PS113 Chapter 7. Impulse and Momentum

Transcription

1 PS113 Chapter 7 Impulse and Momentum 1 The impulse-momentum theorem There are many situations in which the force acting on a object is not constant, but varies with time. The resulting motion can be simply analyzed by introducing a new concept called the impulse. Definition of impulse: The impulse of a force is the product of the average force F and the time interval during which the force acts: Impulse = F where the SI unit of impulse is: newton second (N s). Using this definition, we hope to describe a physical quantity using a constant force instead of a time varying force. From Newton s 2 nd law, we have for a constant force: F = m (a ) = m(v v o ) = mv }{{} mv }{{ o } final momentum initial momentum We define a new physical quantity called the momentum mv, a quantity that changes in proportion to the impulse F. The definition of linear momentum: The linear momentum p of an object is the product of the object s mass m and velocity v: p = mv The linear momentum is a vector quantity that points in the same direction as the velocity. The SI unit of momentum is: kilgoram meter/second (kg m/s). 1

2 The Impulse-Momentum theorem states that when a net force acts on an object, the impulse of the net force is equal to the change in momentum of the object: F = mv mv o (1) Impulse = Change in momentum The impulse-momentum theorem also gives us another way to write Newton s 2 nd Law. We can write it as: Problem 2: F = p The average force is the time rate of change of the momentum. A model rocket is constructed with a motor that can provide a total impulse of 29.0 N s. The mass of the rocket is kg. What is the speed that this rocket achieves when launched from rest? Neglect the effects of gravity and air resistance. (2) 2 The principle of conservation of linear momentum Let s imagine two objects with masses m 1 and m 2 and initial velocities v 01 and v o2 collide and exert a force on each other. They leave the collision with velocities v f1 and v f2 respectively. The total momentum before the collision can be written as: P o = m 1 v o1 + m 2 v o2 The total momentum after the collision can be written as: P f = m 1 v f1 + m 2 v f2 The total change in momentum can be written as: P = P f P o = (m 1 v f1 + m 2 v f2 ) (m 1 v o1 + m 2 v o2 ) 2

3 Rearranging terms, we can write the total change in momentum as: P = P f P o = (m 1 v f1 m 1 v 01 ) + (m 2 v f2 m 2 v o2 ) = ( p 1 ) + ( p 2 ) If we divide both sides by the elapsed time, we have: Using Eq. 2, we can identify: P = p 1 + p 2 (3) p 1 = F 12 as the force on 1 due to 2, and p 2 = F 21 as the force on 2 due 1 Rewriting Eq. 3, we find that: P = F 12 + F 21 (4) 3

4 However, we know from Newton s 3 rd Law, that F 12 = F 21 So, Eq. 4 becomes: P = 0 (5) What does Eq. 5 mean? 1. The change in the total momentum P for a closed system of particles interacting with each other, is zero. The total momentum P does not change. 2. The Principle of conservation of linear momentum states: The total linear momentum of an isolated system remains constant (is conserved). An isolated system is one for which the vector sum of the external forces acting on the system is zero, that is, P/ = 0 Problem 16: In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer space. From stationary positions they push against each other. Bonzo flies off with a velocity of +1.5 m/s, while Ender recoils with a velocity of -2.5 m/s. (a) Without doing any calculations, decide which person has the greater mass. Give your reasoning. (b) Determine the ratio m Bonzo /m Ender of the masses of these two enemies. 4

5 3 Collisions in one dimension As mentioned in the previous section, the total linear momentum is conserved when two objects collide, provided they constitute an isolated system (e.g., no external forces) When objects collide, their collisions are classified according to whether the total kinetic energy (KE) changes. 1. Elastic collisions One in which the total kinetic energy of the system after the collision is equal to the total kinetic energy before the collision. If m 2 is initially at rest (v 02 = 0), then, v f1 = v f2 = ( m1 m 2 ( 2m1 ) ) v 01 v Inelastic collisions One in which the total kinetic energy of the system is not the same before and after the collision; if the objects stick together after colliding, the collision is said to be completely inelastic. If m 2 is initially at rest (v 02 = 0), then, v f = ( ) m 1 v 01 Problem 29: A 2.50-g bullet, traveling at a speed of 425 m/s, strikes the wooden block of a ballistic pendulum, such as that in Figure The block has a mass of 215 g. (a) Find the speed of the bullet/block combination immediately after the collision. (b) How high does the combination rise above its initial position? 5

6 4 Collisions in two dimensions We are not covering this section 5 Center of mass In the previous section, we observed the changes of momentum for particles 1 and 2 ( p 1 and p 2 ) inside a closed system. The result of these interactions led to p 1 + p 2 = 0. Now, we would like to observe the motion of these two particles as an observer outside the closed system and treat the two particles (m 1 and m 2 ) as if they were one particle whose mass (M) is concentrated at a single point. To do this, we introduce the concept of center-of-mass. 6

7 The position of the center of mass for two particles (m 1 and m 2 ) along the x-axis at (x 1 and x 2 ) is defined to be: x cm = m 1x 1 + m 2 x 2 If the center of mass appears to be constant in time, we can say that, We are in the center-of-mass reference frame. (i.e., x cm is unchanged). The particles may be moving individually, but there center of mass is not moving. The other possibility is that two-particles are moving in such a way that their center of mass appears to be moving. In this case, the change in center-of-mass can be written as: x cm = m 1 x 1 + m 2 x 2 Furthermore, we can write the velocity of the center mass as V cm = x cm / : V cm = x cm = m 1( x 1 /) + m 2 ( x 2 /) = m 1v 1 + m 2 v 2 = P cm M Finally, we can write that the momentum of the two-particle system is: P cm = MV cm (6) which is the same as our definition for a one-particle system. We can extend Eq. 6 to include more than two particles where: P cm = p 1 +p 2 +p 3 + and M = m 1 +m 2 +m 3 + (7) 7

8 Problem 50: John s mass is 86 kg, and Barbara s is 55 kg. He is standing on the x axis at x J = +9.0 m, while she is standing on the x axis at x B = +2.0 m. They switch positions. How far and in which direction does their center of mass move as a result of the switch? 8