 1 - Continued**Person 1 Person 2**Example 2**A truck must exert 3750 N of force to pull a trailer at a speed of 90.0 km/h. What power output (in horsepower) is required of the truck? Solution:**Example 2**New Equation**Check Your Learning**A person is pulling a sled that has a mass of 93 kg. The sled starts from rest and you can assume that there is no friction. The person is pulling with a horizontal force of 225 N over a distance of 12.0 m. What was the average power developed by the person? We must first find the work done by the person:**Check Your Learning**We must now use the rest of the information to find the time so that we can get the power:**Kinetic Energy**• Energy = ability to do work • Kinetic Energy is energy of motion • Consider the case where a car with an initial velocity is being pushed horizontally along a table with a net force. Since there is a net force in the direction of motion, there is net work being done.**Kinetic Energy**• Since • If we define kinetic energy to be**Example 1**A 0.410 kg rugby ball is thrown with a horizontal speed of 15.0 m/s. • What is its kinetic energy? • What was the work done by the person throwing the ball? Solution: