Momentum and the Flow of Mass Challenge Problems Solutions

Transcription

1 Poblem 1: Steam Bouncing off Wall Momentum and the Flow of Ma Challenge Poblem Solution A team of paticle of ma m and epaation d hit a pependicula uface with peed v. The team ebound along the oiginal line of motion with the ame peed. The ma pe unit length of the incident team i! = m / d. What i the magnitude of the foce on the uface? Solution: We begin by defining ou ytem a a tube of wate of co ectional aea A and length v!t. Ma Conevation: The ma of the wate in thi tube i given by!m i = "Av!t = #v!t (1.1) - 1 -

2 whee! = " A = m / d i the ma pe unit length. Thi mean that all the wate inide thi tube will hit the wall duing the inteval [ t, t +!t] and eflect. Note that if the colliion with the wall i elatic, the eflected wate will have the ame peed a the incident wate. Othewie to allow fo a lightly moe geneal cae, we hall aume that the eflected wate tavel backwad with peed v!. We aume that the wate i incompeible o that the denity ha not changed. The ma of the wate that ha been eflected i given by!m = "A#v#!t = $#v#!t (1.2) whee!# = " A# = m / d# i the ma pe unit length of the eflected wate. We aume that the wate i incompeible o that the denity ha not changed. We hall alo aume fo implicity that all the wate incident on the wall eflect with the peed v!. Thu Thu in the time inteval!m i =!m. (1.3) Thi implie that!m i!m =. (1.4)!t!t!v =!"v". (1.5) Note that the intantaneou ate that ma i deliveed to the uface and leave i "m dm dm dm i "m i! = lim = lim = = $v = $%v%. (1.6) "t#0 "t "t#0 "t Momentum Pinciple: The momentum pinciple tate that the impule fom the uface change the momentum of the wate. The momentum of the incident wate i given by! p = "m v i ˆ = (!v" ) ˆ =!v 2 i i t v i "t i ˆ. (1.7) The x-component of the momentum of the eflected wate i given by! p = #$ m v i ˆ = #(!"v" $ t) v" i ˆ = #!" v" 2 $t i ˆ. (1.8)!! ave ave The uface exet an aveage foce on the wate, F w, hence an impule F w!t. So the momentum pinciple become! ave! F w!t =!p. (1.9) - 2 -

3 Baed on ou analyi of the momentum incident and eflected the momentum pinciple become! ave!! 2 2 F #t = p $ p = $!"v" #t i ˆ $!v #t i ˆ = $(!"v" v " +!vv)#t i ˆ. (1.10) w i Theefoe the foce of the uface on the wate i! ave Fw = #(!"v" v " +!vv) î. (1.11) We can apply ou condition fo ma conevation, Eq. (1.5) and find that! ave Fw = #!v( v " + v) î. (1.12) By Newton Thid Law the aveage foce of the wate on the uface i!! ave ave F = #F =!v( v " + v) î. (1.13) w w In ode to poduce the maximum poible aveage foce of wate on the uface, the colliion mut be elatic ince then the eflected peed i equal to the incident peed and! ave 2 ) max = 2!v i ˆ. (1.14) (F w Note that the uing the intantaneou ate that ma aive o leave the uface, we have that So the magnitude of the foce on the uface i! ave ˆ dm (F ) = 2!vv i = 2 v i ˆ w max. (1.15)! ave (Fw ) max = 2!v 2. (1.16) - 3 -

4 Poblem 2 A ocket ha a dy ma (empty of fuel) m,0 = 2! 10 7 kg, and initially caie fuel with ma m f,0 = 5! 10 7 kg. The fuel i ejected at a peed u = 2.0! 10 3 m " -1 elative to the ocket. What i the final peed of the ocket afte all the fuel ha buned? Solution The initial ma of the ocket included the fuel i m = m + m = 2! 10 7 kg + 5! 10 7 kg = 7! 10 7 kg (2.1),i,0 f,0 The atio of the initial ma of the ocket (including the ma of the fuel) to the final dy ma of the ocket (empty of fuel) i The final peed of the ocket i then m,i 7! 10 7 kg R = = = 3.5 (2.2) m,d 2! 10 7 kg v = u ln R = (2.0! 10 3 m " -1 )ln3.5 = 2.5! 10 3 m " -1. (2.3), f - 4 -

5 Poblem 3: Coal Ca An empty coal ca of ma m 0 tat fom et unde an applied foce of magnitude F. At the ame time coal begin to un into the ca at a teady ate b fom a coal hoppe at et along the tack. Find the peed when a ma m c of coal ha been tanfeed. Solution: We hall analyze the momentum change in the hoizontal diection which we call the x-diection.. Since the coal doe not have any hoizontal velocity, the falling coal i not tanfeing any momentum to the coal ca. So we hall take a ou ytem the empty coal ca and a ma m c of coal that ha been tanfeed. Ou initial tate at t = 0 i when the coal ca i empty and at et befoe any coal ha been tanfeed. The x- component of the momentum of thi initial tate i zeo, p x (0) = 0. (3.1) Ou final tate at t = t i when all the coal of ma m = bt ha been tanfeed into the f c f ca which i now moving at peed v f. The x-component of the momentum of thi final tate i p x (t f ) = (m 0 + m c )v f = (m 0 + bt f )v f. (3.2) Thee i an extenal contant foce F x = F applied though the tanfe. The momentum pinciple applied to the x-diection i t f F = "p = p ) # p (0). (3.3)! x x x (t f x 0 Since the foce i contant, the integal i imple and the momentum pinciple become Ft f = (m 0 + bt f )v f. (3.4) So the final peed i - 5 -

6 Ft f v f =. (3.5) (m 0 + bt f ) - 6 -

7 Poblem 4: Emptying a Feight Ca A feight ca of ma m c contain a ma of and m. At t = 0 a contant hoizontal foce of magnitude F i applied in the diection of olling and at the ame time a pot in the bottom i opened to let the and flow out at the contant ate b = dm /. Find the peed of the feight ca when all the and i gone. Aume that the feight ca i at et at t = 0. Solution: Chooe the poitive-diection to point in the diection that the ca i moving. Let take a ou ytem the amount of and of ma!m that leave the feight ca duing the time inteval [ t, t +!t], and the feight ca and whateve and i in it at time t. At the beginning of the inteval the ca and and i moving with peed v o the x- component of the momentum at time t i given by p (t) = (!m + m (t))v), (4.1) x c whee m c ( t ) i the ma of the ca and and in it at time t. Denote by m c,0 = m c + m whee the m c i the ma of the ca and m i the ma of the and in the ca at t = 0, and m ( t ) = bt i the ma of the and that ha left the ca at time t ince Thu t t dm m (t) = = b = bt. (4.2)!! 0 0 m (t) = m! bt = m + m! bt. (4.3) c c,0 c Duing the inteval [ t, t +!t], the mall amount of and of ma!m leave the ca with the peed of the ca at the end of the inteval v +!v. So the x-component of the momentum at time t +!t i given by - 7 -

8 p (t +!t) = (!m + m (t))(v +!v). (4.4) x c Thoughout the inteval a contant foce F i applied to the ca o p x (t +!t) # p x ( t) F = lim. (4.5)!t"0!t Fom ou analyi above Eq. (4.5) become Eq. (4.6) implifie to (m c ( t ) +!m )( v +!v) # (m c ( t ) +!m )v F = lim. (4.6)!t"0!t!v!m!v F = lim m c ( t ) + lim. (4.7)!t"0!t! t " 0!t The econd tem vanihe when we take the!t " 0 becaue it i of econd ode in the infiniteimal quantitie (in thi cae!m!v ) and o when dividing by!t the quantity i of fit ode and hence vanihe ince both!m " 0 and!v " 0. So Eq. (4.7) become We now ue the definition of the deivative: in Eq. (4.8) to fund the diffeential equation Uing Eq. (4.3) we have!v F = lim m c ( t ). (4.8)! t " 0!t!v dv lim = (4.9)!t"0!t dv F = m c ( t ). (4.10) dv F = (m c + m! bt ). (4.11) (b) We can integate thi equation though the epaation of vaiable technique. Rewite Eq. (4.11) a - 8 -

9 F dv = (mc + m! bt ). (4.12) We can then integate both ide of Eq. (4.12) with the limit a hown v( t ) t F " dv = " m + m! bt. (4.13) v=0 0 c Integation yield the velocity of the ca a a function of time F t F! m c + m # bt " v( t ) = # ln( m c + m # bt ) %. (4.14) b 0 = # ln $ b & m c + m ' - 9 -

10 Poblem 5: Falling Chain A chain of ma m and length l i upended vetically with it lowet end touching a cale. The chain i eleaed and fall onto the cale. What i the eading of the cale when a length of chain, y, ha fallen? (Neglect the ize of the individual link.) Let g denote the gavitational contant. Solution: Suppoe the chain i eleaed at time t = 0. At time t, a length y of the chain ha fallen onto the cale and the et of the chain ha fallen a ditance y. (Note that we have choen downwad a the poitive y-diection.) We can ue ou enegy condition, applied to the potion of the chain that i feely falling (ma m( t ) ), to find that 2 (1/ 2) m( t) v y =!m( t ) g "h = m( t ) gy. (5.1) Hence at time t, the entie chain i falling with a y-component of the velocity given by v y = 2gy. (5.2) Note at the bottom of the chain, a mall incement of chain, (length!y and ma "m =!"y = (m / l )"y ), i moving with peed v y = 2gy. Afte a mall inteval of time!t ha elaped (we hall hotly conide the limit a!t " 0 ), thi mall piece of chain ha come to et on the cale. Note that we have aumed that the ma pe unit length! = m / l i unifom

11 Let take a ou ytem thi mall piece of chain. Thee i a contact foce between the!! chain and the cale, F ( t ), that act on thi piece. Thi foce F ( t ) vaie in time ince c, c, the peed of the chain inceae in time a the chain fall. It alo vaie due to the uneven colliion between the chain link and the cale. We will aveage thi foce ove the uneven colliion in ode to mooth out thee mall vaiation, and o the aveage! extenal foce on ou ytem i F ave ( t ). c, The y-component of the momentum of the ytem at time t +!t i at time i p y (t +!t) = 0. (5.3) The y-component of the momentum of the ytem at time t i at time i p ( t ) =!mv = (m / l )!y v, (5.4) y y y The change in the y-component of the momentum i theefoe!p y = p y (t +!t) " p y ( t ) = "(m / l )!y v y, (5.5) Since the extenal foce i equal to the change in momentum we have that.!!p ave y!y dy 2 F c, ( t ) = lim ĵ = #(m / l ) v y lim ĵ = #(m / l ) v y ĵ = #(m / l ) v y ĵ, (5.6)!t"0!t! t " 0!t whee we have ued the fact that dy / = v y. Note that we have choen downwad a the poitive y-diection o the foce on the chain i upwad a we expect in ode to top the 2 chain. Recall fom the enegy condition that v y = 2gy. So the magnitude of the extenal foce i given by! F ave ( t ) = 2( m / l ) gy. (5.7) c, The cale eading i equal to the magnitude of thi foce plu the weight of the chain, (m / l ) yg, that i aleady on the cale

12 ! Scale eading = F ave ( t ) + (m / l ) yg = 2( m / l ) yg + (m / l ) yg = 3( m / l ) gy. (5.8) c, Note that at time t, the amount of chain that i eted on the cale i 1 2 So the cale eading a a function of time i given by y( t ) = gt. (5.9) Scale eading = (m / l ) g t. (5.10) 2 We note that when the entie chain ha jut come to et on the cale, y = l, the cale ead Scale eading = 3mg. (5.11) Afte the colliion ha ended the cale eading dop to the weight of the chain. Scale eading = mg

13 Poblem 6 A pacecaft i launched fom an ateoid by being bombaded by a team of ock dut. The team of dut i ejected fom the dut gun at a contant ate dm e / = b at a peed u with epect to the ateoid, which we take to be tationay. Aume that the dut come momentaily to et at the pacecaft and then lip away ideway; the effect i to keep the pacecaft ma m contant. a) Deive an equation fo the acceleation dv / of the pacecaft at time t, in tem of the ate that the dut ma hit the uface of the pacecaft dm d /, the peed of the dut elative to the ateoid u, the ma of the pacecaft m, and the velocity of the pacecaft v. Show you momentum flow diagam at time t and time t +!t. Clealy identify you ytem and label all the object in you ytem. What i the teminal velocity of the pacecaft? Hint: dm d /! b. b) Uing conevation of ma, at time t, find an expeion fo the ate that the dut ma hit the pacecaft, dm d /, a a function of the peed of the pacecaft v, the ate that the dut ma i hot fom the ateoid dm e / = b, and the peed u of the dut elative to the ateoid. Hint: dm d /! b. c) Ue you eult fom pat b) in pat a) to find the peed v (t) of the pacecaft a a function of time, auming v (t = 0) = 0. (If you get an integal that you ae not ue how to integate, you can leave you anwe in integal fom.) Solution: We chooe a ou ytem, the mall amount of dut!m d that hit the pacecaft duing the inteval [t,t +!t] and the pacecaft of ma m. Becaue we aumed that the dut come momentaily to et at the pacecaft and then lip away ideway; at the end of the inteval the dut ha the ame peed a the pacecaft and the ma m of the pacecaft emain contant. We how the momentum diagam fo the ytem at time t and t +!t below

14 Becaue we ae ignoing the gavitational foce (it i vey mall), the x-component of the momentum of the ytem i contant duing the inteval [t,t +!t] and o p x (t +!t) # p x (t) 0 = lim. (6.1)!t"0!t Uing the infomation fom the figue above, Eq. (4.5) become Eq. (4.6) implifie to (m +!m d )(v +!v ) # (!m d u + m v ) 0 = lim. (6.2)!t"0!t!v!m!v d!m d!m 0 = lim m + lim v + lim # lim d u. (6.3)!t"0!t!t"0!t!t"0!t!t"0!t The thid tem vanihe when we take the!t " 0 becaue it i of econd ode in the infiniteimal quantitie (in thi cae!m d!v ) and o when dividing by!t the quantity i of fit ode and hence vanihe ince both!m d " 0 and!v " 0. So Eq. (4.7) become!v!m d!m 0 = lim m + lim v # lim d u. (6.4)!t"0!t!t"0!t!t"0!t We now ue the definition of the deivative:!v dv!m d dm d lim = ; lim =. (6.5)!t"0!t!t"0!t

15 in Eq. (4.8) to fund the diffeential equation decibing the elation between the acceleation of the pacecaft and the time ate of change of the ma of dut tiking the pacecaft dv dm 0 = m + d (v! u). (6.6) In the limit a t! ", dv /! 0, hence v (!) = u the pacecaft can go no fate than the peed of the dut. In ode to olve the above diffeential equation, we begin by obeving that dm d dm! e = b (6.7) The eaon i that b i the ate that the dut i ejected fom the ateoid. In the figue below a column of ejected dut of length!l = u!t ha ma!m ejected = "u!t whee! i the ma pe unit length and i aumed to be contant. The ate that the ma i ejected fom the ateoid i then dm!m b = e = lim e = #u. (6.8)!t"0!t In the figue below we conide a column of dut of length u!t that i jut behind a pependicula uface of the pacecaft at time t

16 Duing the time inteval [t,t +!t], the pacecaft i moving a ditance v!t o the entie column of dut doe not hit the pacecaft. Only a faction of the column hit the pacecaft with the ma of the dut that tike the pacecaft given by b = "(u # v )!t = (u # v )!t (6.9)!m d u whee we ued! = b / u. Dividing the equation though by!t and taking limit we have that dm d!m d b = lim = (u # v ). (6.10)!t"0!t u Subtituting into Eq. (4.10) and afte ome eaanging yield dv b m = (v! u) 2. (6.11) u We can integate thi equation by epaating vaiable to find an integal expeion fo the ma of the pacecaft a a function of time v! = v (t ) t!=t dv! b =!. (6.12) (v! " u) 2 um # # v! =0 t!=0 We can eaily integate both ide of the equation, yielding v" =v (t ) 1 b! = t. (6.13) (v "! u) um v" =0-16 -

17 Evaluating the endpoint of the integal yield A little algebaic eaangement yield Inveting yield So the peed of the pacecaft a a function of time i then 1 1 b t. (6.14)!! = u (v (t)! u) um u m + bt! =. (6.15) (v (t)! u) m m v (t)! u =!u. (6.16) m + bt ubt v (t) = m + bt. (6.17) Note that in the limit a t! ", v (!) = u in ageement with what expect, that the pacecaft can go no fate than the peed of the dut

18 Poblem 7 Space Junk A pacecaft of co-ectional aea A, poceeding along the poitive x-diection, ente an ateoid tom at time t = 0, in which the mean ma denity of the ateoid tom i!! and the aveage ateoid velocity i u =!u î in the negative x-diection. A the pacecaft poceed though the tom, all of the ateoid that hit the pacecaft tick to it.! a) Suppoe that at time t the velocity of the pacecaft i v = v î in the poitive x- diection, and it ma i m. Futhe, uppoe that in an inteval! t, the ma of the pacecaft inceae by an amount! m. Given that thee ae no extenal foce, uing conevation of momentum find an equation fo the change of the pacecaft velocity!v, in tem of! m, u, and v? b) When the pacecaft ente the ateoid tom, the magnitude of it velocity and ma ae v 0 and m 0, epectively. Integate you diffeential equation in pat a) to find the velocity v of the pacecaft when the ma i m. c) Find an expeion fo the ma of the ateoid! m that tick to the pacecaft within the time inteval! t? (Hint: conide the volume of ateoid wept up by the pacecaft in time! t). d) When the pacecaft ente the ateoid tom, the magnitude of it velocity and ma ae v 0 and m 0, epectively. What i the ma of the pacecaft at time t? (Ue you eult fom pat c) and b).) Solution: Let chooe a ou ytem a mall element of the ateoid cloud! m that i abobed by the ocket duing the inteval [ t, t +! t] and the ocket itelf. Chooe poitive x-diection a the diection the ocket i moving. The momentum diagam at the beginning of the inteval at time t i hown in the figue below. The momentum diagam at time t +! t i alo hown below

19 Becaue we ae auming that thee ae no extenal foce acting on ou ytem, p x (t +!t) # p x ( t) 0 = lim. (7.1)! t " 0! t Uing the infomation fom the figue above, Eq. (7.1) become (m +!m)( v +!v) # (#! mu + mv ) 0 = lim. (7.2)! t " 0!t Eq. (7.2) implifie to!v!m!m!v!m 0 = lim m + lim v + lim + lim u. (7.3)! t " 0!t! t " 0!t! t " 0!t! t " 0!t The thid tem vanihe when we take the!t " 0 becaue it i of econd ode in the infiniteimal quantitie (in thi cae!m!v ) and o when dividing by!t the quantity i of fit ode and hence vanihe ince both!m " 0 and!v " 0. So Eq. (7.3) become We now ue the definition of the deivative:!v!m!m 0 = lim m + lim v + lim u. (7.4)! t " 0!t!t"0!t! t " 0!t!v dv!m dm lim = ; lim =. (7.5)!t"0!t! t " 0!t in Eq. (7.5) to find the diffeential equation decibing the elation between the acceleation of the ocket and the time ate of change of the ma of the ocket dv dm 0 = m + (v + u). (7.6) b) We can integate thi equation though the epaation of vaiable technique. Rewite Eq. (7.6) a (cancel the common facto ) dv dm! =. (7.7) u + v m We can then integate both ide of Eq. (7.7) with the limit a hown

20 Integation yield v!=v( m ) m!=m dv! dm! " # = # (7.8) u + v! m! v!=v0 m!=m0! u + v( m ) "! m " # ln $ % = ln $ % (7.9) & u + v0 ' & m0 ' Recall that ln( a / b ) =! ln( b / a ) o Eq. (7.9) become! u + v 0 "! m " ln # $ = ln # $ (7.10) % u + v & % m 0 & Alo ecall that exp(ln( a / b )) = a / b and o exponentiating both ide of Eq. (7.10) yield u + v0 m = (7.11) u + v( m ) m 0 Afte ome eaanging, the peed v(m) of the ocket a a function of m can be expeed a m( u + v( m )) = m 0 (u + v 0 ) m u + ( ) 0 (u + v 0 ) (7.12) v m = m m v( m ) = 0 (u + v 0 )! u m In the above expeion fo the peed v( m ) of the ocket i a function of the ma of the ocket. We can detemine how the ma of the ocket behave a a function of time by conideing ma conevation. In the inteval [ t, t +! t] the ocket weep out a tube of length!l = v! t. All the ateoid inide thi tube ae collected by the ocket adding an amount of ma to the ocket "m 1 =! Av " t. In addition, becaue the ateoid ae moving towad the ocket an additional amount of ma "m 2 =! Au " t alo i collected by the ocket (ee figue below)

21 Theefoe the total amount of ma collected by the ocket duing the inteval i!m =!m 1 +!m 2 = " Av! t + " Au!t = " A(v + u)!t. (7.13) Dividing Eq. though by!t and taking limit we have that dm "m = lim =! A( v + u). (7.14) " t # 0 "t We now ubtitute the econd line in Eq. (7.12) into Eq. (7.14) and find that dm m =! A 0 (u + v0 ). (7.15) m We can integate thi equation by epaating vaiable to find an integal expeion fo the ma of the ocket a a function of time m" =m( t ) t"=t # m " dm " b = #! Am (u + v 0 0 )". (7.16) m" =m t" =0 0 We can eaily integate both ide of the equation yielding So the ma of the ocket a a function of time i then 1 (m(t) 2! m 0 2 ) = "Am 0 (u + v 0 )t. (7.17) 2 2! A( u + v 0 )t m( t ) = m (7.18) We now ubtitute Eq. (7.18) into the thid line of Eq. (7.12) yielding the peed of the ocket a a function of time m 0 v ( t ) = u + v 2! A ( u + v 0 ) t 1 + m 0 0 " u. (7.19)

22 Poblem 8 Continuou Ma Tanpot: falling aindop A aindop of initial ma m 0 tat falling fom et unde the influence of gavity. Aume that the aindop gain ma fom the cloud at a ate popotional to the momentum of the aindop, dm / = km v, whee m i the intantaneou ma of the aindop, v i the intantaneou velocity of the aindop, and k i a contant with unit [m -1 ]. You may neglect ai eitance. a) Deive a diffeential equation fo the velocity of the aindop. b) Show that the peed of the dop eventually become effectively contant and give an expeion fo the teminal peed. Solution: At time t chooe the aindop with ma m (t) and a mall ma element!m c in the cloud that will be added to the aindop duing the time inteval!t. Chooe the poitive y diection downwad. Then the momentum flow diagam i hown in the figue below. At time t, the mall ma element!m c i at et in the cloud at time t. The aindop i moving downwad with velocity! v ˆ ( t ) = v j. (8.1) So the momentum of the dop i equal to the momentum of the ytem! p ( t ) = m v ˆj (8.2) y At time t +!t, the aindop ha added ome ma!m c fom the cloud and i moving downwad with velocity! ˆ v (t +!t) = (v +!v ) j. (8.3)

23 whee!v i the infiniteimal change in the y-component of the velocity of the aindop. So the momentum of the dop i! ˆ p (t +!t) = (m +!m ) ( v +!v ) j. (8.4) y c Duing thi inteval thee i an extenal gavitational foce acting on the ytem and a buoyant foce in the cloud that i keeping the ma element!m c fom falling. Since the total foce on the ma element!m c i zeo, the total extenal foce acting on the ytem i jut the gavitational foce acting on the aindop,! F total ˆ ext = m g j. (8.5) Newton Second Law and Thid Law fo the ytem tate that he total extenal foce i equal to the deivative of the momentum!! total dp y F ext =. (8.6) Recall the definition of the deivative of the ytem momentum,!!! dpy p y (t +!t) # p y ( t) = lim. (8.7)! t " 0!t Subtitute Eq.(8.4), Eq.(8.2), and Eq.(8.5) into Eq.(8.6) yield Expanding the numeato in Eq.(8.8), (m ˆ ˆ ˆ +!m c ) ( v +!v ) j # m v j m g j = lim. (8.8)! t " 0! t (m!v +!m v +!m!v ) ˆ ˆ c c j m g j = lim. (8.9)! t " 0!t The thid tem in the numeato on the ight hand ide of Eq.(8.9) i a econd oede tem in infiniteimal and hence in the limit a!t " 0 vanihe, Thu Eq.(8.9) become!m ˆ c!v j lim " 0. (8.10)! t " 0! t

24 !v!m c dv dm m g ˆ j = m lim ˆ j + v lim ˆ j = m ˆ j + v ˆ j (8.11)! t " 0!t! t " 0!t The y-component of Eq.(8.11) ae equal, dv dm m g = m + v (8.12) Fom the tatement of the poblem, we have modeled the ate that the ate that aindop add ma i popotional to the momentum of the aindop Thu Eq.(8.12) become dm = km v. (8.13) Divide out the ma of the aindop fom Eq.(8.14) giving dv 2 m g = m + km v. (8.14) dv 2 g = + kv (8.15) Thu the acceleation of the aindop i thu non-unifom and given by dv = g! kv 2. (8.16) The aindop will each a teminal velocity when the acceleation i zeo, So the teminal velocity i 0 = g! kv,te 2. (8.17) g v =. (8.18),te k

25 Poblem 9: Rocket Poblem A ocket acend fom et in a unifom gavitational field by ejecting exhaut with contant peed u elative to the ocket. Aume that the ate at which ma i expelled i given by dm f / =! m, whee m i the intantaneou ma of the ocket and! i a contant. The ocket i etaded by ai eitance with a foce F = bm v contant and v time. popotional to the intantaneou momentum of the ocket whee b i a i the peed of the ocket. Find the peed of the ocket a a function of Solution: To find the ocket velocity a a function of time, we fit need to find how the velocity change with epect to time, the diffeential equation mentioned in the poblem tatement. Take the poitive y-diection to be upwad and aume that the ocket velocity i upwad a well. Fo pupoe of pace and claity, the poitive upwad diection i hown a being to the ight in the figue. The gavitational foce and ai eitance foce ae then in the negative diection, to the left in the figue. A uggeted by the figue above, take the ytem to be the ocket and fuel combination. The malle quae epeent the mall amount of fuel of ma!m f,out that i ejected duing the inteval [t,t +!t]. In the above figue m (t) = m + m (t) (9.1),0 f,in i the combined ma of the ocket whee m,0 i the dy ma of the ocket and m f,in ( t ) i the ma of fuel inide the ocket at time t that doe not leave the ocket duing the inteval. Note that diffeentiating the above equation yield Denote by m f,0 the fuel in the ocket at t = 0. Then dm (t) dm f,in (t) =. (9.2)

26 m f,0 = m f,in (t) + m f,out (t) (9.3) whee m f,out ( t ) i the ma of the fuel that ha been ejected duing the inteval [0,t ]. Since m f,0 i contant, we can diffeentiate the above equation yielding Thu combining equation Eq. (9.2) and Eq. (9.4) yield 0 = dm f,0 dm f,in (t) dm f,out (t) = +. (9.4) (t) dm f,out dm (t) =!. (9.5) Fom the tatement of the poblem, the bun ate of fuel i then equal to the negative of the ate that the ma of the ocket i deceaing and fom the tatement of the poblem i given by dm f,out (t) dm (t)! m (t)= = ". (9.6) Retuning to ou momentum diagam we ee that at time t, the y-component of the y momentum of the ytem p y ( t ) i given by y p y ( t ) = (m ( t ) +!m f,out ) v (9.7) Duing the inteval the fuel i ejected backwad elative to the ocket with peed u. Afte the inteval ha ended the momentum diagam of the ytem i hown below. At time t +!t, the y-component of the momentum of the ytem p y y (t +!t) i given by p y (t t)= m t v +!v ) m v +!v - u) (9.8) y +! ( )( +! f,out (

27 and o the change in y-component of the momentum of the ytem duing the inteval [t,t +!t] i y y y!p y = p y (t +!t) " p y ( t ) = m ( t )( v +!v ) +!m f,out (v +!v " u) " (m ( t ) +!m f,out ) v. (9.9) = m ( t )!v +!m f,out!v "! m f,out u The extenal foce acting on the ytem i given by The momentum pinciple F ext =!m (t)g! bm (t)v. (9.10) y!p y ext y F = lim (9.11) y!t"0!t then become We note that m ( t )!v +!m f,out!v #! m f,out u #m ( t) g # bm ( t) v = lim. (9.12)! t " 0!t!m f,out!v lim = 0. (9.13)!t"0!t Then uing the definition of the deivative, we find that the diffeential equation decibing the motion of the ytem i given by dv d m f,out!m (t)g! bm (t)v = m (t)! u (9.14) We can now ubtitute Eq. (9.6) into the above equation and find that dv!m (t)g! bm (t)v = m (t)! " m (t)u. (9.15) We can divide though by the ma of the ocket in the above equation yielding

28 dv =! u " g " bv (9.16) On to the Diffeential Equation: We ve done the phyic, but thee ae eveal inteeting apect to the calculation of the velocity a a function of time. Ou diffeential equation i epaable and o Rewite Eq. (9.16) a We can integate both ide dv =. (9.17)! u " g " bv v " =v ( t) t"=t dv " = ". (9.18)! u # g # bv " $ $ v" =0 t" =0 Integation yield $ 1 ln "! u $ g $ bv ( t ) # % & b! u $ g = t. (9.19) ' ( Afte a light eaangement, thi expeion can be exponentiated yielding! u " g " bv (t) "bt = e. (9.20)! u " g Afte ome eaangement we find that the peed of the ocket i given by 1 " bt v (t) = (! u " g)(1" e ). (9.21) b Thee ae eveal apect of the eult in Eq. (9.21) that ae woth conideing. Fit, unle! u > g, the ocket doen t get off the gound. Next, the ocket continue to acceleate, only eaching teminal peed v teminal = (! u " g )/ b in the limit t! " ; the ai eitance foce eem to have wimped out. The key point in conideing the eult of Eq. (9.21) i in the model fo the ate at which the fuel i exhauted

29 dm f,out dm = " =! m ( t ), (9.22) Solving Equation (9.22) would yield that the ma of the ocket and emaining fuel i m (t) = m 0 e!" t (9.23) whee m 0 = m,0 + m f,0 The ma being acceleated deceae exponentially, the gavitation foce and the ai eitance foce, being popotional to the ma, deceae a well, and o the ocket can continue to acceleate indefinitely. The poduct u( dm f,out / ) i ometime called the thut (ak you Coue XVI fiend); check to ee that the thut ha dimenion of foce. In the model fo thi poblem, we would have that the thut foce i dm f,out F = u =! u m, (9.24) thut an unlikely featue of ealitic ocket deign. At bet, even if Equation (9.24) wee an appoximate model, the thut would have to vanih when the fuel un out. The bottom line i that the model fo the fuel bun ate wa given in the fom it wa in ode to make olving fo the velocity a a function of time poible. Maybe a neat math poblem, but the phyic wa done once we found the ate of change of momentum

30 MIT OpenCoueWae SC Phyic I: Claical Mechanic Fo infomation about citing thee mateial o ou Tem of Ue, viit: