111010

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course phy 201

10/17 12:45am

Class Physics I 111010Common errors from major quiz

Units calculations

What do you get if you divide 14 kg by 7 (m^2 / s^3)?

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2 kg m^2/s^3

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Find (5 m/s + 9 m/s) * 14 kg. Does this result have units of energy?

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(13m/s)14kg=182 kg m/s

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Find ( (3 m/s) / 5 s ) * 12 kg. What quantity has the same units as the result?

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(.6m/s^2)12kg=7.2kg m/s^2

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Are the units of vf^2 the same as the units of v_0 * `dt?

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Are the units of `ds the same as the units of v0 * `dt * a?

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Applying definitions

Define the following:

average rate of change

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the average change wrt clock time

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average velocity

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the average chang in velocity wrt clocktime

(vf+vo)/2

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average acceleration

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the average acceleration for the given time or distance

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@& You don't have all the words in any of these three definitions.

(v0 + vf) / 2 is not the definition of average rate of change of anything with respect to anything else; if acceleration is constant it is equal to the average velocity, but the average velocity is the quantity defined by the definition.*@

force

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work

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f*dS

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kinetic energy

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1/2MV^2

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potential energy

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net force

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the total combined forces on an object

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Identifying quantities

Give possible units for each of the quantities v0, vf, `ds, a, `dt, vAve, `dv.

Identify which of the quantities v0, vf, `ds, a, `dt, vAve, `dv are given in each of the following situations:

A car travels 40 meters down a hill, starting at velocity 5 m/s and increasing velocity by 12 m/s.

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A ball rolls 60 cm down an incline, starting from rest and changing velocity at a constant rate of 10 cm/s^2.

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vo 0

vf 34.64 cm/s

vave 17.32 cm/s

dv 34.64 cm/s

ds 60cm

dt 3.46s

a 10cm/s^2

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@& You were asked to identify which quantities are given. The only given quantites are `ds, v0 and a.

Clearly you knew this because you calculated the quantities correctly, so no problem, but be sure you know the meaning of 'given quantities'.*@

A runner starts at the 50-yard line. 8 seconds later, after having traveled 70 yards, he arrives at the goal line.

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vo 0

vf 17.5 y/s

vave 8.75 y/s

dv 17.5 y/s

ds 70 y

dt 8s

a 2.18y/s^2

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An arrow leaves the bow traveling at 60 meters / second. 3 second later it strikes its target, traveling at 55 meters / second.

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vo 60 m/s

vf 55 m/s

vave 57.5 m/s

dv -5 m/s

ds 172.5m

dt 3s

a -1.67m/s^2

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A swimmer pushes off the wall at 3 meters / second, and begins her stroke after traveling 5 meters, at which point her velocity is 2 meters / second.

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vo 3m/s

vf 2 m/s

vave 2.5 m/s

dv -1 m/s

ds 5m

dt 2s

a -.5m/s^2

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Two masses are suspended from a pulley. The lesser mass is given an upward velocity of 20 cm/s, and its acceleration is -6 m/s^2. When does it reach the point 50 meters above its original position?

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vo .2m/s

vf -7.74 m/s

vave -3.77 m/s

dv -7.55 m/s

ds 50m

dt -13.26s

a -.6m/s^2

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A ball is released from rest at a height of 3 meters above the floor, and falls freely to the floor.

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vo 0

vf 7.75 m/s

vave 3.87 m/s

dv 7.75 m/s

ds 3m

dt .78s

a 10m/s^2

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Projectiles

A 70 gram ball is shot from a spring gun resting on a level tabletop, clears the edge of the table and falls to the floor. Its velocity when leaving the gun is 5 meters / second.

Is that initial velocity in the horizontal or the vertical direction?

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both

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@& When leaving the gun the ball is not moving in the vertical direction, only the horizontal.*@

What is the nature of the net force on the ball after losing contact with the gun? Specify the source of this force.

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downward because of gravity

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What is the ball's acceleration after leaving the gun? Specify the magnitude and direction of this acceleration. Assume air resistance to be negligible.

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vertical and 9.8m/s^2 because of the acceleration of gravity

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Is the acceleration of the ball in the same direction as its initial velocity?

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Is the acceleration of the ball in the same direction as its horizontal displacement?

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Is the acceleration of the ball in the same direction as its vertical displacement?

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yes

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Knowing equations

List the four equations of uniformly accelerated motion, and state which two come directly from the v0, vf, `dt trapezoid.

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*** ds=(vo+vf)/s)dt****

***vf=vo+adt*****

ds=vodt+1/2adt^2

vf^2=vo^2+2ads

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List the four equations of uniformly accelerated motion, and state which two come directly from the definitions of average velocity and average acceleration.

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*** ds=(vo+vf)/s)dt****

vf=vo+adt

****ds=vodt+1/2adt^2****

vf^2=vo^2+2ads

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@& The second equation results from the definition of acceleration.

The third and fourth come from combining the first two in different ways.*@

Solving equations

Solve the first equation of motion for vf.

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dt=((vo+vf)/2)dt=2(ds/dt)-vo=vf

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@& dt=((vo+vf)/2)dt=2(ds/dt)-vo=vf

says, among other things, that

`dt = vf.

This is clearly not so.

Solving

dt=((vo+vf)/2)dt

for vf you do get

2(ds/dt)-vo=vf.

Don't use equal signs for anything but equality. To do otherwise is to invite unnecessary confusion.*@

Solve the second equation of motion for a.

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vf=vo+adt=(vf-vo0/dt=a

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@& vf = a is not so.

See my previous note.

Typo understood, you do have a correct solution for a.*@

Solve the third equation of motion for a.

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ds=vodt+1/2adt^2

sqrt(dt-vodt)/(1/2dt)=a

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@& a is not squared, `dt is.

a = (`ds - v0 `dt) * 2 / `dt^2,

which can be put into a more standard order as

a = 2 ( `ds - v0 `dt) / `dt^2.*@

Solve the fourth equation of motion for `ds.

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vf^2=vo^2+2ads

(vf^2-vo^2)/2A=ds

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Force and work\energy

What is the speed of a 60 gram arrow whose kinetic energy is 3 million gram cm / sec^2?

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3000000g cm/s^2=1/2(60g)v^2

100000cm/s^2=v^2

316.23 cm/s

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If a 60 kg runner speeds up from 3 m/s to 5 m/s, how much work is done by the net force acting on her?

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60kg(2m/s^2)=120kg m/s^2

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Masses of 5 kg and 6 kg are suspended from a pulley. How much force does gravity exert on each, and what is the net force on this 11-kg system?

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5kg+6kg-6kg-5kg=0

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@& 5 kg and 6 kg are not forces. Those are masses.

If gravity is the only force present, then the two opposing unequal forces do not add up to zero.*@

How much work does gravity do on a 15 kg block as it drops from a height of 140 cm to a height of 20 cm?

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w=(15kg(9.8m/s^2)(140cm-20cm)

147kgm/s^2(1.2m)

176.4 kg m^2/s^2

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How much work does gravity do on the same block if it rises from 20 cm to a height of 80 cm?

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w=(15kg(9.8m/s^2)(20cm-80cm)

147kgm/s^2(-.6m)

-88.2 kg m^2/s^2

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To load a spring gun I exert an average force of 50 Newtons through a distance of 20 cm. How much work do I do?

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50n(20cm)

100n cm

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A 70 gram ball is shot from the spring gun in the preceding question. How fast will that ball be traveling as it leaves the gun, if the energy imparted to it by the spring is equal to the work I did to compress the spring?

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100n/m=1/2(70)v^2

2.8n/cm)/g=v^2

1.69=v

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111010

@& You were asked to identify which quantities are given. The only given quantites are `ds, v0 and a.

Clearly you knew this because you calculated the quantities correctly, so no problem, but be sure you know the meaning of 'given quantities'.*@

@& When leaving the gun the ball is not moving in the vertical direction, only the horizontal.*@

@& The second equation results from the definition of acceleration.

The third and fourth come from combining the first two in different ways.*@

@& dt=((vo+vf)/2)dt=2(ds/dt)-vo=vf

says, among other things, that

`dt = vf.

This is clearly not so.

Solving

dt=((vo+vf)/2)dt

for vf you do get

2(ds/dt)-vo=vf.

Don't use equal signs for anything but equality. To do otherwise is to invite unnecessary confusion.*@

@& vf = a is not so.

See my previous note.

Typo understood, you do have a correct solution for a.*@

@& a is not squared, `dt is.

a = (`ds - v0 `dt) * 2 / `dt^2,

which can be put into a more standard order as

a = 2 ( `ds - v0 `dt) / `dt^2.*@

@& 5 kg and 6 kg are not forces. Those are masses.

If gravity is the only force present, then the two opposing unequal forces do not add up to zero.*@

@& 100 N cm doesn't change into 100 n / m.

Except for units your calculation is correct, but it's not clear what the units of v would be.*@

@& Very good overall, but you do have a few errors so check my notes.*@

111010

@& You were asked to identify which quantities are given. The only given quantites are `ds, v0 and a. Clearly you knew this because you calculated the quantities correctly, so no problem, but be sure you know the meaning of 'given quantities'.*@

@& When leaving the gun the ball is not moving in the vertical direction, only the horizontal.*@

@& The second equation results from the definition of acceleration. The third and fourth come from combining the first two in different ways.*@

@& dt=((vo+vf)/2)dt=2(ds/dt)-vo=vf says, among other things, that `dt = vf. This is clearly not so. Solving dt=((vo+vf)/2)dt for vf you do get 2(ds/dt)-vo=vf. Don't use equal signs for anything but equality. To do otherwise is to invite unnecessary confusion.*@

@& vf = a is not so. See my previous note. Typo understood, you do have a correct solution for a.*@

@& a is not squared, `dt is. a = (`ds - v0 `dt) * 2 / `dt^2, which can be put into a more standard order as a = 2 ( `ds - v0 `dt) / `dt^2.*@

@& 5 kg and 6 kg are not forces. Those are masses. If gravity is the only force present, then the two opposing unequal forces do not add up to zero.*@

@& 100 N cm doesn't change into 100 n / m. Except for units your calculation is correct, but it's not clear what the units of v would be.*@

@& Very good overall, but you do have a few errors so check my notes.*@

@& You were asked to identify which quantities are given. The only given quantites are `ds, v0 and a.

Clearly you knew this because you calculated the quantities correctly, so no problem, but be sure you know the meaning of 'given quantities'.*@

@& The second equation results from the definition of acceleration.

The third and fourth come from combining the first two in different ways.*@

@& dt=((vo+vf)/2)dt=2(ds/dt)-vo=vf

says, among other things, that

`dt = vf.

This is clearly not so.

Solving

dt=((vo+vf)/2)dt

for vf you do get

2(ds/dt)-vo=vf.

Don't use equal signs for anything but equality. To do otherwise is to invite unnecessary confusion.*@

@& vf = a is not so.

See my previous note.

Typo understood, you do have a correct solution for a.*@

@& a is not squared, `dt is.

a = (`ds - v0 `dt) * 2 / `dt^2,

which can be put into a more standard order as

a = 2 ( `ds - v0 `dt) / `dt^2.*@

@& 5 kg and 6 kg are not forces. Those are masses.

If gravity is the only force present, then the two opposing unequal forces do not add up to zero.*@

@& 100 N cm doesn't change into 100 n / m.

Except for units your calculation is correct, but it's not clear what the units of v would be.*@