Physics 11. Unit 5 Momentum and Impulse

Transcription

1 Physics 11 Unit 5 Momentum and Impulse 1

2 1. Momentum It is always amazing to see karate experts chopping woods or blocks. They look so extraordinary and powerful! How could they do that? 2

3 Let s consider two situations. (a) A BMW 5-series (about 3800 pounds!) and a bicycle (about 30 pounds) are moving along an alley in a residential area at the same speed (say 7 km/h; about 2 m/s). By coincidence, they are all running to you! You are so freaked out that you try to stop them with your hands. Can you do that? 3

4 (b) You are in Stanley Park enjoying a sunny-day walk. All of a sudden, two bikes lose their control somehow and run to you. (Again?! What a pity!) They are both 2016 Litespeed T1SL. (Wow! Over $4000 each!) One cyclist is too nervous to break (that means he is heading to you at high speed like 10 m/s!) while the other, as he may be more experienced and more calm, is trying hard to break (so his speed is about 1 m/s only). Who would be easier for you to stop? 4

5 From these examples, we see that both mass and velocity are critical to determine how difficult an object can be stopped. A quantity called momentum is therefore defined to describe the extent of motion of an object. By definition: Since velocity is a vector quantity, momentum is also a vector quantity. Its direction is parallel to the direction of velocity vector. The unit of momentum is kg m/s. p = mv 5

6 Example: A 1000 kg car is moving at 10 km/h. Determine the momentum of the car. [ kgm/s] Example: A meteor moving through the Earth s atmosphere has a momentum of kgm/s. As it falls, friction with the atmosphere slows it to ¼ its original speed, as its mass shrinks to 2/9 of its original mass. Determine its new momentum. [639 kgm/s] 6

7 Aside: Influence of the frame of reference In previous chapters, we have learned the concept of relative velocity, which tells us that the velocity of a moving object measured by one observer may be different from that measured by another observer if they are not in the same frame of reference. Since momentum depends on velocity, it is also subjected to the choice of reference frame. Object has zero momentum in one frame but not in another. 7

8 2. Impulse In the previous examples, we see that objects with a large momentum are hard to stop. In order to change the momentum (or the extent of motion) of an object, we either (i) change its mass, or (ii) change its velocity. 8

9 In most cases, the momentum is changed by altering the velocity of the object, since changing mass means mass is either removed from or added to the object, which is quite difficult to do. It is therefore assumed usually that mass remains the same when momentum is changed. The change of momentum is called impulse, which is defined as p = m v = mv f mv i The unit of impulse is the same as the unit of momentum. 9

10 Example: A box of tic tacs (15 g) is sliding along the table at 5.0 m/s. I try to stop it, but only slow it down to 1.6 m/s. Determine the impulse I impart to the box. By definition p = m v = m v f v i = (0.015)( ) = kg m/s The negative sign of p means the impulse is in opposite direction to the motion of the box. 10

11 Since changing the momentum of a moving object requires a change in velocity, it implies acceleration, and therefore force must be involved. Recall from kinematics a = v. Hence v = a t. So t p = m v = ma t = F net t That means, momentum can be changed only if a force is applied over a period of time. Note that F net is the constant net force that is applied to the object. But in collision processes, the applied force is rarely constant. If that s the case, what can we do? 11

12 Usually, calculation of impulse makes use of the average net force that is applied to the object. Alternatively, if the force-time diagram of an object is given, impulse can be determined by measuring the area under the curve. The same amount of impulse exerted by an average force over the same period of time p = F av t The impulse exerted by the varying force F on the object over the time t 12

13 Example: A tennis player hits a ball with a racket, giving it a speed of 43 m/s. If the ball has a mass of kg and is in contact with the racket for s, what is the average force exerted on the ball by the racket? The change of momentum is p = m v = = 3.7 kgm/s Since impulse is the change of momentum, we have F = p t = = 820 N Hence, the average force is F av = 820 N 13

14 The concept of impulse and average force is very important to engineering. Consider the situation where a person is involved in a car accident. Does the crumple zone in front of the car make a difference in the outcome? 14

15 3. Conservation of momentum Consider the collision of two balls, object 1 and object 2, with the masses of m 1 and m 2, and initial velocities of v 1 and v 2 respectively. v 1 v 2 m 1 m 2 During the collision, forces are acting on these objects. They are equal in magnitude but opposite in directions, according to the Newton s third law. F 1 F 2 m 1 m 2 F 1 = F 2 15

16 Since the time in which the force F 1 is acting on object 1 should be equal to the time in which the force F 2 is acting on object 2, we can denote the contact time by t, and write F 1 t = F 2 t This is actually the impulse experienced by each of these objects. It can be rewritten in terms of momentum: p 1 = p 2 m 1 v 1 = m 2 v 2 m 1 v 1 v 1 = m 2 (v 2 v 2 ) 16

17 Rearranging the equation yields the following relationship: m 1 v 1 + m 2 v 2 = m 1 v 1 + m 2 v 2 It tells us that the two balls experience the equal yet opposite change of impulse and, in turns, momentum. This can be generalized into the law of conservation of momentum which states that: For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. 17

18 A short yet important reminder The law of conservation can only be applied to isolated system, in which all forces are conservative (i.e., the forces which are not applied by outsiders)! If other forces such as friction are involved, then the system is no longer isolated, and the law of conservation will not be applicable. Consider the following cases: What s the difference? How can we explain it? 18

19 Example: Someone throws a heavy ball to you when you are standing on a skateboard. The ball has a mass of 6.0 kg and is thrown at a speed of 3.4 m/s. The combined mass of the skateboard and you is 68 kg, and you are initially at rest. What is your velocity on the skateboard after catching the ball? By the conservation of momentum, the initial momentum of the ball is equal to the final sum of the momenta of the ball, the skateboard and you. Hence, m b v b = m b + m s + m y v t = ( )v t v t = (6.0)(3.4) (74) = 0.28 m/s 19

20 4. One-dimensional collisions The formula for the law of conservation of momentum can be applied to linear (or one-dimensional) collisions. In fact in Physics 12 you will see that the formula can be extended to describe two-dimensional collisions (that is, collisions that happen on a plane); however, some modification is needed. There are three types of collisions that we will discuss in this chapter, namely: (1) Collisions in which objects get bounced apart (2) Collisions in which objects stick together (3) Explosion 20

21 (a) Collision in which objects bounce apart The sum of individual momenta of objects is equal to the sum of individual momenta after collision. The two objects move at different velocities after collision. Example: A 0.15 kg blue ball moving at 8.0 m/s to the right hits a 0.10 kg red billiard ball at rest. If the blue ball continues to move to the right at 2.5 m/s, determine the velocity of the red ball. Due to the conservation of momentum p b + p r = p b + p r = v r v r = 8.3 m/s The red ball moves at 8.3 m/s to the right after collision. 21

22 (b) Collision in which objects stick together Momentum during the collision is conserved, yet the objects stick together and move at the same speed after the collision. Example: Two balls of clay, a blue one being 2.3 kg and the second red one being 5.6 kg, hit each other and stick together. If the blue one is moving to the right at 12 m/s, and the red is moving at 8.1 m/s to the left, determine their final velocity. Due to the conservation of momentum, p r + p b = p r + p b = ( )v r+b v r+b = 2. 2 m/s The balls move at 2.2 m/s to the left together. 22

23 (c) Explosion Explosion is a process in which two or more objects are moving apart from one another due to the forces that are exerted on one another. Strictly speaking, explosion is not a collision of objects but is somehow related to it. During the process, momentum is conserved. Each object may have its own individual, non-zero momentum value after explosion, but the sum of these values must be equal to the initial momentum value of the original object. In one-dimensional situation, the above-mentioned constraint implies that some pieces may carry a negative momentum after explosion. 23

24 Example: 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? Since the initial momentum is zero, the sum of momentum after the person throwing the ball must still be zero. p i = p p + p b = 68.0 v p + (6.0)(4.2) 0 = 68.0 v p + (6.0)(4.2) v p = 0.37 m/s The negative velocity of the person means he is sliding to the left. 24

25 Practice: A 5.0 g pellet is compressed against a spring in a gun of mass 300 g. The spring is released and the gun allowed to recoil with no friction as the pellet leaves the gun. If the speed of the recoiling gun is 8.0 m/s, what is the speed of the pellet? [480 m/s] Practice: James Bond is skiing along being pursued by Goldfinger, also on skis. Assume no friction. Mr. Bond, at 100 kg, fires backward a 40 g bullet at 800 m/s. Goldfinger, at 120 kg, fires forward at Bond with a similar weapon. What is the relative velocity change after the exchange of 6 shots each? No bullet hit Bond or Goldfinger. [3.48 m/s] 25

26 Practice: A freight train is being assembled in a switching yard. Car 1 has a mass of 65,000 kg and moves at a velocity of m/s. Car 2, with a mass of 92,000 kg and a velocity of +1.3 m/s, overtakes car 1 and couples to it. Neglecting friction, find the common velocity of the cars after they become coupled. [+1.1 m/s] 26

27 Challenging question: A 1.2 kg croquet ball moving at 2.0 m/s is struck from behind by the impulse force as shown. What is the final velocity of the croquet ball? Force/N [10.3 m/s] Time/s 27