Linear Momentum and Collisions

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1 Chapter 7 Linear Momentum and Colliion 7.1 The Important Stuff Linear Momentum The linear momentum of a particle with ma m moving with velocity v i defined a p = mv (7.1) Linear momentum i a vector. When giving the linear momentum of a particle you mut pecify it magnitude and direction. We can ee from the definition that it unit mut be. Oddly enough, thi combination of SI unit doe not have a commonly ued named o we leave it a! The momentum of a particle i related to the net force on that particle in a imple way; ince the ma of a particle remain contant, if we take the time derivative of a particle momentum we find dp dt = mdv dt = ma = F net o that Impule, Average Force F net = dp dt (7.2) When a particle move freely then interact with another ytem for a (brief) period and then move freely again, it ha a definite change in momentum; we define thi change a the impule I of the interaction force: I = p f p i = p Impule i a vector and ha the ame unit a momentum. When we integrate Eq. 7.2 we can how: I = tf t i Fdt = p 155

2 156 CHAPTER 7. LINEAR MOMENTUM AND COLLISIONS We can now define the average force which act on a particle during a time interval t. It i: F = p t = I t The value of the average force depend on the time interval choen Conervation of Linear Momentum Linear momentum i a ueful quantity for cae where we have a few particle (object) which interact with each other but not with the ret of the world. Such a ytem i called an iolated ytem. We often have reaon to tudy ytem where a few particle interact with each other very briefly, with force that are trong compared to the other force in the world that they may experience. In thoe ituation, and for that brief period of time, we can treat the particle a if they were iolated. We can how that when two particle interact only with each other (i.e. they are iolated) then their total momentum remain contant: or, in term of the mae and velocitie, p 1i + p 2i = p 1f + p 2f (7.3) m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f (7.4) Or, abbreviating p 1 + p 2 = P (total momentum), thi i: P i = P f. It i important to undertand that Eq. 7.3 i a vector equation; it tell u that the total x component of the momentum i conerved, and the total y component of the momentum i conerved Colliion When we talk about a colliion in phyic (between two particle, ay) we mean that two particle are moving freely through pace until they get cloe to one another; then, for a hort period of time they exert trong force on each other until they move apart and are again moving freely. For uch an event, the two particle have well-defined momenta p 1i and p 2i before the colliion event and p 1f and p 2f afterward. But the um of the momenta before and after the colliion i conerved, a written in Eq While the total momentum i conerved for a ytem of iolated colliding particle, the mechanical energy may or may not be conerved. If the mechanical energy (uually meaning the total kinetic energy) i the ame before and after a colliion, we ay that the colliion i elatic. Otherwie we ay the colliion i inelatic. If two object collide, tick together, and move off a a combined ma, we call thi a perfectly inelatic colliion. One can how that in uch a colliion more kinetic energy i lot than if the object were to bounce off one another and move off eparately.

3 7.1. THE IMPORTANT STUFF 157 When two particle undergo an elatic colliion then we alo know that 1 m 2 1v1i 2 + 1m 2 2v2i 2 = 1m 2 1v1f 2 + 1m 2 2v2f 2. In the pecial cae of a one-dimenional elatic colliion between mae m 1 and m 2 we can relate the final velocitie to the initial velocitie. The reult i v 1f = v 2f = ( m1 m 2 m 1 + m 2 ( 2m1 m 1 + m 2 ) ( 2m2 v 1i + ) v 1i + m 1 + m 2 ( m2 m 1 m 1 + m 2 ) v 2i (7.5) ) v 2i (7.6) Thi reult can be ueful in olving a problem where uch a colliion occur, but it i not a fundamental equation. So don t memorize it The Center of Ma For a ytem of particle (that i, lot of em) there i a pecial point in pace known a the center of ma which i of great importance in decribing the overall motion of the ytem. Thi point i a weighted average of the poition of all the ma point. If the particle in the ytem have mae m 1, m 2,... m N, with total ma N m i = m 1 + m m N M i and repective poition r 1, r 2,...,r N, then the center of ma r CM i: r CM = 1 M N m i r i (7.7) i which mean that the x, y and z coordinate of the center of ma are x CM = 1 M N m i x i i y CM = 1 M N m i y i i z CM = 1 M N m i z i (7.8) i For an extended object (i.e. a continuou ditribution of ma) the definition of r CM i given by an integral over the ma element of the object: r CM = 1 M rdm (7.9) which mean that the x, y and z coordinate of the center of ma are now: x CM = 1 M xdm y CM = 1 M y dm z CM = 1 M z dm (7.10)

4 158 CHAPTER 7. LINEAR MOMENTUM AND COLLISIONS When the particle of a ytem are in motion then in general their center of ma i alo in motion. The velocity of the center of ma i a imilar weighted average of the individual velocitie: v CM = dr CM = 1 N m i v i (7.11) dt M i In general the center of ma will accelerate; it acceleration i given by a CM = dv CM dt = 1 M N m i a i (7.12) i If P i the total momentum of the ytem and M i the total ma of the ytem, then the motion of the center of ma i related to P by: v CM = P M and a CM = 1 M dp dt The Motion of a Sytem of Particle A ytem of many particle (or an extended object) in general ha a motion for which the decription i very complicated, but it i poible to make a imple tatement about the motion of it center of ma. Each of the particle in the ytem may feel force from the other particle in the ytem, but it may alo experience a net force from the (external) environment; we will denote thi force by F ext. We find that when we add up all the external force acting on all the particle in a ytem, it give the acceleration of the center of ma according to: N F ext,i = Ma CM = dp (7.13) dt i Here, M i the total ma of the ytem; F ext,i i the external force acting on particle i. In word, we can expre thi reult in the following way: For a ytem of particle, the center of ma move a if it were a ingle particle of ma M moving under the influence of the um of the external force. 7.2 Worked Example Linear Momentum 1. A 3.00kg particle ha a velocity of (3.0i 4.0j) m. Find it x and y component of momentum and the magnitude of it total momentum. Uing the definition of momentum and the given value of m and v we have: p = mv = (3.00kg)(3.0i 4.0j) m = (9.0i 12.j)

5 7.2. WORKED EXAMPLES kg 10 m/ 10 m/ y 60 o 60 o x Figure 7.1: Ball bounce off wall in Example 3. So the particle ha momentum component The magnitude of it momentum i p x = +9.0 and p y = 12.. p = p 2 x + p 2 y = Impule, Average Force (9.0) 2 + ( 12.) 2 = A child bounce a uperball on the idewalk. The linear impule delivered by the idewalk i 2.00N during the 1 of contact. What i the magnitude of the 800 average force exerted on the ball by the idewalk. The magnitude of the change in momentum of (impule delivered to) the ball i p = I = 2.00 N. (The direction of the impule i upward, ince the initial momentum of the ball wa downward and the final momentum i upward.) Since the time over which the force wa acting wa t = = then from the definition of average force we get: F = I t = 2.00N = N 3. A 3.0kg teel ball trike a wall with a peed of 10 m at an angle of 60 with the urface. It bounce off with the ame peed and angle, a hown in Fig If the ball i in contact with the wall for 0.20, what i the average force exerted on the wall by the ball?

6 160 CHAPTER 7. LINEAR MOMENTUM AND COLLISIONS x 750 m/ m m m m m = 35 g Figure 7.2: Simplified picture of a machine gun pewing out bullet. An external force i neceary to hold the gun in place! The average force i defined a F = p/ t, o firt find the change in momentum of the ball. Since the ball ha the ame peed before and after bouncing from the wall, it i clear that it x velocity (ee the coordinate ytem in Fig. 7.1) tay the ame and o the x momentum tay the ame. But the y momentum doe change. The initial y velocity i and the final y velocity i o the change in y momentum i v iy = (10 m )in60 = 8.7 m v fy = +(10 m )in 60 = +8.7 m p y = mv fy mv iy = m(v fy v iy ) = (3.0kg)(8.7 m ( 8.7 m The average y force on the ball i F y = p y t = I y t = (52 ) (0.20) = N )) = 52 Since F ha no x component, the average force ha magnitude N and point in the y direction (away from the wall). 4. A machine gun fire 35.0g bullet at a peed of m. If the gun can fire 200bullet/min, what i the average force the hooter mut exert to keep the gun from moving? Whoa! Lot of thing happening here. Let draw a diagram and try to ort thing out. Such a picture i given in Fig The gun interact with the bullet; it exert a brief, trong force on each of the bullet which in turn exert an equal and oppoite force on the gun. The gun force change the bullet momentum from zero (a they are initially at ret) to the final value of p f = mv = (0.0350kg)(750 m ) = 26.2.

7 7.2. WORKED EXAMPLES 161 o thi i alo the change in momentum for each bullet. Now, ince 200 bullet are fired every minute (60 ), we hould count the interaction time a the time to fire one bullet, t = = 0.30 becaue every 0.30, a firing occur again, and the average force that we compute will be valid for a length of time for which many bullet are fired. So the average force of the gun on the bullet i F x = p x t 26.2 = 0.30 = 87.5N From Newton Third Law, there mut an average backward force of the bullet on the gun of magnitude 87.5N. If there were no other force acting on the gun, it would accelerate backward! To keep the gun in place, the hooter (or the gun mechanical upport) mut exert a force of 87.5N in the forward direction. We can alo work with the number a follow: In one minute, 200 bullet were fired, and a total momentum of P = (200)(26.2 ) = wa imparted to them. So during thi time period (60 econd!) the average force on the whole et of bullet wa F x = P t = = 87.5N A before, thi i alo the average backward force of the bullet on the gun and the force required to keep the gun in place Colliion 5. A 10.0g bullet i topped in a block of wood (m = 5.00kg). The peed of the bullet plu wood combination immediately after the colliion i m. What wa the original peed of the bullet? A picture of the colliion jut before and after the bullet (quickly) embed itelf in the wood i given in Fig The bullet ha ome initial peed v 0 (we don t know what it i.) The colliion (and embedding of the bullet) take place very rapidly; for that brief time the bullet and block eentially form an iolated ytem becaue any external force (ay, from friction from the urface) will be of no importance compared to the enormou force between the bullet and the block. So the total momentum of the ytem will be conerved; it i the ame before and after the colliion. In thi problem there i only motion along the x axi, o we only need the condition that the total x momentum (P x ) i conerved.

8 162 CHAPTER 7. LINEAR MOMENTUM AND COLLISIONS m = 10.0 g m/ v 0 m 5.00 kg (a) (b) Figure 7.3: Colliion in Example 5. (a) Jut before the colliion. (b) Jut after. m l v M v/2 Figure 7.4: Bullet pae through a pendulum bob and emerge with half it original peed; the bob barely wing through a complete circle! Jut before the colliion, only the bullet (with ma m) i in motion and it x velocity i v 0. So the initial momentum i P i,x = mv 0 = ( kg)v 0 Jut after the colliion, the bullet block combination, with it ma of M + m ha an x velocity of m. So the final momentum i P f,x = (M + m)v = (5.00kg kg)(0.600 m Since P i,x = P f,x, we get: ) = 3.01 ( kg)v 0 = 3.01 = v 0 = 301 m The initial peed of the bullet wa 301 m. 6. A hown in Fig. 7.4, a bullet of ma m and peed v pae completely through a pendulum bob of ma M. The bullet emerge with a peed v/2. The pendulum bob i upended by a tiff rod of length l and negligible ma. What

9 7.2. WORKED EXAMPLES 163 v v' v/2 (a) (b) Figure 7.5: Colliion of the bullet with the pendulum bob. (a) Jut before the colliion. (b) Jut after. The bullet ha gone through the bob, which ha acquired a velocity v. i the minimum value of v uch that the pendulum bob will barely wing through a complete vertical circle? Whoa! There a hell of a lot of thing going on in thi problem. Let try to ort them out. We break thing down into a equence of event: Firt, the bullet ha a very rapid, very trong interaction with the pendulum bob, where i quickly pae through, imparting a velocity to the bob which at firt will have a horizontal motion. Secondly, the bob wing upward and, a we are told, will get up to the top of the vertical circle. We how the colliion in Fig In thi rapid interaction there are no net external force acting on the ytem that we need to worry about. So it total momentum will be conerved. The total horizontal momentum before the colliion i P i,x = mv + 0 = mv If after the colliion the bob ha velocity v, then the total momentum i ( v P f,x = m + Mv 2) Conervation of momentum, P i,x = P f,x give ( ( v v mv = m + Mv 2) = Mv = m 2) and o: v = 1 mv M 2 = mv (7.14) 2M Now conider the trip of the pendulum bob up to the top of the circle (it mut get to the top, by aumption). There are no friction type force acting on the ytem a M move, o mechanical energy i conerved. If we meaure height from the bottom of the wing, then the initial potential energy i zero while the initial kinetic energy i K i = 1 2 M (v ) 2.

10 164 CHAPTER 7. LINEAR MOMENTUM AND COLLISIONS T v top Mg Figure 7.6: Force acting on pendulum bob M at the top of the wing. Now uppoe at the top of the wing ma M ha peed v top. It height i 2l and it potential energy i Mg(2l) o that it final energy i o that conervation of energy give: E f = 1 2 Mv2 top + 2Mgl 1 M 2 (v ) 2 = 1 2 Mv2 top + 2Mgl (7.15) What do we know about v top? A drawing of the force acting on M at the top of the wing i hown in Fig Gravity pull down with a force Mg. There may be a force from the upending rod; here, I ve happened to draw it pointing upward. Can thi force point upward? Ye it can... we need to read the problem carefully. It aid the bob wa upended by a tiff rod and uch an object can exert a force (till called the tenion T) inward or outward along it length. (A tring can only pull inward.) The bob i moving on a circular path with (intantaneou) peed v top o the net force on it point downward and ha magnitude Mv 2 top/l: Mg T = Mv2 top. l Since T can be poitive or negative, v top can take on any value. It could be zero. What condition are we looking for which correpond to the mallet value of the bullet peed v? We note that a v get bigger, o doe v (the bob initial peed). A v increae, o doe v top, a we ee from conervation of energy. But it i entirely poible for v top to be zero, and that will give the mallet poible value of v. That would correpond to the cae where M picked up enough peed to jut barely make it to the top of the wing. (And when the bob goe pat the top point then gravity move it along through the full wing.) So with v top = 0 then Eq give u 1 2 M (v ) 2 = 2Mgl = v 2 = 4gl and: v = 4gl

11 7.2. WORKED EXAMPLES 165 A m m m 2 B C Figure 7.7: Frictionle track, for Example 7. Ma m 1 i releaed and collide elatically with ma m 2. and putting thi reult back into Eq. 7.15, we have 4gl = 1 mv M 2 = mv 2M. Finally, olve for v: v = 2M 4gl = 4M gl m m The minimum value of v required to do the job i v = 4M gl/m. 7. Conider a frictionle track ABC a hown in Fig A block of ma m 1 = 5.001kg i releaed from A. It make a head on elatic colliion with a block of ma m 2 = 10.0kg at B, initially at ret. Calculate the maximum height to which m 1 rie after the colliion. Whoa! What i thi problem talking about?? We releae ma m 1 ; it lide down to the lope, picking up peed, until it reache B. At B it make a colliion with ma m 2, and we are told it i an elatic collion. The lat entence in the problem implie that in thi colliion m 1 will revere it direction of motion and head back up the lope to ome maximum height. We would alo gue that m 2 will be given a forward velocity. Thi equence i hown in Fig Firt we think about the intant of time jut before the colliion. Ma m 1 ha velocity v 1i and ma m 2 i till tationary. How can we find v 1i? We can ue the fact that energy i conerved a m 1 lide down the mooth (frictionle) lope. At the top of the lope m 1 had ome potential energy, U = m 1 gh (with h = 5.00m) which i changed to kinetic energy, K = 1m 2 1v1i 2 when it reache the bottom. Conervation of energy give u: o that m 1 gh = 1 2 mv2 1i = v 2 1i = 2gh = 2(9.80 m 2 )(5.00m) = 98.0 m2 2 v 1i = m

12 166 CHAPTER 7. LINEAR MOMENTUM AND COLLISIONS v 1i v 2i = 0 v 1f v 2f m 1 m 2 m1 m 2 (a) (b) Figure 7.8: (a) Jut before the colliion; m 1 ha acquired a velocity of v 1i from liding down the lope. (b) Jut after the colliion; ma m 1 ha velocity v 1f and ma m 2 ha velocity v 2f. Dir. of motion h v 1f (a) (b) Figure 7.9: (a) After the colliion, m 1 goe to the left and will move back up the lope. (b) After moving back up the lope, m 1 reache ome maximum height h. We choe the poitive value here ince m 1 i obviouly moving forward at the bottom of the lope. So m 1 velocity jut before triking m 2 i m. Now m 1 make an elatic (one dimenional) colliion with m 2. What are the final velocitie of the mae? For thi we can ue the reult given in Eq. 7.5 and 7.6, uing v 2i = 0. We get: ( ) ( ) m1 m kg 10.0kg v 1f = v 1i = (+9.90 m m 1 + m kg kg ) = 3.30 m ( ) 2m1 v 2f = v 1i = m 1 + m 2 ( 2(5.001 kg) 5.001kg kg ) (+9.90 m ) = m So after the colliion, m 1 ha a velocity of 3.30 m; that i, it ha peed 3.30 m and it i now moving to the left. After the colliion, m 2 ha velocity m, o that it i moving to the right with peed 6.60 m. Since m 1 i now moving to the left, it will head back up the lope. (See Fig. 7.9.) How high will it go? Once again, we can ue energy conervation to give u the anwer. For the trip back up the lope, the initial energy (all kinetic) i E i = K i = 1 2 m(3.30 m )2

13 7.2. WORKED EXAMPLES 167 v =? v=0 v=? (a) (b) Figure 7.10: (a) Cat leap from left led to the right led. What i new velocity of left led? (b) Cat ha landed on the right led. What i it velocity now? and when it reache maximum height (h) it peed i zero, o it energy i the potential energy, E f = U f = mgh Conervation of energy, E i = E f give u: 1 m(3.30 m 2 )2 = mgh = h = (3.30 m )2 2g Ma m 1 will travel back up the lope to a height of 0.556m. = 0.556m 8. Two 22.7kg ice led are placed a hort ditance apart, one directly behind the other, a hown in Fig (a). A 3.63 kg cat, tanding on one led, jump acro to the other and immediately back to the firt. Both jump are made at a peed on 3.05 m relative to the ice. Find the final peed of the two led. We will let the x axi point to the right. In the initial picture (not hown) the cat i itting on the left led and both are motionle. Taking our ytem of interacting particle to be the cat and the left led, the initial momentum of the ytem i P = 0. After the cat ha made it firt jump, the velocity of thi led will be v L,x, and the (final) total momentum of the ytem will be P f = (22.7kg)v L,x + (3.63kg)(+3.05 m ) Note, we are uing velocitie with repect to the ice, and that i how we were given the velocity of the cat. Now a there are no net external force, the momentum of thi ytem i conerved. Thi give u: with which we eaily olve for v L,x : 0 = (22.7kg)v L,x + (3.63kg)(+3.05 m ) v L,x = (3.63kg)(+3.05 m ) (22.7 kg) = m o that the left led move at a peed of m to the left after the cat firt jump.

14 168 CHAPTER 7. LINEAR MOMENTUM AND COLLISIONS v =? v =? (a) (b) Figure 7.11: (a) Cat leap from right led back to left led. What i new velocity of right led? (b) Cat ha landed back on left led. What i it velocity now? Note that thi i baically the mirror image of the previou figure, o I only had to draw it once! Hah! The cat land on the right led and after landing it move with the ame velocity a that led; the colliion here i completely inelatic. For thi part of the problem, the ytem of interacting particle we conider i the cat and the right led. (The left led doe not interact with thi ytem.) The initial momentum of thi ytem i jut that of the cat, P i = (3.63kg)(+3.05 m ) = 11.1 If the final velocity of both cat and led i v R,x then the final momentum i P f = (22.7kg kg)v R,x = (26.3kg)v R,x (The cat and led move a one ma, o we can jut add their individual mae.) Conervation of momentum of thi ytem, P i = P f give 11.1 = (26.3kg)v R,x o v R,x = ( ) 11.1 (26.3 kg) = m Now we have the velocitie of both led a they are pictured in Fig (b). And now the cat make a jump back to the left led, a hown in Fig (a). Again, we take the ytem to be the cat and the right led. It initial momentum i P i = (22.7kg kg)(0.422 m ) = 11.1 Now after the cat leap, the velocity of the cat (with repect to the ice) i 3.05 m, a pecified in the problem. If the velocity of the right led after the leap i v R,x then the final momentum of the ytem i P f = (3.63kg)( 3.05 m ) + (22.7kg)v R,x Conervation of momentum for the ytem, P i = P f, give 11.1 = (3.63kg)( 3.05 m ) + (22.7kg)v R,x

15 7.2. WORKED EXAMPLES 169 o that we can olve for v R,x: v R,x = ( ) (22.7 kg) = m o during it econd leap the cat make the right led go fater! Finally, for the cat landing on the left led we conider the (iolated) ytem of the cat and the left led. We already have the velocitie of the cat and led at thi time; it initial momentum i P i = (22.7kg)( m ) + (3.63kg)( 3.05 m ) = After the cat ha landed on the led, it i moving with the ame velocity a the led, which we will call v L,x. Then the final momentum of the ytem i P f = (22.7kg kg)v L,x = (26.3kg)v L,x And momentum conervation for thi colliion give 22.1 = (26.3kg)v L,x and then v L,x = ( 22.1 ) (26.3 kg) = m Summing up, the final velocitie of the led (after the cat i done jumping) are: Left Sled: Right Sled: v L,x = m v r,x = m Two Dimenional Colliion 9. An untable nucleu of ma kg initially at ret diintegrate into three particle. One of the particle, of ma kg, move along the y axi with a peed of m. Another particle of ma kg, move along the x axi with a peed of m. Find (a) the velocity of the third particle and (b) the total energy given off in the proce. (a) Firt, draw a picture of what i happening! Such a picture i given in Fig In the mot general ene of the word, thi i indeed a colliion, ince it involve the rapid interaction of a few iolated particle. The are no external force acting on the particle involved in the diintegration; the total momentum of the ytem i conerved. The parent nucleu i at ret, o that the total momentum wa (and remain) zero: P i = 0.

16 170 CHAPTER 7. LINEAR MOMENTUM AND COLLISIONS y y 6.0 x 10 8 m/ 5.0 x kg 8.4 x kg x 4.0 x 10 8 m/ x (v x, v y ) (a) (b) Figure 7.12: Nucleu diintegrate in Example 9. (a) Before the plit; nucleu i at ret. (b) Afterward; three piece fly off in different direction. Afterward, the ytem conit of three particle; for two of thee particle, we are given the mae and velocitie. We are not given the ma of the third piece, but ince we were given the ma of the parent nucleu, we might think that we can ue the fact that the mae mut um up to the ame value before and after the reaction to find it. In fact relativity tell u that mae don t really add in thi way and when nuclei break up there i a meaurable ma difference, but it i mall enough that we can afely ignore it in thi problem. So we would ay that ma i conerved, and if m i the ma of the unknown fragment, we get: kg = kg kg + m o that m = kg. We will let the velocity component of the third fragment be v x and v y. Then the total x momentum after the colliion i P f,x = ( kg)( m) + ( kg)v x Uing P i,x = 0 = P f,x, we find: ( kg)( m ) + ( kg)v x = 0 which eaily give: v x = m Similarly, the total y momentum after the colliion i: and uing P i,y = 0 = P f,y, we have: P f,y = ( kg)( m ) + ( kg)v y ( kg)( m ) + ( kg)v y = 0

17 7.2. WORKED EXAMPLES 171 which give v y = m Thi really doe pecify the velocity of the third fragment (a requeted), but it i alo ueful to expre it a a magnitude and direction. The peed of the third fragment i v = ( m )2 + ( m )2 = m and it direction θ (meaured counterclockwie from the x axi) i given by tan θ = ( ) 8.33 = Realizing that θ mut lie in the third quadrant, we find: θ = tan 1 (0.893) 180 = 138. (b) What i the gain in energy by the ytem for thi diintegration? By thi we mean the gain in kinetic energy. Initially, the ytem ha no kinetic energy. After the breakup, the kinetic energy i the um of 1 2 mv2 for all the particle, namely K f = 1 ( kg)( m )2 + 1 ( kg)( m )2 + 1( kg)( m )2 = J We might ay that the proce give off J of energy. 10. A billiard ball moving at 5.00 m trike a tationary ball of the ame ma. After the colliion, the firt ball move at 4.33 m at an angle of 30.0 with repect to the original line of motion. Auming an elatic colliion (and ignoring friction and rotational motion), find the truck ball velocity. The colliion i diagrammed in Fig We don t know the final peed of the truck ball; we will call it v, a in Fig. 7.13(b). We don t know the final direction of motion of the truck ball; we will let it be ome angle θ, meaured below the x axi, alo a hown in Fig. 7.13(b). Since we are dealing with a colliion between the two object, we know that the total momentum of the ytem i conerved. So the x and y component of the total momentum i the ame before and after the colliion. Suppoe we let the x and y component of the truck ball final velocity be v x and v y, repectively. Then the condition that the total x momentum be conerved give u: m(5.00 m ) + 0 = m(4.33 m )co mv x (The truck ball ha no momentum initially; after the colliion, the incident ball ha an x velocity of (4.33 m )co 30.0.)

18 172 CHAPTER 7. LINEAR MOMENTUM AND COLLISIONS 5.00 m/ y y 4.33 m/ m m x q 30 o x (v x, v y ) v (a) (b) Figure 7.13: Colliion in Example 10. (a) Before the colliion. (b) After the colliion. Luckily, the m cancel out of thi equation and we can olve for v x : and then: (5.00 m ) = (4.33 m )co v x v x = (5.00 m ) (4.33 m )co 30.0 = 1.25 m We can imilarly find v y by uing the condition that the total y momentum be conerved in the colliion. Thi give u: which give = m(4.33 m )in mv y v y = (4.33 m )in30.0 = 2.16 m Now that we have the component of the final velocity we can find the peed and direction of motion. The peed i: v = v 2 x + v 2 y = and the direction i found from (1.25 m )2 + ( 2.16 m )2 = 2.50 m tan θ = v y v x = 2.16 m 1.25 m = 1.73 = θ = tan 1 ( 1.73) = 60 So the truck ball move off with a peed of 2.50 m at an angle of 60 downward from the x axi. Thi really complete the problem but we notice omething trange here: We were given more information about the colliion than we ued. We were alo told that the colliion wa elatic, meaning that the total kinetic energy of the ytem wa the ame before and after the colliion. Since we now have all of the peed, we can check thi.

19 7.2. WORKED EXAMPLES kg 4.00 kg Figure 7.14: Mae and poition for Example 11. x, m The total kinetic energy before the colliion wa K i = 1 2 m(5.00 m )2 = m(12.5 m2 2 ). The total kinetic energy after the colliion wa K f = 1 2 m(4.33 m ) m(2.50 m )2 = m(12.5 m2 2 ) o that K i i the ame a K f ; the tatement that the colliion i elatic i conitent with the other data; but in thi cae we were given too much information in the problem! The Center of Ma 11. A 3.00kg particle i located on the x axi at x = 5.00m and a 4.00kg particle i on the x axi at x = 3.00m. Find the center of ma of thi two particle ytem. The mae are hown in Fig There i only one coordinate (x) and two ma point to conider here; uing the definition of x CM, we find: x CM = m 1x 1 + m 2 x 2 m 1 + m 2 (3.00kg)( 5.00m) + (4.00kg)(3.00m) = (3.00kg kg) = m The center of ma i located at x = 0.429m. 12. An old Chryler with ma 2400 kg i moving along a traight tretch of road at 80km/h. It i followed by a Ford with ma 1600kg moving at 60km/h. How fat i the center of ma of the two car moving? The car here have 1 dimenional motion along (ay) the x axi. From Eq we ee that the velocity of their center of ma i given by: v CM, x = 1 M N i m i v i,x = m 1v 1x + m 2 v 2x m 1 + m 2

20 174 CHAPTER 7. LINEAR MOMENTUM AND COLLISIONS Plugging in the mae and velocitie of the two car, we find v CM, x = (2400kg) ( ) ( ) 80 km h + (1600kg) 60 km h = 72 km h (2400kg kg) In unit of m thi i 72 km h ( 10 3 ) ( ) m 1h = 20 m 1km The Motion of a Sytem of Particle 13. A 2.0kg particle ha a velocity of v 1 = (2.0i 3.0j) m, and a 3.0kg particle ha a velocity (1.0i + 6.0j) m. Find (a) the velocity of the center of ma and (b) the total momentum of the ytem. (a) We are given the mae and velocity component for the two particle. Then writing out the x and y component of Eq we find: v CM,x = (m 1v 1x + m 2 v 2x ) (m 1 + m 2 ) = (2.0kg)(2.0 m ) + (3.0kg)(1.0 m ) (2.0kg + 3.0kg) = 1.4 m v CM,y = (m 1v 1y + m 2 v 2y ) (m 1 + m 2 ) = (2.0kg)( 3.0 m ) + (3.0kg)(6.0 m ) (2.0kg + 3.0kg) = 2.4 m The velocity of the center of ma of the two particle ytem i v CM = (1.4i + 2.4j) m (b) The total momentum of a ytem of particle i related to the velocity of the center of ma by P = Mv CM o we can ue the anwer from part (a) to get: P = Mv CM = (2.0kg + 3.0kg)((1.4i + 2.4j) m ) = (7.00i j)

21 Appendix A: Ueful Number Converion Factor: Length cm meter km in ft mi 1cm = m = km = in = ft = mi = Ma g kg lug u 1g = kg = lug = u = An object with a weight of 1lb ha a ma of kg. Earth Data Sun Ma kg Mean Radiu m Orbital Period Mean Ditance from Sun m Mean Denity kg/m 3 Surface gravity 9.81 m 2 Ma kg Mean Radiu m Mean Ditance from Galactic Center m Mean Denity kg/m 3 Surface Temperature K 175

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