When solving problems involving changing momentum in a system, we shall employ our general problem solving strategy involving four basic steps:

Transcription

1 10.9 Worked Examles Problem Solving Strategies When solving roblems involving changing momentum in a system, we shall emloy our general roblem solving strategy involving four basic stes: 1. Understand get a concetual gras of the roblem. 2. Devise a Plan - set u a rocedure to obtain the desired solution. 3. Carry our your lan solve the roblem! 4. Look Back check your solution and method of solution. We shall develo a set of guiding ideas for the first two stes. 1. Understand get a concetual gras of the roblem The first question you should ask is whether or not momentum is constant in some system that is changing its state after undergoing an interaction. First you must identify the objects that comose the system and how they are changing their state due to the interaction. As a guide, try to determine which objects change their momentum in the course of interaction. You must kee track of the momentum of these objects before and after any interaction. Second, momentum is a vector quantity so the question of whether momentum is constant or not must be answered in each relevant direction. In order to determine this, there are two imortant considerations. You should identify any external forces acting on the system. Remember that a non-zero external force will cause the momentum of the system to change, (Equation (10.4.9) above), F ext = d sys. (10.9.1) dt Equation (10.9.1) is a vector equation; if the external force in some direction is zero, then the change of momentum in that direction is zero. In some cases, external forces may act but the time interval during which the interaction takes lace is so small that the imulse is small in magnitude comared to the momentum and might be negligible. Recall that the average external imulse changes the momentum of the system I = F ext Δt int = Δ sys. (10.9.2) If the interaction time is small enough, the momentum of the system is constant, Δ 0. If the momentum is not constant then you must aly either Equation (10.9.1) or Equation (10.9.2). If the momentum of the system is constant, then you can aly Equation (10.7.5), =. (10.9.3) s ys, i sys, f 10-1

2 If there is no net external force in some direction, for examle the x -direction, the comonent of momentum is constant in that direction, and you must aly = (10.9.4) sys, x,i sys, x, f 2. Devise a Plan - set u a rocedure to obtain the desired solution Draw diagrams of all the elements of your system for the two states immediately before and after the system changes its state. Choose symbols to identify each mass and velocity in the system. Identify a set of ositive directions and unit vectors for each state. Choose your symbols to corresond to the state and motion (this facilitates an easy interretation, for examle (v,i ) 1 reresents the x -comonent of the velocity of object 1 in the initial x state and (v x, f ) 1 reresents the x -comonent of the velocity of object 1 in the final state). Decide whether you are using comonents or magnitudes for your velocity symbols. Since momentum is a vector quantity, identify the initial and final vector comonents of the momentum. We shall refer to these diagrams as momentum flow diagrams. Based on your model you can now write exressions for the initial and final momentum of your system. As an examle in which two objects are moving only in the x -direction, the initial x -comonent of the momentum is The final x -comonent of the momentum is If the x -comonent of the momentum is constant then = m (v ) + m (v ) +. (10.9.5) sys, x,i 1 x,i 1 2 x,i 2 = m ) + s ys, x, f 1 (v x, f 1 m 2 (v x, ) +. ( ) f 2 sys, x,i = sys, x, f. (10.9.7) We can now substitute Equations (10.9.5) and (10.9.6) into Equation (10.9.7), yielding m 1 (v x,i ) 1 + m (v ) + = m (v ) + m (v ) +. (10.9.8) 2 x,i 2 1 x, f 1 2 x, f 2 Equation (10.9.8) can now be used for any further analysis required by a articular roblem. For examle, you may have enough information to calculate the final velocities of the objects after the interaction. If so then carry out your lan and check your solution, esecially dimensions or units and any relevant vector directions. 10-2

3 Examle 10.5 Exloding Projectile An instrument-carrying rojectile of mass m 1 accidentally exlodes at the to of its trajectory. The horizontal distance between launch oint and the exlosion is x i. The rojectile breaks into two ieces that fly aart horizontally. The larger iece, m 3, has three times the mass of the smaller iece, m 2. To the surrise of the scientist in charge, the smaller iece returns to earth at the launching station. Neglect air resistance and effects due to the earth s curvature. How far away, x 3, f, from the original launching oint does the larger iece land? m 1 v 1 arabolic orbit î x = 0 x i v 2 v 3 m 2 m 3 x = 0 x i x = 0 x i x 3, f Figure 10.8 Exloding rojectile trajectories Solution: We can solve this roblem two different ways. The easiest aroach is utilizes the fact that the external force is the gravitational force and therefore the center of mass of the system follows a arabolic trajectory. From the information given in the roblem m 2 = m 1 / 4 and m 3 = 3m 1 / 4. Thus when the two objects return to the ground the center of mass of the system has traveled a distance R cm = 2x i. We now use the definition of center of mass to find where the object with the greater mass hits the ground. Choose an origin at the starting oint. The center of mass of the system is given by m 2 r + m r R = m 2 + m 3 cm. 10-3

4 ! So when the objects hit the ground R cm = 2x i î, the object with the smaller mass returns to the origin, r 2 = 0, and the osition vector of the other object is r! = x î. So using 3 3, f the definition of the center of mass, Therefore (3m 1 / 4)x 3, f î (3m 1 / 4)x 3, f î 3 2x i î = = = x 3, f î. m1 / 4 + 3m 1 / 4 m = x i. x 3, f 3 Note that the neither the vertical height above ground nor the gravitational acceleration g entered into our solution. Alternatively, we can use conservation of momentum and kinematics to find the distance traveled. Because the smaller iece returns to the starting oint after the collision, the velocity of the smaller iece immediately after the exlosion is equal to the negative of the velocity of original object immediately before the exlosion. Because the collision is instantaneous, the horizontal comonent of the momentum is constant during the collision. We can use this to determine the seed of the larger iece after the collision. The larger iece takes the same amount of time to return to the ground as the rojectile originally takes to reach the to of the flight. We can therefore determine how far the larger iece traveled horizontally. We begin by identifying various states in the roblem. Initial state, time t 0 = 0 : the rojectile is launched. State 1 time t 1 : the rojectile is at the to of its flight trajectory immediately before the exlosion. The mass is m 1 and the velocity of the rojectile is v! ˆ 1 = v 1 i. State 2 time t 2 : immediately after the exlosion, the rojectile has broken into two ieces, one of mass m 2 moving backwards (in the negative x -direction) with velocity!! v 2 = v 1. The other iece of mass m 3 is moving in the ositive x -direction with velocity! v 3 = v ˆ i, (Figure 10.8)

5 The initial momentum at time t 1 immediately before the exlosion is The momentum at time t 2 immediately after the exlosion is! sys (t 1! ) = m 1 v 1. (10.9.9)! sys!! 1!! (t ) = m v + m v = m v + 3 m v ( ) During the duration of the instantaneous exlosion, imulse due to the external gravitational force may be neglected and therefore the momentum of the system is constant. In the horizontal direction, we have that! m v 1! 3! = m v + m v. ( ) Equation ( ) can now be solved for the velocity of the larger iece immediately after the collision,! 5! v 3 = v 1. ( ) 3 The larger iece travels a distance 5 5 x = v t = v t = x. ( ) 3, f i 3 3 Therefore the total distance the larger iece traveled from the launching station is in agreement with our revious aroach. 5 8 x f = x + x = x, ( ) i i i 3 3 Examle 10.6 Landing Plane and Sandbag Figure 10.9 Plane and sandbag 10-5

6 A light lane of mass 1000 kg makes an emergency landing on a short runway. With its engine off, it lands on the runway at a seed of 40 m s -1. A hook on the lane snags a cable attached to a 120 kg sandbag and drags the sandbag along. If the coefficient of friction between the sandbag and the runway is µ k = 0.4, and if the lane s brakes give an additional retarding force of magnitude 1400 N, how far does the lane go before it comes to a sto? Solution: We shall assume that when the lane snags the sandbag, the collision is instantaneous so the momentum in the horizontal direction remains constant, =. ( ) x,i x,1 We then know the seed of the lane and the sandbag immediately after the collision. After the collision, there are two external forces acting on the system of the lane and sandbag, the friction between the sandbag and the ground and the braking force of the runway on the lane. So we can use the Newton s Second Law to determine the acceleration and then one-dimensional kinematics to find the distance the lane traveled since we can determine the change in kinetic energy. The momentum of the lane immediately before the collision is = i m ˆ v,i i ( ) The momentum of the lane and sandbag immediately after the collision is = (m + m )v î ( ) 1 s,1 Because the x - comonent of the momentum is constant, we can substitute Eqs. ( ) and ( ) into Eq. ( ) yielding m,i s )v v = (m + m,1. ( ) The seed of the lane and sandbag immediately after the collision is m v,i v = ( ),1 m + m s The forces acting on the system consisting of the lane and the sandbag are the normal force on the sandbag, N = N ˆ j, ( ) g,s g,s 10-6

7 the frictional force between the sandbag and the ground the braking force on the lane and the gravitational force on the system, f = f î = µ N î, ( ) k k k g,s F = F î, ( ) g, g, (m + m ) g = ( m + m s s )gˆ j. ( ) Newton s Second Law in the î -direction becomes F g, f k = (m + m s )a x. ( ) If we just look at the vertical forces on the sandbag alone then Newton s Second Law in the ĵ-direction becomes N m g = s 0. The frictional force on the sandbag is then f = µ N ˆ i k k g,s = µ m gi ˆ. ( ) k s Newton s Second Law in the î -direction becomes F g, µ k m s g = (m + m s )a x. The x -comonent of the acceleration of the lane and the sand bag is then F µ g, k m s g a x = ( ) m + m s We choose our origin at the location of the lane immediately after the collision, x (0) = 0. Set t = 0 immediately after the collision. The x -comonent of the velocity of the lane immediately after the collision is v x,0 = v,1. Set t = t when the lane just f comes to a sto. Because the acceleration is constant, the kinematic equations for the change in velocity is v x, f (t f ) v = a t f.,1 x 10-7

8 We can solve this equation for t = t f, where v x, f (t f ) = 0 t f = v / a t.,1 x Then the osition of the lane when it first comes to rest is v 2,1 x (t ) x (0) = v t + a t =. ( ) f,1 f x f 2 2 a x Then using x (0) = 0 and substituting Eq. ( ) into Eq. ( ) yields 2 1 (m x (t ) = + m )v s,1. ( ) f 2 (F + µ m g) g, k s We now use the condition from conservation of the momentum law during the collision, Eq. ( ) in Eq. ( ) yielding m 2 v 2,i x (t ) =. ( ) f 2(m + m s )(F g, + µ k m s g) Substituting the given values into Eq. ( ) yields (1000 kg) 2 (40 m s -1 ) 2 2 x (t f ) = = m. ( ) -2 2(1000 kg kg)(1400 N + (0.4)(120 kg)(9.8m s )) 10-8

9 MIT OenCourseWare htts://ocw.mit.edu 8.01 Classical Mechanics Fall 2016 For Information about citing these materials or our Terms of Use, visit: htts://ocw.mit.edu/terms.