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Chapter 7 LINEAR MOMENTUM Conceptual Questions 1. The likelihood of injury resulting from jumping from a second floor window is primarily determined by the average force acting to decelerate the body. (a) The deceleration time interval for a person landing stiff legged on pavement is very short. The impulsemomentum theorem tells us that the average force acting on the person’s feet must therefore be very large— such a person is likely to incur injuries. (b) Jumping into a privet hedge increases the time interval over which the body decelerates. This decreases the average force on the person’s limbs and therefore decreases the likelihood of injury. (c) Jumping into a firefighter’s net is the best option of the three. The net stretches downward, gradually bringing the person to rest. Additionally, the firefighters lower the net with their hands as the person lands to further lengthen the time interval during which the person is brought to rest. 2. (a) A body’s momentum change is equal to the impulse that has acted on it. Impulse is defined as the product of the average force acting on a body and the time interval over which it acts—the bodies therefore experience the same impulse and so have equal momentum changes. (b) The change in a body’s velocity is defined as the ratio of the change in its momentum to its mass—the less massive body therefore incurs a larger velocity change. (c) The acceleration of a body is defined as the ratio of the force acting on it to its mass—the less massive body therefore has the larger acceleration. 3. The muzzle speed is determined by the change in the bullet’s momentum. The impulse-momentum theorem tells us that this momentum change is determined by the impulse acting on the bullet. The force acting on the bullet due to the expanding hot gases is roughly constant throughout the muzzle. A shorter muzzle produces a shorter time interval over which the bullet is accelerated by the firing force. This results in a smaller impulse and therefore a smaller momentum change—thus producing lower bullet velocities. 4. After the explosion, each piece of the firecracker has a momentum vector associated with it that points in the direction of its motion. The law of conservation of linear momentum tells us that the vector sum of the momentum of all the pieces of the firecracker must equal the initial momentum of the whole firecracker—in this case, both the initial and the final net momentum vectors equal zero. 5. The law of the conservation of linear momentum states that in the absence of external interactions, the linear momentum of a closed system is constant. Floating in free space, the astronaut and the wrench form a closed system free from interactions with other bodies. If the astronaut throws the wrench in the direction opposite the ship, conservation of momentum dictates that he must in turn move toward the ship. 6. The horizontal component of the golf ball’s momentum is conserved since no external force acts on the ball in the horizontal direction. The vertical component of the ball’s momentum is not conserved however because the Moon’s gravitational force interacts with it and changes its momentum. 371 Chapter 7: Linear Momentum Physics 7. In an elastic collision between the hammer and nail, the kinetic energy of the system is conserved while in a perfectly inelastic collision, the greatest percentage of the kinetic energy is lost. The energy lost by the system in a perfectly inelastic collision is used to do the work required to bring the hammer and nail together. In an elastic collision, this work is available to drive the nail into the wood—the total work available to drive the nail is therefore greater for an elastic collision. Thus, for equal applied forces, the hammer will drive the nail further into the wood if the collision is elastic. 8. The momentum of the squid (including the water inside its cavity) comprises a system for which momentum is conserved. The means the momentum of the squid (plus water) must be the same before and after some of the water has been ejected. When the squid expels some water, the water gains momentum in the direction it is being expelled. To conserve momentum, the squid must gain an equal amount of momentum in the opposite direction, propelling it forward. Similarly, a rocket engine expels exhaust from burning fuel to propel itself forward. 9. First law: The momentum of an object is constant unless acted upon by an external force. Second law: The net force acting on an object is equal to the rate of change of the object’s momentum. Third law: When two objects interact, the changes in momentum that each imparts to the other are equal in magnitude and opposite in direction. 10. Noting that the (translational) kinetic energy can be written as p 2 /(2m), and that both objects have the same kinetic energy, it is evident that the object with the greater mass has the larger magnitude of momentum. 11. The woman’s center of mass is not necessarily 0.80 m above the floor, because her mass is not necessarily distributed uniformly with height. Normally, the upper body of a person is more massive than the lower body and thus we would expect the woman’s center of mass to be slightly higher than 0.80 m. 12. The frictional force of the road on the tires supplies the external force to change the bicycle’s momentum. Changes in the bicycle’s kinetic energy do not require an external force. For example, the rider could throw her helmet away hard, increasing both her and the helmet’s speed. The kinetic energy of the system (bicycle, rider, and helmet) would increase, while the momentum would remain the same. Note that the work-energy theorem (total work done equals change in kinetic energy) cannot be used here, because the internal structure of the system cannot be ignored. 13. An impulse must be supplied to the egg to change its momentum and bring it to rest. A good strategy is to make the time interval over which the stopping force is applied as large as possible. This will reduce the magnitude of the force required to stop the egg. One should therefore attempt to catch the egg with a swinging motion, moving the hand backwards as it is being caught, to bring it to rest as slowly and gently as possible. 14. The collisions of the balls in the “executive toy” are nearly perfectly elastic. The kinetic energy of the system just before and after a collision must therefore be the same. This is the reason we never see three balls moving away after a collision in which two balls were initially pulled back and released—such an event would not conserve kinetic energy. 15. According to the impulse-momentum theorem, the change in momentum of the baseball is equal to the impulse it receives from the bat. Impulse is equal to the average force times the time interval over which the force is applied. To give the ball the greatest possible momentum, one should attempt to maximize the amount of time during which the force is being applied. 16. Jeremy has it right. By momentum conservation, Micah needs to throw the balls forward if he wants to propel himself backward, but the balls need not strike any surface. You can also consider Newton’s third law and see that it is the force by the balls on Micah’s hand that pushes Micah backward. 17. Daryl has done his homework. If he falls when rock climbing, his rope will stretch and stop him more gradually than the rope Mary wants to buy. In a fall, the climber’s momentum must go from some initial value to zero. If the time over which the momentum is decreased to zero is longer, the average force delivered by the rope is smaller. 372 Physics Chapter 7: Linear Momentum Problems 1. Strategy Use the definition of linear momentum. Solution Find the magnitude of the total momentum of the system. p total p1 p 2 mv1 mv 2 m( v1 v 2 ) m[ v1 ( v1 )] 0, so the magnitude is 0 . 2. Strategy Use the definition of linear momentum. Solution Find the momentum of the automobile. 9800 N W p mv v (35 m s south) 3.5 104 kg m s south g 9.80 m s 2 3. Strategy and Solution Impulse F t , so the SI unit is N s kg m s 2 s kg m s. p mv, so the SI unit is kg m s. Therefore, the SI unit of impulse is the same as the SI unit of momentum. 4. Strategy Use the impulse-momentum theorem. Solution Find the final speed of the cue ball. F t (24 N)(0.028 s) p pf pi mvf m(0) Fav t , so vf av 4.2 m s . m 0.16 kg 5. Strategy Add the momenta of the three particles. Solution Find the total momentum of the system. p tot p1 p 2 p3 m1v1 m2 v 2 m3 v3 m1v1 north m2 v2 south m3v3 north (m1v1 m2 v2 m3v3 ) north (3.0 kg)(3.0 m s) (4.0 kg)(5.0 m s) (7.0 kg)(2.0 m s) north 3 kg m s north 6. (a) Strategy Form a ratio of the magnitudes of the final and initial momenta. Solution Compute the ratio. pf mvf vf 60.0 mi h 3.00 pi mvi vi 20.0 mi h (b) Strategy Form a ratio of the final and initial kinetic energies. Solution Compute the ratio. Kf Ki 1 mv 2 f 2 1 mv 2 i 2 v f vi 2 2 3.00 9.00 7. Strategy The initial momentum is toward the wall and the final momentum is away from the wall. Solution Find the change in momentum. p pf pi mvf mvi m(vf vi ) (5.0 kg)(2.0 m s 2.0 m s) 20 kg m s , so p 20 kg m s in the x-direction . 373 Chapter 7: Linear Momentum Physics 8. Strategy The final and initial velocities are the same, since air resistance is ignored. Use the definition of the 1 linear momentum. Use Eq. y viy t g (t )2 to find the initial speed. Let up be the positive direction. 2 Solution Find the initial speed. 1 1 y 0 viy t g (t ) 2 , so viy g t. 2 2 Find p. 1 p pfy piy m(vfy viy ) m(viy viy ) 2m g t mg t (3.0 kg)(9.80 m s 2 )(3.4 s) 2 1.0 102 kg m s So, p 1.0 102 kg m s downward . 9. Strategy Use the definition of linear momentum. Let up be the positive direction. Solution vf vfy viy g t g t , since the object starts from rest. Find p. p pf pi m(vf vi ) m( g t 0) mg t (3.0 kg)(9.80 m s 2 )(3.4 s) 1.0 102 kg m s, so p 1.0 102 kg m s downward . 10. Strategy Use the impulse-momentum theorem. Solution Find the average force. p mv (50.0 kg)(3.0 m s 0) 7.5 N Fav t t 20.0 s The force necessary is 7.5 N in the direction of the sled’s velocity. 11. Strategy Use the impulse-momentum theorem. Let the forward direction be positive. Solution Find the time interval for which the engine must be fired. p mv (3800 kg)(1.1 104 m s 2.6 104 m s) t 320 s Fav Fav 1.8 105 N 374 Physics Chapter 7: Linear Momentum 12. (a) Strategy Use the component method of subtracting vectors. y Solution Compute the magnitude of the change in momentum. p x mvx and p y mv y . p m (vx )2 (v y ) 2 (0.15 kg) [0 (20 m s)]2 (15 m s 0)2 3.8 kg m s x pf pi Find the angle of the change in momentum. p y v y 15 tan 1 tan 1 tan 1 37 p x vx 20 The change in momentum of the baseball is 3.8 kg m s at 37 above the horizontal direction opposite vi . (b) Strategy Use the impulse-momentum theorem. Solution Find the average force of the bat on the ball. p 3.75 kg m s Fav 75 N, so F 75 N in the same direction as p . 0.050 s t 13. Strategy Use the impulse-momentum theorem. Let the positive direction be in the direction of motion. Solution Find the average horizontal force exerted on the automobile during breaking. p m(vf vi ) (1.0 103 kg)(0 30.0 m s) 6.0 103 N Fav t t 5.0 s So, Fav 6.0 103 N opposite the car’s direction of motion . 14. Strategy Use the impulse-momentum theorem. Solution (a) Compute the changes in momenta for each direction. pnorth 0 and peast Fav t mveast mveast . Find the magnitude and direction of v f . 15 m/s N 15 N 2 2 (15 N)(4.0 s) F t vf vnorth 2 veast 2 vnorth 2 av (15 m s)2 25 m s and m 3.0 kg v 15 m s tan 1 north tan 1 37 north of east, so v f 25 m s at 37 north of east . veast 20 m s (b) Let +y be north and +x be east. Compute the change in momentum. p Fav t (15 N)(4.0 s) 60 kg m s. The entire change in momentum is due to the force, so p 60 kg m s east . 375 Δp pi pf y x Chapter 7: Linear Momentum Physics 15. (a) Strategy Use the definition of linear momentum. Use vf2y viy 2 2a y y to find the speed after the fall. Solution Find the initial speed, which is the final speed after the fall. vf2y viy 2 vf2y 0 2 g y 2 gh, so vfy 2 gh . If up is positive, v y 2 gh down 2 gh up vi . p m(vf vi ) m 0 2 gh m 2 gh (60.0 kg) 2(9.80 m s 2 )(8.0 m) 750 kg m s , so p 750 kg m s upward . (b) Strategy The impulse on the net is equal to the boy’s weight times t plus the change in momentum of the boy due to the net, p. Solution Find the impulse on the net. mg t downward p (60.0 kg)(9.80 N kg)(0.40 s) downward 750 kg m/s downward 990 N s downward (c) Strategy Use Eq. (7-3). Solution Find the average on the net due to the boy. p 990 kg m/s downward 2500 N downward Fav t 0.40 s 16. Strategy The impulse is equal to the area under the graph. Use the impulse-momentum theorem. Let the positive direction be to the right. Solution Each rectangle of the grid is equal to (100 N)(0.0010 s) 0.10 kg m s. The area can be divided easily into three right triangles and one rectangle. Thus, there are 12 (7)(4) 12 (6)(2) 12 (8)(6) (6)(4) 68 rectangles under the graph and the magnitude of the impulse is p 68(0.10 kg m s) 6.8 kg m s mv. The impulse is opposite the direction of motion of the initial velocity. Compute the final speed. p p 6.8 kg m s v vf vi , so vf vi 30 m s 29 m s . m m 0.115 kg 17. (a) Strategy Use conservation of energy to find the speed with which the pole-vaulter lands on the padding. Solution 1 K mv 2 U mgh, so v 2 gh 2(9.80 m s 2 )(6.0 m) 10.84 m/s 11 m s . 2 (b) Strategy The padding exerts an upward force on the pole-vaulter while gravity continues to exert a downward force. Use the impulse-momentum theorem and let Fav represent the average force exerted by the padding. Solution p Fnet t , and Fnet Fav mg , so m(vf vi ) m[0 ( 2 gh )] p Fav Fnet mg mg mg mg t t t 2 gh 10.84 m/s Fav m g 60.0 kg 9.80 m s 2 1900 N . t 0.50 s 376 Physics Chapter 7: Linear Momentum 18. Strategy Use conservation of momentum. Solution Find the recoil speed of the rifle. m 0.0100 kg p rf p bf mr v rf mb v bf p ri p bi 0 0, so v rf b v bf (820 m s) 1.8 m s . mr 4.5 kg 19. (a) Strategy Right after the collision, the bullet and baseball combination must have the same momentum as the bullet had just before it stuck the baseball. Solution Before the collision, the momentum of the bullet is pi mbullet vbullet . After the collision, the momentum of the bullet and baseball combination is pf (mbullet mbaseball )vf pi . Thus, the speed of the bullet and baseball combination right after the collision was mbullet vi pi (0.030 kg)(200 m s) vf 33 m s . mbullet mbaseball mbullet mbaseball 0.030 kg 0.15 kg (b) Strategy Use conservation of energy to determine the work done by air resistance on the bullet and baseball combination. Solution Determine the work done by air resistance. Let up be the positive direction. Wtotal Wc Wnc U Wair K , so 1 1 Wair K U 0 mv 2 mg y (0.18 kg) (33.333 m s) 2 (9.80 m s 2 )(37 m) 34.73 J. 2 2 W 34.73 J Since Wair Fair, av y, Fair, av air 0.94 N. Therefore, the average force of air resistance 37 m y was 0.94 N down . 20. Strategy Use conservation of momentum. Solution Find the recoil speed of the submarine. m 250 kg (100.0 m s) 0.010 m s . psf p tf ms vsf mt v tf psi p ti 0 0, so vsf t v tf ms 2.5 106 kg 21. Strategy Use conservation of momentum. Solution Find the recoil speed of the thorium nucleus. pi 0 p f , so if n = nucleus and p = particle, mp 4.0 u 0.050(2.998 108 m s) 2.6 105 m s . p n p p mn v n mp v p 0, so v n vp 234 u mn 22. Strategy Use the law of conservation of linear momentum to determine the speed Dash must throw the balls. Solution According to the law of conservation of linear momentum, Dash and his skateboard will move backward with linear momentum equal in magnitude to the magnitude of the combined momentum of the balls. Find the speed of the balls. m v (60 kg)(0.50 m s) 100 m s (224 mph) . pb 3mb vb mD vD pD , so vb D D 3mb 3(0.10 kg) Since 224 mph is faster than any human can throw a ball, Dash will not succeed . 377 Chapter 7: Linear Momentum Physics 23. Strategy Use the law of conservation of linear momentum to determine the astronaut’s speed. Solution According to the law of conservation of linear momentum, the astronaut will move toward the ship with linear momentum equal in magnitude to the magnitude of the combined momentum of the objects thrown. Find the speed of the astronaut after he throws the mallet. pobjects pw ps pm mw vw ms vs mm vm pA mA vA , so vA mw vw ms vs mm vm mA (0.72 kg)(5.0 m s) (0.80 kg)(8.0 m s) (1.2 kg)(6.0 m s) 0.30 m s . 58 kg 24. Strategy Use conservation of energy to determine the skier’s speed and momentum just before he grabs the backpack. Then, use conservation of linear momentum to find his new speed after he grabs the backpack. Finally, from Chapter 4, use the equations for motion with a changing velocity. Solution Use conservation of energy to find the speed of the skier just before he grabs the backpack. 1 1 E K U K f Ki U f U i ms vs 2 0 0 ms gh ms vs 2 ms gh 0, so vs 2 gh . 2 2 Use conservation of linear momentum to find his new speed after he grabs the backpack. pi ms vs pf (ms mb )vx , so vx ms vs ms mb ms 2 gh ms mb . Now, find the time it takes for an object to fall 2.0 m from rest. 1 1 1 2 y y viy t g (t )2 (0)t g (t )2 g (t )2 , so t . g 2 2 2 The skier will travel a horizontal distance of m 2 gh 2y 2ms yh 2(65 kg) (2.0 m)(5.0 m) x vx t s 4.8 m . ms mb g ms mb 65 kg 20 kg 25. Strategy Use conservation of momentum. Solution Find the recoil speed of the railroad car. pi 0 p f , and since we are only concerned with the horizontal direction, we have: m 98 kg (105 m s) cos 60.0 0.10 m s . mc vcx ms vsx , so vcx s vsx mc 5.0 104 kg 26. Strategy Use conservation of momentum. Solution Find the mass of the man and the car. pi 0 p f , and since we are only concerned with the horizontal direction, we have: mmc vmc mb vb , so mmc vb vmc mb (173 m s) cos 30.0 1.0 103 m s 378 (0.010 kg) 1500 kg . Physics Chapter 7: Linear Momentum 39. Strategy Use conservation of momentum. Let the positive direction be to the right. Solution Find the final velocity of the helium atom. mHe vHef mO vOf mHe vHei mO vOi , so vHef mO (vOi vOf ) mHe vHei mHe mO mHe (vOi vOf ) vHei 32.0 u 412 m s 456 m s 618 m s 270 m s. 4.00 u Thus, the velocity of the helium atom after the collision is 270 m s to the right . 40. Strategy Linear momentum is conserved, so pf pi . Solution Find the change in speed of the car. pf (mcar mclay )vf pi mcar vi , so vf v vf vi mcar mcar mclay mcar mcar mclay vi , and mcar 120 g vi vi 1 vi 1 (0.75 m s) 0.15 m s . mcar mclay 120 g 30.0 g 41. Strategy Use conservation of momentum. Solution (a) The collision is perfectly inelastic, so v1f v2f vf . Find the speed of the two cars after the collision. v 1.0 m s m1v1i m2 v2i mv1i 4.0m(0) m1v1f m2 v2f mvf 4.0mvf , so vf 1i 0.20 m s . 5.0 5.0 (b) The cars are at rest after the collision, so v1f v2f 0. mv1i 4.0mv2i 0, so v2i v1i 4.0 1.0 m s 0.25 m s. The initial speed was 0.25 m s . 4.0 42. Strategy Use conservation of momentum. The block is initially at rest, so v2i 0. Let east be in the +xdirection. Solution Find the final velocity of the block. m1v1f m2 v2f m1v1i m2 v2i m1v1i m2 (0), so m (v v ) 0.020 kg v2f 1 1i 1f 200.0 m s (100.0 m s) 3.0 m s. m2 2.0 kg Thus, v block 3.0 m s east . 43. Strategy Use conservation of momentum. The collision is perfectly inelastic, so v1f v2f vf . Also, the block is initially at rest, so v2i 0. 379 Chapter 7: Linear Momentum Physics Solution Find the speed of the block of wood and the bullet just after the collision. m1v1f m2 v2f (m1 m2 )vf m1v1i m2 v2i m1v1i m2 (0), so m1 0.050 kg vf v (100.0 m s) 5.0 m s . m1 m2 1i 0.050 kg 0.95 kg 44. Strategy The collision is perfectly inelastic since the bullet embeds in the wood. Friction does negative work on the block and bullet combination. Use Newton’s second law and conservation of momentum. Solution Let the +x-direction be in the direction of motion. Find the acceleration of the block and bullet due to friction. Fy N (mbullet mblock ) g 0, so N (mbullet mblock ) g. Fx f k k N k (mbullet mblock ) g (mbullet mblock )ax , so a x k g . Find the initial speed of the block and bullet (just after the collision). N x fk (mbullet + mblock)g vfx vix 0 vix 2a x x 2 k g x, so vix 2k g x v. Use conservation of momentum to find the speed of the bullet just before its collision with the block. mbullet vbullet (mbullet mblock )v (mbullet mblock ) 2k g x , so 2 2 vbullet 2 (mbullet mblock ) 2 k g x mbullet (2.02 kg) 2(0.400)(9.80 m s 2 )(1.50 m) 0.020 kg 350 m s . 45. Strategy Use conservation of momentum. Solution Find the total momentum of the two blocks after the collision. p2 p1 p2f p2i p1i p1f p1f p2f p1i p2i (m1 m2 )vf m1v1i m2 v2i pf (2.0 kg) 1.0 m s (1.0 kg)(0) 2.0 kg m s p1i Since p1i was directed to the right, and pf p1i , the total momentum of the two blocks after the collision is 2.0 kg m s to the right . 46. Strategy The collision is perfectly inelastic, so v1f v2f vf . Use conservation of momentum. Let the positive direction be the initial direction of motion. Solution Find the speed of the man (1) just after he catches the ball (2). m1v1f m2 v2f m1vf m2 vf m1v1i m2 v2i m1 (0) m2 v2i , so m2 0.20 kg vf v (25 m s) 0.066 m s . m1 m2 2i 75 kg 0.20 kg 47. Strategy Use conservation of momentum. Let the positive direction be in the initial direction of motion. Solution Find the speed of the Volkswagen after the collision. pV mV vVf mV vVi pB mB vB , so vVf mV vVi mB vB mV (1.0 103 kg) 25 m s (2.0 103 kg)(33 m s 42 m s) 1.0 103 kg 380 43 m s . Physics Chapter 7: Linear Momentum 48. Strategy Use conservation of linear momentum. Since the collision is perfectly elastic, kinetic energy is conserved. Solution The 100-g ball is (1) and the 300-g ball is (2). Note that m2 3m1. m1v1i m2 v2i m1v1i m2 (0) m1v1i m1v1f m2 v2f , so v1i v1f m2 m1 v2f v1f 3v2f . 1 1 1 1 1 1 1 m1v1i 2 m2 (0)2 m1v1i 2 m1v1f 2 m2 v2f 2 m1v1f 2 (3m1 )v2f 2 , so v1i 2 v1f 2 3v2f 2 . 2 2 2 2 2 2 2 Substitute for v1i . v1f 3v2f 2 v1f 2 6v1f v2f 9v2f 2 v1f 2 3v2f 2 , so 6v1f v2f 6v2f 2 , or v1f v2f . Find the final velocities of each ball. v1i v1f 3v2f v2f 3v2f 2v2f , so v2f 1 1 v (5.00 m s) 2.50 m s. 2 1i 2 Since v1f v2f , v1f 2.50 m s. So, the 300-g ball moves at 2.50 m s in the +x -direction and the 100-g ball moves at 2.50 m s in the x-direction . 49. Strategy Use conservation of momentum. Let the positive direction be the initial direction of motion. Solution Find the speed of the 5.0-kg body after the collision. m1v1f m2 v2f m1v1i m2 v2i , so m (v v ) m2 v2i (1.0 kg) 10.0 m s ( 5.0 m s) (5.0 kg)(0) 3.0 m s . v2f 1 1i 1f 5.0 kg m2 50. Strategy The collision is perfectly inelastic, so v1f v2f vf . Use conservation of momentum. Let the positive direction be the initial direction of motion. Solution Find the speed of the combination. m (0) m2 v2i 3.0 kg m1vf m2 vf m1v1i m2 v2i , so vf 1 (8.0 m s) 4.8 m s . m1 m2 2.0 kg 3.0 kg 51. Strategy The spring imparts the same (in magnitude) impulse to each block. (The same magnitude force is exerted on each block by the ends of the spring for the same amount of time.) So, each block has the same final magnitude of momentum. (The initial momentum is zero.) Solution Find the mass of block B. mB vB mA vA , so v d t d 1.0 m mB A mA A mA A mA (0.60 kg) 0.20 kg . 3.0 m vB d B t dB 381 −p p A B Chapter 7: Linear Momentum Physics 52. (a) Strategy Use conservation of momentum. Let the +x-direction be to the right. Solution Find the final velocity of the other glider. mv1f mv2f mvi mvi , so v2f vi vi v1f 0.50 m s 0.50 m s 1.30 m s 0.30 m s. So, the velocity of the other glider is 0.30 m s to the left . (b) Strategy Form a ratio of the final to the initial kinetic energies. Solution Compute the ratio. 1 mv 2 1 mv 2 Kf v 2 v2f 2 (1.30 m s)2 (0.30 m s) 2 1f 2f 2 2 1f 3.6 Ki 2vi 2 2(0.50 m s)2 2 1 mvi 2 2 The final kinetic energy is greater than the initial kinetic energy. The extra kinetic energy comes from the elastic potential energy stored in the spring. 53. Strategy The collision is perfectly inelastic, so v1f v2f v. The block is initially at rest, so v2i 0 and v1i vi . Use conservation of momentum. Solution Find the speed of the bullet and block system. mbul (mbul mblk )v mbul vi mblk (0), so v v. mbul mblk i Determine the time it takes the system to hit the floor. 1 1 2h y h viy t g (t )2 0 g (t ) 2 , so t . 2 2 g h = 1.2 m Δx Find the horizontal distance traveled. x vix t vt mbul 2h 0.010 kg 2(1.2 m) 0.49 m vi 400.0 m s 0.010 kg 4.0 kg mbul mblk g 9.80 m s 2 54. Strategy Use conservation of momentum. K i Kf , since the collision is elastic. Solution Show that the final speed of each object is the same as the initial speed. m1v1f m2 v2f m1v1i m2 v2i and p1i p2i. So, m1v1i m2 v2i and m1v1f m2 v2f 0, or m1v1f m2 v2f . 1 1 1 1 m v 2 m2 v2i 2 m1v1f 2 m2 v2f 2 , so m1v1i 2 m2 v2i 2 m1v1f 2 m2 v2f 2 . 2 1 1i 2 2 2 Eliminate v1i and v1f . 2 2 m m m1 2 v2i m2 v2i 2 m1 2 v2f m2 v2f 2 m1 m1 m22 2 m22 2 m2 v2i m2 v2f m m 1 1 2 2 v2i v2f 382 Physics Chapter 7: Linear Momentum Therefore, the initial and final speeds of object 2 are the same. Eliminate v2i and v2f . 2 2 m m m1v1i m2 1 v1i m1v1f 2 m2 1 v1f m2 m2 2 2 m m m1 1 v1i 2 m1 1 v1f 2 m2 m2 v1i 2 v1f 2 Therefore, the initial and final speeds of object 1 are the same. 2 55. Strategy Use conservation of linear momentum. Since the collision is perfectly elastic, kinetic energy is conserved as well. Solution Let the +x-direction be in the original direction of motion of the 2.0-kg object. The 2.0-kg object is (1) and the 6.0-kg object is (2). Note that m2 3m1. m m1v1i m2 v2i m1v1i m2 (0) m1v1i m1v1f m2 v2f , so v1i v1f 2 v2f v1f 3v2f . m1 1 1 1 1 1 1 1 m1v1i 2 m2 (0)2 m1v1i 2 m1v1f 2 m2 v2f 2 m1v1f 2 (3m1 )v2f 2 , so v1i 2 v1f 2 3v2f 2 . 2 2 2 2 2 2 2 Substitute for v1i . v1f 3v2f 2 v1f 2 6v1f v2f 9v2f 2 v1f 2 3v2f 2 , so 6v1f v2f 6v2f 2 , or v1f v2f . Find the final speed of the 6.0-kg object. 1 1 v1i v1f 3v2f v2f 3v2f 2v2f , so v2f v1i (10 m s) 5.0 m s . 2 2 56. Strategy Look at the collision in its center of mass frame. Assume a one-dimensional collision. The initial velocities are v1ix and v2ix . The masses are m1 and m2. Solution Transform the initial velocities to the CM frame by subtracting vCMx from each. v1ix v1ix vCMx v2ix v2ix vCMx According to the result from Problem 54, the final speeds of the objects must be the same as the initial speeds, but the final and initial velocities are oppositely directed since the objects rebound after colliding. Therefore, v1fx v1ix v1ix vCMx v2fx v2ix v2ix vCMx Transform back to the original frame of reference. v1fx v1fx vCMx v1ix 2vCMx v2fx v2fx vCMx v2ix 2vCMx The relative speed after the collision is v1fx v2fx v1ix 2vCMx v2ix 2vCMx v1ix v2ix , which is the relative speed before the collision. 57. Strategy Use conservation of momentum. Let each of the first two pieces be 45 from the positive x-axis (one CW, one CCW). Solution Find the speed of the third piece. Find v3 x . 383 Chapter 7: Linear Momentum Physics p1x p2 x p3 x mv1x mv2 x mv3 x 0, so v3 x v1x v2 x v cos 45 v cos(45) v 2 v 2 v 2. Similarly, v3 y v1 y v2 y v sin 45 v sin( 45) v 2 v 2 0, so v3 v3 x v 2 120 m s 2 170 m s . 58. Strategy Use conservation of momentum. Solution Find vBfx . pix MvAix pfx MvAfx MvBfx , so vBfx vAix vAfx . Find vBfy . piy 0 pfy MvAfy MvBfy , so vBfy vAfy . Calculate vBf . vBf (vAix vAfx ) 2 (vAfy )2 6.0 m s 1.0 m s 2.0 m s 5.4 m s 2 2 59. Strategy Use conservation of momentum. Refer to Practice Problem 7.11. Solution (a) Find the momentum change of the ball of mass m1. 1 p1x p2 x m2 v2ix m2 v2fx m2 (0 v2fx ) m2 v2fx 5m1 vi cos( 36.9) 1.00m1vi 4 1 p1 y p2 y m2 v2iy m2 v2fy 5m1 (0 v2fy ) 5m1v2fy 5m1 vi sin(36.9) 0.751m1vi 4 (b) Find the momentum change of the ball of mass m2 . p2 x p1x m1 (v1ix v1fx ) m1 (vi 0) m1vi p2 y p1 y m1 (v1iy v1fy ) m1 (0 v1 ) m1v1 m1 (0.751vi ) 0.751m1vi The momentum changes for each mass are equal and opposite. 60. Strategy Use conservation of momentum. Let right be +x and +y be in the initial direction of the puck. y Solution Find v2fx . mv1fx mv2fx mv1ix mv2ix 0 0, so v2fx v1fx . Find v2fy . mv1fy mv2fy mv1iy mv2iy v1iy 0, so v2fy v1iy v1fy . 384 v1i v1f θ1 = 37° x Physics Chapter 7: Linear Momentum Calculate v2f . v2f v2fx 2 v2fy 2 (v1fx ) 2 (v1iy v1fy ) 2 (v1f sin 1 ) 2 (v1i v1f cos 1 ) 2 0.36 m s sin 37 0.45 m s 0.36 m s cos 37 0.27 m s Calculate the direction of the second puck. 0.36 m s sin 37 v1fx v tan 1 53 to the left tan 1 2fx tan 1 v2fy v1iy v1fy 0.45 m s 0.36 m s cos 37 Thus, v 2f 0.27 m s at 53 to the left . 2 2 61. Strategy Use conservation of momentum. Solution Find v2f in terms of v1f . mv1fy mv2fy mv1f sin 1 mv2f sin 2 mv1iy mv2iy 0 0, so sin 1 sin 60.0 v2f v v 1.73v1f . sin 2 1f sin(30.0) 1f y v1i v1f 60.0° 30.0° x v2f 62. Strategy The collision is perfectly inelastic, so the final velocities of the blocks are identical. Use conservation of momentum. Solution Find the initial speed of block B. pix mA vAix mB vBix 0 mB vBix mB vBix pfx (mA mB )vfx , (m mB )vfx (mA mB )vf cos so vBix A vBi . Compute vBi . mB mB vBi y vf vBi 42.5° (220 g 300 g)(3.13 m s) cos(180 42.5) 4.0 m s 300 g N x vAi Thus the initial speed of block B was 4.0 m s . 63. Strategy Use conservation of momentum. Let +x be along the initial direction of the projectile. Solution Find the magnitude of the momentum of the target body after the collision. Find p2fx . p2 x p2fx p2ix p2fx 0 p1x p1ix p1fx mvi mvfx m(vi vf cos ), so p2fx m(vi vf cos ). Find p2fy . p2 y p2fy p2iy p2fy 0 p1 y p1iy p1fy 0 mvf sin , so p2fy mvf sin . Calculate p2f . p2f p2fx 2 p2fy 2 m (vi vf cos )2 vf 2 sin 2 (2.0 kg) 5.0 m s 3.0 m s cos 60.0 3.0 m s sin 2 60.0 8.7 kg m s 2 2 385 Chapter 7: Linear Momentum Physics 64. (a) Strategy The collision is perfectly inelastic, so the final velocities of the cars are identical. Use conservation of momentum. y Solution Let the 1500-kg car be (1) and the 1800-kg car be (2). m1 pix m1v1ix m2 v2ix m1v1ix 0 pfx (m1 m2 )vfx , so vfx v . m1 m2 1ix m2 piy m1v1iy m2 v2iy 0 m2 v2iy pfy (m1 m2 )vfy , so vfy v . m1 m2 2iy v2i v1i N x Compute the final speed. 2 2 m1v1ix m2 v2iy [(1500 kg)(17 m s)]2 [(1800 kg)(15 m s)]2 vf vfx vfy 1500 kg 1800 kg m1 m2 m1 m2 11 m s Compute the direction. vfy (1800 kg)(15 m s) tan 1 tan 1 47 (1500 kg)(17 m s) vfx 2 2 Thus, the final velocity of the cars is 11 m s at 47 S of E . (b) Strategy Find the change in kinetic energy. Solution 1 1 1 K K f Ki (m1 m2 )vf 2 m1v1i 2 m2 v2i 2 2 2 2 1 1 2 1 (1500 kg 1800 kg)(11.254 m s) (1500 kg)(17 m s) 2 (1800 kg)(15 m s) 2 210 kJ 2 2 2 Thus, 210 kJ of the initial kinetic energy was converted to another form of energy during the collision. 65. Strategy The collision is perfectly inelastic, so the final velocities of the cars are identical. Use conservation of momentum. 386 Physics Chapter 7: Linear Momentum y Solution Let the 1700-kg car be (1) and the 1300-kg car be (2). m1 pix m1v1ix m2 v2ix m1v1ix 0 pfx (m1 m2 )vfx , so vfx v . m1 m2 1ix piy m1v1iy m2 v2iy pfy (m1 m2 )vfy , so vfy N m1v1iy m2 v2iy . m1 m2 v1i Compute the final speed and the direction. mv vf vfx 2 vfy 2 1 1ix m1 m2 2 m1v1iy m2 v2iy m1 m2 2 [(1700 kg)(14 m s) cos 45]2 [(1700 kg)(14 m s)sin 45 (1300 kg)(18 m s)]2 tan 1 1700 kg 1300 kg vfy vfx tan 1 6.0 m s (1700 kg)(14 m s) sin 45 (1300 kg)(18 m s) 21 (1700 kg)(14 m s) cos 45 Thus, the final velocity of the cars is 6.0 m s at 21 S of E . 66. Strategy Use conservation of momentum. y Solution Find the components of the deuteron’s velocity after the collision. Find vdfx . mn vnix md vdix mn vi 0 mn vnfx md vdfx 0 md vdfx , so vdfx vi / 3 m n vi . md vi x Find vdfy . mn vniy md vdiy 0 0 mn vnfy md vdfy mn vi 3 md vdfy , so vdfy 1 3 vi mn . md Find the components, vdfx and vdfy . m m m m 1 1 (vdfx , vdfy ) n vi , vi n n vi , vi n 3 md 2mn 3 2mn md vi vi , 2 3 2 67. Strategy Use conservation of linear momentum. y Solution Find vBfx . pix mvAix pfx mvAfx mvBfx , so vBfx vAix vAfx . Find vBfy . piy mvAiy 0 pfy mvAfy mvBfy , so vBfy vAfy . Compute the final speed of puck B. vBf (vBfx )2 (vBfy )2 (vAix vAfx ) 2 (vAfy )2 [2.0 m s (1.0 m s) cos 60]2 [(1.0 m s) sin 60]2 1.7 m s Compute the direction of puck B. vBfy (1.0 m s)sin 60 tan 1 30 tan 1 2.0 m s (1.0 m s) cos 60 vBfx 387 vAf vAi 60° x v2i x Chapter 7: Linear Momentum Physics Thus, the speed and direction of puck B after the collision is 1.7 m s at 30 below the x-axis . 68. Strategy The collision is perfectly inelastic, so the final velocities of the acrobats are identical. Use conservation of momentum. Solution Let the first acrobat be (1) and the second acrobat be (2). pix m1v1ix m2 v2ix pfx (m1 m2 )vfx , so vfx (m1v1ix m2 v2ix ) (m1 m2 ). piy m1v1iy m2 v2iy pfy (m1 m2 )vfy , so vfy (m1v1iy m2 v2iy ) (m1 m2 ). y 2.0 m/s 20° 3.0 m/s 10° Compute the final speed. vf vfx 2 vfy 2 (m1v1ix m2 v2ix )2 (m1v1iy m2 v2iy )2 m1 m2 [(60)(3.0 m s) cos10 (80)(2.0 m s) cos160]2 [(60)(3.0 m s) sin10 (80)(2.0 m s) sin160]2 60 80 0.64 m s Compute the direction. vfy (60)(3.0 m s) sin10 (80)(2.0 m s) sin160 tan 1 73 tan 1 vfx (60)(3.0 m s) cos10 (80)(2.0 m s) cos160 Thus, the final velocity of the acrobats is 0.64 m s at 73 above the +x-axis . 69. Strategy Use conservation of momentum. 388 x Physics Chapter 7: Linear Momentum Solution Let swallow 1 and its coconut be (1) and swallow 2 and its coconut be (2) (before the collision). After the collision, let swallow 1’s coconut be (3), swallow 2’s coconut be (4), and the tangled-up swallows be (5). m v m4 v4 x pix m1v1x m2 v2 x 0 0 pfx m3v3 x m4 v4 x m5 v5 x , so v5 x 3 3 x . m5 piy m1v1 y m2 v2 y m1v1 m2 v2 pfy m3v3 y m4 v4 y m5v5 y , so m1v1 m2 v2 m3v3 y m4 v4 y v5 y . m5 y v2 30° N x 10° v3 v4 v1 Compute the final speed of the tangled swallows, v5 . v5 v5x 2 v5y 2 1 [ (m3v3 x m4 v4 x )]2 (m1v1 m2 v2 m3v3 y m4 v4 y )2 m5 [(0.80 kg)(13 m s) cos 260 (0.70 kg)(14 m s) cos 60]2 [(1.07 kg)(20 m s) (0.92 kg)( 15 m s) (0.80 kg)(13 m s) sin 260 (0.70 kg)(14 m s) sin 60]2 0.270 kg 0.220 kg 20 m s Compute the direction. v5 y tan 1 v5 x (1.07 kg)(20 m s) (0.92 kg)(15 m s) (0.80 kg)(13 m s) sin 260 (0.70 kg)(14 m s) sin 60 tan 1 (0.80 kg)(13 m s) cos 260 (0.70 kg)(14 m s) cos 60 72 Since v5 x 0 and v5 y 0, the velocity vector is located in the second quadrant, so the angle is 180 72 108 from the positive x-axis or 18° west of north. Thus, the velocity of the birds immediately after the collision is 20 m s at 18 W of N . 70. Strategy Use conservation of momentum. The collision is perfectly inelastic, so v1f v2f vf . Solution Find the speed of the sled once the book is on it. m1v1f m2 v2f (m1 m2 )vf m1v1i m2 v2i m1vi 0, so m1 5.0 kg vf vi (1.0 m s) 0.83 m s . m1 m2 5.0 kg 1.0 kg 71. Strategy Use conservation of momentum. The collision is perfectly inelastic, so v1f v2f vf . Solution Find the speed of the cars just after the collision. m1v1f m2 v2f (m1 m2 )vf m1v1i m2 v2i m1vi 0, so m1 g 13.6 kN vf v 17.0 m s 10.2 m s . (m1 m2 ) g i 13.6 kN 9.0 kN 72. Strategy Use Eqs. (7-10) and (7-11). Solution Find the velocity of the center of mass of the system. p CM Mv CM m1v1 m2 v 2 m3 v3 , so 389 Chapter 7: Linear Momentum Physics (3.0 kg)(290 m s) (5.0 kg)(120 m s) (2.0 kg)(52 m s) 37 m s 3.0 kg 5.0 kg 2.0 kg 0, so v CM 37 m s in the +x-direction . vCMx vCMy 73. Strategy Use the definition of linear momentum. Solution Find the magnitude of the total momentum of the ship and the crew. ptot mtot v (2.0 103 kg 4.8 104 kg)(1.0 105 m s) 5.0 109 kg m s 74. Strategy Use the definition of linear momentum and the impulse-momentum theorem. Solution (a) Compute the magnitude of the change in momentum of the ball. p pf pi mvf mvi m(vf vi ) (0.145 kg) 37 m s 41 m s 11 kg m s (b) Compute the impulse delivered to the ball by the bat. Impulse p 11 kg m s (c) Compute the magnitude of the average force exerted on the ball by the bat. p 11.31 kg m s Fav 3.8 kN t 3.0 103 s 75. Strategy Use the impulse-momentum theorem. y Solution Find the average force exerted by the ground on the ball. p x Fav t 54 m/s 53 m/s m (vx )2 (v y ) 2 18° 22° t 0.060 kg 2 2 53 m s cos18 54 m s cos(22) 53 m s sin18 54 m s sin(22) 34 N 0.065 s 76. Strategy The center of each length is its center of mass. Use the component form of the definition of center of mass. Solution Find the location of the center of mass of the rod. mx mx mx 1 1 3 2 3 3 xCM 3 (0 5.0 cm 10.0 cm) 5.00 cm 3 m 390 Physics Chapter 7: Linear Momentum my my my 1 1 3 2 3 3 yCM 3 (5.0 cm 10.0 cm 5.0 cm) 6.67 cm 3 m Thus, ( xCM , yCM ) (5.00 cm, 6.67 cm) . 77. Strategy The center of mass of each block is its center. Add up the individual center of mass components to find the components of the center of mass of the block structure. Solution mx 2mx2 5mx3 4mx4 x1 2 x2 5 x3 4 x4 0 2(1.0 in) 5(2.0 in) 4(3.0 in) 2.0 in xCM 1 12m 12 12 6my1 4my2 my3 my4 6 y1 4 y2 y3 y4 6(0) 4(1.0 in) 2.0 in 3.0 in 0.75 in yCM 12m 12 12 9mz1 3mz2 9 z1 3 z2 9(0) 3(1.0 in) 0.25 in zCM 12m 12 12 The center of mass of the block structure is located at (2.0 in, 0.75 in, 0.25 in). 78. Strategy Use the impulse-momentum theorem. Solution Compute the force exerted by the stream on a person in the crowd. p mv m v (24 kg s)(17 m s) 410 N Fav t t t 79. Strategy Use the impulse-momentum theorem. Solution Compute the average forces imparted to the two gloved hands during the catches. Inexperienced: Fav 130 km h 10kmm p mv (0.14 kg) t t 103 s Experienced: Fav (0.14 kg) 130 km h 103 m km 3 10 10 1h 3600 s 3 1h 3600 s 5000 N 500 N s 80. Strategy The fly splatters on the windshield, so the collision is perfectly inelastic (vfly, final vcar, final vf ). Use conservation of momentum. Let the positive direction be along the velocity of the automobile. Solution (a) Compute the change in momentum. pcar pfly mfly (vfly, f vfly, i ) mfly (vcar, i 0) (0.1103 kg)(100 km h) 0.01 kg km h So, the change in the car’s momentum due to the fly is 0.01 kg km/h opposite the car’s motion. (b) Compute the change in momentum. pfly pcar 0.01 kg km h , or 0.01 kg km h along the car’s velocity. (c) Compute the number of flies N required to slow the car. (1000 kg) 1 km h m v 105 flies . N pfly mcar vcar , so N car car pfly 0.01 kg km h 391 Chapter 7: Linear Momentum Physics 81. (a) Strategy The initial momentum of the baseball is pi mvi . The final momentum is zero. Solution Compute the change in momentum. p pf pi 0 mvi mvi (0.15 kg)(35 m s) 5.3 kg m s Thus, the change in momentum was 5.3 kg m s opposite the ball’s direction of motion . (b) Strategy and Solution According to the impulse-momentum theorem, the impulse applied to the ball is equal to the change in the momentum of the ball, or 5.3 kg m s opposite the ball’s direction of motion . (c) Strategy Use the impulse momentum theorem. Solution Since the acceleration is assumed constant, the time it takes for the ball to come to a complete stop is t x vav . Compute the average force applied to the ball by the catcher’s glove. p p 35 m s 5.25 kg m s opposite the direction of motion Fav vav t x 2 0.050 m 1.8 kN opposite the ball’s direction of motion 82. Strategy Use conservation of momentum. The collision is perfectly elastic, so Ki K f . Also, v1ix v1i and v1fy v1f . Solution Find the speed of the target body after the collision. x-direction: m1v1fx m2 v2fx 0 m2 v2fx m1v1ix m2 v2ix m1v1ix 0 y-direction: m1v1fy m2 v2fy m1v1iy m2 v2iy 0 0, so m2 v2fy m1v1fy . y v1f = 6.0 m/s v1i = 8.0 m/s x Square the results and add. m22 v2fx 2 m22 v2fy 2 m2 2 v2f 2 m12 v1ix 2 m12 v1fy 2 m12 v1i 2 m12 v1f 2 , so m2 v2f 2 Calculate the kinetic energies. 1 1 1 1 1 m v 2 m2 v2f 2 m1v1i 2 m2 v2i 2 m1v1i 2 0, so m1v1f 2 m2 v2f 2 m1v1i 2 . 2 1 1f 2 2 2 2 392 m12 m2 (v1i 2 v1f 2 ). Physics Chapter 7: Linear Momentum Thus, m2 v2f 2 m1 (v1i 2 v1f 2 ) m12 m2 (v1i 2 v1f 2 ) and m2 m1 v1i 2 v1f 2 v1i 2 v1f 2 . From the kinetic energies, v2f m1 m2 (v1i 2 v1f 2 ) m1 (v1i 2 v1f 2 ) v 2 v 2 m1 1i 2 1f 2 v1i v1f (v1i 2 v1f 2 )2 v1i 2 v1f 2 [(8.0 m s) 2 (6.0 m s) 2 ]2 (8.0 m s)2 (6.0 m s) 2 2.8 m s . 83. Strategy Use conservation of momentum. Let e = electron, neutrino, and n = nucleus. Solution (a) Find the direction of motion of the recoiling daughter nucleus. pe p p n 0, so pnx pex p x pex 0 pe and pny pey p y 0 p y p ( p 0). Find the angle with respect to the electron’s direction. p 5.00 1019 tan 1 tan 1 31.4 180 148.6 CCW from the electron’s direction pe 8.20 1019 (b) Find the momentum of the recoiling daughter nucleus. pn pnx 2 pny 2 ( pe ) 2 ( p ) 2 (8.20 1019 kg m s)2 (5.00 1019 kg m s)2 9.60 1019 kg m s Thus, p n 9.60 1019 kg m/s in the direction found in (a) . 84. Strategy Use conservation of momentum and the definition of center of mass. Let the pier be to the left of the raft and woman at x 0. Solution (a) Since p CM 0, as the woman walks toward the pier, the raft moves away from the pier, and the center of mass does not change. So, xCM mw xwi mr xri mw mr mw xwf mr xrf . mw mr Initially, xCM is to the right of xri . When the woman has walked to the other end of the raft, xCM is to the left of xrf . By symmetry, the distance xCM xri equals the distance xrf xCM , thus xrf xCM xCM xri , so xrf 2 xCM xri . The final distance of the raft from the dock, df , is equal to the difference between xrf and half its length, 3.0 m. 393 Chapter 7: Linear Momentum Physics df xrf 3.0 m 2 xCM xri 3.0 m 2 xCM (3.0 m 0.50 m) 3.0 m 2 xCM 6.5 m Calculate xCM . xCM (60.0 kg)(6.5 m) (120 kg)(3.5 m) 4.5 m, so df 2(4.5 m) 6.5 m 2.5 m . 60.0 kg 120 kg (b) Find the distance the woman walked relative to the pier. xw xwf xwi df xwi 2.5 m 6.5 m 4.0 m 85. Strategy We must determine the initial speeds of the two cars. The collision is perfectly inelastic, so the final velocities of the cars are identical. Use conservation of momentum and the work-kinetic energy theorem. y Solution Let the 1100-kg car be (1) and the 1300-kg car be (2). Use the work-kinetic energy theorem to determine the kinetic energy and, thus, the initial speed of the wrecked vf cars, which is the final speed of the collision. v1i 30° 1 W F r f k r k mg r K 0 mvi 2 , so vi 2k g r . v2i 2 Thus, the final speed of the collision is vf 2k g r . Find the initial speeds. pix m1v1ix m2 v2ix m1v1i 0 pfx (m1 m2 )vfx , so m m2 m m2 1 km h 2400 kg v1i 1 vfx 1 2k g r cos150 2(0.80)(9.80 m s 2 )(17 m) cos150 m1 m1 1100 kg 0.2778 m 110 km h . N x s piy m1v1iy m2 v2iy 0 m2 v2i pfy (m1 m2 )vfy , so m1 m2 m m2 vfy 1 m2 m2 54 km h . v2i 2 k g r sin150 1 km h 2400 kg 2(0.80)(9.80 m s 2 )(17 m) sin150 1300 kg 0.2778 m Since 110 > 70, the lighter car was speeding . 86. Strategy Use the impulse-momentum theorem. Solution Find the speed of the expelled gas relative to the ground. Fav p (m)vgas m 6.0 104 N Fav vgas , so vgas 740 m s . t t t m t 81 kg s 87. Strategy Use the definition of linear momentum and the impulse-momentum theorem. Solution Find the force the kinesin molecule needs to deliver in order to accelerate the organelle. p mv (0.01 1015 kg)(1 106 m s 0) Fav 1018 N t t 10 106 s 88. Strategy Use conservation of momentum. The collision is perfectly inelastic, so vAf vBf vf . 394 s Physics Chapter 7: Linear Momentum Solution Find the final speed in terms of the initial speed. 1 1 1 mA vAf mB vBf (mA mB )vf m m vf mA vAi mB vBi mA vi 0 mvi , so vf vi . 2 3 2 Calculate the ratio of the final kinetic energy to the initial kinetic energy. Kf Ki 1 ( m m )v 2 A B f 2 1m v2 2 A i 1 m m 13 vi 2 1 mv 2 i 2 2 3m 1v2 2 9 i 1 mv 2 i 2 1 3 89. Strategy Use conservation of energy and momentum. Let 2m mB 2mA . Solution Find the maximum kinetic energy of A alone and, thus, its speed just before it strikes B. 1 K mv12 0 U mgh 0, so v1 2 gh . 2 Use conservation of momentum to find the speed of the combined bobs just after impact. The collision is perfectly inelastic, so vAf vBf v2 . 1 mA vAf mB vBf (m 2m)v2 mA vAi mB vBi mv1 0, so v2 v1. 3 Find the maximum height. 2 1 1 1 1 K 0 mv2 2 m h . 2 gh U 0 mgh2 , so h2 2 2 3 9 90. Strategy The center of mass of the disk prior to drilling is ( xCM , yCM ) (0, 0). Let S stand for the small circle removed and L stand for the large circle that remains. Solution Find the center of mass of the metal disk after the hole has been drilled. m x mS xS 0, so xCM L L mL mS xL mS mL xS AS AL xS rS2 rL 2 rS2 xS (1.5 cm)2 (3.0 cm) 2 (1.5 cm)2 (1.5 cm) 0.50 cm. By symmetry, yL yS yCM 0, so ( xL , yL ) (0.50 cm, 0) . 91. Strategy Use conservation of momentum and energy. The collision is elastic, so kinetic energy is conserved. Solution Find the speed of bob B immediately after the collision. Momentum conservation: mvAf mvBf mvAi mvBi mvAi 0, so vBf vAi vAf . Perfectly elastic collision ( Ki K f ): 1 1 1 1 1 mvAf 2 mvBf 2 mvAi 2 mvBi 2 mvAi 2 0, so vAf 2 vBf 2 vAi 2 . 2 2 2 2 2 Energy conservation: 1 mv 2 mgh, so vAi 2 gh . 2 Ai Find vAf in terms of vAi . 395 Chapter 7: Linear Momentum Physics vAi 2 vAf 2 vBf 2 vAf 2 (vAi vAf ) 2 vAf 2 vAi 2 2vAi vAf vAf 2 , so vAf (vAf vAi ) 0. Thus, vAf 0 or vAi . The only way vAf could equal vAi is if bob B didn’t exist, so vAf 0. Calculate vBf . vBf vAi vAf 2 gh 0 2(9.80 m s 2 )(5.1 m) 10 m s 92. Strategy Use conservation of momentum and energy. The collision is elastic, so kinetic energy is conserved. Let the positive direction be to the right. Solution Find the velocities of the gliders after the collision. Momentum conservation: mv1f mv2f mv1i mv2i mv1i 0, so v2f v1i v1f . Perfectly elastic collision due to bumpers ( K i K f ): 1 1 1 1 1 mv 2 mv2f 2 mv1i 2 mv2i 2 mv1i 2 0, so v1f 2 v2f 2 v1i 2 . 2 1f 2 2 2 2 Find v1f in terms of v1i . v1f 2 (v1i v1f ) 2 v1f 2 v1i 2 2v1i v1f v1f 2 v1i 2 , so v1f (v1f v1i ) 0. So, v1f 0 or v1i . The only way v1f could equal v1i is if glider 2 didn’t exist, so v1f 0. Calculate v2f . v2f v1i v1f 0.20 m s 0 0.20 m s After the collision, glider 1 is stationary and glider 2 has a velocity of 0.20 m s in the direction of glider 1’s initial velocity. 93. Strategy Use conservation of momentum and Eq. (6-6) for the kinetic energies. Since the radium nucleus is at rest, pi p Ra 0. Solution (a) Find the ratio of the speed of the alpha particle to the speed of the radon nucleus. pf mRn vRn m v pi 0, so m v mRn vRn . Therefore, v vRn mRn m 222 u 222 111 , where the negative was dropped because speed is nonnegative. 4u 4 2 396 Physics Chapter 7: Linear Momentum (b) Since the initial momentum is zero, p Rn p ; therefore, (c) Find the ratio of the kinetic energies. 1m v 2 m 2 K Rn 1 mRn vRn 2 mRn 2 K v vRn 2 2 4 u 111 111 222 u 2 2 397 p p 1 . pRn p Rn