ps08sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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- 1 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 Fall Term 2010 Problem Set 8: Continuous Mass Transfer Solutions Problem 1 A rocket has a dry mass (empty of fuel) m r ,0 = 2 ! 10 7 kg , and initially carries fuel with mass m f ,0 = 5 ! 10 7 kg . The fuel is ejected at a speed u = 2.0 ! 10 3 m " s -1 relative to the rocket. What is the final speed of the rocket after all the fuel has burned? Solution The initial mass of the rocket included the fuel is m r , i = m r ,0 + m f ,0 = 2 ! 10 7 kg + 5 ! 10 7 kg = 7 ! 10 7 kg (1) The ratio of the initial mass of the rocket (including the mass of the fuel) to the final dry mass of the rocket (empty of fuel) is R = m r , i m r , d = 7 ! 10 7 kg 2 ! 10 7 kg = 3.5 (0.0.2) The final speed of the rocket is then v r , f = u ln R = (2.0 ! 10 3 m " s -1 )ln3.5 = 2.5 ! 10 3 m " s -1 . (0.0.3)
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- 2 - Problem 2: An ice skater of mass m is holding a bag of sand of mass m s that is leaking sand at a constant rate b . The ice skater is pushed with a constant force of magnitude F on a frictionless ice surface. At the instant that all the sand has leaked out, what is the speed of the skater? Solution: Choose the positive x-direction to point in the direction that the ice skater is moving. Let’s take as our system the amount of sand of mass s m ! that leaves the bag during the time interval [ , ] t t t + ! , and the ice skater with bag of remaining sand. At the beginning of the interval the ice skater with bag sand is moving with speed v so the x-component of the momentum at time t is given by p x ( t ) = ( ! m s + m ( t )) v ) , (0.0.4) where m ( t ) is the mass of the car and sand in it at time t . Denote by m 0 = m 1 + m s where m is the mass of the skater and m s ,0 is the mass of the sand in the bag at 0 t = , and ( ) s m t bt = is the mass of the sand that has left the car at time t since m s ( t ) = dm s dt dt 0 t ! = bdt 0 t ! = bt . (0.0.5) Thus m ( t ) = m 0 ! bt = m 1 + m s ,0 ! bt . (0.0.6) During the interval [ , ] t t t + ! , the small amount of sand of mass s m ! leaves the bag with the speed of the skater at the end of the interval v v + ! . So the x-component of the momentum at time t t + ! is given by p x ( t + ! t ) = ( ! m s + m ( t ))( v + ! v ) . (0.0.7) Throughout the interval a constant force F is applied to the skater so
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- 3 - 0 ( ) ( ) lim x x t p t t p t F t ! " + ! # = ! . (0.0.8) From our analysis above Eq. (0.0.8) becomes F = lim ! t " 0 ( m ( t ) + ! m s )( v + ! v ) # ( m ( t ) + ! m s ) v ! t . (0.0.9) Eq. (0.0.9) simplifies to F = lim ! t " 0 m ( t ) ! v ! t + lim ! t " 0 ! m s ! v ! t . (0.0.10) The second term vanishes when we take the 0 t ! " because it is of second order in the infinitesimal quantities (in this case s m v ! ! ) and so when dividing by t ! the quantity is of first order and hence vanishes since both 0 s m ! " and 0 v ! " . So Eq. (0.0.10) becomes F = lim ! t " 0 m ( t ) ! v ! t . (0.0.11) We now use the definition of the derivative: 0 lim t v dv t dt ! " ! = ! (0.0.12) in Eq. (0.0.11) which then becomes the differential equation F = m ( t ) dv dt . (0.0.13) Using Eq. (0.0.6) we have F = ( m 1 + m s ,0 ! bt ) dv dt . (0.0.14) (b) We can integrate this equation through the separation of variable technique. Rewrite Eq. (0.0.14) as dv = Fdt ( m 1 + m s ,0 ! bt ) . (0.0.15)
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