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    Velocity, conservation of momentum, force, force of gravity

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    1. A ball is thrown in the air with enough force so that it goes straight up for several seconds.
    a. What is the velocity of the ball 1 sec before it's highest point?
    b. What is the change in velocity during this 1 sec interval?
    c. What is the velocity 1 sec after it reaches it's highest point?
    d. what is the change in velocity during this 1 sec interval?
    e. What is the change in velocity during the 2 sec interval?
    f. What is the acceleration of the ball during any of these time intervals and at the moment the ball has a zero velocity?

    2. An airplane is flying horizontally with speed 1000 km / h ( 280 m/s) when the engine falls off. Neglecting air resistance, if it takes 30 sec for the engine to hit the ground:
    a. How high is the plane?
    b. How far horizontally will the engine travel while it falls?
    c. If the airplane somehow continues to fly as if nothing happened, where is the engine relative to the airplane at the moment the engine hits the ground?

    3. A bicycle has wheels with a circumference of 2m. What is the linear speed of the bicycle when the wheels rotate at 1 revolution per second?

    4. Before going into orbit, an astronaut has a mass of 55kg, While in orbit, a measurement determines that a force of 100N causes her to move with an acceleration of 1.90 m/s2. to regain her original weight, should she diet, or eat more?

    5. A railroad car weighs four times as much as a freight car. If the railroad car coasts at 5 km/h into the freight car that is initially at rest, how fast do the two coast together after they are coupled?

    6. A 60kg skydiver moving at terminal speed falls 50m in 1 sec. What power is the skydiver expending on the air?

    7. To tighten a bolt, you push with a force of 80N at the end of a wrench handle that is 0.25m from the axis of the bolt:
    a. What torque are you exerting?
    b. If you move your hand to 0.10 from the bolt, what force do you have to exert to achieve the same torque?
    c. Do your answers depend on the direction of your push relative to the direction of the wrench handle?

    8. The value of g at the earth's surface is about 9.8m/s2. What is the value of g at a distance from the earth's center that is 4 times the earth's radius?

    9. A 3 kg newborn at the earth's surface is gravitationally attracted to earth with a force of about 30N.
    a. Calculate the force of gravity with which the baby on earth is attracted to the planet Mars, when Mars is closest to earth. (The mass of Mars is 6.4 x 1023 kg, and it's closest distance is 5.6 x 1010 meters)
    b. Calculate the force of gravity between the baby and the physician who delivers it. Assume the physician has a mass of 100kg and is 0.5 m from the baby.
    c. How do the forces compare?

    10. Does the speed of a falling object depend on its mass? Does the speed of a satellite in orbit depend on its mass? Defend your answer.

    © BrainMass Inc. brainmass.com December 24, 2021, 4:42 pm ad1c9bdddf
    https://brainmass.com/physics/applied-physics/velocity-conservation-of-momentum-force-force-of-gravity-2581

    SOLUTION This solution is FREE courtesy of BrainMass!

    See the attached file for the fully formatted responses.

    1. A ball is thrown in the air with enough force so that it goes straight up for several seconds.
    a. What is the velocity of the ball 1 sec before it's highest point?
    b. What is the change in velocity during this 1 sec interval?
    c. What is the velocity 1 sec after it reaches it's highest point?
    d. what is the change in velocity during this 1 sec interval?
    e. What is the change in velocity during the 2 sec interval?
    f. What is the acceleration of the ball during any of these time intervals and at the moment the ball has a zero velocity?

    Answer:

    A. We know that at the highest point, the ball will have a vertical velocity of zero. Knowing that after it is released the only force acting on it is gravity, we can work backwards from this zero-velocity point to find the velocity a second earlier. Use the formula for constant acceleration:

    V2 = V1 + at

    0 = V1 - 9.8(1)
    V1 = 9.8 m/s
    Note that gravity acts downwards, while the velocity is upwards, hence the minus sign.

    B. The change in velocity is obvious: the difference between the final (0 m/s) and initial (9.8 m/s) velocities.

    V = -9.8 m/s (the minus sign indicates that the upward velocity decreased)

    C. Following the same equation in part A. (or just looking at the same event in reverse, if you like approaching problems from a symmetry perspective), the velocity after one second will be -9.8 m/s.

    D. Following part B., the change in velocity would be -9.8 m/s. Note how they are both minus - that's because gravity is acting downwards the whole time.

    E. The change in velocity for the 2-second interval is simply the difference between the initial velocity (9.8 m/s upwards) and final velocity (9.8 m/s downwards, or -9.8 m/s), giving us a difference of -19.6 m/s.

    2. An airplane is flying horizontally with speed 1000 km / h ( 280 m/s) when the engine falls off. Neglecting air resistance, if it takes 30 sec for the engine to hit the ground:
    a. How high is the plane?
    b. How far horizontally will the engine travel while it falls?
    c. If the airplane somehow continues to fly as if nothing happened, where is the engine relative to the airplane at the moment the engine hits the ground?

    Answer:

    If the aircraft is flying horizontally, then when the engine falls off it has zero initial velocity in the vertical direction. It then takes 30 seconds to hit the ground. Using this information you can find the height of the aircraft from the ground with the formula:

    d = v1t + .5at2

    And since v1 is zero, it is simply:

    D = .5(9.8)(302) = 4410 m

    Then using the same formula you can find the horizontal travel of the engine (8400 m), with the initial speed being the speed of the aircraft in flight (280 m/s), but with no acceleration (since we are ignoring air resistance).

    Now you should see that if we ignore air resistance and any effects on the airplane, the engine will follow directly beneath the plane as it falls, so that when it hits the ground, it is 4410 m directly below the plane! Remember: the engine has inertia.

    3. A bicycle has wheels with a circumference of 2m. What is the linear speed of the bicycle when the wheels rotate at 1 revolution per second?

    Answer:

    Well, if there is no slip between the wheels and the ground, then every time the wheels rotate once, the bike will have advanced by the circumference of the wheel - 2 m. If there is one revolution per second, then the bike should advance at a pace of 2 m/s.

    4. Before going into orbit, an astronaut has a mass of 55kg, While in orbit, a measurement determines that a force of 100N causes her to move with an acceleration of 1.90 m/s2. to regain her original weight, should she diet, or eat more?

    Answer:

    Remember, F = ma.
    Find her mass based on the force and acceleration given:
    m = F/a = 52.6 kg. She needs to eat more.

    5. A railroad car weighs four times as much as a freight car. If the railroad car coasts at 5 km/h into the freight car that is initially at rest, how fast do the two coast together after they are coupled?

    Answer:

    This is a conservation of momentum question. Remember, momentum (p) is:

    p = mv (mass times velocity).

    In a conservative system (one without loss of energy to friction, etc.), momentum is conserved, so that

    p1 = p2

    Now, the initial momentum is the mass of the railroad car (mr) times its velocity (5 km/h)

    p1 = 5mr

    The momentum after the two are coupled is equal to their combined weight times their velocity. The question says that the freight car is ¼ the mass of the railroad car, so their combined mass is 1.25mr. Therefore we have:

    5mr = 1.25mrv2

    Notice that the mr's cancel - you don't need to know the exact mass of the railroad cars to find their velocities after a collision.

    v2 = 4 km/h.

    6. A 60kg skydiver moving at terminal speed falls 50m in 1 sec. What power is the skydiver expending on the air?

    Answer:

    This question is a little tricky: you have to find the energy expended in one second due to friction with the air. To find that, you must first find the force exerted... however, since the skydiver is falling at terminal velocity we know that the force of air resistance must exactly equal the force of gravity:

    F = mg = 60(9.8) = 588 N.

    The energy expended in one second is then equal to the work done by the skydiver on the air in one second, which we know is falling through a distance of 50 m.

    W = Fd = 588(50)

    Since we dealt with a 1-second time interval, the power expended on the air is simply:

    P = W/t = 29400 W (J/s)

    7. To tighten a bolt, you push with a force of 80N at the end of a wrench handle that is 0.25m from the axis of the bolt:
    a. What torque are you exerting?
    b. If you move your hand to 0.10 from the bolt, what force do you have to exert to achieve the same torque?
    c. Do your answers depend on the direction of your push relative to the direction of the wrench handle?

    Answer:

    Torque is the rotational (angular) equivalent of force. It absolutely does depend on the angle between the force and the wrench - if you push directly along the wrench, there will be no twisting force on the bolt.

    The torque exerted is the cross product between the vector from the axis of rotation to the application point and the force vector.

     = F x r

    Or, in scalar terms,  = Frsin-- the dependence of torque on the direction of push is explicit in the sin term.

    So, assume that in part a and b you will be as efficient as possible, and apply your force at 90° to the wrench handle. Then you have:

    a.  = .25*80 = 20 Nm
    b. 20 = .1*F
    F = 200 N

    8. The value of g at the earth's surface is about 9.8m/s2. What is the value of g at a distance from the earth's center that is 4 times the earth's radius?

    Answer:

    All you need to know for this question is that gravity is a 1/r2 law, so that at a point 4 times farther away, the force of gravity will be 1/(4)2 = 1/16th as strong.

    Don't be afraid to express answers in terms of other variables (in this case, g) when very little information is given. These types of relational questions can sometimes be more valuable and insightful to a physicist than ones involving precise numerical answers.

    9. A 3 kg newborn at the earth's surface is gravitationally attracted to earth with a force of about 30N.
    a. Calculate the force of gravity with which the baby on earth is attracted to the planet Mars, when Mars is closest to earth. (The mass of Mars is 6.4 x 1023 kg, and it's closest distance is 5.6 x 1010 meters)
    b. Calculate the force of gravity between the baby and the physician who delivers it. Assume the physician has a mass of 100kg and is 0.5 m from the baby.
    c. How do the forces compare?

    Answer:

    The force of gravity is given by:

    F = GmM/r2 where G is the gravitational constant, equal to 6.67 x 10-11 Nm2/kg2

    Simply plug into the formula the known values to find that F = 4.09 x 10-8 N for the gravitational attraction between the newborn and the planet Mars, and F = 8.01 x 10-8 N for the gravitational attraction between the newborn and the doctor. After doing this question you may have some reservations about astrology...

    10. Does the speed of a falling object depend on its mass? Does the speed of a satellite in orbit depend on its mass? Defend your answer.

    Answer:

    No. Remember, the force of gravity is:

    F = GmM/r2, which does depend on the mass, but that the acceleration due to gravity (which is what determines the speed) just simplifies to "g" for all objects, no matter what their mass is. This holds for baseballs as well as it does for satellites, though the value of "g" may change slightly.

    There are two exceptions to this:
    1. The object is SO massive that it pulls the Earth towards itself, in which case the speeds relative to the centre of mass of the system would be unchanged, but the speeds relative to the Earth would be.
    2. You take air resistance into account. Air resistance depends on many things, such as the density of an object and its cross-sectional area (thought not, strictly speaking, it's mass).

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 4:42 pm ad1c9bdddf>
    https://brainmass.com/physics/applied-physics/velocity-conservation-of-momentum-force-force-of-gravity-2581

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