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Chapter 5 Chapter 5 Center of Mass and Linear Momentum Center of Mass and Linear Momentum Basic Requirements: 1. Understand the concept of the center of mass 2. Master Newton's second law for a system of particles 3. Master the law of conservation of linear momentum 4. Understand the motion of a system with varying mass 5. Understand the external forces and internal energy changes Review and Summary Center of Mass The center of mass of a system of n particles defined to be the point whose coordinates are given by xcom 1 M n mi xi i 1 1 rcom M or ycom , n 1 n mi yi M i 1 , zcom 1 n mi zi M i 1 m r i 1 (5-5) (5-8) i i where M is the total mass of the system. If the mass is continuously distributed, the center of mass is given by xcom 1 xdm M , ycom 1 M ydm , zcom 1 zdm M (5-9) If the density (mass per unit volume) is uniform, the Eq.5-9 can be written as xcom 1 xdV V , ycom 1 ydV V , zcom 1 zdV V (5-11) where V is the volume occupied by M. Newton’s Second Law for a System of Particles The motion of the center mass of any system of particles is governed by Newton’s second law for a system of particles, which is Fnet Macom - 111 - (5-14) Chapter 5 Center of Mass and Linear Momentum Here Fnet is the net force of all the external force acting on the system, M is the total mass of the system, and acom is the acceleration of the system’s center of mass. Linear Momentum and Newton’s Second Law For a single particle, we define a quantity p called its linear momentum as p mv (5-22) and can write Newton’s second law in terms of this momentum: dp Fnet dt (5-23) For a system of particles these relations become dP P Mvc o m and Fnet dt Conservation of Linear Momentum (5-25, 5-27) If a system is isolated so that no net external force acts on the system, the linear momentum P of the system remains constant: P =constant (closed, isolated system) (5-42) This can also be written as Pi Pf (closed, isolated system) (5-43) where the subscripts refer to the value of P at some initial time and at a later time. Eqs. 5-42 and 5-43 are equivalent statements of the law of conservation of linear momentum. Collisions In a collision, two bodies exert strong forces on each other for a relatively short time. These forces are internal to the two-body system and are significantly larger than any external force during the collision. - 112 - Chapter 5 Center of Mass and Linear Momentum Impulse and Linear Momentum Applying Newton’s second law in momentum form to a particle-like body involved in a collision leads to the impulse-linear momentum theorem: p f pi p J where p f pi p is the change in the body’s linear momentum, and J is the impulse due to the force F t exerted on the body by the other body in the collision: tf J F t dt ti If Favg is the average magnitude of F t during the collision and t is the duration of the collision, the for one-dimensional motion J Favg t When a steady stream of bodies, each with mass m and speed v, collides with a body whose position is fixed, the average force on the fixed body is Favg n n p mv t t where n / t is the rate at which the bodies collide with the fixed body, and v is the change in velocity of each colliding body. This average force can also be written as Favg m v t where m / t is the rate at which mass collides with the fixed body. In Eqs.5-15 and 5-16, v v if the bodies stop upon impact, or v 2v if they bounce directly backward with no change in their speed. Inelastic Collision—One Dimension In an inelastic collision of two bodies, the kinetic energy of the two-body system is not conserved. If the system is closed and isolated, then the total linear momentum of the system must be conserved, which we can write in vector form as p1i p2i p1 f p2 f - 113 - (5-77) Chapter 5 Center of Mass and Linear Momentum where subscripts i and f refer to values just before and just after the collision, respectively. If the motion of the bodies is along a single axis, the collision is one-dimensional and we can write Eq.5-77 in terms of velocity components along the axis: mi v1i m2v2i m1v1 f m2v2 f If the bodies stick together, the collision is a completely inelastic collision and the bodies have the same final velocity V (because they are stuck together). Motion of the Center Mass The center of mass of a closed, isolated system of two colliding bodies is not affected by the collision. In particular, the velocity vcom of the center of mass cannot be changed by the collision and its related to the constant total momentum P of the system by vcom P p1i p2i m1 m2 m1 m2 Elastic Collisions—One Dimension An elastic collision is a special type of collision in which the kinetic energy of the system of colliding bodies is conserved. Some collisions in the everyday world can be approximated as being elastic collisions. If the system is closed and isolated, its linear momentum is also conserved. For a one-dimensional collision in which body 2 is a target and body 1 is an incoming projectile, conservation of kinetic energy and linear momentum yield the following expressions for the velocities immediately after the collisions: and v1 f m1 m2 v1i m1 m2 v2 f 2m1 v1i m1 m2 If both bodies are moving prior to the collision, their velocities immediately after the collision are given by - 114 - Chapter 5 Center of Mass and Linear Momentum v1 f and v2 f m1 m2 2m2 v1i v2i m1 m2 m1 m2 2m1 m m1 v1i 2 v2i m1 m2 m1 m2 Note the symmetry of subscripts 1 and 2 in Eqs.5-22 and 5-23. Collisions in Two Dimensions If two bodies collide and their motion is not along a single axis (the collision is not head-on), then the collision is two-dimensional. If the two-body system is closed and isolated, then the law of conservation of momentum applies to the collision and can be written as P1i P2i P1 f P2 f In component form, the law gives two equations that describe the collision (one equation for each of the two dimensions). If the collision is also elastic (a special case), then the conservation of kinetic energy during the collision gives a third equation: K1i K 2i K1 f K 2 f Examples Example 1 Assume that a cannon-shot with mass of 2m is projected out from the ground, when it reaches the maximum height it explodes into two fragments with equal masses (Fig.5-1), among them one piece falls vertically freely, and the other piece horizontally projects out, they reach the ground at the same time. Where does the second fragment land? Solution: Consider the cannon-shot as a system and ignore the air resistance. Before and after the explosion the center of mass of the cannon-shot moves on the same parabola. That is to say, after the explosion the trajectory of the center of mass of the two fragments still follows the parabolic trajectory of the cannon-shot before the explosion. Fig.5-1 Example 1 If we choose where the first fragment lands as the origin O of the coordinates system - 115 - Chapter 5 Center of Mass and Linear Momentum and the rightward horizontal axis as the positive direction of the Ox axis, and we assume m1 and m2 as the masses of the two fragments, we have as given by the question m1 m2 m ; assume x1 and x 2 are the distances between where the fragments land and the origin O. From the figure we know that x1 0 , and we have xC m1 x1 m2 x2 m1 m2 Since m1 m2 m , we have x2 2 xC (Answer) i.e., the horizontal distance between the landing point of the second fragment and the landing point of the first fragment is twice the horizontal distance between the center of mass of the fragments and the landing point of the first fragment. Although this problem can also be solved by the kinetics method, it would be much more complicated than the above method. Example 2 There is soft chain with length l and uniform mass density, the mass per unit is , we put it in a pile on the ground, if we hold on one end of the chain and pull it up with a constant speed v , when the hand-holding end of the chain reaches a height y above the ground, what is the pulling force of the hand? Solution: From the figure we can see that the coordinate ycom of the mass center of the part of the chain already above the ground changes as the chain is pulled up, select the coordinates system as shown in the figure and the center of mass is at ycom my m i i i y y l y 0 y2 2 l 2l (1) Fig.5-2 Example 2 - 116 - Chapter 5 Where Center of Mass and Linear Momentum is the mass per unit length of the chain. Since the combined d 2 ycom external force on the chain if F yg l dt 2 i.e., F yg ˆj l d 2 ycom ˆ j dt 2 (2) Taking the second derivative with respect to time t on Eq.(1), we have 2 d 2 ycom 1 dy d2y y dt 2 l dt dt 2 Considering v=dy/dt, v=0, we have yd y / d t 0 . The above becomes 2 2 d 2 y com v 2 l dt 2 Substituting the above into Eq.(2), we have F yg ˆj v 2 ˆj F v 2 yg and Example 3 (Answer) Assume that a nucleus at rest decays and emits an electron and a neutrino and becomes a new nucleus. It is known that the directions of motion of the electron and the neutrino are perpendicular to each other, the momentum of the 1.2 10 22 kg m / s and the momentum of the neutrino is 6.4 10 23 kg m / s , what is the magnitude and direction of the momentum of the electron is new nucleus? Solution: Let pe , pv and p N denote the momentum of the electron, the neutrino, and the new nucleus respectively, and pe and pv are perpendicular as shown in the figure. In the short time period of the nuclear decay, the internal force between the particles are much greater than the external forces acting on the particle system, therefore the momentum of the particle system is conserved before and after the decay - 117 - Chapter 5 Center of Mass and Linear Momentum process. Consider that the nucleus was at rest before the decay, then after the decay the sum of the momentum of the electron, the neutrino, and the new nucleus is also equal to 0, i.e., pe pv pN 0 Since pe and pv are perpendicular, we have pN pe2 pv2 1/ 2 Substituting the known data in it we, have 2 pN 1.2 10 22 6.4 10 23 2 1/ 2 kg m s 1 1.36 10 22 kg m s 1 The angle α in the figure is: arctan pe 1.2 1022 arctan 61.9 pv 6.4 1023 Or the angle between the momentum p N of the new nucleus and the momentum pv of the neutrino is 180 61.9 118.1 (Answer) Example 4 A returning rocket flies at 2.5 10 m / s along the horizontal direction 3 with respect to the ground, ignore the air resistance. The control system separates the rocket into two parts, the front part is the instrumental module with mass of 100kg and the back part is the rocket container with mass of 200kg If the speed of the instrumental module with respect to the rocket container is 1.0103 m/s, what are the speeds of the instrumental module and the rocket container with respect to the ground? Solution: As shown in the figure, take the ground as an inertial frame S (O xyz ) , let v be the velocity of the rocket with respect to the inertial frame S before the separation, v1 and v2 be the velocities of the instrumental module and the rocket - 118 - Chapter 5 Center of Mass and Linear Momentum container respectively, with respect to the inertial frame S after the separation, v ' be the velocity of the instrumental module with respect to the rocket container after the separation, and take the rocket container as another inertial frame S (O x y z ) , the S system moves with velocity v2 along the xx’ axis respect to the S frame. With the velocity formula of the relative motions, we have v1 v2 v ' Since v1 , v2 and v ' are all in the same horizontal direction, the above equation becomes v1 v2 v' Before and after the rocket separation, it is only subject to the gravity in the vertical direction, so the momentum component in the horizontal direction is conserved, we have m1 m2 v m1v1 m2v2 Solving the above two equations, we have v2 v m1 v' m1 m2 Substituting the data in it, we have 1 v2 2.5 103 1.0 103 m / s 2.17 103 m / s 3 3 3 v1 2.17 10 1.0 10 m / s 3.17 103 m / s (Answer) Both v1 and v2 are positive numbers, indicating that their velocities are in the same direction as v , only that the instrumental module increased its speed by the thrust of the rocket, while the speed of the rocket container got slowed down, therefore the momentum transfer is realized. Example 5 Assume there are dusts with density ρ in the universe, such dusts are at rest with respect to the inertial reference frame. A spaceship with mass m0 penetrates - 119 - Chapter 5 Center of Mass and Linear Momentum through the dusts with the initial velocity v0 , since the dusts get stuck on the spaceship, it causes the speed of the spaceship to change, what is the relationship between the speed of the spaceship and its time of flight in the dusts? For convenience, assume the external shape of the spaceship is a cylinder with the cross-sectional area S. Solution: According to the conditions given in the problem, we may consider the collisions between the dusts and the spaceship are complete inelastic collisions, we treat the dusts and the spaceship as a single system, also consider the spaceship flies in free space and there are no external forces on this system, therefore the momentum of the system in conserved. Before entering the dusts (i.e., t=0) the mass and the speed of the spaceship are m0 and v0 respectively, and when the spaceship is in the dusts (at time t) the mass and the speed are m and v respectively. Therefore, according to momentum conservation, we have Fig.5-3 Example 5 m0 v0 mv (1) Also, in the time interval t t dt , since the collision between the spaceship and the dusts are complete inelastic collisions, the mass of the dusts stuck on the spaceship, i.e., the increase of the mass of the spaceship, is dm Svdt (2) From (1) we have dm m0 v0 dv v2 Therefore, we have Svdt m0 v 0 dv v2 From the given conditions, integrating the above, we have dv S 3 v0 v m0 v 0 v We obtain - 120 - dt 0 dt Chapter 5 Center of Mass and Linear Momentum 1 1 1 S 2 2 t 2v v0 m0 v0 We have v m0 v0 2Sv0t m0 (Answer) Obviously, the longer the spaceship flies in the dusts, the lower its speed. Example 6 As shown in the figure, assume two elastic small balls with masses m1 and m2 and velocities v10 and v 20 collide in such a way that their centers are aligned in a straight line, the directions of the two velocities are the same. If the collision is m1v10 m2 v20 m1v1 m2 v2 (1) completely elastic, what are the velocities v1 and v2 after the collision? Solution: From the theorem of momentum conservation, we have From the law of conservation of mechanical energy we have 1 1 1 1 2 2 m1v10 m2 v20 m1v12 m2 v22 2 2 2 2 (2) Fig.5-4 Example 6 Eq.(1) can be written as m1 v10 v1 m2 v2 v20 Eq.(2) can be written as m1 v10 v1 m2 v2 v20 2 2 2 2 (3) (4) From Eqs.(3) and (4) we have v10 v1 v2 v20 Or v10 v20 v2 v1 - 121 - (5) Chapter 5 Center of Mass and Linear Momentum Eq.(5) indicates that before the collision the relative velocity v10 v20 of the two balls approaching each other is equal to the relative velocity v2 v1 of the two balls separating each other after the collision. From Eq.(3) and Eq.(5), we obtain: v1 v2 m1 m2 v10 2m2 v20 m1 m2 (Answer) m m1 v20 2m1v10 2 m1 m2 Problem Solving 1 (2) In the ammonia ( NH 3 ) molecule of Fig.5-5, three hydrogen ( H ) atoms form an equilateral triangle, with the center of the triangle at distance d 9.40 10 11 m from each hydrogen atom. The nitrogen ( N ) atom is at the apex of a pyramid, with the three hydrogen atoms forming the base. The nitrogen-to-hydrogen atomic mass ratio is 13.9 , and the nitrogen-to-hydrogen distance is L 10.14 10 are the (a) x and (b) y coordinates of the molecule’s center of mass? Solution: The position of the molecule’s center of mass can be found out by the center of mass equations of a system of n particles. But here we can first get the 11 m . What position of the center of mass of the three hydrogen atoms on the base, then the center of mass of the molecule can be determined by the coordinates of the center of mass of the base and the nitrogen atom. Fig.5-5 Problem 1 Obviously, the coordinates of the center of mass of equilateral triangle of hydrogen atoms has xbase d , ybase 0 The x and y coordinates of the nitrogen atom are x 0, y L2 d 2 10.14 10 9.40 10 11 2 11 2 3.80 1011 m Then, the x and y coordinates of center of mass of the molecule are - 122 - Chapter 5 Center of Mass and Linear Momentum xcom 0 y com 3m H ybase m N y N 13.9m H y N 3m H m N 3m H 13.9m H (Answer) 13.9 3.80 10 11 3.13 10 11 m 16.9 2 (4) A metal soda can of uniform composition has a mass of 0.140kg and is 12.0cm tall (Fig.5-6). The can is filled with 1.31kg of soda. Then small holes are drilled in the top and bottom (with negligible loss of metal) to drain the soda. What is the height h of the center of mass of the can and the contents (a) initially and (b) after the can loss all the soda? (c) What happens to h as the soda drains out? (d) If x is the height of the remaining soda at any given instant, find x when the center of mass reaches its lowest point. Solution: (a) Since the can is uniform and filled with soda, the center of mass initially is right at the center of the can: hi 12.0 6.0cm 2 (Answer) (b) After the can loss all the soda, the center of mass the system is that of the can: hf 12.0 6.0cm 2 (Answer) Fig.5-6 Problem 2 (c,d) The system consists of the can and the remained soda, so 0.14kg6cm 1.31kg x x cm h 12 2 1.31kg 0.14kg x 12 dh 0 , from the calculus, x has the minimum value that means dx the center of mass reaches its lowest point is x 2.84cm So h decreases first, after the center of mass reaches its lowest point when We know when - 123 - Chapter 5 Center of Mass and Linear Momentum x 2.84cm , it begins to increase until the can loss all the soda. 3 (5) In Figure 5-7, two particles are launched from the origin of the coordinate system at time t 0 . Particle 1 of mass m1 5.00g is shot directly along the x axis (on a frictionless floor), where it moves with a constant speed of 10.0m/ s . Particle 2 of mass m2 3.00 g is shot with a velocity of magnitude 20.0m/ s at an upward angle such that it always stays directly above particle 1 during its flight. (a) What is the maximum height H max reached by the com of the two-particle system? In unit-vector notation, what are the (b) velocity and (c) acceleration of the com when the com reaches H max ? Solution: (a) Suppose the launching angle of Particle 2 is . To make Particle 2 always stay directly above particle 1 v2 x v2 cos v1 (1) Fig.5-7 Problem 3 Substituting v1 10.0m / s, v2 20m / s into Eq.(1) yields 1 m2 v2 sin t gt 2 m2 y2 2 So ycom m1 m2 m1 m2 60 (2) Substituting m1 5.00 g , m2 3.00 g , g 9.8m / s into Eq.(2) and suppose 2 dycom 0 dt yields t 3s so H max ycom (t 3s) 5.625m (Answer) (b) As the com reaches Hmax, v2 y 0 , so vcom, x 10m / s, vcom, y 0 - 124 - (3) Chapter 5 And Center of Mass and Linear Momentum vcom vcom, xiˆ vcom, y ˆj 10iˆm / s (Answer) (c) From the equation Fnet Ma com , we get m2 g ˆ a a y ˆj j 3.75 ˆjm / s 2 m1 m2 (Answer) 4 (6) Ricardo, of mass 80kg , and Carmelita, who is lighter, are enjoying Lake Merced at dusk in a 30kg canoe. When the canoe is at rest in the placid water, they exchange seats, which are 3.0m apart and symmetrically located with respect to the canoe’s center. Ricardo notices that the canoe moves 40cm horizontally relative to a pier post during the exchange and calculates Carmelita’s mss. What is it? Solution: The key idea here is that during the exchange, there is no external force on the two-people and the canoe system, and the total linear momentum is conserved. So the center of mass of the system will keep Ricardo Carmelita unchanged. Choose the initial position of the center of mass of the canoe as origin and construct x x axis as shown. As the diagrammatic sketch O shown, Ricardo notices that the canoe Carmelita Ricardo moves 40cm relative to a pier post, that means the canoe moves 20cm relative to the unchanged center of mass of the system. 40 cm (The initial and final position of the canoe is symmetric about the center of mass of the Center of Mass system.) Since Fig. 5-8 Problem 4 xcom mRicardox Ricardo mCarmelitaxCarmelita mcanoe xcanoe mRicardo mCarmelita mcanoe we have - 125 - Chapter 5 Center of Mass and Linear Momentum 80kg 3m mCarmelita 3m 30kg 0 2 2 3m 3m 80kg 20cm mCarmelita 20cm 30kg 20cm 2 2 yields mCarmelita 68.125kg (Answer) 5 (9) A 0.30kg softball has velocity of 15m / s at an angle of 35 below the horizontal just before making contact with the bat. What is the magnitude of the change in momentum of the ball while it is in contact with the bat if the ball leaves the bat with a velocity of (a) 20m/ s , vertically downward, and (b) 20m/ s , horizontally back toward the pitcher? Solution: (a) Construct the coordinates as shown in Fig. 5-9. The change in momentum p mv f mvi O 0.30kg 20 ˆj 15 cos 35iˆ 15 sin 35 ˆj m / s 3.69iˆ 3.42 ˆj kg m / s Magnitude p 3.692 3.422 5.03kg m / s (Answer) (b) The change in momentum of the ball is 35º vi x y Fig. 5-9 Problem 5 p 0.30kg 20iˆ 15 cos 35iˆ 15 sin 35 ˆj m / s 9.69iˆ 2.58 ˆj kg m / s p 9.692 2.582 10.03kg m / s (Answer) 6 (13) Figure 5-10 shows a two-ended “rocket” that is initially stationary on a frictionless floor, with its center at the origin of an x axis. The rocket consists of a central block C (of mass M 6.00kg ) and blocks L and R (each of mass m 2.00kg ) on the left and right sides. Small explosions can shoot either of the side blocks away from block C and along the x axis. Here is the sequence: (1) At time t 0 , block L is shot to the left with a speed of 3.00m/ s relative to the - 126 - Chapter 5 Center of Mass and Linear Momentum velocity that the explosion gives the rest of the rocket. (2) Next, at time t 0.80s , block R is shot to the right with a speed of 3.00m/ s relative to the velocity that block C then has. At t 2.80s , what are (a) the velocity of block C and (b) the position of its center? Solution: (a) From time t 0s to t 0.80s , the velocity of block C is v C 1 , as the linear momentum is conserved: Fig.5-10 Problem 6 2.00kgvC1 3.00m / s 6.00kg 2.00kgvC1 0 yields vC1 0.6m / s (Answer) From time t 0.80s , the velocity of block C is vC 2 , and 2.00kgvC 2 3.00m / s 6.00kgvC 2 6.00kg 2.00kgvC1 yields vC 2 0.15m / s At t 2.80s , the velocity of block C is vC 2 0.15m / s . (Answer) (b) The position of center of block C: x vC1 0.80s vC 2 2.80s 0.80s 0.18m (Answer) 7 (19) In Fig.5-11, a 10 g bullet moving directly upward at 1000m/ s strikes and passes through the center of mass of a 5.0kg block initially at rest. The bullet emerges from the block moving directly upward at 400m/ s . To what maximum height does the block then rise above its initial position? Solution: Choose the initial position of the block as origin and the upward direction as positive. The linear momentum of the bullet-block system is conserved during the interaction: mbulletvbullet,i mbulletvbullet, f mblock vblock - 127 - Fig.5-11 Problem 7 Chapter 5 Center of Mass and Linear Momentum 10 g 1000m / s 10 g 400m / s 5.0kgvblock vb l o ck 1.2m / s yields The maximum height of the block: vblock 1.2m / s 2 0.073m 7.3cm 2g 2 9.8m / s 2 2 y max (Answer) 8 (20) A completely inelastic collision occurs between two balls of wet putty that move directly toward each other along a vertical axis. Just before the collision, one ball, of mass 3.0kg , is moving upward at 20m/ s and the other ball, of mass 2.0kg , is moving downward at 12m / s . How high do the combined two balls of putty rise above the collision point? (Neglect air drag) Solution: Choose the collision point as origin and the upward direction as positive. The velocity of the combined balls right after the collision is v . The linear momentmum of the system is conservedin the collision, so we get 3.0kg20m / s 2.0kg 12m / s 3.0kg 2.0kgv Solving for v , yields v 7.2m / s Then the maximum height that the combined balls rise is 7.2m / s 2.64m v2 2 g 2 9.8m / s 2 2 ymax (Answer) 9 (21) In Fig.5-12, block 1 (mass 2.0kg ) is moving rightward at 10 m/s and block 2 (mass 5.0kg ) is moving rightward at 3.0m/ s . The surface is frictionless, and a spring with a spring constant of 1120N / m is fixed to block 2. When the blocks collide, the compression of the spring is maximum at the instant the blocks have the same velocity. Find the maximum compression. Fig.5-12 Problem 9 - 128 - Chapter 5 Center of Mass and Linear Momentum Solution: The linear momentum of the two blocks and the spring system is conserved, and the mass of the spring is negligible. Choose rightward as positive direction and suppose when the compression of the spring is maximum, the velocity of the blocks is v , we have 2.0kg10m / s 5.0kg3.0m / s 2.0kg 5.0kgv Solving for v , yields v 5m / s Because of the compression of the spring, the total kinetic energy of the system decreased, gives us 1 1 1 1 2 2 2 2 kxmax 2.0kg 10m / s 5.0kg 3m / s 2.0kg 5.0kg 5m / s 2 2 2 2 Substituting k=1120N/m into the above equation yields the maximum compression of the spring: xmax 0.25m (Answer) 10 (25) In Fig.5-13, puck 1 of mass m1 0.20kg is sent sliding across a frictionless lab bench, to undergo a one-dimensional elastic collision with stationary puck 2. Puck 2 then slides off the bench and lands a distance d from the base of the bench. Puck 1 rebounds from the collision and slides off the opposite edge of the bench, landing a distance 2d from the base of the bench. What is the mass of puck 2? (Hint: Be careful with signs.) Fig.5-13 Problem 10 Solution: Choose rightward as positive direction. The linear momentum of the two-puck system is conserved during the collision, so m1v1i m1v1 f m2v2 With the landing distance of the two pucks, we have - 129 - (1) Chapter 5 Center of Mass and Linear Momentum v1 f v2 2d 2 d (2) During one-dimensional elastic collision, the total kinetic energy is conserved: 1 1 1 2 2 2 m1v1i m1v1 f m2v2 2 2 2 (3) From Eqs.(1)~(3), we get the mass of puck 2: m2 5m1 5 0.20kg 1.00kg (Answer) 11 (33) Figure 5-14 shows a uniform square plate of edge length 6d 6.0m from which a square piece of edge length 2d has been removed. What are (a) the x coordinate and (b) the y coordinate of the center of mass of the remaining piece? Solution: The square plate of edge length 6d consists of the square piece of edge length 2d and the remaining piece, and obviously the center of mass of the uniform square plate is at the origin. So we have xcom, plate ycom, plate m piece xcom, piece mremainxcom, remian m plate m piece ycom, piece mremain ycom, remian m plate Fig.5-14 Problem 11 0 (1) 0 where xcom, piece 2d , ycom, piece 0 (2) The uniform square gives us m p i e ce 2d 2 m 1 m , 6d 2 p l a t e 9 p l a t e 8 mr e m a i nm p l a m p l a t e (3) t e m p i e ce 9 Eqs.(1)~(3) give us the coordinates of the center of mass of the remaining piece: 1 xcom, remian d 0.25m , 4 12 (44) ycom, remain 0 (Answer) Two identical coins are initially held at height h 11.0m . Coin 1 is - 130 - Chapter 5 Center of Mass and Linear Momentum dropped at time t 0 and then lands on a muddy field where it sticks. Coin 2 is dropped at t 0.500s and then lands on the field. What is the acceleration acom of the center of mass (com) of the two-coin system (a) between t 0 and t 0.500s , (b) between t 0.500s and time t1 when coin hits and sticks, and (c) between t1 and time t 2 when coin 2 hits and sticks? What is the speed of the center of mass when t is (d) 0.250s , (e) 0.750s , (f) 1.75s ? Solution: Choose the dropping point as origin and downward as positive direction,construct x coordinate. t1 2h 2 11m 2h 1.50s , t 2 0.500s 2.00s 2 g 9.8m / s g (a) between t 0 and t 0.500s , coin 1 is free falling and coin 2 is at rest, so m g ˆ 1 ˆ acom i gi 4.9iˆ m / s 2 2m 2 (Answer) (b) between t 0.500s and time t1 1.50s , coin 1 and 2 are both free falling, and m g m g ˆ ac o m i giˆ 9.8iˆ m / s 2 2m (Answer) (c) between t1 1.50s and time t 2 2.00s , coin 1 is at rest and coin 2 is free falling, and m g ˆ 1 ˆ acom i gi 4.9iˆ m / s 2 (Answer) 2m 2 (d) When t 0.250s , coin 1 is free falling and coin 2 is at rest, and the speed of com is vcom mv1 1 1 gt 9.8m / s 2 0.250s 1.225m / s 2m 2 2 (Answer) (e) When t 0.750s , 0.500s t t1 , coin 1 and 2 are both free falling, and mv1 mv2 1 1 gt g t 0.500s g 2t 0.500s 2m 2 2 1 9.8m / s 2 2 0.750s 0.500s 2 4.9m / s vcom - 131 - (Answer) Chapter 5 Center of Mass and Linear Momentum t1 t t 2 , coin 1 is at rest and coin 2 is free falling, (f) When t 1.75s , and vcom mv2 1 1 g t 0.500s 9.8m / s 2 1.75s 0.500s 6.125m / s 2m 2 2 (Answer) 13 (47) In Fig.5-15, a 3.2kg box of running shoes slides on a horizontal frictionless table and collides with a 2.0kg box of ballet slippers initially at rest on the edge of the table, at height h 0.40m . The speed of the 3.2kg box is 3.0m/ s just before the collision. If the two boxes stick together because of packing tape on their sides, what is their kinetic energy just before they strike the floor? Fig.5-15 Problem 13 Solution: Choose the edge of the table as origin, and construct x y coordinates as shown. Just before the boxes strike the floor, their velocity v vxiˆ v y ˆj During the collision, the linear momentum of the two-box system is conserved: 3.2kg3.0m / s 3.2kg 2.0kgvx vx 1.85m / s Solving for v x , yields So vy 2 gh 2 9.8m / s 2 0.40m 2.8m / s Along the y axis v 2 vx v y 1.85m / s 2.80m / s 11.2625m / s 2 2 2 2 2 The kinetic energy of the two boxes just before they strike the floor is K 1 3.2kg 2.0kg v 2 1 5.2kg 11.2625m2 / s 2 29.28 J 2 2 (Answer) 14 (49) In Fig.5-16, block 1 slides along an x axis on a frictionless floor with a speed of 0.75m/ s . When it reaches stationary block 2, the two blocks undergo an elastic collision. The following table gives the mass and length of the (uniform) blocks - 132 - Chapter 5 Center of Mass and Linear Momentum and also the locations of their centers at time t 0 . Where is the center of mass of the two-block system located (a) at t 0 , (b) when the two blocks first touch, and (c) at t 4.0s ? Block Mass (kg) Length (cm) Center at t=0 1 2 0.25 0.50 5.0 6.0 x=-1.50m x=0 Fig.5-16 Problem 14 Solution: (a) At t 0 , the coordinate of the center of mass: xcom 0.25kg 1.50m 0.50kg0 0.5m 0.25kg 0.50kg (Answer) (b) When the two blocks first touch, the coordinate of com: 0.25kg 5.0cm 0.50kg0 xcom 2 0.25kg 0.50kg 0.83cm (c) Suppose block 1reaches stationary block 2 at time t1 , then 5.0cm 6.0cm 1.50m 2 2 t1 1.93s 0.75m / s After collision, the velocity of com is that of the two boxes: v 0.25kg0.75m / s 0.25m / s 0.25kg 0.50kg At t 4.0s , the position of the com - 133 - (Answer) Chapter 5 Center of Mass and Linear Momentum xcom 0.83cm vt t1 0.83cm 0.25m / s 4.0s 1.93s 50.92cm (Answer) 15 (52) Two particles P and Q are released from rest 1.0m apart, P has a mass of 0.10kg , and Q a mass of 0.30kg . P and Q attract each other with a 2 constant force of 1.0 10 N . No external forces act on the system. (a) What is the speed of the center of mass of P and Q when the separation is 0.50m ? (b) At what distance from P' s original position do the particles collide? Solution: (a) Choose P' s original position as origin and P' s velocity as positive. Since no external forces act on the system, the linear momentum of the system is conserved, and the speed of the center mass will keep unchanged as zero. vcom 0 (Answer) (b) As the speed of the com is zero, the position of the com is at rest, and the two particles will collide at that point, that is xcom 0.10kg0 0.30kg1.0m 0.75m 0.10kg 0.30kg So the particles collide 0.75m from P' s original position. - 134 - (Answer)