Class 091021

Collision Demonstration

In class we observed a steel ball rolling down an incline with slope approximately .03, colliding with a stationary marble, with both objects then falling 90 cm to the floor.

• The steel ball, when unimpeded by the marble, traveled 8.5 cm while falling.
• The marble (which was initially stationary), after being hit by the ball, traveled 16 cm while falling.
• The steel ball after hitting the marble traveled 5.5 cm while falling.
• You should be able to figure out the time of fall, assuming the initial vertical velocity to be close to zero.

Approximating the time of fall by .4 sec we conclude that the ball reached the end of the ramp at a speed of about 20 cm/s, the marble traveled at 40 cm/s after being struck, and the steel ball was moving at about 12 cm/s after striking the marble.

`q001.  Using the above figures:

• What is the change in the ball's velocity between the instant before and the instant after it strikes the marble?
• What is the change in the marble's velocity the instant before and the instant after the collision?
• What do you therefore conjecture to be the ratio of the ball's mass to that of the marble?

****

The ball's velocity just before collision is its velocity 20 cm/s at the end of the ramp.  After collision its velocity is 12 cm/s, so its velocity changed by -8 cm/s.

The marble was initially at rest.  As a result of the collision it velocity changed to 40 cm/s.

The two objects exert equal and opposite forces on one another during the time interval during which they are in contact.  So they have equal and opposite momentum changes.

For these two objects we therefore have

m1 `dv1 = -m2 `dv2 so that

m2 / m1 = - `dv1 / `dv2 and, letting object 1 be the marble and object 2 the steel ball,

m2 / m1 = - (40 cm/s) / (-8 cm/s) = 5.</h3>

&&&&

`q002.  Using your own data for the same experiment, what are the following:

• the unimpeded steel ball's velocity at the end of the ramp
• the velocity of the glass marble the instant after being struck by the ball and
• the velocity of the ball immediately after striking the marble?

What therefore should be the ratio of the ball's mass to that of the marble?

Be sure to include your raw data and explain how you get from the raw data to your conclusions.

****

<h3>Your analysis will be similar to that of the example.</h3>

&&&&

Let's assume that we can neglect forces exerted by the ramp on the glass marble and the steel ball, and that we can in fact neglect all forces except the forces exerted by each ball on the other.  We can therefore regard the two balls, during the time of collision, as a closed system.  We have draw our conclusions about the velocities involved in the collision from the projectile behavior of the objects after collision.

From F_net `dt = `dp and the fact that the forces exerted by the two objects are equal and opposite, we can draw the conclusion that the momentum change `dp of one object should be equal and opposite to that of the other.  (recall that p = m v, so `dp = p_f - p_0 = m vf - m v0)

Since momentum changes are equal and opposite, we conclude that the total momentum of the system is the same immediately after collision as it was immediately before.

`q003.  Let's let m1 and m2 be the two masses, let v1 and v2 be their velocities before the collision, and let u1 and u2 be their velocities after collision.

• What are the values of v1, v2, u1 and u2 for your trial?

****

For the example, we have

• v1 = 0
• v2 = 20 cm/s
• u1 = 40 cm/s
• u2 = -8 cm/s.

&&&&

`q004.  Using the symbols m1, m2, v1, v2, u1 and u2, give the symbolic expression for each of the following:

• The momentum of the steel ball before collision.
• The momentum of the glass marble before collision.
• The momentum of the steel ball after collision.
• The momentum of the glass marble after collision.
• The total momentum of the two objects before collision.
• The total momentum of the two objects after collision.

****

<h3>

• The momentum of the steel ball before collision is m2 v2.
• The momentum of the glass marble before collision is m1 v1.
• The momentum of the steel ball after collision is m2 u2.
• The momentum of the glass marble after collision is m1 u1.
• The total momentum of the two objects before collision is m1 v1 + m2 v2.
• The total momentum of the two objects after collision is m1 u1 + m2 u2.

</h3>

&&&&

`q005.  Write the symbolic equation which states that the total momentum of the two objects before collision is equal to their total momentum after collision.  (your equation will set your answers the last two expressions in the preceding question equal to one another)

****

<h3>The equation is

m1 v1 + m2 v2 = m1 u1 + m2 u2.</h3>

&&&&

`q006.  Substitute the values you observed for v1, v2, u1 and u2 into your equation.  Your equation will still have the symbols m1 and m2, but the other symbols will be replaced by the various velocities you observed.

****

<h3>For the given example the equation would be

m1 * 20 cm/s + m2 * 0 = m1 * 12 cm/s + m2 * 40 cm/s.</h3>

&&&&

`q007.  Rearrange your equation so that all the quantities containing m1 are on the left-hand side and all the quantities containing m2 are on the right-hand side (your equation should have been m1 v1 + m2 v2 = m1 u1 + m2 u2).

Factor m1 out of the left-hand side and m2 out of the right-hand side.

Solve the equation for m1.  The left-hand side should end up just as m1; the right-hand side will include the symbol m2.

Give the steps of your solution:

****

<h3>m1 * 20 cm/s + m2 * 0 = m1 * 12 cm/s + m2 * 40 cm/s simplifies to

m1 * 20 cm/s = m1 * 12 cm/s + m2 * 40 cm/s.  Dividing both sides by the unit cm/s we get

m1 * 20 = m1 * 12 + m2 * 40, or

20 m1 = 12 m1 + 40 m2.  Rearranging we get

20 m1 - 12 m1 = 40 m2 so that

8 m1 = 40 m2 and

m1 = 40 m2 / 8 or

m1 = 5 m2.</h3>

&&&&

The foam wheel was observed to rotate through 8 revolutions, coming to rest after about 20 seconds.  We want to find its acceleration.  We follow the usual procedure for this situation:  find the average velocity, use this along with the final velocity to to determine the initial velocity, then find the rate of change of velocity with respect to clock time.

Its position is measured in revolutions.

The average rate of change of position with respect to clock time is therefore

• ave rate = change in position / change in clock time = 8 rev / 20 sec = .4 rev / sec.

Revolutions are a measure of angle or angular displacement.  The 8 revolutions therefore represent a change in angular position.  The rate of change of angular position will be called angular velocity.  Angular velocity is represented by the symbol omega (the Greek symbol that looks something like a W but should never be called 'double-yew' or written as 'w'; write it as 'omega').

So we say that

• ave angular velocity = .4 rev / sec, or
• omega_Ave = .4 rev / sec.

This is the 'average velocity'.  To find the acceleration we have to find the initial velocity.

A linear velocity vs. t trapezoid tells us that since final vel is zero, the initial vel is double the average. 

Our conclusion:

• the initial angular velocity is .8 rev / sec, double the average angular velocity
• the average rate of change of angular velocity with respect to clock time is therefore
• ave rate = (0 rev / sec - .8 rev/sec) / (20 sec) = -.04 rev / sec^2.

This is what we call acceleration.  Since it's the average rate of change of angular velocity, we call this the angular acceleration.  Angular acceleration is represented by the symbol alpha.

Thus our conclusion is that

• angular acceleration = alpha = `dOmega / `dt = -.04 rev/s^2.

Our conclusions:

• omega_Ave = .4 rev / sec
• omega_f = .8 rev / sec
• alpha = -.04 rev / s^2.

We could express these quantities using degrees instead of revolutions.  Since 1 revolution = 360 degrees, we have

• omega_Ave = .4 rev / sec = .4 (360 deg) / sec = 144 deg / sec
• alpha = -.04 rev / s^2 = -.04 * 360 deg / s^2 = -14.4 deg/s^2.

Or we could express these quantities in terms of radians.  You might not be familiar with radians, and if not you should note and remember that 1 revolution = 2 pi radians.  Using this fact

• omega_Ave = .4 rev / sec = .4 * 2 pi rad / sec = .8 *pi rad / sec and
• alpha = -.04 rev / s^2 = -.04 * 2 pi rad / s^2 = -.08 pi rad / sec.

How would someone who measured the motion in cycles of a 9 cm pendulum calculate these quantities?  Let's see:

• We know that for this length 1 cycle = .2 sqrt(9) sec = .6 sec.
• We know omega_Ave in rev / sec.  To get this quantity in rev/cycle we need to know what to substitute for 'sec'.
• If 1 cycle = .6 sec, then 1 sec = 1 / .6 cycle = 1.67 cycle.  So we can substitute 1.67 cycle for 'sec', obtaining
• omega_Ave = .4 rev / sec = .4 rev / (1.67 cycle) = .24 rev / cycle.
• alpha = -.04 rev / sec^2 = -.04 rev / (1.67 cycle)^2 = -.014 rev / cycle^2.

`q008.  If someone observed that the wheel went through 10 revolutions in 50 cycles of a pendulum of unspecified length, then in units of revolutions and cycles, reason out the values of omega_Ave and alpha.

****

omega_Ave is ave. angular velocity, which is ave rate of change of angular position with respect to clock time, so

• omega_Ave = (change in angular position) / (change in clock time) = 10 revolutions / (50 cycles) = .2 rev / cycle.

Assuming uniform angular acceleration, since the system ends up at rest its initial angular velocity will be double its average angular velocity so

• omega_0 = .4 rev / cycle.

Angular acceleration is ave rate of change of angular velocity with respect to clock time, so

• alpha = (change in angular velocity) / (change in clock time) = (0 - .4 rev/sec) / (50 cycles) = -.008 rev / cycle^2.

&&&&

`q009.  If we later find that the pendulum in the preceding has length .8 cm, then what would be the values of omega_Ave and alpha in terms of revolutions and seconds, in terms of degrees and seconds, and in terms of radians and seconds?

****

A pendulum of length .8 cm has period .2 sqrt(.8) sec = .18 sec.

A revolution corresponds to 360 degrees, or to 2 pi radians.

So the average angular velocity is

omega_Ave = .2 rev / cycle = .2 rev / (.18 sec) = 1.1 rev / sec, which in turn can be written

1.1 rev / sec = 1.1 (360 deg) / sec = 400 deg / sec, or as

1.1 rev / sec = 1.1 ( 2 pi rad) / sec = 2.2 pi rad / sec. 

2.2 pi rad / sec can be approximated as 2.2 * 3.14 rad / sec = 6.9 rad / sec.

&&&&

An Atwood machine consists of a 20 kg mass on one side and a 22 kg mass on the other.  Let the positive direction be that in which the 20 kg mass descends.  The system therefore experiences forces of 20 kg * 9.8 m/s^2 = 200 N (approx.) in the positive direction, and 22 kg * 9.8 m/s^2 = 220 N in the negative direction, so the net force is 200 N - 220 N = -20 N.  The total mass is 20 kg + 22 kg = 42 kg, so the acceleration of the system is

• a = F_net / m = -20 N / (42 kg) = -.48 m/s^2.

We come to this conclusion without considering the tension in the string connecting the two masses.  To figure out this tension we could consider just the 20 kg mass.  The forces on this mass are the tension T, acting upward, and the weight (200 N) of the 20 kg mass, acting downward.  Letting downward be the positive direction (consistent with our choice in the original problem) we see that

• F_net = 200 N - T.

We know that F_net = m a, so we can say

• m a = 200 N - T.

We also know that a = -0.48 m/s^2.  Of course we know that m = 20 kg.  So we can solve the equation for T:

• T = 200 N - m a = 200 N - 20 kg * (-0.48 m/s^2) = 200 N - (-10 N) = 210 N.

That is, the tension is 210 N.  We have already assumed that the tension acts upward (we did this by saying that the tension force is -T), so the tension is 210 N in the upward direction.

This makes sense.  The net force on the 20 kg mass is 200 N - 210 N = -10 N, i.e., 10 N in the upward direction.  Its acceleration is therefore -10 N / (20 kg) = -.5 m/s^2, within approximation error of the more accurate -.48 m/s^2.

`q010.  We have seen that the acceleration of the system is -.48 m/s^2.  Use this fact to find the tension acting on the 22 kg mass.  Your reasoning will be similar to that used above, but some of the details will be different.

****

For this mass, F_net = m a = 22 kg * .48 m/s^2 = 10 N, approx..

The weight of the 22 kg mass is 220 N, approx.. 

• net force = weight + tension, so that
• 10 N = 220 N + tension.  We easily rearrange this to obtain
• tension = 10 N - 220 N = -210 N.

&&&&

A 20 kg mass on a 10 degree incline, with the incline running down and to the right, experiences a gravitational force of about 200 N.

If an x-y coordinate system is imposed on this system, with the x axis down and to the right parallel to the incline, then the downward gravitational force will lie at angle 280 deg, measured counterclockwise from the positive x axis.

From the figure on the board we estimated that the y component of the 200 N weight is about -.95% of the weight, while the x component is about +30% of the weight.  According to our estimates we thus estimated

• wt_x = .30 * 200 N = 60 N
• wt_y = -.95 * 200 N = -190 N
• (ball-park estimated values).

Using the sine and cosine we can make these calculations more accurate:

• wt_x = 200 N * cos(280 deg)
• wt_y = 200 N * sin(280 deg)

`q011.  Using the sine and cosine what do you get for the x and y components of the weight?  How far off were our ballpark estimates?

****

We get

• wt_x = 200 N * cos(280 deg) = 200 N * .17 = 34 N
• wt_y = 200 N * sin(280 deg) = 200 N * (-.98) = -196 N.

&&&&

`q012.  What are the x and y components of the weight of a 40 kg mass on a 20 degree incline? 

Sketch and estimate, give your estimated percents, and the resulting x and y components of the weight.

Use sines and cosines to get the precise values of the components.

****

The weight is 40 kg * 9.8 m/s^2 = 400 N, approx..

If the figure is sketched in a manner analogous to the preceding, the weight vector makes angle 270 deg + 20 deg = 290 deg.  The components are therefore

• wt_x = 400 N * cos(290 deg) =  400 N * .34 = 140 N  and
• wt_y = 400 N * sin(290 deg) = .400 N * (-.94) = - 380 N (both final results very approximate)

&&&&

`q013.  Continuing the preceding, if the normal force is equal to the y component of the weight and the frictional force resisting motion is 15% of the normal force, then what is the net force on the block?  What therefore is its acceleration?

****

The y component of the weight is -380 N so the normal force will be + 380 N in the positive y direction.

The normal force is therefore 380 N.

15% of the normal force is 380 N * .15 = 57 N, and is directed opposite the motion.

Assuming the motion to be down the incline, then frictional force is in the negative x direction and the net force, which is is in the x direction, will be

• F_net = wt_x + f_frict = 140 N + (-57 N) = 90 N, approx.

The acceleration will be

• a = F_net / m = 90 N / (40 kg) = 2.25 m/s^2.

&&&&

Homework:

Your label for this assignment: 

ic_class_091021

Copy and paste this label into the form.

Report your results from today's class using the Submit Work Form.  Answer the questions posed above.

Work through and submit q_A_15 and q_A_16.  These qa's are concerned primarily with vectors, and you should find them straightforward.

URL's of qa's 10-19:

http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_10.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_11.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_12.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_13.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_14.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_15.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_16.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_17.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_18.htm
http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/ph1/ph1_qa_19.htm
 

Read through Chapter 2 of your text to familiarize yourself with the author's notation, and to familiarize yourself with his examples and methods of solution.  Just about everything you see should be familiar to you.  The author does a good job explaining a lot of the things we've seen in class, and you'll find the text to be a very useful reference.

You should take a good look at Introductory Problem Sets 3 and 4.  The Phy 121 test, probably to be assigned the week after next, will consist mostly of problems from these sets.  A number of these problems will also appear on the Phy 201 test.