Linear Momentum Collisions

Transcription

1 Linear Momentum Collisions Lana Sheridan De Anza College Nov 8, 2017

2 Last time applying the rocket equation conservation of momentum in isolated systems nonisolated systems impulse average force

3 Overview collisions elastic vs. inelastic collisions

4 Collisions A major application of momentum conservation is studying collisions. This is not just useful for mechanics but also for statistical mechanics, subatomic physics, etc. For our purposes, there are two main kinds of collision: elastic inelastic

5 Collisions If two objects collide and there are no external forces, then the only forces each object experiences are internal forces. Internal forces obey Newton s third law Momentum is conserved.

6 Collisions If two objects collide and there are no external forces, then the only forces each object experiences are internal forces. Internal forces obey Newton s third law Momentum is conserved. This is true for both elastic and inelastic collisions. (So long as there is no external net force.)

7 Collisions 9.4 Colli Collisions can occur in macroscopic systems through contact. In this section happens when S S which two part F 21 F 12 interaction for m 1 m 2 so we can use t a A collision described in F generalized be hence meaning scale (Fig. 9.5b p of a helium at ++ each other due 4 and never com He When two p b impulsive force 9.3. Regardless

8 Collisions 9.4 Colli 9.4 Collisio Collisions can occur in macroscopic systems through contact. In this section happens when In this section, we S S which two part F 21 F 12 happens when two interaction for S S m 1 which two particle F 21 m 2 F 12 interaction so we can forces use ta m a 1 m 2 so we A can collision use the i And collisions can occur through purely repulsive forces, described even if in F a A collision may two particles never make contact. described generalized in Figur be generalized hence meaning becau scale (Fig. 9.5b p hence meaningles scale of (Fig. a helium 9.5b) su p at ++ of each a helium other atom) due ++ each 4 and other never due com to He 4 and never When come two in He p When two part b impulsive force b impulsive forces m 9.3. Regardless

9 Collisions The internal force on each particle can vary over a collision.

10 Collisions The internal force on each particle can vary over a collision. In particular, this is clearly the case for repulsive force collisions. Let the charge on a proton be q. The the force between a proton and an alpha particle (Helium nucleus) will vary with the separation distance r : F = k(q)(2q) r 2 The force increases as the two particles approach one another, then decreases as they move apart again.

11 Types of Collision Elastic collisions are collisions in which kinetic energy is conserved.

12 Types of Collision Elastic collisions are collisions in which kinetic energy is conserved. K i = K f truly elastic collisions do not occur at macroscopic scales many collisions are close to elastic, so we can model them as elastic

13 Types of Collision Elastic collisions are collisions in which kinetic energy is conserved. K i = K f truly elastic collisions do not occur at macroscopic scales many collisions are close to elastic, so we can model them as elastic Inelastic collisions do not conserve kinetic energy.

14 Types of Collision Elastic collisions are collisions in which kinetic energy is conserved. K i = K f truly elastic collisions do not occur at macroscopic scales many collisions are close to elastic, so we can model them as elastic Inelastic collisions do not conserve kinetic energy. energy is lost as sound, heat, or in deformations all macroscopic collisions are somewhat inelastic

15 Types of Collision Elastic collisions are collisions in which kinetic energy is conserved. K i = K f truly elastic collisions do not occur at macroscopic scales many collisions are close to elastic, so we can model them as elastic Inelastic collisions do not conserve kinetic energy. energy is lost as sound, heat, or in deformations all macroscopic collisions are somewhat inelastic when the colliding objects stick together afterwards the collision is perfectly inelastic

16 Because all veloci a Before the collision, the alternative app particles move separately. represented tion of Equatio by th direction. factor 1 We sha S S 2 in Equ v tive if it moves to 1i v2i After the collision, the left and 2 on th In a typical pro particles continue to move m 1 m 2 ties, and Equatio separately with new velocities. a alternative approa S S tion Factoring of Equation both v1f v2f factor 1 2 in Equati After the collision, the left and 2 on the r b particles continue to move For two particles involved separately in with an elastic new velocities. collision, we can write Next, two let us equations: Figure 9.7 Schematic representation of van 1f elastic head-on v2fcolli- p i = p f sion m 1 between v 1i + m 2 vtwo 2i = particles. m 1 v 1f + m 2 v lar way to obta S S Factoring both sid 2f Elastic Collisions K i = K f 1 b To Next, obtain let us our 2 m 1(v 1i ) m 2(v 2i ) 2 = 1 2 m 1(v 1f ) m 2(v 2f ) 2 sep Figure 9.7 Schematic represen- lar way to obtain (Assume the masses of the two particles remain unchanged.) tation of an elastic head-on colli-

17 Elastic Collisions p i = p f m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f 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 These are independent equations can solve for multiple unknowns.

18 Elastic Collisions p i = p f m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f 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 These are independent equations can solve for multiple unknowns. These equations can be solved however you like.

19 Elastic Collisions One convenient trick might be to remove the quadratic terms. This only works in 1-dimensional collisions. m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f 1 2 m 1(v 1i ) m 2(v 2i ) 2 = 1 2 m 1(v 1f ) m 2(v 2f ) 2

20 Elastic Collisions One convenient trick might be to remove the quadratic terms. This only works in 1-dimensional collisions. m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f 1 2 m 1(v 1i ) m 2(v 2i ) 2 = 1 2 m 1(v 1f ) m 2(v 2f ) 2 From those two, this equation can be derived: (v 1i + v 1f ) = (v 2f + v 2i ) This only applies to 1-dimensional collisions! (The v s are assumed to lie along a single direction and can be positive or negative) Try to derive it; use the textbook if you get stuck.

21 the front car and gives it a big push, increasing its speed to 4.00 m/s southward. The remaining three cars continue moving south, now at 2.00 m/s. (a) Find the initial speed of the four cars. (b) By how much did the potential energy within the body of the actor change? (c) State the relationship between the process described here and the process in Problem 25. Example - Elastic Particle Collision Page 285, # A neutron in a nuclear reactor makes an elastic, headon collision with the nucleus of a carbon atom initially M at rest. (a) What fraction of the neutron s kinetic energy is transferred to the carbon nucleus? (b) The initial kinetic energy of the neutron is J. Find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision. (The mass of the carbon nucleus is nearly 12.0 times the mass of the neutron.) 28. A 7.00-g bullet, when fired from a gun into a 1.00-kg block of wood held in a vise, penetrates the block to a depth of 8.00 cm. This block of wood is next placed on a frictionless horizontal surface, and a second 7.00-g bullet is fired from the gun into the block. To what depth will the bullet penetrate the block in this case? AMT W w m a P st ti o n m h 34. (a Q/C a

22 Example - Elastic Particle Collision (a) Fraction of neutron s kinetic energy transferred to carbon nucleus?

23 Example - Elastic Particle Collision (a) Fraction of neutron s kinetic energy transferred to carbon nucleus? K n K n,i

24 Example - Elastic Particle Collision (a) Fraction of neutron s kinetic energy transferred to carbon nucleus? K n K n,i p tot = 0 ; K tot = 0

25 Example - Elastic Particle Collision (a) Fraction of neutron s kinetic energy transferred to carbon nucleus? K n K n,i p tot = 0 ; K tot = 0 K n K n,i = (b) Final KE of neutron and carbon, given K n,i = J.

26 Example - Elastic Particle Collision (a) Fraction of neutron s kinetic energy transferred to carbon nucleus? K n K n,i p tot = 0 ; K tot = 0 K n K n,i = (b) Final KE of neutron and carbon, given K n,i = J. K n,f = J K c,f = J

27 Question Quick Quiz 9.6 A table-tennis ball is thrown at a stationary bowling ball. The table-tennis ball makes a one-dimensional elastic collision and bounces back along the same line. Compared with the bowling ball after the collision, the table-tennis ball has which of the following? (A) a larger magnitude of momentum and more kinetic energy (B) a smaller magnitude of momentum and more kinetic energy (C) a smaller magnitude of momentum and less kinetic energy (D) the same magnitude of momentum and the same kinetic energy 1 Serway & Jewett, page 259.

28 Question Quick Quiz 9.6 A table-tennis ball is thrown at a stationary bowling ball. The table-tennis ball makes a one-dimensional elastic collision and bounces back along the same line. Compared with the bowling ball after the collision, the table-tennis ball has which of the following? (A) a larger magnitude of momentum and more kinetic energy (B) a smaller magnitude of momentum and more kinetic energy (C) a smaller magnitude of momentum and less kinetic energy (D) the same magnitude of momentum and the same kinetic energy 1 Serway & Jewett, page 259.

29 Inelastic Collisions For general inelastic collisions, some kinetic energy is lost.

30 Inelastic Collisions For general inelastic collisions, some kinetic energy is lost. We have only the conservation of momentum to work from: p i = p f

31 Inelastic Collisions For general inelastic collisions, some kinetic energy is lost. We have only the conservation of momentum to work from: p i = p f m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f

32 Inelastic Collisions For general inelastic collisions, some kinetic energy is lost. We have only the conservation of momentum to work from: p i = p f m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f This means we can only solve for 1 unknown. (Or 2 unknowns if we have a 2-D collision, 2 component equations.) However, there is a special case where we have more information: perfectly inelastic collisions.

33 sion, we can ntum Perfectly of the Inelastic Collisions a (9.15) (9.14) 1 m Before the collision, the 2 particles move separately. After the S m vcollision, S 1i vthe 2i particles 1 move together. and S v a 2i (9.15) articles S v 1f and v After the collision, f the system particles m 1 move m 2 together. s tion S v 1i and in S v 2i b wo particles velocity! ties, S S v 1f and Figure 9.6 Schematic representation of a perfectly m 1 m 2 inelastic v f f the (9.16) system direction in head-on collision between two b (9.17) particles. Now the two particles stick together after colliding same final m2 p i = p f m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f

34 Perfectly Inelastic Collisions In this case it is straightforward to find an expression for the final velocity: So, m 1 v 1i + m 2 v 2i = (m 1 + m 2 )v f v f = m 1v 1i + m 2 v 2i m 1 + m 2

35 Summary collisions elastic collisions 3rd Collected Homework due Mon, Nov 20. (Uncollected) Homework Serway & Jewett, Ch 9, onward from page 285. Probs: 23, 25, 27, 29, 31