course phy 121
10/31 3pm
`q001. ** One-step-at-a-time questions:•If the velocity of a 5-kg object changes from +3 m/s to -2 m/s, what is the change in its velocity, and what is the change in its momentum?
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`dV = -2 m/s – 3 m/2 = -5 m/s
dp (1) = 5 kg * 3 m/s = 15 kg*m/s dp (2) = 5 kg * -2 m/s = -10 kg*m/s
Those are values of the momentum of a single object, which could be designated p_0 and p_f. Those values are not changes in momentum.
`dp = -10 kg*m/s – 15 kg*m/s = -25 kg*m/s
&#See the appended document for details and/or discussion of the solution.
&#
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• A 12-kg object experiences a velocity change of +2 m/s during a time interval of .01 second. What average force was exerted on it during that interval?
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F = w*a
F = 12 kg * (2m/s/.01 s)
F = 24 kg*m/s^2 or 24 N
it doesn't look like you divided by .01 s
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• What is the resultant of two displacement vectors, the first having magnitude 8 meters directed at angle 53 degrees, the second with x component -3 meters and y component -4 meters?
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1) I guess the first is just 8 m at 53 deg, with a x-comp = .60 m= 4.8m, y-comp = .79m = 6.3 m
2) the second is 5 m at 233 deg
the resultant is what you get when you add the two (see also Intro Prob Set 5).
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• Two objects collide, one of mass 5 kg and the other of mass 12 kg. The momentum of the 5 kg object changes by -40 kg m/s. What is the momentum change of the 12 kg object, and what is the change in its velocity?
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This is my thought process, the 5 kg object is moving -40 kg m/s because the 12 kg object hit it. So I would divide the -40 kg m/s by 12 kg and that gives me -3.33 m/s, and I would multiply that vel by 5 and get -16 kg m/s.
right general idea, but if the momentum change of one object in a collision is positive, the momentum change of the other is negative
And to get the vel of the 12 kg object I would divide -16 kg m/s by 12 kg = -1.4 m/s
For the 5 kg object, -40 kg m/s by 5 kg = -8 m/s
&#See the appended document for details and/or discussion of the solution.
&#
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• What is the direction of a displacement vector whose x and y components are +5 m and - 4 m? What are the components of a force vector of magnitude 80 N in the same direction?
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A positive x direction and a negative y direction
the direction of the vector is specified by its angle theta as measured in the counterlockwise direction from the positive x axis
from your subsequent answer it looks like you probably figured this out, but you didn't mention the angle in your answer to this first question
For 80 N, x-comp = .78 N = 62 N y-comp = -.62 N = -50 N
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• A rubber band chain exerts force 4 N when its length is 50 cm, and 12 N when its length is 80 cm. As the chain is extended from the first length to the second, what average force is exerted by the rubber band (assuming that the force vs. length is linear)? How much work is done by the chain during this interval?
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Ave is (12 N + 4 N) / 2 = 8 N
Work = 8 N * .3 m = 2.4 J
good
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• A 200 kg mass is subjected to a net force of 1000 Newtons for half a second. What is its acceleration? What therefore is its change in velocity? What therefore is the change in its momentum? What is the impulse of the force on this interval?
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Accel = 5 m/s^2
vel 2.5 m/s
`dp = 500 kg m/s
impulse?
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`q002. Give your data for the experiment in which you attempted to release the pendulum the point that would maximize its horizontal range. Include all data that would be necessary to replicate your trials.
Attempt to organize your data in such a way that it is easy to read at a glance.
Include a description how you conducted the experiment.
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We used the pendulum length (20cm) and pull back (14cm) constant
Four trials distance let go deg (estimate)
1 43 cm 25 deg
2 51 cm 20 deg
3 59 cm 12 deg
4 45 cm 5 deg
good data report
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`q003. How much force would it take to accelerate an Atwood machine consisting of two 50 kg masses at 9.8 m/s^2?
• What is the total weight of these two masses?
• If the addition of a single dollar coin to one side of the system results in an acceleration of 5 cm/s^2, then what is the mass of the coin? What is the weight of the coin?
• How many coins would it therefore take to match the mass of the system?
• (Assume that the friction in the pulley is negligible).
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• Would it be 0 because one side is the opposite weight of the other?
they are opposite so it would take some force to accelerate them
• How could you figure out the mass of the coin?
&#See the appended document for details and/or discussion of the solution.
&#
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`q004. Three rubber bands are attached to a small ring at the origin of an x-y coordinate system. The opposite ends of the rubber bands are respectively at the points (8 cm, 6 cm), (-10 cm, 0) and (0, -12 cm).
� For each rubber band, determine the angle of the displacement vector from the origin to its other end.
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Respectfully:
A = 10 cm at 37 deg
B = 10 cm at 180 deg
C = 12 cm at 270 deg
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Using a graph of force vs. length for each rubber band, we find that they exert forces of 5.0 N, 3.8 N and 3.2 N, respectively.
� Find the x and y components of each each force.
� Use your results to find the x and y components of the net force, and
� find the magnitude and direction of the net force.
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A: x-comp = 80 % = 4 N, y-comp = 60 % = 3 N
B: x-comp = 100% = 3.8 N, y-comp = 0
C: x-comp = 0, y-comp = 100% = 3.2 N
some of those components would be negative
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`q005. A ball starts rolling along the x axis at point x = x0, with velocity v0. Its acceleration is a. This continues through time interval `dt.
• By how much does the velocity of the ball change?
• What is the ball's displacement during the interval?
• What therefore is its final position along the x axis?
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• `dV = a * `dt
• `ds = (V0 + Vf)/2 * `dt
• `ds = (Vf – V0)/2 * `dt
xf – x0 = (Vf – V0)/2 * `dt
xf = x0 + (Vf – V0)/2 * `dt
right idea, but vf isn't a given quantity
the equation needs to be expressed in terms of the given quantities x0, v0, a, `dt
however it's clear you could have done this easily enough; just be sure you understand why you should have
&#See the appended document for details and/or discussion of the solution.
&#
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&#Your work looks good. See my notes. Let me know if you have any questions. &#