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course Phy 201
121022`q001. Describe briefly what you did in lab today and show you data.
We balanced dominos on both sides of a ramp that was balanced on a dice to get numerous variations of data. There was data that consisted of a single domino on both sides or a single on one side and 2 dominos on the other. There was also some data taken that there was 2 dominos on both sides. There was no set distance from the edge of the strap or the balancing point they were just randomly balanced.
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What did you learn from doing this brief experiment?
The location of the domino in comparison to the balancing point determines how much weight it can balance on the other side of the strap. A single domino can balance double dominos if it is in the right position.
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Pick four of your trials, choosing the most varied among the situations you set up.
Assume the weight of a domino is 1 dominoWeight. If you really want to use a weight in Newtons, you can use .16 Newtons, but dominoWeight is fine.
For your first chosen trial, what was the net torque exerted by the dominoes? Be sure to give the details of your calculations.
-mg*d
-(-1dw)(9.8m/s^2)*.21m
Domino 1 = -2.058
-(-1dw)(9.8m/s^2)*.22m
Domino 2 = -2.156
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What was the net torque for this trial, as a percent of the sum of the magnitudes (i.e., absolute values) of the torques?
M1g*d1-m2g*d2=0
-2.058—2.156= 0.098Nm
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Report the net torque and the net torque as a sum of the magnitudes of the torques for the second chosen trial. You have already explained the details of how you do the calculations, and don't need to include the details here or for the next two trials.
2dw(9.8m/s^2)(0.09m) = -1.764 What units should these measurements be in?
1dw(9.8m/s^2)(0.18cm) = -1.764
Would net torque = 0??
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The net torque on any system in equilibrium, as measured relative to any point, is zero.
In this case you are measuring relative to the balancing point for the foam strip, so the strip itself contributes nothing to the net torque. So the sum of the torques of the dominoes would be zero.
However it's unlikely that your most accurate measurements would give you exactly zero. The positions of the dominoes can be fairly easily be measured to the within about a millimeter. Rounding measurements off to the nearest centimeter isn't appropriate to this experiment.
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Report the net torque and the net torque as a sum of the magnitudes of the torques for the third chosen trial.
1dw(9.8m/s^2)(0.2m) = 0.196
1dw(9.8m/s^2)(0.2m) = 0.196
0.196+0.196= 0.392
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Report the net torque and the net torque as a sum of the magnitudes of the torques for the fourth chosen trial.
2dw(9.8m/s^2)*(0.06m) = 1.18
1dw(9.8m/s^2)(0.205m) = .02
1.18+0.02 = 1.20
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`q002. A projectile has initial upward vertical velocity 5 meters / second. It is released at a height of 2 meters above the ground. Its acceleration is -9.8 m/s^2, as is the case for all ideal projectiles. How long does it take to reach the ground?
Vf^2 = vo^2+2ads
Vf^2 =( 5m/s)^2+2(-9.8m/s^2)(2m)
V=3.76m/s
Vave = ds/dt
4.38 = 2m/dt
dt= 0.46s
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The acceleration and the displacement are both in the downward direction, so your 9.8 m/s^2 acceleration and your 2 m displacement would have the same sign.
This would change your results fairly significantly.
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`q003. A projectile is released from a height of 4 meters and requires 1.2 seconds to reach the ground. Its initial horizontal velocity is 12 meters / second. How far does it travel in the horizontal direction before hitting the ground?
12m/s*1.2s
It travels 14.4 m in the horizontal direction before it hits the ground
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`q004. What would be the initial vertical velocity of a projectile which, when released from a height of 4 meters, requires 1.2 seconds to reach the ground?
12m/scos(theta)
The angle was about 70degrees
It initial vwould be about 11.3 m/s
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The angle isn't mentioned so isn't relevant to this question.
For the vertical motion you know `ds, a and `dt. From this information you can find v0 and vf.
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`q005. Just before striking the ground a projectile has a vertical velocity of 10 m/s, downward, and a horizontal velocity of 15 m/s. What are the magnitude and direction of its velocity at that instant, with direction measured on an x-y coordinate system with a vertical y axis?
a^2 +b^2=c^2
10^2+15^2 = c^2
c =18
the direction is right and downward
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The direction can be specified in degrees.
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If the mass of the projectile is 5 kg, what is its kinetic energy at the instant it strikes the ground?
KE = 1/2mv^2
KE = ½(5kg)(5m/s)^2
KE = 62.5J
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You determined previously that the velocity was c = 18. You didn't include units with that result or in the calculation; one reason for doing so is that it helps you identify the quantity and its meaning. Had you done this you would probably have been more likely to apply that result to the present question.
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What is its kinetic energy in the x direction (i.e., its kinetic energy calculated just on the basis of its horizontal velocity?)
KE = 1/2mv^2
KE = ½(5kg)(15m/s)^2
KE = 562.5J
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What is its kinetic energy in the y direction (i.e., its kinetic energy calculated just on the basis of its vertical velocity?)
KE = 1/2mv^2
KE = ½(5kg)(-10m/s)^2
KE = 250J
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What is the sum of its x and y kinetic energies?
250J+562.5J = 812.5J
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Using the 18 m/s result you would get a result close to this. You did some rounding with that result, so you won't get exactly the same answer, but any discrepancy should be explainable by the rounding.
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`q006. The washer I tossed into the trash can appears to have had an initial velocity of about 5 meters / second and left my hand at an angle of about 20 degrees above horizontal. The trash can was 5 meters away, the washer was released at a height of 170 cm and entered the trash can at a height of 80 cm.
Are these quantities consistent (remember that our original estimate of 80 cm/s at 30 degrees was consistent with hitting a trash can about a foot away)? Specifically, would the given initial velocity, angle, height of release, and height of the trash can result in reaching the 80 cm height after traveling 5 meters? If not, at what distance would this 'shot' reach the 80 cm height?
Suggested procedure (if your solution above is complete you don't have to answer these individual questions; if not, you should):
Find the initial vertical and horizontal velocities.
Vx = 50cm/s(cos)(160degrees) = -47cm/s
Vy = 50cm/s(sin)(160degrees) = 17.1cm/s
ds= -90m
dt = 5s
down2s
right 3s
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Between two given events, the time for the horizontal and the vertical motion will be identical. There is only one time interval between two given events.
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5 meters is 500 cm, not 50 cm. Otherwise your calculations of the initial horizontal and vertical velocities are correct.
When you correct those calculations, you will know the initial vertical velocity of the washer. You know its vertical displacement, and its acceleration is that of gravity. So you can analyze the vertical velocity and calculate the final vertical velocity, average vertical velocity and time interval.
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I think that it will reach the 80cm height after traveling 5meters.
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Figure out what other two vertical quantities are known from the given information.
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Analyze the vertical motion to find the time required to get to the 80 cm height.
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Apply this time to the horizontal motion.
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`q007. Give a brief synopsis of what you learned by using the PHeT program.
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&#Good responses. See my notes and let me know if you have questions. &#