LINEAR MOMENTUM. Contents. (A. Savas ARAPO GLU) July 16, Introduction 2. 2 Linear Momentum 2

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1 LINEAR MOMENTUM (A. Savas ARAPO GLU) July 16, 2017 Contents 1 Introduction 2 2 Linear Momentum 2 3 Impulse & Collisions Collison of Particles Collison Types Center of Mass of a System of Particles 7 5 Motion of Systems of Particles 9 6 The Systems of Variable Mass 10 7 Symmetries and the Conservation Principles 14 1

2 1 Introduction In this part of the course we will introduce a new quantity (momentum) and write Newton's Second Law in terms of this quantity. The procedure is very similar to the one in the Work & Energy part. This new quantity will be especially useful, together with its conservation, in considering the collision problems, motion of rigid bodies, variable mass problems. 2 Linear Momentum For a point particle of mass m, the linear momentum is dened as Note that p = m v (1) it is a vectorial quantity (and we will see that it is really a very useful quantity, like W, K, and U, in analyzing some mechanical systems), its SI unit is kg.m/s = N.s. The Newton's Second Law can be written in terms of linear momentum: F net = m a = m d v dt = }{{} m:cnst d dt (m v) F net = d p dt. (2) Although we have derived Eq.(2) for constant mass, this form of Newton's Second Law is applied in variable mass problems (i.e. M = M(t)). Another important point about the linear momentum is its conservation, which simply follows from Eq.(2): If F net = 0 d p dt = 0 p is constant in time p is conserved, i.e. p i = p f, (3) that is, if the net force acting on the system is zero, the momentum is conserved. 2

3 Example: Two blocks of masses m and 3m are placed on a frictionless, horizontal surface. A light spring is attached to the more massive block, and the blocks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after that happens, the block of mass 3m moves to the right with a speed of 2.0 m/s. (a) What is the velocity of the block of mass m? (b) Find the system's original elestic potential energy, taking m = kg. Example: A small block of mass m 1 = kg is released from rest at the top of a frictionless, curveshaped wedge of mass m 2 = 3.00 kg, which sits on a frictionless, horizontal surface as shown in the gure. When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right as shown. (a) What is the velocity of the wedge after the block reaches the horizontal surface, i.e. v 2 =? (b) What is the height h of the wedge? 3

4 3 Impulse & Collisions F~net d~p = dt Z p ~f Z tf d~p = F~net dt, (4) ti p ~i and de ning J~net : the impulse of the force F~net, J~net Z tf F~net dt, (5) ti we can re-write the previous integral equality as J~net = ~p. (6) This is called as the Impulse-Momentum Theorem, which is especially important to analyze the collison processes. Obviously, the are under the F t graph is the impulse transferred to/from the object: The average force concept is especially useful for the case of impulsive collisions (a large for acting in a very small interval of time); since the time interval t is very small, it is di cult to model the force by an analytic function in this time interval, and the real impulsive force is generally represented by its (approximate) average value: 4

5 It is very interesting and illuminating to compare the construction above with the similar construction in the Work & Energy part: W net xf x i Fnet d x and W net = K. (7) A short review on the approaches to analyze the motion of a single particle: There are (one can say) three coequal theories of motion for a single particle: Newton's second law, stating that the total force on the particle causes its acceleration; the workenergy theorem, stating that the total work on the particle causes its change in kinetic energy; and the impulsemomentum theorem, stating that the total impulse on the particle causes its change in momentum. They are coequal in the sense that the Work-Energy Theorem and Impulse-Momentum Theorem actually is the reformulation of Newton's Second Law, F net = m a, in terms of some other quantites, and each form is appropriate for a specic class of problems through the use of some conservation laws. Example: After a kg rubber ball is dropped from a height of 1.75 m, it bounces o a concrete oor and rebounds to a height of 1.50 m. (a) Determine the magnitude and direction of the impulse delivered to the ball by the oor. (b) Estimate the time the ball is in contact with the oor and use this estimate to calculate the average force the oor exerts on the ball. 3.1 Collison of Particles For simplicity we will consider the collision of two particles but what we will derive in the following will also be valid for the collision of more than two particles. For each particle, dene the change in the momentum as p p p, (8) where p is the momentum of the particle just after the collision and p is the momentum of the particle just before the collision. Then the total momentum change of the two particles during the collision is p = p 1 + p 2, (9) = tf t i Fnet,1 dt + tf t i tf Fcollision,1 dt + }{{} = F collision,2 t i Fnet,2 dt, (10) tf t i Fcollision,2 dt, (11) p = 0 p i = p f in collisions. (12) 3.2 Collison Types 5

6 In the previous section we have seen that: provided that the magnitude of the impulsive collision forces exceeds the magnitude of all the other forces present in the system, the linear momentum during the collision is conserved. The following classication of collisions is very useful: Elastic Collision: p and K are conserved, Inelastic Collision: only p is conserved, Completely Inelastic Collision: Particles stick together after the collision and move as a single particle, and only p is conserved. Example: A kg puck, initially at rest on a horizontal, frictionless surface, is struck by a kg puck moving initially along the x-axis with a speed of 2.00 m/s. After the collision, the kg puck has a speed of 2.00 m/s at an angle of θ = 53 o to the positive x-axis. (a) Determine the velocity of the kg puck after the collision. (b) Is this collision elastic, or inelastic? 6

7 4 Center of Mass of a System of Particles The center of mass concept is very useful in understanding the motion of rigid/extended objects. But we start with the simplest cases as usual to introduce the rudiments of the concept. First consider two point masses m 1 and m 2 located on the x-axis at points x 1 and x 2, respectively. The center of mass coordinate of this two-particle system is dened by the average of the coordinates weighted by masses (called `weighted mean') x cm m 1x 1 + m 2 x 2 m 1 + m 2. (13) Note that x 1 < x cm < x 2. The generalization of this averaging process to the case of N-particles (m 1, m 2,, m N ) in 3-dimensional space is r cm i=1 m i r i. (14) i=1 m i Note that this equation has 3 components, for x cm, y cm, and z cm. 7

8 Finally, we will generalize to the case of rigid bodies which are non-deformable solid objects. In this case, we simply make the following replacement mi r i r dm, (15) where dm is the innitesimal mass! Then r cm size r dm M, (16) where M is the total mass of the rigid body. Example: The center of mass of a rod of (M, L): (a) Show that the center of mass of a uniform rod of (M, L) lies midway between its ends. (b) Assume that the rod is not uniform and its mass density is given as λ(x) = αx where α is a constant. If this rod has total mass M and length L, then rst calculate α in terms of M and L, and then nd the position of the center of mass. Example: A semicircular wire of (M, R): Find the coordinates of the center of mass of the wire, i.e. (x cm, y cm ). 8

9 5 Motion of Systems of Particles In this section we will learn the physical signicance and utility of the CM concept. We can shortly expalin it like this: For example, for a system of N point particles, m 1, m 2,, m N, } { } to describe the motion we can determine the motion we can represent the whole system OR by the CM and describe of a system of particles of each particle as a fn of time the motion of this special point Supposing that all the particles of the system are in motion, we can dene the center of mass velocity, v cm, v cm d r cm = 1 d r i mi dt M dt = 1 mi v i, (17) M then M v cm = m i v i. (18) Note that ths equation is the mathematical realization of the idea of describing the system either by a single particle of mass M or by N point particles, m 1, m 2,, m N. Similarly, we can also dene the center of mass acceleration, a cm, a cm d v cm dt = = 1 M mi a i, (19) = 1 M Fi, (20) = 1 M ( F ext i + F int i ), but F int i = 0 why? (21) M a cm = F ext i = F net. (22) A very important result! It says that the CM of a system of particles moves as though all the mass of the system were concentrated at the CM and all external forces were applied at the CM. Example: The gure on the right shows an overhead view of the initial conguration of two pucks of mass m on frictionless ice. The pucks are tied together with a string of length l, and negligible mass. At time t = 5.0 s, a constant force of magnitude F begins to pull to the right on the center point of the string. At time t, the moving pucks strike each other and stick together. At this time, the force has moved through a distance d, and the pucks have attained a speed v. (a) What is v in terms of F, d, l, and m? (b) How much of the energy transferred into the system by work done by the force has been transformed to internal energy? 9

10 6 The Systems of Variable Mass If the mass of a system is changing continuously with time, i.e. if m = m(t), the Newton's Second Law in the form F net = m a can not be applied to these systems. The problem of this type is formulated for the motion of a rocket and it is called as rocket motion. In the gure, v is the velocity of the rocket relative to the ground, u is the velocity of the exhaust gas relative to the ground; then the change in momentum between t and t + t is: p = p f p i, (23) and the time rate of change of momentum is, = {M( v + v) + m u} (M + m) v, (24) = M v + m( u v); (25) p lim t 0 t d p dt { = lim M v t 0 t + m t = F ext = M(t) d v dt where the conservation of mass implies that M + m = constant dm dt we get F ext = M(t) d v dt + ( v u)dm = M(t) d v dt dt u dm dt } ( u v), (26) + ( u v)dm dt, (27) = dm, and nally dt, (28) where u is the velocity of the exhaust gas relative to the rocket 1. This equation is generically known as Rocket Equation. 1 Generally u is a constant characteristic of the rocket engine. 10

11 Example: A rocket moving in space far from all other objects, has a speed of m/s relative to the earth. Its engines are turned on, and fuel is ejected in the direction opposite to the rocket's motion at a speed of m/s relative to the rocket. (a) What is the speed of the rocket relative to the earth once the rocket's mass is reduced to the half its mass befor ignition? (b) What is the thrust on the rocket if it burns fuel at a rate of 50 kg/s? (Thrust is the force exerted on the rocket by the ejected exhaust gas.) Example: A garden hose is held as shown in the gure. The hose is originally full of motionless water. What additional force is necessary to hold the nozzle stationary after the ow is turned on if the discharge rate is 0, 6 kg/s with a speed of 25.0 m/s? 11

12 Example: As shown in the gure, a bullet of mass m and speed v passes completely through a pendulum bob of mass M. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a sti rod (not a string) of length l and negligible mass. What is the minimum value of v such that the pendulum bob will barely swing through a complete vertical circle? Example: A cannon is rigidly attached to a carriage, which can move along horizontal rails but is connected to a post by a large spring, initially unstretched and with force constant k = N/m as shown in the gure. The cannon res a 200 kg projectile at a velocity of 125 m/s directed 45.0 o above the horizontal. (a): Assuming that the mass of the cannon and its carriage is 5000 kg, nd the recoil speed of the cannon. (b): Determine the maximum extension of the spring. (c): Find the maximum force the spring exerts on the carriage. (d): Consider the system consisting of the cannon, carriage, and projectile. Is the momentum of this system conserved during the ring? Why or why not? Example: A mass m 1 = 1 kg sliding with a speed of v 1 = 5 m/s on a horizontal oor collides with another mass m 2 = 4 kg sliding with a speed of v 2 = 3 m/s on the same oor as shown in the Figure 1. The speed of m 1 just after the collision is again 5 m/s and it is moving in the direction shown in the Figure 2. (a) What is the direction of motion of mass m 2 after the collision, i.e. θ =? (b) Calculate the change in the momentum during the collision for each mass. (c) Find the velocity of the center of mass of the system before and after the collision. (d) Is this collision elastic, or inelastic? 12

13 7 Symmetries and the Conservation Principles Newton's Laws have led us to two conservation principles up to now: the conservation of (mechanical) energy and the conservation of (linear) momentum. (There is a third conservation principle, the conservation of angular momentum, which we will consider in the Rotations part.) The classical mechanics is not always valid: for objects as small as atoms, quantum mechanics replaces classical mechanics; and for objects traveling near the speed of light, Einstein's special theory of relativity is applicable. However, even when these modern theories replace Newton's classical theory, the conservation principles remain valid - the conservation principles transcend the mechanical theories. Thus the question naturally emerging at this point is do the conservation principles have a more fundamental basis than the mechanical theories? YES! The conservation principles are connected to certain symmetries in the Universe: 1. The uniformity of time (temporal symmetry) energy conservation, 2. The homogeneity of space (spatial symmetry) momentum conservation, 3. The isotropy of space (spatial symmetry) angular momentum conservation. 13