#$&*
course phy 201
11/4 2:00 pm
`q001. Describe briefly what you did in lab today and show you data.****
We balanced dominos from different positions on top of a piece of Styrofoam, trying to reach an equilibrium each time.
No dominos: 23 cm on each side of equilibrium
One domino on each side: 21.5 cm on left, 22.5 cm on right
2 dominos on each side: 21.5 cm on left, 22.5 cm on right
#$&*
What did you learn from doing this brief experiment?
****
when you have more weight on one side than the other, the greater weight must be moved closer to the middle to achieve equilibrium.
#$&*
Pick four of your trials, choosing the most varied among the situations you set up.
Assume the weight of a domino is 1 dominoWeight. If you really want to use a weight in Newtons, you can use .16 Newtons, but dominoWeight is fine.
For your first chosen trial, what was the net torque exerted by the dominoes? Be sure to give the details of your calculations.
****
first trial: zero torque; mass = 0 domino weights, total torque on each side: 0, net torque: 0
#$&*
@&
You can't assume that the torques exerted by the dominoes are equal and opposite. The foam might also be exerting a torque. You need to calculate the torques based on your observations, not on your expectations.
*@
@&
Having done that, of course, you would then compare results with the expectation of zero net torque.
*@
What was the net torque for this trial, as a percent of the sum of the magnitudes (i.e., absolute values) of the torques?
****
100%
#$&*
@&
If the net torque was zero, that would be 0% of the sum of the magnitudes of the individual torques.
However torques should have been calculated based on observations, per my previous note.
*@
Report the net torque and the net torque as a sum of the magnitudes of the torques for the second chosen trial. You have already explained the details of how you do the calculations, and don't need to include the details here or for the next two trials.
****
-210.7 domino weights m^2/s^2, 220.5 domino weights m^2/s^2
net torque: 9.8 domino weights m^2/s^2
percentage: 2.27%
#$&*
@&
The domino weight wouldn't be multiplied by the acceleration of gravity; the domino weight is equal to the domino mass multiplied by the acceleration of gravity, so the domino weight already includes the acceleration of gravity.
*@
@&
Since you multiplied both domino weights by the acceleration of gravity, when you divide them in the process of calculating percents the result isn't affected, so you have the right percent.
*@
Report the net torque and the net torque as a sum of the magnitudes of the torques for the third chosen trial.
****
421.4 torque, 441 torque
net torque: 19.6 torque
percentage: 2.27%
#$&*
Report the net torque and the net torque as a sum of the magnitudes of the torques for the fourth chosen trial.
****
no fourth trial
#$&*
@&
You had the opportunity to do many more trials with varied configurations, and should have done so.
However your analysis was mostly good.
*@
`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?
****
initial velocity: 5 m/s; final velocity: ?; vf^2 = v0^2 + 2a * ds; vf^2 = 2^2 + 2(-9.8) * -2; vf^2 = 4 + -19.6 *-2; vf^2 =43.2; vf = 6.57 m/s; vAve: 5.78 m/s; dt: 0.34 seconds;
#$&*
@&
You haven't actually declared a positive direction, but it's implicit from the signs you used for the quantities in the equation that you are using upward as positive.
The initial and final velocities have different signs, which will very significantly affect the average velocity and the time interval.
What you have calculated would be the correct time interval if the ball had been given a downward initial velocity of 5 m/s. Your `dt is not correct for the upward initial velocity.
*@
`q003. A projectile is released from a height of 4 meters and requires 1.2 seconds to reach the ground. Its initial horizontal velocity is 12 meters / second. How far does it travel in the horizontal direction before hitting the ground?
****
vertical velocity does not affect horizontal velocity; 12 m/s * 1.2 seconds; 14.4 meters;
#$&*
`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?
****
ds = -4 meters; dt = 1.2 seconds; a = - 9.8 m/s^2; vAve = -3.33 m/s; 9.8 m/s^2 * 1.2 seconds = 11.76 m/s; dv = -11.76 m/s; STUMPED.
#$&*
@&
Good reasoning. Just one step left.
vAve = (vf + v0) / 2.
Solve the equation for v0, then plug in your values for vAve and `dv (which are both correct).
*@
`q005. Just before striking the ground a projectile has a vertical velocity of 10 m/s, downward, and a horizontal velocity of 15 m/s. What are the magnitude and direction of its velocity at that instant, with direction measured on an x-y coordinate system with a vertical y axis?
****
magnitude: 18 m/s (10^2 + 15^2 = magnitude^2)
angle: arctan(rise/run), rise is negative, so add 180; 146.3 degrees;
#$&*
@&
Good, but you add 180 degrees only if the x component (which corresponds to the run) is negative. So the angle is about -34 degrees.
*@
If the mass of the projectile is 5 kg, what is its kinetic energy at the instant it strikes the ground?
****
½ m * v^2; 810 Joules
#$&*
What is its kinetic energy in the x direction (i.e., its kinetic energy calculated just on the basis of its horizontal velocity?)
****
562.5 Joules
#$&*
What is its kinetic energy in the y direction (i.e., its kinetic energy calculated just on the basis of its vertical velocity?)
****
250 Joules
#$&*
What is the sum of its x and y kinetic energies?
812.5 Joules
#$&*
`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?
****
velocity: 5 m/s; direction: 20 degrees; horizontal velocity: 4.7 m/s; vertical velocity: 1.7 m/s; vertical acceleration: -9.8 m/s^2; time to reach trash can: 1.06 seconds; dv: -10.39 m/s; final velocity: -8.68 m/s; average velocity:-3.5 m/s; distance covered in 1.06 seconds: -3.71 meters; predicted distance: 90 cm; very clearly not consistent. Actual height: -2 meters, 2 meters below the floor. Not valid.
#$&*
@&
Good.
*@
Suggested procedure (if your solution above is complete you don't have to answer these individual questions; if not, you should):
Find the initial vertical and horizontal velocities.
****
#$&*
Figure out what other two vertical quantities are known from the given information.
****
#$&*
Analyze the vertical motion to find the time required to get to the 80 cm height.
****
#$&*
Apply this time to the horizontal motion.
****
#$&*
`q007. Give a brief synopsis of what you learned by using the PHeT program.
****
From what I could tell, usually the faster object left with less energy than the slower object upon collision. Of course, size played a factor in this as well (the definition of momentum is of course Mass * velocity).
#$&*
"
&#Good responses. See my notes and let me know if you have questions. &#