Chapter 07 impulse and momentum


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  • J = 8.00 Nοƒ—s
  • Ξ”t = 0.0041666.. Or0.00417 s
  • F = 500 N
  • Chapter 07 impulse and momentum

    1. 1. Impulse and Momentum<br />
    2. 2. ASTRONAUT Edward H. White IIis in the zero gravity of space. By firing the gas-powered gun, he gains momentum and maneuverability. Credit: NASA<br />
    3. 3. Momentum Defined<br />Momentum is defined as the product of mass and velocity. (vector quantity)Units:kg m/s; g cm/s; slug ft/s<br />Momentum<br />m = 1000.0 kg<br />𝒑=𝟏𝟎𝟎𝟎.πŸŽΒ π€π πŸπŸ”Β π¦/𝐬<br />Β <br />𝒑=πŸπŸ”πŸŽπŸŽπŸŽΒ π€π Β π¦/𝐬<br />Β <br />𝒗=πŸπŸ”Β π¦/𝐬<br />Β <br />
    4. 4. Dt<br />Impulse𝒋 is a force 𝐹 acting for a small time interval Dt.<br />Β <br />IMPULSE<br />Impulse:<br />
    5. 5. Dt<br />Example 1:The face of a golf club exerts an average force of 4000 N for 0.002 s. What is the impulse imparted to the ball?<br />Impulse:<br />The unit for impulse is the newton-second(N-s)Other units are dyne-second and pound-second (lb-s)<br />
    6. 6. F<br />time, t<br />Impulse from a Varying Force<br />Normally, a force acting for a short interval is not constant. It may be large initially and then play off to zero as shown in the graph.<br />In the absence of calculus, we use the average force Favg.<br />
    7. 7. B<br />A<br />Example 2: Two flexible balls collide. The ball B exerts an average force of 1200 N on ball A. How long were the balls in contact if the impulse is 5 N s?<br />Dt = 0.00417 s<br />The impulse is negative; the force on ball A is to the left. <br />
    8. 8. In a Β½ sheet of pad paper, discuss briefly the movie clip Spiderman 2 in the application of impulse and momentum (their relationship):<br />Notice:<br />Train and its motion <br />How spiderman wants to stop the train using his legs, single shot of spiderweb by each hand; and multiple shots of spiderwebs; <br />β€œstopper” in the railroad. <br />
    9. 9. Impulse Changes Velocity and Momentum<br />Consider a mallet hitting a ball:<br />Impulse = Change in β€œmv”<br />
    10. 10. Impulse and Momentum<br />F<br />Dt<br />mv<br />Impulse = Change in momentum<br />Newton’s Second Law of Motion <br />A force F acting on a ball for a time Dtincreases its momentum mv.<br />Conversion: 1.00 N.s = 1.00 kg m/s<br />
    11. 11. F<br />+<br />Dt<br />0<br />mv<br />Example 3: A 50-g golf ball leaves the face of the club at 20 m/s. If the club is in contact for 0.002 s, what average force acted on the ball?<br />Given: m = 0.05 kg; vo = 0; <br />Dt = 0.002 s; v= 20 m/s<br />Choose right as positive.<br />Average Force:<br />
    12. 12. +<br />v<br />vo<br />Vector Nature of Momentum<br />Consider the change in momentum of a ball that is dropped onto a rigid plate:<br />A 2-kg ball strikes the plate with a speed of 20 m/s and rebounds with a speed of 15 m/s. What is the change in momentum?<br />Dp = mv- mvo<br />Dp = (2 kg)(15 m/s) - (2 kg)(-20 m/s)<br />Dp= 70 kg m/s<br />
    13. 13. Momentum<br />Impulse<br />𝒋=𝑭.βˆ†π’•<br />Β <br />Impulse = Change in momentum<br />𝒋=βˆ†π’‘;      𝑭=βˆ†π’‘βˆ†π’•<br />Β <br />
    14. 14. More Sample Problems:<br />A 0.10-kg ball is thrown straight up into the air with an initial sped of 15 m/s. Find (a) the ball’s initial momentum; (b) its momentum at its maximum height; and (c) impulse of the weight on the ball from initial to maximum height.<br />The force shown in the force vs. time diagram acts on a 1.5 kg object. Find (a) the impulse of the force, (b) the final velocity of the object if it is initially at rest.<br />Fx (N)<br />𝑨𝒓=π’π’˜<br />Β <br />𝑨𝒕=πŸπŸπ’ƒπ’‰<br />Β <br />t (s)<br />
    15. 15. Conservation of Linear Momentum<br />Momentum is conserved in this rocket launch. The velocity of the rocket and its payload is determined by the mass and velocity of the expelled gases. Photo: NASA<br />
    16. 16. Conservation of Momentum <br />The total momentum of an isolated system of bodies remains constant.<br />system = set of objects that interact with each other.<br />An isolated system is one in which the only force present are those between the objects of the system; that is there is no net external force.<br />𝐹=0 therefore 𝑗=0<br />βˆ†π‘=0<br />Β <br />
    17. 17. Conservation of Momentum<br />Since 𝐹21=βˆ’πΉ12<br />Β <br />π‘­πŸπŸ<br />Β <br />βˆ΄Β Β Β Β Β Β Β Β Β π‘—21=βˆ’π‘—12<br />Β <br />𝑗21=𝐹21.βˆ†π‘‘=π‘š1𝑣1β€²βˆ’π‘š1𝑣1<br />Β <br />𝑗12=𝐹12.βˆ†π‘‘=π‘š2𝑣2β€²βˆ’π‘š2𝑣2<br />Β <br />π‘š1𝑣1β€²βˆ’π‘š1𝑣1=βˆ’π‘š2𝑣2β€²βˆ’π‘š2𝑣2<br />Β <br />π‘­πŸπŸ<br />Β <br />π‘š1𝑣1+π‘š2𝑣2=π‘š1𝑣1β€²+π‘š2𝑣2Β Β β€²<br />Β <br />Initial momenta = final momenta<br />
    18. 18. Conservation of Momentum Revisited: Into Details<br />Sample Illustrations:<br />Initially, the system is at rest:<br />Rocket: π‘£π‘Ÿ=0<br />Gas: 𝑣𝑔=0<br />Β <br />π‘šπ‘Ÿπ‘£π‘Ÿβ€²<br />Β <br />π‘šπ‘Ÿπ‘£π‘Ÿ+π‘šπ‘”π‘£π‘”=π‘šπ‘Ÿπ‘£π‘Ÿβ€²+π‘šπ‘”π‘£π‘”β€²<br />Β <br />π‘šπ‘”π‘£π‘”β€²<br />Β <br />𝑣𝑔′=βˆ’π‘šπ‘Ÿπ‘£π‘Ÿβ€²π‘šπ‘”<br />Β <br />
    19. 19. π’ŽπŸπ’—πŸ<br />Β <br />π’ŽπŸπ’—πŸ<br />Β <br />APPLICATION: COLLISIONS and Conservation of Momentum<br />2<br />1<br />π‘­πŸπŸ<br />Β <br />π‘­πŸπŸ<br />Β <br />2<br />1<br />π’ŽπŸπ’—πŸβ€™<br />Β <br />π’ŽπŸπ’—πŸβ€²<br />Β <br />1<br />2<br />π‘š1𝑣1+π‘š2𝑣2=π‘š1𝑣1β€²+π‘š2𝑣2Β Β Β Β β€²<br />Β <br />
    20. 20. Example 01 A 730-N man stands in the middle of a frozen pond of radius 5.0 m. He is unable to get to the other side because of a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 1.2-kg physics book horizontally toward the north shore at a speed of 5.0 m/s. How long does it take him to reach the south shore?<br />Simple Collisions (Head-on Collisions)<br />A car with a mass of 9.00 x 102 kg is traveling at +15.0 m/s while an SUV with a mass of 1.80 x 103 kg is traveling at -15.0 m/s. (a) If the cars collide head-on, becoming entangled, find the velocity of the entangled cars after the collision. (b) If the SUV stops after collision and not entangled, what is the velocity of the car after collision?<br />
    21. 21. Assignment:<br />What is the momentum of a 15-g sparrow flying with an initial speed speed of 12 m/s to the east? If the sparrow increases its speed to 14 m/s in the same direction, what is the impulse of its force?<br />Calculate the force exerted on a rocket, given that the propelling gases are expelled at a rate of 1000 kg/s with a speed of 60,000 m/s (at take off)<br />A 900-kg box car traveling at +20 m/s strikes a stationary second car. After collision, the box car stops and the second car travels with a velocity of +30 m/s. What is the mass of the second car?<br />
    22. 22. Summary of Formulas:<br />Momentum<br />Impulse<br />𝒋=𝑭.βˆ†π’•<br />Β <br />Impulse = Change in momentum<br />𝒋=βˆ†π’‘;      𝑭=βˆ†π’‘βˆ†π’•<br />Β <br />Conservation of Momentum<br />βˆ†π’‘=𝟎;Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β π’ŽπŸπ’—πŸ+π’ŽπŸπ’—πŸ=π’ŽπŸπ’—πŸβ€²+π’ŽπŸβ€²<br />Β <br />