query asst 20

course Phy 201

LvnybǦcž֒nyassignment #020

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

020. `query 20

Physics I

11-18-2007

......!!!!!!!!...................................

00:03:51

Explain how we get the components of the resultant of two vectors from the components of the original vectors.

......!!!!!!!!...................................

RESPONSE -->

x component=L*cos theta

ycomponent=L*sin theta

x^2+y^2=z^2

Solve for z which is the hypotenouse

confidence assessment: 1

.................................................

......!!!!!!!!...................................

00:04:56

** If we add the x components of the original two vectors we get the x component of the resultant.

If we add the y components of the original two vectors we get the y component of the resultant. **

......!!!!!!!!...................................

RESPONSE -->

Misunderstood the question.

Add the x components of the two vectors to get the resulting x component.

Add the y components of the two vectors to get the resulting y component.

self critique assessment: 2

.................................................

......!!!!!!!!...................................

00:06:21

Explain how we get the components of a vector from its angle and magnitude.

......!!!!!!!!...................................

RESPONSE -->

x component= magnitude *cos angle

y component=magnitude *sin angle

x^2+y^2=z^2

Solve for z

confidence assessment: 2

.................................................

......!!!!!!!!...................................

00:06:37

** To get the y component we multiply the magnitude by the sine of the angle of the vector (as measured counterclockwise from the positive x axis).

To get the x component we multiply the magnitude by the cosine of the angle of the vector (as measured counterclockwise from the positive x axis). **

......!!!!!!!!...................................

RESPONSE -->

ok

self critique assessment: 3

.................................................

......!!!!!!!!...................................

00:10:54

prin phy, gen phy 7.02. Const frict force of 25 N on a 65 kg skiier for 20 sec; what is change in vel?

......!!!!!!!!...................................

RESPONSE -->

F=ma

65kg*9.8m/s^2=637kgm/s^2=637N

637N-25N=612N

612N/65kg=9.42m/s^2

aave=`dv/`dt

9.42m/s^2*20s=188.4m/s

confidence assessment: 2

.................................................

......!!!!!!!!...................................

00:14:31

If the direction of the velocity is taken to be positive, then the directio of the frictional force is negative. A constant frictional force of -25 N for 20 sec delivers an impulse of -25 N * 20 sec = -500 N sec in the direction opposite the velocity.

By the impulse-momentum theorem we have impulse = change in momentum, so the change in momentum must be -500 N sec.

The change in momentum is m * `dv, so we have

m `dv = impulse and

`dv = impulse / m = -500 N s / (65 kg) = -7.7 N s / kg = -7.7 kg m/s^2 * s / kg = -7.7 m/s.

......!!!!!!!!...................................

RESPONSE -->

Frictional force of -25N for 20s=-500Nsec=impulse

impulse=change in momentum

change in momentum=-500N

Change in momentum is m*`dv

`dv=impulse/m=-500Nsec/65kg=-7.7N sec/kg=-7.7m/s

Am confused as to when you should use the impulse momentum theory versus using the mass *acceleration.

self critique assessment: 2

Impulse and momentum are very useful concepts relating force, time interval and momentum without getting into the detail of the kinematics. They are time-saving concepts, they simplify our thinking in a wide range of situations, they provide us with another important tool and they lead to the law of conservation of momentum.

.................................................

......!!!!!!!!...................................

00:25:19

gen phy #7.12 23 g bullet 230 m/s 2-kg block emerges at 170 m/s speed of block

......!!!!!!!!...................................

RESPONSE -->

23g=.023kg

momentum 1= .023kg*230m/s=5.29kgm/s

momentum 2=2kg*170m/s=340kgm/s

confidence assessment: 0

.................................................

......!!!!!!!!...................................

00:27:43

**STUDENT SOLUTION: Momentum conservation gives us

m1 v1 + m2 v2 = m1 v1' + m2 v2' so if we let v2' = v, we have

(.023kg)(230m/s)+(2kg)(0m/s) = (.023kg)(170m/s)+(2kg)(v). Solving for v:

(5.29kg m/s)+0 = (3.91 kg m/s)+(2kg)(v)

.78kg m/s = 2kg * v

v = 1.38 kg m/s / (2 kg) = .69 m/s.

INSTRUCTOR COMMENT:

It's probably easier to solve for the variable v2 ':

Starting with m1 v1 + m2 v2 = m1 v1 ' + m2 v2 ' we add -m1 v1 ' to both sides to get

m1 v1 + m2 v2 - m1 v1 ' = m2 v2 ', then multiply both sides by 1 / m2 to get

v2 ` = (m1 v1 + m2 v2 - m1 v1 ' ) / m2.

Substituting for m1, v1, m2, v2 we will get the result you obtained.**

......!!!!!!!!...................................

RESPONSE -->

m1v1+m2v2=m1v1` +m2v2`

Should have used 0m/s for v2

.023kg*230m/s +2kg(0m/s)=.023kg*170m/s +2kg(v)

Solve for v

5.29kgm/s=3.91kgm/s + 2kg(v)

.78kgm/s=2kg*v

v=1.38kgm/s/2kg=.69m/s

self critique assessment: 2

.................................................

......!!!!!!!!...................................

00:27:52

**** Univ. 8.70 (8.68 10th edition). 8 g bullet into .992 kg block, compresses spring 15 cm. .75 N force compresses .25 cm. Vel of block just after impact, vel of bullet?

......!!!!!!!!...................................

RESPONSE -->

confidence assessment:

.................................................

......!!!!!!!!...................................

00:27:57

** The spring ideally obeys Hook's Law F = k x. It follows that k = .75 N / .25 cm = 3 N / cm.

At compression 15 cm the potential energy of the system is PE = .5 k x^2 = .5 * 3 N/cm * (15 cm)^2 = 337.5 N cm, or 3.375 N m = 3.375 Joules, which we round to three significant figures to get 3.38 J.

The KE of the 1 kg mass (block + bullet) just after impact is in the ideal case all converted to this PE so the velocity of the block was v such that .5 m v^2 = PE, or v = sqrt(2 PE / m) = sqrt(2 * 3.38 J / ( 1 kg) ) = 2.6 m/s, approx.

The momentum of the 1 kg mass was therefore 2.6 m/s * .992 kg = 2.6 kg m/s, approx., just after collision with the bullet. Just before collision the momentum of the block was zero so by conservation of momentum the momentum of the bullet was 2.6 kg m/s. So we have

mBullet * vBullet = 2.6 kg m/s and vBullet = 2.6 kg m/s / (.008 kg) = 330 m/s, approx. **

......!!!!!!!!...................................

RESPONSE -->

self critique assessment:

.................................................

&#

Your work looks very good. Let me know if you have any questions. &#