Lesson 2-Impulse and Momentum PDF

Lesson 2 Momentum and Impulse Advanced General Physics Physics 1/2 Science, Third Quarter Technology, Engineering, and Mathematics Shown in the figure is a device called “Newton’s cradle.” Each ball contains an equal mass as with the rest of the balls....

Lesson 2 Momentum and Impulse Advanced General Physics Physics 1/2 Science, Third Quarter Technology, Engineering, and Mathematics Shown in the figure is a device called “Newton’s cradle.” Each ball contains an equal mass as with the rest of the balls. 2 When you pull away one ball and release it, the entire system will “gain momentum.” 3  Learning Objectives At the end of the lesson, you should be able to do the following: Explain the concept of momentum and impulse. Determine how momentum affects the motion of a system’s center of mass. Explain how internal and external forces affect the momentum in a given system. Solve problems related to momentum and impulse. 4  Momentum and the Second Law of Motion Linear Momentum (Momentum) It is the product of the object’s mass and velocity. It is symbolized by lowercase p, with a unit of kg*m/s. It is expressed as: 5  Momentum and the Second Law of Motion Linear Momentum (Momentum) It is a vector quantity, and has a direction similar to the velocity. It is proportional to both mass (m) and speed (v). 6  Conservation of Momentum Consider two skaters exerting forces (of equal magnitude and opposite directions) on each other. How will you illustrate the action-reaction pair of forces exerted by each other? 12  Conservation of Momentum The internal forces exerted by each skater on each other are of equal magnitudes but opposite in direction. The total momentum in the system is conserved. 13  Conservation of Momentum The total momentum of a system is constant when the summation of all the external forces acting upon that system amounts to zero. 15  Momentum and Center of Mass The total momentum of a system amounts to the product of the mass and the velocity of the center of mass. This can be expressed as: 16  Let’s Practice! Determine the momentum of a child riding a wagon with a velocity of 1.75 m/s if the wagon-child system has a mass of 35.5 kg. 𝑝റ = 𝑚𝑣 𝑝റ = 35.5(1.75) 𝒑 = 𝟔𝟐. 𝟏𝟐𝟓 𝒌𝒈 ∙ 𝒎/𝒔 17  Let’s Practice! A 1350-kg vehicle’s parking brake has malfunctioned when it reached a momentum of 6 500 kg-m/s. What is the vehicle’s velocity? 𝑝റ = 𝑚𝑣 6500 = 1350(v) 𝒗 = 𝟒. 𝟖𝟏𝟓 𝒎/𝒔 18  Let’s Practice! A dart gun produces a 1.56-kg-m/s momentum and an 11.41-mph velocity when it thrusts its darts. Determine the mass of each dart. 𝑝റ = 𝑚𝑣 𝑚 𝑚𝑖 1.609 𝑘𝑚 1000 𝑚 1 ℎ𝑟 𝑚 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 𝑚𝑝ℎ 𝑡𝑜 : 11.41 = 5.1 𝑠 ℎ𝑟 1 𝑚𝑖 1 𝑘𝑚 3600 𝑠 𝑠 1.56 = 𝑚(5.1) 𝒎 = 𝟎. 𝟑𝟎𝟔 𝒌𝒈 19  Change in Momentum We can note that the change in velocity (Δv) of an object is given by the product between the acceleration (a) and the time elapsed (Δt): We can recall that the equation for the second law of motion is: 25 Change in Momentum Combining and we can get a working equation for the change in the momentum of the object given by: 26  Impulse The impulse of a force Δp is also a vector quantity and is denoted by either Imp or. The SI unit of impulse is N-s (newton-second). It can also be expressed as. 27  Impulse of a Constant Force The impulse of a net force is given by the equation: 28  Let’s Practice! A child with a mass of 23.75 kg runs with a velocity of 3.6 m/s eastward and heads back to her initial position with a velocity of 2.5 m/s. What is her impulse? 𝐽റ = 𝑚 ∆𝑣 = 𝑚(𝑣2 − 𝑣1 ) 𝐽റ = 23.75(−2.5 − 3.6) 𝑱റ = −𝟏𝟒𝟒. 𝟖𝟕𝟓 𝑵 ∙ 𝒔 𝒐𝒓 𝟏𝟒𝟒. 𝟖𝟕𝟓 𝑵 ∙ 𝒔 𝒆𝒂𝒔𝒕𝒘𝒂𝒓𝒅  Let’s Practice! A 0.45-kg ball has been flung against a concrete wall. It is moving in a straight path to the west at 43.2 m/s when it hits the wall, and rebounds to the right at 35.45 m/s. Determine the (a) impulse of the net force exerted on the ball when it hit the wall and the (b) force that the wall exerted on the ball during the collision if the ball is in contact with the wall for 0.02 seconds. 𝑎) 𝐼𝑚𝑝𝑢𝑙𝑠𝑒: 𝑏) 𝑁𝑒𝑡 𝐹𝑜𝑟𝑐𝑒: 𝐽റ = 𝑚 ∆𝑣 = 𝑚(𝑣2 − 𝑣1 ) 𝐽റ = Σ𝐹റ ∆𝑡 𝐽റ = 0.45 35.45 − (−43.2) റ 35.393 = Σ𝐹(0.02) 𝑱റ = 𝟑𝟓. 𝟑𝟗𝟑 𝑵 ∙ 𝒔 𝜮𝑭 = 𝟏𝟕𝟔𝟗. 𝟔𝟓 𝑵  Let’s Practice! A 0.66-kg ball is initially moving to the west at 19.5 m/s when it was kicked 49° upward to the right at 25 m/s. Calculate the magnitude and direction of the average net force assuming a collision time of 0.01 s. 𝐽റ = Σ𝐹റ ∆𝑡 𝑎𝑙𝑜𝑛𝑔 𝑥 − 𝑎𝑥𝑖𝑠: 𝐽𝑥 = 𝑚 ∆𝑣𝑥 = 0.66[ 25𝑐𝑜𝑠49 − −19.5 ] 𝐽𝑥 = 23.695 𝑁 ∙ 𝑠 𝑎𝑙𝑜𝑛𝑔 𝑦 − 𝑎𝑥𝑖𝑠: 𝐽𝑦 = 𝑚 ∆𝑣𝑦 = 0.66[ 25𝑠𝑖𝑛49 − 0] 𝐽𝑦 = 12.453 𝑁 ∙ 𝑠 𝑓𝑜𝑟 𝑡ℎ𝑒 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛: 2 2 𝐽റ = 𝐽𝑥 + 𝐽𝑦 = 23.6952 + 12.4532 = 26.768 𝑁 ∙ 𝑠 −1 12.453 𝜃 = 𝑡𝑎𝑛 23.695 𝐽റ = Σ𝐹റ ∆𝑡 → 26.768 = Σ𝐹റ 0.01 ∴ 𝜮𝑭 = 𝟐𝟔𝟕𝟔. 𝟖𝟎𝟖 𝑵 𝜽 = 𝟐𝟕. 𝟕𝟐°  WORKSHEET NO. 2 MOMENTUM AND IMPULSE (15 POINTS) CN LN, FN MI. 10-__________ 1 WHOLE INTERMEDIATE PAPER ONLY  Check Your Understanding INSTRUCTIONS: Solve the following problems with complete solution. Use GRESA Format. Box the final answer with the correct unit. 2 points per required answer. 1 point for the solution. 1. An object with m = 150,000 g moves to the right with a velocity of 20.808 kph and then heads left with a velocity of 3.154 mph. What is its impulse in N·s? 2. The momentum of a 3000 kg truck is 6.36 x 104 kg·m/s. At what speed, in kph, is the truck traveling? 3. Suppose another ball, with mass equal to 0.59 kg hit the wall after having traversed horizontally westward at 54.0 m/s. It then rebounds at 40.96 m/s. Calculate the (a) impulse of the net force exerted on the ball when it hit the wall and the (b) force that the wall exerted on the ball during the collision if the ball is in contact with the wall for 5143 milliseconds (ms). 4. A 0.84-kg ball is initially moving to the west at 16.7 m/s when it was kicked 38.6° upward to the right at 28.14 m/s. Calculate the magnitude and direction of the average net force, assuming a time of collision of 0.05 s.