Module 3 Impulse Momentum Bello [PDF]

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MODULE 3 IMPULSE AND MOMENTUM



Learning Outcomes: At the end of this module, you should be able to: 1. Define impulse and momentum 2. State the relationship between impulse and momentum 3. Apply mathematical expressions for impulse and momentum in problem solving 4. State the law of conservation of momentum Introduction: This module aims to broaden your knowledge about impulse and momentum, the principle behind it, the relationship between them and the laws of conservation of momentum. Force can be observed anywhere by their motion. In collision between objects the force involves may be extremely large. Force maybe great during impact that acts for short time like when you bat a ball or when you hit the pin balls in bowling. Every object has a momentum when is it in motion and impulse produces the change in the motion. Momentum and impulse are both vector quantities. An object has momentum whether or not it interacts with another object. However, an impulse can only be provided when two or more objects interact. Impulse is equivalent to the change in momentum of an object, so impulse depends on time while momentum is time independent. The module also includes the conservation of momentum. The conservation of momentum and the conservation of energy are universal laws because they apply throughout the universe and have no known exceptions to it. Lesson: Momentum Momentum, P, is the product of the mass, m, of the body and the velocity, v, of the object. Momentum is a vector quantity. Its direction is being the direction of the velocity. When momentum is applied to an extended body, the velocity used is the velocity at the center of the mass. Momentum is directly proportional to mass and speed. A net force is required to change a body’s momentum.



In terms of momentum, Newton’s second law states that a net force acting on an object is equal to how fast the momentum of the object is changing. Formula: Momentum = mass x velocity; where: P, momentum P = mv m, mass of the object v, velocity of the object Units System of Measuremen t SI - MKS CGS Eng. -FPS



m. mass



V, velocity



kg g slug



P, Momentum



m/s cm/s ft/s



kg- m/s g-m/s slug-ft/s



Example. A car, 1200 kg mass, moves at the speed of 55 m/s due East. Determine its momentum. N W



m = 1200 kg



E S



v = 55 m/s



Given: m = 1200 kg V = 55 m/s Required: P Solution: To find the momentum, P, defined as the product of mass and velocity of object Formula: P = mv P = (1200 kg) (55m/s) = 66,000 k-m/s, East Example. A bowling ball moves at a velocity of 4 m/s and a momentum of 30 kgm/s North. Find its mass of the ball. P = 30 kg-m/s W S V = 4m/s



N E



Solution:



P = mv m = P/v m = 30 kg-m/s 4 m/s m = 7.5 kg



Collision 1. During the collision, the object ball gains momentum and the cue ball loses the same amount of momentum 2. The momentum of each ball changes during the collision but total momentum remains constant 3. Momentum is conserved during collisions and when objects push away from each other 4. Since the momentum of the two balls remains constant after the collisions, we say momentum is conserved ΔP = Pf - Pi ΔPA = PAf - PAi = mAvAf - mAvAi = mA (vAf - vAi) ΔPB = PBf - PBi = mBvBf -- mBvBi = mB(vBf - vBi) In,



ΔP = mΔv Δv = ΔP m



I = ΔP = Ft



Example. Two balls roll toward each other having unequal velocity, Ball A, 10 m/s and Ball B, 8 m/s. Both balls weigh 2.0 kg. Find the momentum of both balls. Given: mA = 2.0 kg vA = 10 m/s Ball A



mB = 2.0 kg v B = 8 m/s



A VA = 10 m/s



B



+



Ball B



-



V B = 8 m/s (left is negative)



Solution: P A = m A vA = (2.0 kg) (10 m/s)



PB = m B v B = (2.0kg) (8 m/s)



PA = 20 kg-m/s



PB =16 kg-m/s



Ball A moves to the right, Ball B moves to the left: ΔP = P A - PB, ΔP = 20 kg-m/s --16 kg-m/s ΔP = 4 kg m/s The momentum of the system is 4.0 kg-m/s, to the left Impulse Impulse, I, is the product of a force and the time its action. The impulse produces the change in motion. It is dependent on the velocity and the change in velocity is from full speed to zero. Impulse, I, is the change in momentum Formula:



I = ΔP, P = Ft, I = Fnet t, but F = ma from Second Newton’s Law I = mat



since a = Δv t I = m Δv (t) t I = m Δv and Δv = (vf – vi) I = mvf - mvo I = m (vf - vo) Types of collisions: 1. Elastic. When two objects collide and bounces off each other 2. Inelastic. When two objects collide and stick to each other 3. Explosion. When one object separates into fragments One-dimensional collision is when two ball bearings colliding on a straight tract. Example: A 45 N force is applied to a 70 kg boulder for 3 seconds. What is the impulse of this force? I = Ft I = (45 N) (3 s) I = 135 N-s



Example. A 1.2 kg ball is coming straight at a 65 kg soccer player at 15 m/s who kicks it in the exact opposite direction at 25 m/s with an average force of 1000 N. How long are his foot and the ball in contact?



Given: mass of the ball, mb = 1.5 kg mass of the player, mp = 65 kg velocity of the ball initial, vbi = 15 m/s velocity of the ball final, vbf = 25 m/s Force average applied = 1000 N



Solution



vf = 25 m/s



vi =15 m/s I = mb (vbf - vbi), Ib = 1.5 kg 25 – (-15) m/s Ib = 60 kg-m/s I = Ft, t = I_ = 48 kg-m/s = 60 kg-m/s = 0.06 s or 6 ms 2 F 1000 N 1000 kg-m/s



Impulse – Momentum Theorem I = FΔt = ΔP = Pf - Pi I = Pf -Pi I = mvf – mvi) 1. This states that a net force, F, applied for a certain time interval, Δt, will cause a change in the object’s momentum equal to the product of the force and time interval. 2. A large constant force will cause a rapid change, P, momentum 3. A small constant force would take a much longer time to cause a change in P. 4. The Impulse-Momentum theorem explains why “follow through” is so important in many sports such as baseball, basketball, and boxing. Remember: If a system is isolated, the sum of the internal forces of the system is zero If the system is not isolated, the system experiences an impulse because a net force act on it and the conservation of momentum does not apply.



Example. A truck mass 2000 kg collide with a wall. The initial and final velocities of the automobile ae 3.5 m/s to 4.2 m/s. If the collision last for 0.5 second, find: a. initial and final momenta b. impulse due to collision c. the average force exerted on the automobile Given: m = 2000 kg Vi = 3.5 m/s Vf = 4.2 m/s t = 0.5 s



Required:



a. Initial momentum, Pi and final momentum, Pf b. I, impulse b. F, average force



Solution: a) Momentum: P = mV Initial Momentum: Pi = mVi = 2000 kg (3.5 m/s) = 7000 kg-m/s Final Momentum: Pf = mVf = 2000 kg (4.2 m/s) = 8400 kg-m/s b) Impulse due to the collision: I = ΔP = Pf – Pi I = 8400 kg-m/s -- 7000 kg-m/s I = 1400 kg-m/s c) Average force exerted on the truck ΔP = FΔt F = ΔP = 1400 kg-m/s = 2800 kg-m/s2 = 2800 N Δt 0. 5 s The Law of Conservation of Momentum The momentum of any closed system does not change. The law of conservation of momentum requires an isolated system because any external net force acting on the system will provide an impulse. That impulse will change the momentum of some or all



of the objects in the system, and the momentum of the system will no longer be conserved.



When two bodies of masses, m1 & m2 collide, Total momentum before impact = Total momentum after impact



m1u1+m2u2 = m1v1 + m2v2 u1, u2 – velocities before impact v1, v2 – velocities after impact m1 – mass of ball no.1 m2 – mass of ball no. 2



v2



v2 u1



m1



m2



u2



Law of action - reaction 1. Collision is an interaction between two bodies which have made in contact with each other. 2. A collision results in a force being applied to the two objects collide. 3. Such collisions are governed by Newton’s laws of motion Two-dimensional collision is two balls colliding off-center and deflecting away from each other after the interaction. Example. The baseball mass 150 grams approaches the bat horizontally at a speed, vi of 35 m/s, but the collision is not head-on, and the ball leaves the bat with the speed vf of 40 m/s at an upward angle of 35 o from the horizontal. What is the magnitude and the direction of the average force, F exerted on the ball if the collision lasts 1.65 ms.



Vf=40m/s



35o Vi =35 m/s



Given: mass of baseball, m1 = 135 g



Initial speed, vi = 35 m/s Final speed, vf = 40 m/s



Solution: Vfy = 40m/s sin 35o Vf = 40 m/s 35 o



Vfx = 40m/s cos 35o Components: Vfx = (40 m/s) (cos 35 o) = 32.77m/s Vfy = 40m/s sin 35o = 22.94 m/s Vix = - 35 m/s (motion to the left, negative)



Vfx +



Ix = m (Vfx – Vix) = 0.15 kg [ 32.77 – (-35)] m/s = 10.17 kg-m/s Iy = m (Vfy - Viy) = 0.15kg [22.94 - 0]m/s = 3.44 kg-m/s Magnitude of impulse: I = √ I x 2+ I y 2 =



√ 10.172 +3.442



= 10.74 kg-m/s



Magnitude of the average force: Fave = I = 10.74 kg-m/s = 6509.09 N Δt 0.00165 s The direction of the impulse angled upward from the horizontal by ϴ is given by: tan ϴ = Iy Ix tan ϴ = 3.44 10.17 ϴ = tan -1(3.44/10.17) ϴ = 18.69o



I =10.74 kg-m/s



ϴ = 18.69 o



Component form:



m1u1x+m2u2x = m1v1x + m2v2x Remember that velocity is a vector quantity, they can be positive or negative.



Vi -



v1



u1



v2



m1



m2



u2



In collision or interaction, momentum is conserved, the total momentum before the collision is equal to the total momentum after the collision. Energy is conserved in the collision. But, as a result of collision, the total kinetic energy never increases it usually decreases. KE2 ≤ KE1, for not perfectly elastic collision From kinetic-energy consideration: u1 – initial velocity of ball 1, u2 – initial velocity of ball 2; v1 – final velocity of ball 1; v2 – final velocity of ball no. 2, m1 -mass of ball no. 1 m2-mass of ball no.2. for perfectly elastic collision: KE2 = KE1 ½ m1v12+ ½ m2v22 = ½m1u12 + ½m2u22



V1F -V2f = Vi1 - Vi2 The relative velocity after collision is less than or equal to the negative relative velocity before collision Coefficient of restitution, e – the relative ratio of velocity after collision to the relative velocity before collision e = ̶ v 1- v 2 u1 – u2



e = V1 – V2



u2 – U1



e = 1, if the collision is perfectly elastic e = 0, If the collision is completely inelastic (the body stick together after collision) e < 1 inelastic collision



Example. Two balls are traveling at the same direction at first. Ball B, 2.0 g has a speed of 25 m/s overtakes a 4.0 kg ball A with the speed of 15m/s. If the coefficient of restitution, e, is 0.70. Find the speeds of ball A and ball B after the collision. Given:



mB = 2.0 kg



m A = 4.0 kg



UB = 25 m/s



UA= 15 m/s



e = 0.70



Require: VA and VB, final velocity of each ball Solution: Using the formula of coefficient of restitution: e = VA – VB



UB - UA e (UB - UA) = VA – VB 0.70 (25 – 15) m/s = VA – VB 7.0 m/s = VA – VB VB = VA – 7.0 m/s ----- eq 1 Using the momentum relationship: Total momentum before impact = Total momentum after Impact



mAuA+ mB uB = mAvA + mBvB 4.0 kg(15m/s) + 2.0kg(25m/s) = 4.0kg(vA) + 2.0kg(vB) 60 kg-m/s + 50 kg-m/s = 4.0kg vA + 2.0kg vB 110 kg-m/s = 4.0 kg vA + 2.0kgvB, substitute eq. 1 110 kg-m/s = 4.0 kg vA + 2.0 kg (vA-7.0 m/s) 110 kg-m/s = 4.0 kg VA + 2.0 kg VA – 14 kg-m/s 110 kg-m/s = 6.0 kg VA – 14 kg-m/s (110 + 14) kg-m/s = 6.0 kg VA vA = 124 kg-m/s 6.0 kg vA = 20.67 m/s, final velocity of ball A Solving for vB, use eq. 1; VB = VA – 7.0 m/s vB = (20.67 – 7.0) m/s vB = 13.67 m/s, final velocity of ball B It means they continue to move at different velocity from the initial Example. From Example above, Find the total kinetic energy of both ball A and ball B before collision. Find the kinetic energy of both ball A and ball B after collision Compare the results Solution: mass of ball A & ball B:



mA = 4.0 kg,



mB = 2.0 kg



initial Velocity of ball A and ball B: U A = 15 m/s,



UB= 25 m/s



final velocity of ball A and ball B: vA = 20.67 m/s, vB = 13.67 m/s Total KE1 before collision =½ mA UA2 + ½ mB UB2 = ½(4.0kg) (15m/s)2 + ½(2.0kg) (25m/s)2



= 450 kgm2/s2 + 625 kgm2/s2 KE1 = 1075 J



Total KE2 after collision = ½ mA VA2 + ½ mBvB2 = ½(4.0kg) (20.67 m/s) 2 + ½ (2.0kg) (13.67m/s)2 = 854.5 kg m2 + 186.87 kg-m2 s2 s2 KE = 1041.37 J KE2 < KE1; therefore, KE after collision is less than KE before collision . It is not perfectly elastic because KE is after collision.



lost Activities:



1. Review the lesson and familiarize with the formulas 2. Practice resolving the problems. 3. Watch https://www.youtube.com/watch?v=E13h1E_Pc00 4. Watch https://www.youtube.com/watch?v=8OB8eIPgEkQ



Assessment Direction: After learning about measurement in this the module; understanding the concept, reviewing the examples and convert some physical quantities, the following are the basis of your grade: 1. assignment 2. quiz 3. experiment 4. major exam Reflection Direction: Now that you have completed the course, you will be writing what you have learned in this module and that includes; a) your experience in learning the topic b) your opinion about the topic c) your reasons for saying so and to back up, you cite some evidences



Assignment: In your assignment the following steps are required 1. Review first the lessons. 2. Rewrite the formulas and know when and why you use the formula. 3. Read each problem well. 4. Draw the figure (if not given) that is being described in the problem. 5. Label your drawing that is matched with the description in the stated problem. 6. Include the FBD to isolate the force acting on the object. 7. Solve, show the units cancellation (if needed). 8. Box your final answers Problems to be solved:



ans.



1. A particle has a mass of 10 kg and a velocity of 5 m/s. a) What is the momentum of the particle? b) An impulse of 20 kg· m/s acts on the particle in the same direction as the velocity. What is the final velocity of the particle? c) If the impulse is delivered for a duration of 0.5 seconds, what is the average force acting on the particle? 2. An elastic collision occurs in one dimension, in which a 10 kg block traveling at 5 m/s collides with a 5 kg block traveling at 3 m/s in the same direction. What are the velocities of the two blocks immediately after the collision? (Hint: 3.67 m/s, 5.67 m/s). 3. The baseball mass 140 g approaches the bat horizontally at a speed of 39 m/s but the collision in not head-on and the ball leaves the bat at 45 m/s at an angle of 30o from the horizontal. What is the average force exerted on the ball if the collision last 1.2 ms (Hint ans. 9466.7N). 4. Two balls collide and bounce off each other, Ball A ,800 g, has an initial speed of 30 m/s and Ball B, 500 g, is with speed of 50 m/s. Find: a. If Ball B final velocity of Ball A is 15 m/s b. Is the collision perfectly elastic Figure:



ϴ 30o



References: https://www.slideshare.net/JPoilek/chapter-6-4140



https://www.slideshare.net/merina_90/as-physics-momentum https://www.pdfdrive.com/schaums-outline-of-theory-and-problems-of-college-physicshttps://www.google.com/search?q=baseball+bat+clipart&tbm Ymas, S. E. (2009) College Physics, Ymas Publishing House Weber, R. L. Manning, K. V., White, M. W., (1977), College Physics McGraw -Hill Inc.