Workbook for Introductory Mechanics Problem-Solving. To the Student

Transcription

1 Workbook for Introductory Mechanics Problem-Solving Daniel M. Smith, Jr. South Carolina State University To the Student The Workbook should help you to learn how to think about physics problems before you try using equations. The format is designed to carefully build each level of your problem-solving ability before moving you to the next level. So instead of working one problem at a time, you will work several problems at once. Here is advice on how to use this Workbook. 1. fter printing, place these materials in a binder before beginning work. (If possible, use double-sided or duplex printing.) 2. ttempt to answer every question by drawing or writing before looking at the answer. 3. Warning: Simply reading the questions, the answers, and the discussion without doing the work is a complete waste of time. Copyright by Daniel M. Smith, Jr., ll rights reserved. Sponsored by FIPSE, U.S. Department of Education

2 cknowledgments Students have provided invaluable assistance in word processing, and in creating graphics for this workbook. Thanks go to ndrick nderson, Kyle Herbert, Tyesia Pompey, Tarryn Reeves, and Keilah Spann. Professors Theodore Hodapp and Dave Maloney are thanked for their criticism and suggestions. Clip art is taken from the rt Explosion collection published by the Nova Development Corporation, Calabasas, C. I have had the pleasure of teaching and tutoring at South Carolina State University and Northeastern University, and in the process I have learned much from students about their difficulties in solving physics problems. Those who wish to further my enlightenment may send comments to dsmith@scsu.edu. Copyright by Daniel M. Smith, Jr. ll rights reserved. Students using College Physics, 4th edition by Jerry Wilson and nthony uffa may print out one copy of the Workbook material for their own use but may not otherwise copy or distribute the material in part or in whole by any means whatsoever, electronic or otherwise, without express written permission from the author and Prentice- Hall, Inc.

3 Chapter 6 MOMENTUM Wilson/uffa Chapter 6: Momentum and Collisions Unit 1 Graphical representation of momentum, and energy equations in a collision; statements of momentum, and energy conservation. Unit 2 Review and completion of problem solutions

4 6-2 Copyright by Daniel M. Smith, Jr. ll rights reserved. Students using College Physics, 4th edition by Jerry Wilson and nthony uffa may print out one copy of the Workbook material for their own use but may not otherwise copy or distribute the material in part or in whole by any means whatsoever, electronic or otherwise, without express written permission from the author and Prentice-Hall, Inc.

5 Momentum - Unit Problem lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.1 r v r v = 0 (a) before collision x(m) (b) during collision x(m) r v = 0 r v x(m) (c) after collision efore the collision, velocities are given by v r (pronounced vector v sub ) and v r. fter the collision, velocities will be denoted v r (pronounced vector v sub prime) and v r. In this chapter the prime symbol ( ) always refers to events after a collision (some textbooks use v r f instead of v r ). Draw vectors on diagram (b) to show the horizontal forces exerted on each block, being especially careful about the vector magnitudes. Describe the vectors in words. Workbook for Introductory Mechanics Problem-Solving Copyright by Daniel M. Smith, Jr. Sponsored by FIPSE (U.S. Department of Education)

6 6 4 Momentum - Unit Problem 6.1 lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.2 force that block exerts on bl ock force that block exerts on block x(m) Do your vectors have equal magnitudes? Yes No If No, then correct your diagram. How do you know that the vector magnitudes should be equal? Write your answer. 6.3 ecause of Newton s 3rd Law, the force exerted by block on block is equal in magnitude to the force exerted by block on block but opposite in direction. lock exerts the force F r on block ; block exerts the force F r on block. Write down the mathematical relationship between F r and F r.

7 Momentum - Unit Problem lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.4 Then We have from Newton s 3rd Law r r F = F. (6.1) r r F + F = 0, (6.2) a statement that the total horizontal force exerted on both blocks equals 0. (ecause the total force exerted on both blocks is 0, Newton s 2nd Law implies that the acceleration of the total mass equals 0.)

8 6 6 Momentum - Unit Problem 6.1 Momentum r is r p= mv lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.5 Newton s 2nd Law tells us that the force exerted on block is r r r v F = ma m t and the force exerted on block is r r r v F = ma m, (6.4) t where the acceleration is approximated by r v. Substituting into t equation (6.2) gives r r v v m + m = 0, (6.5) t t r r m v + m v = 0, (6.6) r r r r m ( v v) + m ( v v) = 0, (6.7) r r r r m v m v + m v m v = 0, (6.8) m r v r + m v = m r r v + m v. (6.9) The quantity mv r is called the momentum and is usually represented by the symbol p: r r r p=mv (6.10) r Then, for example, m v is the momentum of block before r the collision, and m v is the momentum of block after the collision. Note that momentum is a vector because velocity is a vector. Write a single sentence to express in words the meaning of equation (6.9).

9 Momentum - Unit Problem lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? Momentum is conserved in all collisions and explosions 6.6 The total momentum before the collision is equal to the total momentum after the collision. More explicitly, the momentum of block plus the momentum of block before the collision is equal to the momentum of block plus the momentum of block after the collision. This statement of the conservation of momentum is an important principle of physics found to hold true whenever two or more masses interact (collisions or explosions). Note from equation (6.2) that momentum is conserved whenever the total external force is 0. Solve part(a) of the problem by using momentum conservation, along with the diagrams below. r v (a) efore collision r v = 0 x(m) r v = 0 r v x(m) (b) fter collision

10 6 8 Momentum - Unit Problem 6.1 lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.7 Total momentum before collision = Total momentum after collision, (6.11) r r r r p + p = p + p, (6.12) s usual for one-dimensional motion, + and will be used to indicate vector direction. Writing equation (6.11) in more detail, p + p = p + p, (6.13) m v + m v = m v + m v. (6.14) For part (a), v = 0, and v = 0: m v + 0 = 0 + m v, (6.15) v = m m v, (6.16) 0.3 kg m = kg ( 008. ), (6.17) s v v m = (6.18) s Part (a) of the problem is solved, but we will understand more about the solution after working part (b).

11 Momentum - Unit Problem lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.8 Part (b) requires that we consider the possible outcomes of a collision. One possibility is that block comes to rest and block gains velocity, considered in part(a). Write out a few other possibilities for what could happen to the two blocks as a result of a collision.

12 6 10 Momentum - Unit Problem 6.1 lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? Kinetic energy is conserved in an elastic collision 6.9 s you will soon discover, there are only three other possibilities for this problem: (1) block bounces backwards off block, causing block to move; (2) block sticks to block, assuming that they carry velcro or glue; (3) block bumps block and they move in the same direction with different speeds. To demonstrate these three possibilities, let us consider an ideal collision, one in which kinetic energy is conserved. Such a collision is called elastic. Write a statement of energy conservation in the case when both blocks are moving after the collision, as shown, then write an equation to express that statement. (a) before the collision x(m) (b) one possibility after the collision x(m)

13 Momentum - Unit Problem lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.10 Kinetic energy of block before the collision is converted to kinetic energy of block and kinetic energy of block after the collision. In equation form, m v = m v + m v (6.19) This collision is ideal, i.e. it is impossible for ordinary objects because energy is lost either in sound or in deforming the objects. Multiply both sides of the equation by 2, then divide both sides by m v 2. The result is v 2 1 = + m v v 2 m v 2 2 (6.20) which can be rewritten as v 2 v 2 v = 1. (6.21) m v m Recall that the equation of an ellipse is x a 2 2 y2 + = 1, (6.22) b 2 therefore equation (6.21) is the equation of an ellipse with x and y replaced by v and v.

14 6 12 Momentum - Unit Problem 6.1 lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.11 Use the given values of m, m, and v to sketch a graph of the ellipse v 2 v 2 v = 1 (6.23) m v m after completing the table. What do the points on the ellipse represent? v v cm s v cm ( ) s v cm ( ) s

15 Momentum - Unit Problem lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? v v 8.0 cm s cm s 4.0 cm s 5.4 cm s v 6.12 cm ( ) s cm ( ) Substituting the given values into v 2 v 2 v = 1 m v m enables us to complete the table as shown, then to make a sketch using the symmetry of the ellipse. Points on the ellipse represent all of the values for block velocities after a collision which are consistent with energy conservation. ll of these velocity values are not physically possible, however. Notice, for example, that the part (a) answer does not lie on the ellipse. Evidently that collision does not conserve kinetic energy. We must also take into account the conservation of momentum. v s

16 6 14 Momentum - Unit Problem 6.1 lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.13 Using the diagrams of frame 6.9, write the conservation of momentum statement and momentum equation when blocks and are both moving after the collision.

17 Momentum - Unit Problem lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.14 Momentum of block before the collision is equal to the momentum of block plus the momentum of block after the collision. In vector equation form, r r r p = p + p, (6.24) and when vector directions are indicated by signs, p = p + p, (6.25) m v = m v + m v. (6.26) lthough the signs in this equation have been chosen according to the frame 6.9 diagrams, both positive and negative values are possible for v and v as will be seen.

18 6 16 Momentum - Unit Problem 6.1 lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.15 gives Solving for v, beginning with equation (6.26) above, m v = m v + m v, (6.27) m v = m v + m m v. (6.28) Graph this equation below, using the given values of m, m, and v. v cm ( ) s v cm ( ) s

19 Momentum - Unit Problem lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? v v cm s 8.0 cm s Equation (6.28) is the equation of a line, therefore only two points on it need to be calculated, as indicated in the table. v cm ( ) s v cm ( ) s Use this graph to get approximate answers for part (b) of the problem. e careful in finding the lowest value for v.

20 6 18 Momentum - Unit Problem 6.1 lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? Two objects stick together in a completely inelastic collision 6.17 Velocity v has its maximum value at the point where the line intersects the ellipse, near the top of the ellipse: v 6.0 cm, v = 2.1 cm s s. The minus sign means that block bounces backwards. This is the case for an elastic collision, where both momentum and kinetic energy are conserved. Realistic collisions are less than ideal, so the line describing momentum conservation is followed for other possible post-collision velocity values. Note, for example, that the part (a) solution (v = 4.8 cm, s v = 0) falls on the line. Possible values for v grow smaller as we follow the line downward and away from the kinetic energy conserving ellipse because more of the energy is expended in deforming the blocks. The smallest value for v, however, occurs for v = v when the two blocks stick together in what is called a completely inelastic collision. Following the line downward to smaller values of v would require block to move faster than block, impossible because that would require block to pass through block. From the graph, v = v for v 3.0 cm. s The material composition of the blocks determines whether the collision is elastic, completely inelastic, or inelastic (somewhere in between). Making the blocks from the same material as billiard balls would give a nearly elastic collision. pplying glue, or velcro to the blocks would make a collision completely inelastic.

21 Momentum - Unit Problem lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.18 Use the previous two frames to answer part (c) of the problem.

22 6 20 Momentum - Unit Problem 6.1 lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.19 The graph shows that after an elastic collision, block has a speed of 6 cm, and a speed of only about 3 cm s s after a completely inelastic collision. Since block is initially at rest (v = 0), its largest momentum change, m v m v, occurs when block collides with it elastically.

23 Momentum - Unit Problem lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.20 Suppose that block moves initially as before with block at rest, but that blocks and are of equal mass. Then the frame 6.14 momentum conservation statement is unchanged, but equation (6.28) becomes (for m /m = 1) v = v + v, (6.29) the equation of a line whose slope is 1. For a perfectly elastic collision, the statement of kinetic energy conservation is unchanged, but equation (6.19) becomes v 2 = v 2 + v 2, (6.30) the equation of a circle of radius v. ccording to the graph below, what are the velocities of the two equal-mass blocks after (a) elastic, and (b) completely inelastic collisions? 10 5 v cm ( ) s v cm ( ) s

24 6 22 Momentum - Unit Problem 6.1 lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.21 The point where the line intersects the top of the circle corresponds to an elastic collision, at v = 0, v = 8.0 cm. The s graph shows that the largest post-collision velocity for block occurs for a completely inelastic collision when block sticks to block : v = v = 4.0 cm. s

25 Momentum - Unit Problem lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.22 Let us change the problem again: this time the block mass is 0.5 kg and the block mass is 0.3 kg. efore the collision block moves at 8.0 cm s toward a stationary block. The momentum conservation statement in frame 6.14 is the same as before, and the momentum conservation equation (starting from equation (6.25) ) is identical to equation (6.28) except that now m > m. The energy conservation statement is the same as in frame 6.10, and the energy conservation equation (starting from equation (6.19)) is identical to equation (6.21) with m > m. ccording to the graph below, what are the velocities of the blocks after (a) elastic, and (b) completely inelastic collisions v cm ( ) s v cm ( ) s

26 6 24 Momentum - Unit Problem 6.1 lock (mass 0.3 kg) moves at a speed of 8.0 cm s towards a stationary block (mass 0.5 kg) on a horizontal, frictionless surface. lock collides head on with lock. (a) What is the velocity of block if block comes to rest after the collision? (b) If block does not come to rest, what are the largest and smallest possible velocities for block after the collision and the corresponding velocities for block? (c) Which collision in part (b) (elastic or completely inelastic) gives the largest momentum change to block? 6.23 For an elastic collision (intersection of the line and ellipse), v 2.5 cm, v 10.0 cm s s. In a completely inelastic collision, v = v 5.0 cm. s Note that as the ratio m /m changes from m /m < 1 to m /m = 1 to m /m > 1 in frames 6.16, 6.20, 6.22, the postcollision velocity for block in elastic collisions changes from negative to zero to positive. lock cannot bounce back to move in the negative direction when its mass is larger than that of block.

27 Momentum - Unit Problem bullet of mass 4.0 g moves horizontally toward a wooden block of mass 2.3 kg initially at rest. fter the bullet strikes and becomes embedded in the block, the combination slides for a distance of 2.1 m on a horizontal surface of µ k = 0.18 before coming to rest. (a) What is the bullet s speed before it strikes the block? (b) If the bullet comes to rest within the block in a time 0.3 ms after striking it, what average force brings the bullet to rest? (c) How much energy is lost by the bullet as it strikes the block? 6.24 Describing the problem events in the language of physics will eventually lead to a solution of part (a). What are the two important events in the problem?

28 6 26 Momentum - Unit Problem 6.2 bullet of mass 4.0 g moves horizontally toward a wooden block of mass 2.3 kg initially at rest. fter the bullet strikes and becomes embedded in the block, the combination slides for a distance of 2.1 m on a horizontal surface of µ k = 0.18 before coming to rest. (a) What is the bullet s speed before it strikes the block? (b) If the bullet comes to rest within the block in a time 0.3 ms after striking it, what average force brings the bullet to rest? (c) How much energy is lost by the bullet as it strikes the block? 6.25 The two important events are (1) the bullet colliding with the block (figures (a) and (b)), and (2) the bullet-block combination sliding to rest (figures (b) and (c)). Describe these two events using principles of physics. single sentence should be written for each event in terms of a conservation law. (a) before collision (b) immediately after collision (c) after sliding to rest

29 Momentum - Unit Problem bullet of mass 4.0 g moves horizontally toward a wooden block of mass 2.3 kg initially at rest. fter the bullet strikes and becomes embedded in the block, the combination slides for a distance of 2.1 m on a horizontal surface of µ k = 0.18 before coming to rest. (a) What is the bullet s speed before it strikes the block? (b) If the bullet comes to rest within the block in a time 0.3 ms after striking it, what average force brings the bullet to rest? (c) How much energy is lost by the bullet as it strikes the block? 6.26 Momentum conservation and energy conservation are the physics principles needed to describe the problem events. Event (1) can be described in terms of momentum conservation: Momentum of the bullet before the collision equals momentum of the block plus bullet after the collision. Energy conservation describes event (2): kinetic energy of block plus bullet after the collision is converted to heat energy. Explain why energy conservation does not describe the collision, then write equations for these two conservation statements.

30 6 28 Momentum - Unit Problem 6.2 bullet of mass 4.0 g moves horizontally toward a wooden block of mass 2.3 kg initially at rest. fter the bullet strikes and becomes embedded in the block, the combination slides for a distance of 2.1 m on a horizontal surface of µ k = 0.18 before coming to rest. (a) What is the bullet s speed before it strikes the block? (b) If the bullet comes to rest within the block in a time 0.3 ms after striking it, what average force brings the bullet to rest? (c) How much energy is lost by the bullet as it strikes the block? 6.27 Remember that kinetic energy is conserved only in an elastic collision (frame 6.9). In this completely inelastic collision (frame 6.17), kinetic energy is partly converted to heat energy as the bullet enters the block. y using signs to indicate direction in r r r p 1 = p 1 + p 2, the momentum conservation statement becomes p 1 = p 1 + p 2, (6.31) m 1 v 1 = m 1 v 1 + m 2 v 2, (6.32) m 1 v 1 = (m 1 + m 2 )v 1, (6.33) where 1 denotes the bullet, 2 denotes the block, and v 1 = v 2 after the collision. The statement of energy conservation becomes 1 2 ( m + m ) v = Fd, (6.34) f (m 1 +m 2 )g where F f is the magnitude of the frictional force on the block, and d is the magnitude of the block s displacement. (Recall from the chapter Introduction to Energy that heat energy is given by F f d.) Recall that the frictional force F f is given by F f = µ k F s (from Introduction to Forces ), and in this situation F s = (m 1 + m 2 )g (see diagram at left). Equation (6.34) can, therefore, be rewritten: F s ( m + m ) v = µ ( m +m ) gd, (6.35) k 1 2 v = µ gd (6.36) Can part (a) of the problem now be solved? Yes No Explain your answer. k

31 Momentum - Unit Problem bullet of mass 4.0 g moves horizontally toward a wooden block of mass 2.3 kg initially at rest. fter the bullet strikes and becomes embedded in the block, the combination slides for a distance of 2.1 m on a horizontal surface of µ k = 0.18 before coming to rest. (a) What is the bullet s speed before it strikes the block? (b) If the bullet comes to rest within the block in a time 0.3 ms after striking it, what average force brings the bullet to rest? (c) How much energy is lost by the bullet as it strikes the block? 6.28 ecause there are two equations, (6.33) and (6.36), and two unknowns, v 1 and v 1, yes, part (a) of the problem can be solved. In the next unit the solution will be completed.

32 6 30 Momentum - Unit Problem 6.2 bullet of mass 4.0 g moves horizontally toward a wooden block of mass 2.3 kg initially at rest. fter the bullet strikes and becomes embedded in the block, the combination slides for a distance of 2.1 m on a horizontal surface of µ k = 0.18 before coming to rest. (a) What is the bullet s speed before it strikes the block? (b) If the bullet comes to rest within the block in a time 0.3 ms after striking it, what average force brings the bullet to rest? (c) How much energy is lost by the bullet as it strikes the block? End Page

33 Momentum - Unit Problem specially designed bomb of mass 2.3 kg, initially at rest, explodes and fragments into three pieces, all of which move along a line on a horizontal frictionless surface. fter the explosion, one fragment of mass 0.4 kg has a velocity of m s, and another fragment of mass 0.8 kg has a velocity of 34.7 m s. What is the velocity of the third fragment? 6.29 ecause the mass fragments interact, the physics principle needed to describe the explosion is conservation of momentum, frame 6.6. What is the bomb s total momentum before the explosion?

34 6 32 Momentum - Unit Problem 6.3 specially designed bomb of mass 2.3 kg, initially at rest, explodes and fragments into three pieces, all of which move along a line on a horizontal frictionless surface. fter the explosion, one fragment of mass 0.4 kg has a velocity of m s, and another fragment of mass 0.8 kg has a velocity of 34.7 m s. What is the velocity of the third fragment? 6.30 ecause the bomb is initially at rest ( v r = 0), p=m r v r implies that the total momentum is 0 before the explosion. Use the drawings below to write a statement of momentum conservation in words, then write an equation to represent that statement. (a) before the explosion x (m) v 2 = 34.7 m/s v 3 v 1 = m/s (b) after the explosion x (m)

35 Momentum - Unit Problem specially designed bomb of mass 2.3 kg, initially at rest, explodes and fragments into three pieces, all of which move along a line on a horizontal frictionless surface. fter the explosion, one fragment of mass 0.4 kg has a velocity of m s, and another fragment of mass 0.8 kg has a velocity of 34.7 m s. What is the velocity of the third fragment? 6.31 Total momentum before the explosion (0) equals the sum of the momenta of fragments 1, 2, and 3 after the explosion. In equation form, r r r 0 = p 1 + p 2 + p 3, (6.37) r r r = m v + m v + m v, (6.38) m 0= (0.4 kg)(+52.1 ) + (0.8 kg)( kg) + m v 3. (6.39) s 3 Can the velocity of the third fragment now be found? Explain.

36 6 34 Momentum - Unit Problem 6.3 specially designed bomb of mass 2.3 kg, initially at rest, explodes and fragments into three pieces, all of which move along a line on a horizontal frictionless surface. fter the explosion, one fragment of mass 0.4 kg has a velocity of m s, and another fragment of mass 0.8 kg has a velocity of 34.7 m s. What is the velocity of the third fragment? 6.32 Yes, v 3 can be found because the mass of the whole bomb is known, as well as the masses of the two fragments. So m 3 = 1.1 kg, and v 3 can be found from equation (6.39). You will complete this calculation in the next unit.

37 Momentum - Unit Problem kg boy is at a distance of 6.5 m from a lake s shore as he stands at the far end of a 2.6 m long, kg raft. What is the boy s distance from the shore after he walks towards the shore to the other end of the raft? The raft is uniform and moves without friction s usual, when solving a problem the first question to answer is What physics principle should be used to describe the problem? Write an answer to this question whether or not you are sure of your answer The physics principle needed to describe and solve the problem is momentum conservation. How could this be? Initially the boy and boat are at rest, and they are at rest after the boy walks to the other end of the boat. So if we write total momentum before the walk equals total momentum after the walk, this becomes 0=0 in equation form, which gets us nowhere. To make progress, answer the question What is the total force exerted on boy plus boat before, during, and after the walk?

38 6 36 Momentum - Unit Problem kg boy is at a distance of 6.5 m from a lake s shore as he stands at the far end of a 2.6 m long, kg raft. What is the boy s distance from the shore after he walks towards the shore to the other end of the raft? The raft is uniform and moves without friction ecause there are no external horizontal forces, the total force exerted on boy plus boat equals 0 at all times. This means that the total acceleration is 0 (because r r F=m a ), and that the total velocity is constant (because a 0 acceleration implies not speeding-up, slowing-down, nor changing the direction of motion). We might guess that the total velocity is a constant 0 because the boy is initially and finally at rest. ut what has a total velocity of 0? fter all, while the boy is walking his velocity is not 0. The answer is that a quantity called the center of mass has a velocity of 0 throughout the motion of boy and boat. In other words, for this problem the center of mass does not move. The center of mass can be understood as a balance point for an object, or collection of objects. For example, the balance point for the uniform raft is at its center. If the masses of the boy and raft were equal, the balance point would be half-way between the raft s center and the boy. ut the masses are not equal. Use the problem data to write the position values for the boy, and for the raft s center of mass on the x-axis below. Estimate where the center of mass of boy plus raft is, and place a at that position, under the raft. x(m) 0

39 Momentum - Unit Problem kg boy is at a distance of 6.5 m from a lake s shore as he stands at the far end of a 2.6 m long, kg raft. What is the boy s distance from the shore after he walks towards the shore to the other end of the raft? The raft is uniform and moves without friction * 5.2 m 6.5 m x(m) ecause the mass of the raft is larger than that of the boy, the center of mass (the balance point) is closer to the raft s center than to the boy. The center of mass position is calculated exactly by using the definition Center of mass definition in onedimension x cm m1x1 + m2x2 = m + m, (6.40) 1 2 where 1 and 2 refer, respectively, to the boy and raft. Use the definition above to calculate the x-position for the center of mass. First check that the definition makes sense by calculating the center of mass as if the masses of the boy and raft were equal (choose any value for that mass). Looking at the diagram, what do you expect for x cm in this case (expected x cm = )? Now calculate the actual x cm using the problem data.

40 6 38 Momentum - Unit Problem kg boy is at a distance of 6.5 m from a lake s shore as he stands at the far end of a 2.6 m long, kg raft. What is the boy s distance from the shore after he walks towards the shore to the other end of the raft? The raft is uniform and moves without friction If the boy and raft had equal masses, we would expect x cm to be half-way between 5.2 m and 6.5 m: half the distance between 5.2 m and 6.5 m is 1 2 (6.5 m 5.2 m) = 0.65 m, corresponding to an expected x cm = 5.2 m m = 5.85 m. Using definition (6.40), and the diagram, we find for m 1 = m 2 that m 1( 65. m) + m 1( 52. m) equal mass calculated x cm = m + m 1 2 (6.41) 1 equal mass calculated x cm = m ( m) 2m 1 = ( m) (6.42) 1 2 equal mass calculated x cm = 5.85 m, (6.43) just as expected. For this problem, definition (6.40) gives x cm = (60.4 kg)(6.5 m) + (105.3 kg)(5.2) 60.4 kg kg, (6.44) x cm = 5.67 m. (6.45) Notice that this value is less than 5.85 m, indicating that x cm is closer to the center of the raft, as predicted in the previous frame.

41 Momentum - Unit Problem kg boy is at a distance of 6.5 m from a lake s shore as he stands at the far end of a 2.6 m long, kg raft. What is the boy s distance from the shore after he walks towards the shore to the other end of the raft? The raft is uniform and moves without friction To show why we have argued that v cm = 0 in frame 6.35, we start with definition (6.40), x cm = m x m + m x + m and find the change in both sides of the equation: x cm = m1 x1 + m2 x2 m + m (6.46) 1 2 keeping in mind that m 1 and m 2 are constants. Dividing both sides of the equation by an elapsed time t, we get x t cm = x x m m 2 2 t t m + m, 1 2 (6.47) v cm = m 1v1 + m2v2 m + m, (6.48) 1 2 using the definition of velocity by taking v = x/ t in the limit t 0. Initially v 1 = v 2 = 0, so according to equation (6.48) v cm = 0 before the boy walks. Total external horizontal force is 0, so v cm = 0 always and the center of mass does not move. Multiply both sides of equation (6.48) by m 1 + m 2 to get the momentum equation (m 1 + m 2 )v cm = m 1 v 1 + m 2 v 2. (6.49) Considering the values of v 1, v 2, and v cm, what does this equation say about the center of mass momentum?

42 6 40 Momentum - Unit Problem kg boy is at a distance of 6.5 m from a lake s shore as he stands at the far end of a 2.6 m long, kg raft. What is the boy s distance from the shore after he walks towards the shore to the other end of the raft? The raft is uniform and moves without friction The equation says that the center of mass momentum equals zero: because the right side of the equation equals 0, the left side equals 0, i.e. p cm = (m 1 + m 2 )v cm = 0. (6.50) Momentum conservation implies that after the boy starts walking, the center of mass momentum remains zero, p cm = (m 1 + m 2)vcm = m 1v1 + m 2v2 = 0, (6.51) even after he stops at the other end of the raft. s the boy walks the full length of the raft, the raft moves by an unknown distance d. Write expressions for the positions of the boy, and the raft s center of mass on the x-axis below. Figuring out what these expressions are is easier if you pretend that the boy and raft did not move at the same time. lso place a under the raft at the center of mass of boy plus raft. x(m) 0

43 Momentum - Unit Problem kg boy is at a distance of 6.5 m from a lake s shore as he stands at the far end of a 2.6 m long, kg raft. What is the boy s distance from the shore after he walks towards the shore to the other end of the raft? The raft is uniform and moves without friction Center of mass momentum equals zero means that as the boy walks toward the shore, the raft moves away from the shore. Let us pretend, however, that the boy completes his walk while the raft is held in place. The boy s new position would be 6.5 m 2.6 m = 3.9 m. In reality, of course, the raft is shifted away from the shore by a distance d. So when the boy completes his walk, his position is actually 3.9 m + d, and the raft s center of mass is at the new position 5.2 m + d. Note the center of mass position, indicated by the. Momentum conservation implies that relative to the shore, its position is unchanged. Its position is changed relative to the raft. 0 * 3.9 m + d 5.2 m + d x(m) Use equation (6.40) and the diagram to write a new expression for the center of mass position, and set it equal to the value in equation (6.45).

44 6 42 Momentum - Unit Problem kg boy is at a distance of 6.5 m from a lake s shore as he stands at the far end of a 2.6 m long, kg raft. What is the boy s distance from the shore after he walks towards the shore to the other end of the raft? The raft is uniform and moves without friction Starting with x cm = m x m + m x + m we have x cm = (60.4 kg)(3.9 m + d) + (105.3 kg)(5.2 m + d) 60.4 kg kg, (6.52) and using equation (6.45) gives 567. m = (60.4 kg)(3.9 m + d) + (105.3 kg)(5.2 m + d) kg. (6.53) Setting up the problem is now finished; its solution will be completed in the next unit.

45 Momentum - Unit Problem Car of mass 1360 kg (weight of 2992 lbs) is moving eastward at a constant speed. Car of mass 1656 kg (weight of 3643 lbs), traveling southward at constant speed, fails to stop at a red light and hits car. fter the collision, the two cars stick together and start moving together at a velocity of 11.2 m mi s (25 hr ), 30 south of east. What were the velocities of cars and before the collision? 6.42 What principle of physics should be used to solve this problem? Write out a single sentence description of the problem events using this principle.

46 6 44 Momentum - Unit Problem 6.5 Car of mass 1360 kg (weight of 2992 lbs) is moving eastward at a constant speed. Car of mass 1656 kg (weight of 3643 lbs), traveling southward at constant speed, fails to stop at a red light and hits car. fter the collision, the two cars stick together and start moving together at a velocity of 11.2 m mi s (25 hr ), 30 south of east. What were the velocities of cars and before the collision? 6.43 Conservation of momentum is the physics principle to be used to solve the problem. In terms of this principle, the problem restatement is as follows: the momentum of car plus the momentum of car before the collision equals the momentum of cars and after the collision. Why can kinetic energy conservation not be used to describe the collision, i.e. why is this not an elastic collision? 6.44 Kinetic energy is lost in crushing metal during the collision, and in creating sound, so kinetic energy is not conserved. The collision is completely inelastic (frame 6.17). Write an equation equivalent to the momentum conservation statement in the previous frame.

47 Momentum - Unit Problem Car of mass 1360 kg (weight of 2992 lbs) is moving eastward at a constant speed. Car of mass 1656 kg (weight of 3643 lbs), traveling southward at constant speed, fails to stop at a red light and hits car. fter the collision, the two cars stick together and start moving together at a velocity of 11.2 m mi s (25 hr ), 30 south of east. What were the velocities of cars and before the collision? 6.45 Did you remember that momentum is a vector quantity in your equation? If not, correct your equation. We write r r r r p + p = p + p, (6.54) but this equation, although correct, does not take into account the fact that the cars stick together after the collision. Take this into account by first substituting p=mv r r into the equation, then factoring.

48 6 46 Momentum - Unit Problem 6.5 Car of mass 1360 kg (weight of 2992 lbs) is moving eastward at a constant speed. Car of mass 1656 kg (weight of 3643 lbs), traveling southward at constant speed, fails to stop at a red light and hits car. fter the collision, the two cars stick together and start moving together at a velocity of 11.2 m mi s (25 hr ), 30 south of east. What were the velocities of cars and before the collision? 6.46 Equation (6.54) becomes r r r r m v + m v = m v + m v, (6.55) and because v r = v r, r r r r m v + m v = m v + m v, (6.56) m r r v + m v = (m + m ) v. (6.57) Recall that a vector equation actually represents two equations in two-dimensions, as is the case for this problem. Write the two equations represented by equation (6.57).

49 Momentum - Unit Problem Car of mass 1360 kg (weight of 2992 lbs) is moving eastward at a constant speed. Car of mass 1656 kg (weight of 3643 lbs), traveling southward at constant speed, fails to stop at a red light and hits car. fter the collision, the two cars stick together and start moving together at a velocity of 11.2 m mi s (25 hr ), 30 south of east. What were the velocities of cars and before the collision? 6.47 The two equations are m v + m v = (m + m ) v, (6.58) x x x m v + m v = (m + m ) v. (6.59) y y y Write the problem data on the diagrams below. y y x x 30 o (a) before collision (b) after collision

50 6 48 Momentum - Unit Problem 6.5 Car of mass 1360 kg (weight of 2992 lbs) is moving eastward at a constant speed. Car of mass 1656 kg (weight of 3643 lbs), traveling southward at constant speed, fails to stop at a red light and hits car. fter the collision, the two cars stick together and start moving together at a velocity of 11.2 m mi s (25 hr ), 30 south of east. What were the velocities of cars and before the collision? 6.48 y y x x v y = 0, v x v x = 0, v y v x = +(11.2 m/s) cos(30 ) v y = (11.2 m/s) sin(30 ) 30 o (a) before collision (b) after collision Can the problem now be solved? Explain

51 Momentum - Unit Problem Car of mass 1360 kg (weight of 2992 lbs) is moving eastward at a constant speed. Car of mass 1656 kg (weight of 3643 lbs), traveling southward at constant speed, fails to stop at a red light and hits car. fter the collision, the two cars stick together and start moving together at a velocity of 11.2 m mi s (25 hr ), 30 south of east. What were the velocities of cars and before the collision? 6.49 Substituting values from the previous frame into equations (6.58) and (6.59), we find that m v x + 0 = (m + m )(11.2 m s ) cos(30 ) ; (6.60) 0 + m v y = (m + m )( 11.2 m s ) sin(30 ) (6.61) can be solved, since v x and v y are the only unknowns. In the next unit, the solution will be completed. Respond to the following question concerning the collision. Why must we consider the velocity of the cars when they first start moving together in order for momentum conservation to be valid?

52 6 50 Momentum - Unit Problem 6.5 Car of mass 1360 kg (weight of 2992 lbs) is moving eastward at a constant speed. Car of mass 1656 kg (weight of 3643 lbs), traveling southward at constant speed, fails to stop at a red light and hits car. fter the collision, the two cars stick together and start moving together at a velocity of 11.2 m mi s (25 hr ), 30 south of east. What were the velocities of cars and before the collision? 6.50 Starting at frame 6.4, equation (6.2), momentum is conserved (equation (6.9)) only when the total external force on the interacting objects is 0. t any time other than immediately after the collision, we must account for the frictional force exerted by the road on the tires. In that case the total external force is no longer 0.

53 Momentum - Unit Problem On a horizontal, frictionless surface, block (mass 0.2 kg) moves at a velocity of +0.8 m s toward block (mass 0.7 kg) which is initially at rest and is compressing a spring of spring constant 10.3 N m. lock is released and after leaving the spring, collides head-on with block. fter the collision, block has a velocity of 0.3 m, and block has a velocity of +0.2 m s s. (a) y how much was the spring initially compressed? (b) Is the collision elastic or inelastic? 6.51 What physics principles are needed to solve this problem? Describe the problem events, illustrated below, by writing two sentences making use of these principles. x(m) (a) spring initially compressed x(m) (b) before collision x(m) (c) after collision