phy 201
See my notes and let me know if I missed anything.
I didn’t get to post this last night so I’m emailing it; it will be posted tonight.
I want to make sure that I did this right before I continue.
On finding my vertical average velocity, I knew 'dS, acceleration, and vnot.
I plugged these into one of my four equations of motion and found that vf=429.23cm/s
Then I found the average velocity to be 214.6cm/s.
I then plugged this is 'dT='dS/avg velocity and this would be 94cm/214.6cm/s
which would give me .438seconds which is pretty close to what you got.
Ok now that I know the 'dT = .438sec, vnot=0cm/s, and 'dS=39cm which is what I measured for the steel ball to roll of the ramp uninterrupted.
Is this right?
`ds = 39 cm is in the horizontal direction.
v0 = 0 is in the vertical direction and does not apply to the horizontal motion of the projectile. So you would not mix v0 = 0 up with `ds = 39 cm.
In the horizontal direction acceleration is 0, so velocity is constant.
Since velocity is constant initial velocity = ave velocity = final velocity.
So what you know is that `ds = 39 cm and `dt = .438 sec.
From `ds and `dt you can find average velocity.
So what is the average horizontal velocity and what is the initial horizontal velocity of the projectile?
Do I need to find the horizontal velocity for the larger ball after it hits the marble or what? Anyways, I just took those things I know and plugged that into 'dS=(vf+vo)/2'dT and got that vf for that one is equal to 178cm/s
Is that vf for horizontal or vertical motion?
The average velocity for horizontal motion is 80cm/s
You don't say how you got 80 cm/s. That is close to the correct horizontal velocity, but not close enough to make me confident in your method.
Is all of this looking correct to you?
" "Then for my data
undeflected ball - 39cm
deflected ball - 35cm
targeted ball - 68cm
vertical fall after collision =94cm
Ball 1 before collision - v1= 39cm/(.438sec)=89cm/s
Ball 1 after collision - v1=35cm/(.438sec)=79.9cm/s
Ball 2 before collision = 0
Ball 2 after collision = 68cm/(.438sec)=155cm/s
Yes, these calculations look good.
The velocity change of the first ball is v1-v1= -9.1cm/s
The velocity change of the second ball is v2-v2=155cm/s
The magnitude of change in v for the second ball is about 17 times as massive then the first, we conclude that the first ball has 17 times the mass of the second.
That would be the correct conclusion.
OK here on the diameters of the steel ball and the marble, I just used to ones in the example you gave us because I did not measure them myself, HOWEVER, I did use a steel ball and marble in my experiment.
I really don't think my ratio makes sense, I think somehow my measurements are off because 17 times bigger sounds a little extreme for the size balls we were working with, especially since you got five times bigger with the same two balls, however, I will proceede with my experiment with my data anyways.
What are the standard deviations of your distances?
How much uncertainty do you therefore think there is in your distances?
How much uncertainty would this imply in your velocities?
How much uncertainty would this imply for your mass ratio?
So (25mm/18mm)^3 is equal to 2.7
If the steel ball has 17 times the mass and 2.7 times the volume, then the density ratio would be
d=m/v
d=17/2.7
d=6.3 times the density
If the mass of the steel ball is about 60 grams(is this number something you knew or that you figured out using the data given, I assumed that it was just something you knew so I used it for my data too), then according to the ratio of 17/1 the marble's mass is about 3.5 grams.
The KE before and after
Ball 1 before collision
v1=39cm/(.438sec)=82cm/s
Ball 1 after collision
v1= 35cm/(.438sec) =79.9cm/s
Ball 2 before collision
=0
Ball 2 after collision
68cm/(.438sec)=155cm/s
KE
Ball 1 before collision 1/2mv^2=
(1/2)(60g)(89cm/s)^2=237630g/cm^2
=.237630 Joules
Ball 1 after collision
(1/2)(60g)(79.9cm/s)=191520.3g/cm^2
=.1915203 Joules
Ball 2 before collision - Is not moving before collision = 0
Ball 2 after collision
(1/2)(3.5g)(155cm/s)=42043.75g/cm^2
=.4204375 Joules
This gives us a .2 Joules before and a .61 after
Have I done this all correctly?
Pretty much so. But your total KE after collision is more than before, which by energy conservation is impossible. And your ratio of 17 to 1 is probably a bit high. This is likely due to uncertainties in your measurements, and perhaps even to measurement errors.
Nevertheless your analysis is excellent, and your results are no further off than many of the others I've seen.
We'll explore this more in class.