From the section CONSERVING EQUATIONS

Note that the solutions are written in the format you would actually submit as a solution, namely as a Maple file. Maple commands appear red. Maple responses are black with cursive type. Text comments appear black with typewriter text.

• 3.7:
  
  • a.)
    We start by considering the monkey and the package as independent systems coupled only by the common tension acting on both of them. Note that we cannot, in this problem, simply consider the monkey and package as part of a single system with only their weights as the external forces since we would have to account for the relative acceleration of the monkey and package in any case.

    For the monkey, the forces acting are the tension and the monkey's weight. For the package, the tension and the weight act, but we also have the additional constraint that the acceleration of the package is zero (i.e. it is just lifted off the ground). Hence, we can write down force equations for the monkey and package and simply ask Maple to solve them for us. In our equations, we choose the up direction to be positive and down to be negative. So,

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    Plugging in values for the monkey and package masses and the acceleration due to gravity yields:
    

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    Hence the acceleration of the monkey must be at least one-half of g or 4.9 m/s^2.

  • b.)
    If the monkey stops climbing, the tension will change, but the force equation for the monkey is the same as derived in part a. For the package, the only difference is that now the package and monkey must have equal magnitude, opposite direction accelerations since both move with the rope. If we term the monkeys acceleration as positive, then the acceleration of the package must be negative. We write down these equations and once again ask Maple to do the grunt work for us:

    

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    Hence, Maple says that the tension in the rope is 117.6 Newtons and the acceleration is 1.96 m/s^2. Since the value for the acceleration is positive, the monkey is accelerated upward and the package is falling.

• 3.8:
  The tensions are found easiest from the force equations for the elevator passenger cage and the counterweight

  

  Now note that the acceleration of the counterweight is equal magnitude, opposite direction to the elevator cage, so

  

  The force provided by the driving mechanism must make up the difference between the tensions, so, in the convention of force to the right being positive, we must have

  Since the force is negative, the driving mechanism is pulling the cable to the left.

• 3.9:
  We note from the figure that m_2 must go down at the half the acceleration that m_1 goes to the right. We can see this since the string on both sides of the pulley attached to m_2 have to extend by the same length as m_2 goes down. Since the string is constant length, the part of the string attached to m_1 must go twice this length. There are also two tensions attached to the pulley connected to m_2. They both pull in the up direction. Therefore, our equation of motion for m_2 is

  

  since down is the negative direction, while for m_1 we get

  

  Maple gives us the simultaneous solution for the two unknowns, T and a2, as follows:

  

  Therefore, the accelerations of m_2 and m_1 are

  

• 3.10:
  Begin by just writing down momentum and energy conservation equations and asking Maple to simultaneously solve them.

  

  We choose the second set of solutions since the first corresponds to no chnage in velocity for either object - an unphysical situation for our head-on collisions.

  

  Find the final velocities in terms of the initial velocity for the specific case of equal masses.

  

  So we see that the velocities exchange. Mass 1 moves in the direction and speed that mass 2 was originally moving and vice versa. Now consider the solution in the case where m1 and m2 are different and u2 = 0, i.e. mass 2 is initially stationary

  

  For the solutions just derived, we see that if m1=m2, then m1 stops and m2 continues in the direction and with the initial speed of m1. This confirms our previous result which said that the velocities should ``exchange'' if m1=m2. Now plot the general dependence as instructed for m2 greater than and less than m1.

  
  

  We see from our plots that for m2 > m1 (i.e. x > 1) m1 rebounds backward while m2 moves forward. The velocity of m2 decreases as its mass gets larger compared to m1. For m2 >> m1, m1 rebounds back with the same magnitude velocity it came in with while m2 hardly moves at all. This is in line with how we think things should happen. If a light object hits a very massive, stationary object, the stationary object stays almost still while the light object rebounds elastically.

  
  

  For m2 < m1 (i.e. x < 1) we find that both objects always move forward. If m2 << m1, then m2 moves forward almost twice as fast as m1 comes in while m1 hardly changes its velocity. Again, this fits our experience. If a massive object collides with a light one, the light object gets thrown forward while the massive one suffers hardly any change in its motion.

• 3.12:
  This problem looks hard, but in fact all we need to do is specify the equations for momentum and energy conservation and let Maple do the rest. It's remarkable!

  First define the momentum and energy conservation equations. The convention is that motion to the right is positive and to the left is negative (i.e. toward the sun is negative).

  

  Let Maple solve it for the unknowns. For this exercise we know the initial velocities, u1 and u2, and the masses m1 and m2, so the final velocities are the unknowns. We must ask Maple to solve for both explicitly or it will choose what is unknown for us and give us unwieldy equations.

  

  Define the initial velocities

  

  Now put in the fact that Jupiter is essentially infinite in mass compared to the mass of the space probe.

  

  So the probe goes away from Jupiter at 38 km/sec. That's more than 3 times the initial speed! We could never pack enough fuel on board a space probe to get it to such a high velocity. The sling shot effect is essential to our exploration of the solar system. The most recent craft to use it was the now-defunct Mars Explorer probe entering Mars orbit this week. In this case, the gravity boost was obtained from the earth and the moon.

• 3.13:
  The beginning of this problem is simple, we just make use of the formula given in the lecture notes
  • a.)
    We'll use Maple to calculate things for us. This time we'll include units to be sure that we're making the correct calculations (and because it's a good thing to do.) First thing is to define a function for the mass of fuel burned for a given time

    

    Next, we define the final velocity as a function of time using the lecture note formula

    

    Now just plug in the constants

    

    And now, the answer for 120 seconds worth of burning is ...

    

    It's so easy with Maple! Now for the plotting. A Maple hint here is to remember that Maple needs to see ALL your variables defined before it will give you a non-empty plot. The units we use are considered variables by Maple (it's not really smart, it just seems that way). So, be sure to set them to a standard value before plotting. We want our graph to be in units of m/s, so set meters and seconds to 1 for plotting purposes. Also, its a mark of quality to label graphs we include for other people's perusal.

    
    

    The plot increases logarithmically with time. What we saw in the plot done in the lecture notes is something that flattened out as the mass of fuel burned approached the total rocket mass. The difference here is that we are burning a CONSTANT rate of fuel. There is no data point after about the 120th second because all the fuel is exhausted. The remaining mass is the payload of the Saturn V first stage (namely all the stages above the first). So the final velocity is strictly limited, but does not show the tapering off that you get when you plot speed vs. mass where the RATE of fuel burned has to dwindle off if you are not to burn it all by the 120th second.

  • b.)
    The formula for more than one stage is simple to derive. Starting from our formula in the lecture notes and remembering that the second stage starts acceleration from the final speed of the first stage, we get the answer desired. The final speed of the first stage is

    

    Since we have only burned the mass of first stage fuel when the first stage is done at time t_1. Now, we drop the first stage, so the starting mass for the second stage is m_0 - m_1 - m_1'. The final mass is m_0 - m_1 - m_1' - m_2, where m_2 is the mass of second stage fuel burned after additional time t_2. Therefore, adding the two velocities together yields