Linear Momentum 2D Collisions Extended or Composite Systems Center of Mass

Transcription

1 Linear Momentum 2D Collisions Extended or Composite Systems Center of Mass Lana Sheridan De Anza College Nov 13, 2017

2 Last time inelastic collisions perfectly inelastic collisions the ballistic pendulum

3 Overview 2D collisions center of mass finding the center of mass

4 Collisions in 2 Dimensions The conservation of momentum equation is a vector equation. It will apply for any number of dimensions that are relevant in a question. p i = p f m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f

5 Collisions in 2 Dimensions The conservation of momentum equation is a vector equation. It will apply for any number of dimensions that are relevant in a question. p i = p f m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f In particular, we can write equations for each component of the momentum. In 2-d, with x and y components: x : m 1 v 1ix + m 2 v 2ix = m 1 v 1fx + m 2 v 2fx y : m 1 v 1iy + m 2 v 2iy = m 1 v 1fy + m 2 v 2fy

6 Collisions in 2 Dimensions The conservation of momentum equation is a vector equation. It will apply for any number of dimensions that are relevant in a question. p i = p f m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f In particular, we can write equations for each component of the momentum. In 2-d, with x and y components: x : m 1 v 1ix + m 2 v 2ix = m 1 v 1fx + m 2 v 2fx y : m 1 v 1iy + m 2 v 2iy = m 1 v 1fy + m 2 v 2fy If it is an elastic collision: K i = K f 1 2 m 1(v 1i ) m 2(v 2i ) 2 = 1 2 m 1(v 1f ) m 2(v 2f ) 2

7 S x m implies that th m important sub 1 Finalize This answer is not the maximum compression of the spring because iar example the two in m each other at the instant shown in Figure 9.10b. Can you determine 2 the surface. maximum For com su a for conservatio Collisions in 2 Dimensions As an example, consider the case of a glancing collision. a Before the collision S v 1i m 1 m 2 After the collision The velocity of particle 1 is in the x-direction. v 1f sin θ φ θ v2f sinφ S v 1f v 1f cos θ v 2f cos φ S v 2f v 1i After the collision 9.5 Collisions in Two S Dimension In Section 9.2, we v 1f sin showed θ that the momentum where the thr of served when the system is isolated. sent, For any respectiv collis implies that the momentum v 1f cos in θeach and of the the directi veloci θ important subset of collisions takes place Let in a us plane. consi iar example involving φ multiple collisions collides of object with p v 2f cos φ surface. For such two-dimensional collisions, (Fig. 9.11b), we pa ob for conservation of momentum: moves at an an v2f sinφ S v 2f sion. Applying m 1 1ix 1 m 2 vthat 2ix 5 the m 1 initial v 1fx 1 b m 1 v 1iy 1 m 2 v 2iy 5 m 1 v 1fy 1Dp Figure 9.11 An elastic, glancing where collision the three between subscripts two particles. on the velocity compo Dp sent, respectively, the identification of the object (1 and the velocity component (x, y). Let us consider a specific two-dimensional proble collides with particle 2 of mass m 2 initially at rest as i (Fig. 9.11b), particle 1 moves at an angle u with respec moves at an angle f with respect to the horizontal. T sion. Applying the law of conservation of momentum v 1f

8 S x m implies that th m important sub 1 Finalize This answer is not the maximum compression of the spring because iar example the two in m each other at the instant shown in Figure 9.10b. Can you determine 2 the surface. maximum For com su a for conservatio Collisions in 2 Dimensions As an example, consider the case of a glancing collision. a Before the collision S v 1i m 1 m 2 After the collision The velocity of particle 1 is in the x-direction. x-components: y-components: v 1f sin θ φ θ v2f sinφ S v 1f v 1f cos θ v 2f cos φ S v 2f v 1i After the collision 9.5 Collisions in Two S Dimension In Section 9.2, we v 1f sin showed θ that the momentum where the thr of served when the system is isolated. sent, For any respectiv collis implies that the momentum v 1f cos in θeach and of the the directi veloci θ important subset of collisions takes place Let in a us plane. consi iar example involving φ multiple collisions collides of object with p v 2f cos φ surface. For such two-dimensional collisions, (Fig. 9.11b), we pa ob for conservation of momentum: moves at an an v2f sinφ S v 2f sion. Applying m 1 1ix 1 m 2 vthat 2ix 5 the m 1 initial v 1fx 1 b m 1 v 1iy 1 m 2 v 2iy 5 m 1 v 1fy 1Dp Figure 9.11 An elastic, glancing where collision the three between subscripts two particles. on the velocity compo Dp sent, respectively, the identification of the object (1 and the velocity component (x, y). Let us consider a specific two-dimensional proble collides with particle 2 of mass m 2 initially at rest as i (Fig. 9.11b), particle 1 moves at an angle u with respec moves at an angle f with respect to the horizontal. T sion. Applying the law of conservation of momentum m 1 v 1i = m 1 v 1f cos θ + m 2 v 2f cos φ 0 = m 1 v 1f sin θ m 2 v 2f sin φ v 1f

9 Example Car collision ollisions A 1500 kg car traveling east with a speed of 25.0 m/s collides at an intersection with a 2500 kg truck traveling north at a speed of 20.0 m/s. Find the direction and magnitude of the velocity of the wreckage after the collision, assuming the vehicles stick together after the collision. ceptualize the situation before long the positive x direction and y S v f ediately before and immediately, we ignore the small effect that and model the two vehicles as an ore the vehicles sizes and model tic because the car and the truck 25.0i ˆ m/s u x 20.0j ˆ m/s ng momentum in the x direction initial momentum of the system the car. Similarly, the total initial f the truck. After the collision, let espect to the x axis with speed v f. 1 Serway & Jewett, page 265. Figure 9.12 (Example 9.8) An eastbound car colliding with a northbound truck.

10 Example Car collision This is an inelastic collision.

11 Example Car collision This is an inelastic collision. x-components: y-components: Dividing (2) by (1): and v f = m 1 v 1i = (m 1 + m 2 )v f cos θ (1) m 2 v 2i = (m 1 + m 2 )v f sin θ (2) m 2 v 2i m 1 v 1i = tan θ θ = tan 1 ( m2 v 2i m 1 v 1i ) = 53.1 m 2 v 2i = 15.6 m/s (m 1 + m 2 ) sin(53.1)

12 Example Exploding Rocket A rocket is fired vertically upward. At the instant it reaches an altitude of 1000 m and a speed of v i = 300 m/s, it explodes into three fragments having equal mass. One fragment moves upward with a speed of v 1 = 450 m/s following the explosion. The second fragment has a speed of v 2 = 240 m/s and is moving east right after the explosion. What is the velocity of the third fragment immediately after the explosion? (What is the change in kinetic energy of the system of the rocket parts?)

13 Example Exploding Rocket p i = p f M v i = M 3 (v 1 + v 2 + v 3 ) Let j point in the the upward vertical direction, and i point east. v 3 = 3v i v 1 v 2 = j 450 j 240 i = ( 240 i j) m/s Or, 510 m/s at an angle of 62 above the horizontal to the west.

14 Example Exploding Rocket p i = p f M v i = M 3 (v 1 + v 2 + v 3 ) Let j point in the the upward vertical direction, and i point east. v 3 = 3v i v 1 v 2 = j 450 j 240 i = ( 240 i j) m/s Or, 510 m/s at an angle of 62 above the horizontal to the west. ( K = K f K i = 1 2 M( ) 1 2 (3M)(3002 ) = M, a positive number)

15 Center of Mass The center of mass of a system is the average position of the system s mass.

16 Center of Mass The center of mass of a system is the average position of the system s mass. If the system is a point-like particle, the center of mass is just the point-particle s location.

17 Center of Mass The center of mass of a system is the average position of the system s mass. If the system is a point-like particle, the center of mass is just the point-particle s location. We modeled many systems that were not point-like as if they were. We pretended they were points located at their center of mass.

18 Center of Mass The center of mass of a system is the average position of the system s mass. If the system is a point-like particle, the center of mass is just the point-particle s location. We modeled many systems that were not point-like as if they were. We pretended they were points located at their center of mass. For translational motion, it is as if all of the system s mass is concentrated at that one point and all external forces are applied at that point.

19 Center of Mass For translational motion, it is as if all of the system s mass is concentrated at that one point and all external forces are applied at that point.

20 Center of Mass For translational motion, it is as if all of the system s mass is concentrated at that one point and all external forces are applied at that point. That model worked up until now, but we could not model vibrations, rotations, or deformations.

21 Center of Mass For translational motion, it is as if all of the system s mass is concentrated at that one point and all external forces are applied at that point. That model worked up until now, but we could not model vibrations, rotations, or deformations. The translational motion is independent of all these other motions.

22 Center of Mass For a solid, rigid object: center of mass the point on an object where we can model all the mass as being, in order to find the object s trajectory; a freely moving object rotates about this point The center of mass of the wrench follows a straight line as the wrench rotates about that point.

23 Center of Mass 1 Figure from

24 Center of Mass c Figure 9.13 A force is applied For that systemto ofa two system particles: of two particles of unequal mass connected x CM = m 1x 1 + m 2 by x 2 a light, rigid rod. m 1 + m 2 r CM where r S i is th y m 1 x CM x 1 x 2 CM m 2 Figure 9.14 The center of mass of two particles of unequal mass on the x axis is located at x CM, a x Although lo what more cu ticles, the basi system contain Because sidered to hav of mass Dm i w mass is approx

25 Center of Mass For that system of two particles: x CM = m 1x 1 + m 2 x 2 m 1 + m 2 For more particles in 1 dimension: x CM = i m ix i or i m i x CM = 1 m i x i M where M is the total mass of all the particles in the system. i

26 Center of Mass This expression also gives us the x coordinate of the center of mass when we have more dimensions. x CM = 1 m i x i M i Likewise for y: y CM = 1 m i y i M i and z: z CM = 1 m i z i M where M is the total mass of all the particles in the system. i

27 Center of Mass Therefore, we can condense all three expressions into a single vector expression. r CM = 1 m i r i M i where r i = x i i + y i j + z i k is the displacement of particle i from the origin.

28 Summary 2D collisions center of mass Quiz tomorrow, in class. 3rd Collected Homework due Monday, Nov 20. (Uncollected) Homework Serway & Jewett, Look at example 9.15 on page 275. PREV: Ch 9, onward from page 275. Probs: 35, 37, 41, 43 Ch 9, onward from page 288. Probs: 67, 71, 77, 81 Read Chapter 9, if you haven t already.