MOMENTUM AND IMPULSE (LIVE) 07 APRIL 2015 Lesson Description In this lesson we revise: Momentum Collisions Impulse Section A: Summary Notes and Examples 1.1. Momentum Momentum is defined as the product of an object’s mass and velocity. It is a vector which means that it has both magnitude and direction. Momentum is calculated using the following equation: p = momentum measured in ∙ ∙ −1 = m = mass measured in kg v = velocity measured in ∙ −1 When an object changes its velocity then it will experience a change in momentum. If the object changes direction when changing velocity it will experience a much greater change in momentum then if its direction doesn’t change. Example 1: -1 A tennis ball of mass 56 g is thrown towards a wall and hits the wall at 45 m.s . The ball then -1 bounces off the wall with a velocity of 40 m.s as shown in the diagram. Calculate the change in momentum of the ball. Solution: Hint 1: If there was no diagram given, then draw a sketch to help understand the question Hint 2: Write a list of all the information given, including chosen positive direction. Also convert given units to SI units. Page 1 Right is positive -1 vi = 45 m.s -1 vf = −40 m.s m = 56 g = 0,056 kg ∆ = ∆ = − = 0,056 −40 − 45 = −4,76 = 4,76 . −1 In order for an object to change its velocity or change its direction a net force needs to be applied to the object. This net force can be calculated as follows: Fnet = net force measured in N ∆ = ∆ Δp = change in momentum measured in kg.m.s -1 Δt = time measured in s This net force is the same net force that can be calculated using Newton’s second law. Therefore Newton’s second law can state in terms of momentum: Newton’s Second Law in terms of Momentum: The net force acting on an object is equal to the rate of change of momentum. Example 2: A remote controlled toy rocket, with mass 0,5 kg, is launched so that is -1 reaches a velocity of 35 m.s in 0,2 seconds. Calculate the net force acting on the rocket. Solution: m = 0,5 kg -1 vi = 0 m.s (the rocket started from rest) vf = 35 m.s -1 Δt = 0,2 s = ∆ ∆ − ∆ 0,5 35 − 0 = 0,2 = = 87,5 Collisions When two objects collide with each other, their momentum will change. However the total momentum of the system will not change if there are no external forces acting on the objects. External forces are forces such as friction or the brakes on a car. Law of Conservation of Momentum: The total linear momentum of an isolated system remains constant (is conserved) Page 2 As an equation: Σ = Σ This equation can be rewritten in several ways, depending on the type of the collision. For example, if two objects collide with each other and then move apart after the collision. 1 1 + 2 2 = 1 1 + 2 2 m1 = mass of object 1 vi1 = initial velocity of object 1 m2 = mass of object 2 vi2 = initial velocity of object 2 vf1 = final velocity of object 1 vf2 = final velocity of object 2 Elastic and inelastic collisions Although momentum is always conserved during collisions, kinetic energy may not always be conserved. If the total kinetic energy of the system before collision isequal to the total kinetic energy of the system after collision, then the collision is said to be elastic. However, if the total kinetic energy before and after the collision is not the same, then the collision is said to be inelastic. The kinetic energy is often transferred to sound, light and internal heat, due to friction. Example 3: -1 A ball of mass 100 g is moving at 10 m.s to the right when it collides with another ball of mass 150 g -1 -1 moving at 10 m.s to the left. After the collision the 100 g ball moves with a velocity of 7 m.s to the left. 1. Calculate the velocity of the 150 g ball after the collision. 2. Is the collision elastic or inelastic? Use appropriate calculations to prove your answer. Solution 1. Take to the right as positive. Σ = Σ m1 = 100 g 1 1 + 2 2 = 1 1 + 2 2 = 0,1 kg vi1 = 10 m.s -1 vf1 = - 7 m.s 0,1 10 + 0,15 −10 = 0,1 −7 + 0,15 2 -1 −0,5 = −0,7 + 0,152 m2 = 150 g 0,2 = 0,152 = 0,15 kg vi2 = -10 m.s ∴ 2 = 1,33 ∙ −1 -1 vf2 = ? Page 3 2. Inelastic 1 1 2 2 Σ = 1 1 + 2 2 2 2 1 1 = (0,1)(10)2 + (0,15)(−10)2 2 2 1 1 2 2 1 1 + 2 2 2 2 1 1 = (0,1)(−7)2 + (0,15)(1,33)2 2 2 Σ = = 5 + 7,5 = 2,45 + 0,13 = 12,5 = 2,58 ≠ 1.2. Impulse Impulse is defined as the product of the net force acting on an object and the contact time that the net force acts. Impulse = FnetΔt. Also, Fnet∆t =Δp (impulse-momentum theorem). Therefore theimpulse equation can be written as follows: FnetΔt = impulse measured in N.s ∆ = ∆ Fnet = net force measured in N Δt = time measured in s Δp = change in momentum measured in kg.m.s -1 Impulse is a vector therefore it has both magnitude and direction. Example Consider the tennis ball in example 1. 1. If the tennis ball is in contact with the wall for 0,1s what is the impulse of the wall on the ball? Solution: = 4,76 . −1 (Calculated in Example 1) FnetΔt = Δp = 4,76 N.s left 2. Calculate the force exerted by the wall on the tennis ball. Solution FnetΔt = Δp Fnet (0,1) = 4,76 Fnet = 47,6 N left If the ball experiences a force of 47,6 N left, then the wall experiences a force of 47,6 N right due to the tennis ball. This is according to Newton’s third law. Both objects experience the same size force but in opposite directions. Page 4 Application Air bags and seat belts are used in cars to increase the time it takes for passengers to stop during a collision. The passengers will experience the same change in momentum but if the time to stop is increased then the force experience decreases as F netΔt = Δp and Δp is a constant. Section B: Practice Questions Question 1 (Taken from NCS November Paper I 2012) -1 The diagram below shows a car of mass m travelling at a velocity of 20 m.s east on a straight level -1 road and a truck of mass 2m travelling at 20 m.s west on the same road. Ignore the effects of friction. The vehicles collide head-on and stick together during the collision. 1.1. State the principle of conservation of linear momentum in words. (2) 1.2. Calculate the velocity of the truck-car system immediately after the collision. (6) 1.3. On impact the car exerts a force of magnitude F on the truck and experiences an acceleration of magnitude a. 1.3.1. Determine, in terms of F, the magnitude of the force that the truck exerts on the car on impact. Give a reason for the answer. (2) 1.3.2. Determine, in terms of a, the acceleration that the truck experiences on impact. Give a reason for the answer. (2) 1.3.3. Both drivers are wearing identical seat belts. Which driver is likely to be more severely injured on impact? Explain the answer by referring to acceleration and velocity. (3) [15] Question 2 (Taken from NCS November Paper I 2011) -1 A patrol car is moving on a straight horizontal road at a velocity of 10 m.s east. At the same time a -1 thief in a car ahead of him is driving at a velocity of 40 m.s in the same direction. -1 While travelling at 40 m.s , the thief’s car of mass 1000 kg, collides head-on with a truck of mass -1 5000 kg moving at 20 m.s . After the collision, the car and the truck move together. Ignore the effects of friction. 2.1. State the law of conservation of linear momentum in words (2) 2.2. Calculate the velocity of the thief’s car immediately after the collision. (6) Page 5 2.3. Research has shown that forces greater than 85 000N during collisions may cause fatal injuries. The collision described above lasts for 0,5 s. Determine, be means of a calculation, whether the collision above could result in a fatal injury. (5) [13] Question 3 (Taken from Gauteng Prelim Paper I 2013) A delivery van having a mass of 5 000 kg is moving towards the right at a constant velocity of -1 10 m.s . The delivery van collides head on with a car of mass 2 000 kg moving at a velocity of 15 -1 -1 m.s towards the left. Immediately after the collision the car moves at a constant velocity of 5 m.s towards the right. 3.1. Calculate the magnitude of the velocity of the delivery van immediately after the collision. (4) 3.2. Calculate the change in the kinetic energy of the system. (4) 3.3. What type of collision is this? Write only ELASTIC or INELASTIC (1) 3.4. If the collision lasts for 0,4 s, calculate the magnitude of the force that the delivery can exerts on the car. (4) [13] Page 6 Section C: Solutions Question 1 1.1. The total linear momentum remains constant in an isolated system 1.2. East positive 1.3.1. F Newton’s Third Law of motion (2) Σ = Σ 1 1 + 2 2 = 1 + 2 20 + 2 −20 = + 2 20 − 40 = 3 −20 = 3 ∴ = −6,67 = 6,67 . −1 (6) (2) 1.3.2. ½a 1 The truck experiences the same Fnet but has twice the mass. Since ∝ then acceleration must be half (2) 1.3.3. Car driver The car has a greater acceleration , due to the smaller mass but same Fnet. Therefore the car experiences a greater change in velocity. (3) Question 2 2.1. The total linear momentum remains constant in an isolated system 2.2. To the right is positive (2) Σ = Σ + = + 1000 40 + 5000 −20 = (1000 + 5000) −60000 = 6000 ∴ = −10 = 10 . −1 (6) 2.3. Force on car: (To the right is positive) ∆ = ∆ = − 0,5 = 1000 −10 − 40 ∴ = −100 000 = 100 000 Fnet > 85 000 N, therefore the collision will be fatal (5) Question 3 3.1. Right Positive Total p before = Total p after m1 vi1 + m2 vi2 = m1 vf1 + m2 vf2 5000 10 + 2000 −15 = 5000 vf1 + 2000 5 vf1 = 2 m. s −1 OR Page 7 (4) Left Positive Total p before = Total p after m1 vi1 + m2 vi2 = m1 vf1 + m2 vf2 5000 −10 + 2000 15 = 5000 vf1 + 2000 −5 vf1 = −2 m. s −1 = 2 m. s −1 3.2. (4) ∆K = Total K after − Total K before 1 1 1 1 2 2 2 2 m1 vf1 + m2 v2f − m1 vi1 + m2 vi2 2 2 2 2 1 1 1 = ( (5000)(2)2 + 2000 5)2 − ( 5000 10 2 2 2 = 2 1 + (2000)(−15)2 ) 2 = −440 000 J (4) OR Total K before = 1 1 m v2 + m v2 2 1 i1 2 2 i2 1 1 = 5000 10 2 + (2000)(−15)2 2 2 = 475 000 J Total K after = 1 1 2 2 m1 vf1 + m2 vf2 2 2 1 = 5000 2 2 2 1 + (2000)(5)2 2 = 35 000 J ∆K = Total K after − Total K before = 35 000 − 475 000 = −440 000 J 3.3. Inelastic 3.4. Right positive (4) (1) Fnet = Left positive ∆p m1 vf1 − m1 vi1 = ∆t ∆t = Fnet = 2000 5 − (2000)(−15) 0,4 ∆p m1 vf1 − m1 vi1 = ∆t ∆t = = 100 000 N 2000 −5 − (2000)(15) 0,4 = − 100 000 N = 100 000 N OR Page 8 (4) Right positive Fnet = Left positive ∆p m1 vf1 − m1 vi1 = ∆t ∆t = Fnet = ∆p m1 vf1 − m1 vi1 = ∆t ∆t 5000 2 − (2000)(10) 0,4 5000 −2 − (5000)(−10) 0,4 = = −100 000 N = 100 000 N = 100 000 N (4) OR Right Positive Left Positive vf = vi + a∆t for both equations 2 vf = vi + a∆t for both equations −15 = 5 + a(0,4) 15 = −5 + a(0,4)2 ∴ a = −50 m. s −2 ∴ a = 50 m. s −2 Fnet = ma Fnet = ma = 2000 −50 = 2000 50 = −100 000 N = 100 000 N = 100 000 N (4) Page 9

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