class 051031
1. Write down the definition of acceleration.
Acceleration is the rate of change of velocity with respect to clock time.
Average acceleration over and interval is change in velocity / change in clock time. Could be expressed as the slope of a v vs. t graph.
2. Write down Newton's Second Law and briefly explain what it says.
Netwon's Second Law says that F_net = m a. This means that the net force acting on a mass, which is the sum of all the forces acting on that mass, is equal to the product of the mass and its acceleration.
3. Write down the definition of PE and explain how to find the change in gravitational PE.
The change in PE between two states of the system is the work done by the system against the net force.
We can define PE relative to a specified state of the system, so we can talk in terms of PE as a function of position and not just PE changes.
Near the surface of the Earth the change in gravitational PE is F_grav_BY * `dy, where `dy is change in upward vertical position. Since F_grav_BY = m g, the PE change is
`dPE = m g * `dy.
The textbook's definition of gravitational potential energy is
PE = m g y,
where y is upward vertical position measured relative to a fixed point you choose. You can choose any point, but some points make more sense of a given situation than others.
Note that if PE = m g y, then `dPE = m g `dy. The definitions of PE and of `dPE are equivalent.
4. Write down the impulse-momentum theorem, and explain how it is related to Newton's Third Law.
Fnet `dt = `d( m v), impulse = change in momentum.
Impulse if F `dt.
Momentum is m v so change in momentum is `d(m v).
If m is constant then `d( m v ) = m `dv.
This law comes from Newton's Third Law, which says that two objects exert equal and opposite forces on one another, and by substituting a = F / m in the second equation of uniformly accelerated motion. (It actually takes calculus to deal with the cases of variable force and variable mass, but we don't worry about this in the algebra-trig-based course).
5. Explain how to use a graph of v vs. t to find average acceleration over an interval and to find displacement over that interval.
If the graph is straight then you can form a perfect trapezoid corresponding to the interval. The average altitude of the trapezoid represents the average velocity and the width represents the time interval so the area represents the displacement. The change in altitude of the trapezoid represents the change in velocity, so rise / run represent change in velocity / change in clock time = average acceleration.
6. Write down the four equations of uniformly acceleration motion.
7. Write down the law of conservation of energy in at least one of its forms, preferably in all the forms we have discussed.
Fnet_ON * `ds = `dKE is the most basic form.
Fnet_ON = F_noncons_ON + F_cons_ON = F_noncons_ON - F_cons_BY, so since `dPE = F_cons_BY * `ds we get
`dW_noncons_ON - `dPE so that:
8. If two objects collide, one much more massive than the other, which one exerts the greater force on the others? Which one experiences the greater magnitude of momentum change?
By Newton's third law the forces are equal and opposite.
Momentum change is equal to F_net * `dt and the net forces exerted by the objects on one another are equal and opposite. So momentum changes are equal and opposite. So neither momentum change has the greater magnitude. The momentum changes are equal and opposite.
9. If you know the magnitude of a vector and its angle with the positive x axis, how do you find its x and y components?
10. If you know the x and y components of a vector then how do you find its magnitude and its angle with the positive x axis?
In Friday's experiment a typical result would be a 35 cm range for the undeflected ball, 25 cm when deflected, and 50 cm for the target ball.
Assume vertical fall after collision to be 92 cm and assume that all initial projectile velocities have zero vertical component.
An analysis that should be second nature tells you that since init vert. vel. is zero, the equations of motion applied to the vertical component easily give us the time interval, which in this case is about .43 second.
Since horizontal acceleration is unchanged the horizontal velocity immediately after collision will be the same as the horizontal velocity just before impact with the floor, and will be equal to the average horizontal velocity over this interval.
We thus easily find the before- and after-collision velocities to be as follows:
The velocity change of the first ball is therefore v1 ' - v1 = -24 cm/s.
The velocity change of the second ball is therefore v2 ' - v2 = 116 cm/s = 0 cm/s = 116 cm/s.
Which ball is therefore the more massive?
Does this mass ratio make sense for a steel ball of 25 mm diameter and a glass ball of 18 mm diameter?
The steel ball has both greater density and greater volume. How much greater is its volume?
(V2 / V1) = (4/3 pi r2^3) / (4/3 pi r1^3) = (r2 / r1)^3, and since the ratio of diameters is the same as the ratio of the radii we see that
(V2 / V1) = (d2 / d1)^3 = (25 mm / (18 mm) )^3 = 2.7, approx..
If the steel ball has 5 times the mass and 2.7 times the volume, then what is the ratio of densities of steel and glass?
If the steel ball has 2.7 times the volume and 5 times the mass then it has 5 / 2.7 = 1.8 times the density.
The density of steel is about 7.5 grams / cm^3. So the mass of the steel ball is about 60 grams.
If the density ratio is 5/1 then the marble has mass about 12 grams.
What then is the total KE before collision and the total KE after?
so
This gives us .20 J before, and .18 J after collision.
Homework:
Write up and submit electronically a description of how you set up and conducted this experiment. Use the submit work form to submit a text copy; if you have pretty formatting you can also email a formatted copy.
Report your data in a clear and concise table.
Show how you obtain the mean and standard deviation your distances.
Repeat the analysis done here with your data.
1. What were the mean and standard deviation of the distances you observed?
| 1st marble undeflected | 1st marble after collision | 2d marble after collision | |
| first setup | |||
| second setup |
How many trials did you make for each setup?
According to your results does a significant difference occur when the entire ramp system is raised 1 mm relative to the target marble?
2.