what therefore are the magnitude and angle of the resultant?

good, but impulse = F_net * `dt = 500 kg m/s; better to make an independent determination to confirm the impulse-momentum theorem than to assume it

The description was good, but I had to read too many words to find the information and can't readily spot that information later when I might want to confirm your analysis.

It would be easy to correct that by adding a concise synopsis:

length 15 cm

pullback 8 cm

equilibrium to floor 90 cm

landing positions (cm from equilibrium line):

early release: 18, 22, 19

middle release: 38, 40, 44

late release: 33, 30

Estimates of the angle with vertical at each release point would also be good.

To find F_net multiply the mass by 5 cm/s^2; the coin won't appreciably affect the mass of the system.

`ds isn't given; it needs to be expressed in terms of the given quantities x0, v0, a, `dt, as appropriate to the situation.

Very good. See my notes.

&#Check your solutions and your reasoning against the solutions and reasoning given in the appended document, and please submit any specific questions you might still have.

Class 091028

STUDENT QUESTION

Could you tell me how to solve for the 'dPE, and could you give it to me in

INSTRUCTOR

The answer to your question

Any time you think about

• `dPE for a system is equal and opposite to the work done
  by the conservative force on the system for the interval in question.

This

• Work is the product of force and displacement in the
  direction of the force.

The definition of work

So if you know the

• Identify the displacement `ds for the interval and
  multiply it by the conservative force. 

• Be careful about the directions of the force and the
  displacement.  The displacement you use is either in the same direction as
  the force (in which case the force and displacement have the same sign and
  the work is positive), or opposite the direction of the force (in which case
  the force and displacement have opposite signs and the work is negative).`

You can also determine `dPE

• The work-energy theorem tells us that `dW_NC_ON = `dPE +
  `dKE.

• It follows immediately that `dPE = `dW_NC_ON - `dKE.

`q001.  ** One-step-at-a-time questions:

• If the velocity of a 5-kg object changes from +3 m/s
  to -2 m/s, what is the change in its velocity, and what is the change in its
  momentum?

****

Initial velocity is +3 m/s, final velocity is -2 m/s.

Change in velocity is

• `dv = vf - v0 = -2 cm/s - 3 m/s
  = -5 m/s.

Initial momentum is 5 kg * 3 m/s = 15 m/s; final

momentum is 5 kg * (-2 m/s) = -10 m/s.

Change in momentum is

• `dp = p_f - p_0 = -10 kg m/s -
  15 kg m/s = -25 kg m/s.

&&&&

• A 12-kg object experiences a velocity change of +2 m/s
  during a time interval of .01 second.  What average force was exerted on it
  during that interval?

****

The impulse-momentum theorem applies to a force

exerted for a given time interval:

F_net `dt = `d(m v); when m is constant this can

be written

F_net `dt = m `dv.

Thus

• F_net = m `dv / `dt = 12 kg * 2 m/s / (.01 s) = 2400
  kg m/s^2 = 2400 N.

 

&&&&

• What is the resultant of two displacement vectors, the
  first having magnitude 8 meters directed at angle 53 degrees, the second
  with x component -3 meters and y component -4 meters?
  ****

  The magnitude of the first displacement vector is 8

  meters, the angle is 53 deg, so its components are

  • x comp of 1st vector = 8 m * cos(53 deg) = 8 m * .6 =
    4.8 m
  • y comp of 1st vector = 8 m * sin(53 deg) = 8 m * .8 =
    6.4 m

  To find the resultant vector R we add the two vectors. 

  To do this we add their components, obtaining

• R_x = 4.8 m + (-3 m) = 1.8 m and
• R_y = 6.4 m + (-4 m) = 2.4 m.

The magnitude of the resultant vector is found using

the Pythagorean Theorem:

• magnitude = sqrt( (R_x)^2 + (R_y)^2 ) = sqrt( (1.8
  m)^2 + (2.4 m)^2 ) = sqrt ( 3.24 m^2 + 5.76 m^2) = sqrt(9 m^2) = 3 m.

The angle of the resultant is

• theta = arcTan(R_y / R_x) = arcTan ( 2.4 m / (1.8 m) )
  = arcTan(1.33) = 53 deg.

&&&&

• Two objects collide, one of mass 5 kg and the other of
  mass 12 kg.  The momentum of the 5 kg object changes by -40 kg m/s.  What is
  the momentum change of the 12 kg object, and what is the change in its
  velocity?

****

The forces exerted on the particles by one another are

equal and opposite, and the act for the same time interval.  So the

values of the impulse F `dt will be equal and opposite.

Since F `dt = `d(m v), the changes in momentum will be

equal and opposite.

The momentum change of the first object is -40 kg m/s,

so the momentum change of the second is +40 kg m/s.

The mass of the second object is 12 kg.  If `dv

is the change in its velocity we have

• m `dv = `dp so that
• `dv = `dp / m = +40 kg m/s / (12 kg) = 3.33 m/s.

&&&&

• What is the direction of a displacement vector whose x
  and y components are +5 m and - 4 m?  What are the components of a force
  vector of magnitude 80 N in the same direction?

****

The direction of the vector is

• theta = arcTan(-4 m / (5 m) ) = arcTan(-.8) = -39 deg,
  approx..

Measured counterclockwise from the positive x axis,

this is 360 deg - 39 deg = 341 deg.

A force vector of magnitude 80 N in the same direction

has the same angle, so the components of this force vector are

• F_x = 80 N cos(341 deg) = 80 N * .78 = 62 N, approx.

• F_y = 80 N sin(341 deg) = 80 N * .62 = 50 N, approx..

&&&&

• A rubber band chain exerts force 4 N when its length
  is 50 cm, and 12 N when its length is 80 cm.  As the chain is extended from
  the first length to the second, what average force is exerted by the rubber
  band (assuming that the force vs. length is linear)?  How much work is done
  by the chain during this interval?

****

The average force, assuming linearity of F vs. length,

is

• F_ave = (4 N + 12 N ) = 8 N.

As the chain is extended, the ends of the rubber band

must be moved in the direction opposite the tension force, so the work done

by the chain is negative.  The force is exerted through a displacement

of 80 cm - 50 cm = 30 cm, and the displacement is along the line of the

force. 

Thus the work done by the rubber band is

• `dW_by = - (8 N * 30 cm) = -240 N * cm = -240 N * (.01
  m) = -2.4 N * m = -2.4 Joules.

&&&&

• A 200 kg mass is subjected to a net force of 1000
  Newtons for half a second.  What is its acceleration?  What therefore is its
  change in velocity?  What therefore is the change in its momentum?  What is
  the impulse of the force on this interval?

****

The acceleration is

• a = F_net / m = 1000 N / (200 kg) = 5 m/s^2.

The change in velocity is

• `dv = a `dt = 5 m/s^2 * .5 s = 2.5 m/s.

The change in momentum is

• `dp = m `dv = 200 kg * 2.5 m/s = 500 kg m/s.

The impulse of the force is

• F_net `dt = 1000 N * .5 s = 500 N * s = 500 kg m
  / s^2 * s = 500 kg m/s.

The impulse is seen to be equal to the change in the

momentum.

&&&&

`q002.  Give your data for the

Attempt to organize your data in such a

Include a description how you conducted

****

You should have received individual feedback on your

data submission.

&&&&

`q003.  How much force would it take to accelerate an

• What is the total weight of these two masses? 
• If the addition of a single dollar coin to one side of
  the system results in an acceleration of 5 cm/s^2, then what is the mass of
  the coin?  What is the weight of the coin? 
• How many coins would it therefore take to match the
  mass of the system? 
• (Assume that the friction in the pulley is
  negligible).

****

The system has a total mass of 100 kg, so this would

require a force of

• F_net = 100 kg * 9.8 m/s^2 = 980 N.

The total weight of the system is therefore 980 N.

When in balance, the system experiences net force zero

in its direction of motion.

When the coin is added, the net force on the system is

• F_net_coin = 100 kg * .05 m/s^2 = 5 N.

This force comes from the gravitational force on the

coin, so we have

• wt_coin = 5 N.

The mass of the coin is its weight divided by the

acceleration of gravity:

• m = wt / g = 5 N / (9.8 m/s^2) = .5 kg, approx..

That's a pretty massive dollar coin.

It's worth noting that this is .5% the mass of the

system, and that the 5 m/s^2 acceleration is .5% the acceleration of

gravity.  This connection is worth thinking about.

&&&&

`q004.  Three rubber bands are attached to a small ring at the origin of an

****

The lengths of the three rubber bands are respectively

L_1 = sqrt( (8 cm)^2 + (6 cm)^2 ) = 10 cm

L_2 = sqrt( (-10 cm)^2 + (0 cm)^2 ) = 10 cm and

L_3 = sqrt( (0 cm)^s + (-12 cm)^2) = 12 cm.

The lengths could be used with force vs. length graphs

for the three rubber bands, to find the force exerted by each.  This

won't be necessary here, since the forces are simply given.  However,

as we say in class, this is what we would ordinarily do.

The three angles are

theta_1 = arcTan(6 cm / (8 cm) ) = arcTan(.75) =

37 degrees.

theta_2 = 180 deg (this vector is clearly directed

along the negative x axis) and

theta_3 = 270 deg (this vector being clearly

directed along the negative y axis).

&&&&

Using a graph of force vs. length for each rubber band, we find that they

F1_x =

L1 cos(theta_1) = 5.0 N cos(37 deg) = 4.0 N

F1_y =

L1 sin(theta) = 5.0 N sin(37 deg) = 3.0 N.

F2_x =

3.8 N cos(180 deg) = -3.8 N

F2_y =

3.8 N sin(180

deg) = 0 N

F3_x = 3.2 N cos(270 deg) = 0 N

F3_y = 3.2 N sin(270 deg) = -3.2 N.

F_net_x

= 4.0 N + (-3.8 N) + 0 N = 0.2 N

F_net_y

= -3.0 N + 0 N + (-3.2 N) = -0.2 N.

F_net =

sqrt( (F_net_x)^2 + F_net_y)^2 ) = sqrt( (0.2 N)^2 + (-0.2 N)^2 ) = sqrt(.08

N^2) = 0.28 N, approx.

theta =

arcTan(-0.2 N / (0.2 N)) = arcTan(-1) = -45 deg, which when measured

counterclockwise from the positive x axis is

theta =

360 deg - 45 deg = 315 deg.

****

&&&&

`q005.  A ball starts rolling along the x axis at point x =

• By how much does the velocity of the ball change? 

• What is the ball's displacement during the interval? 

• What therefore is its final position along the x axis?

****

During the time interval the ball's velocity changes

by

• `dv = a `dt,

ending up with velocity

• vf = v0 + `dv = v0 + a `dt.

Its average velocity, assuming linear v vs. t (i.e.,

constant acceleration) is

• vAve = (vf + v0) / 2 = (v0 + a `dt + v0) / 2 = (2
  v0 + a `dt) / 2 = v0 + 1/2 a `dt.

Its displacement is therefore

• `ds = vAve `dt = (v0 + 1/2 a `dt) * `dt = v0 `dt
  + 1/2 a `dt^2.

Its final position along the x axis is therefore

• x_f = x_0 + `ds = x_0 + v0 `dt + 1/2 a `dt^2.

Our solutions are therefore

vf = v0 + a `dt

`ds = v0 `dt + 1/2 a `dt^2 and

x_f = x0 + v0 `dt + 1/2 a `dt^2.

You should recognize these expressions as being

similar to expressions given in your text.  If we assume that x0 and v0

are position and velocity at clock time t_0 = 0, then the time interval up

to t = t_f is just `dt = t_f - t_0 = t_f = 0 = t_f.  Our equations

would then become

vf = v0 + a * t_f

`ds = v0 * t_f^2 + 1/2 a * t_f^2 and

x_f = x0 + v0 * t_f + 1/2 a t_f^2.

If we agree that vf, x_f and t_f can just be

represented by v, x and t, our equations are

v = v0 + a t

`ds = v0 t + 1/2 a t^2 and

x = x0 + v0 t + 1/2 a t^2.

These equations specify the velocity v and position x

at clock time t. 

v and x can now be understood as functions of t.

The authors of your text present the four equations of

uniformly accelerated motion as

v = v0 + a t

x = x0 + v0 t + 1/2 a t^2

(these two are identical to those given above)

vAve = (v + v0) / 2 (the familiar statement that

avererage velocity is the average of initial and final velocity,

assuming acceleration to be constant) and

v^2 = v0^2 + 2 a ( x - x0 ).

The last is identical to our equation vf^2 = v0^2 + 2

a `ds, except that `ds is expressed as x - x0.

The author's equations have the advantage that v and x

are now expressed as functions of t.

They have the disadvantage of having six symbols (v0,

v, x0, x, a, t) rather than our five (`ds, v0, vf, a, `dt), and we can no

longer say that 'given three of the five quantities, we can find the other

two'.

&&&&