endassign20

course Phy 231

ؑyMɱ˯assignment #020

020. `query 20

Physics I

02-26-2009

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10:41:31

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

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RESPONSE -->

We have to add the x components and then the y components of the original vectors and the result will be our single resultant vector.

confidence assessment: 2

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10:41:40

** 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. **

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RESPONSE -->

OK

self critique assessment: 3

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10:42:24

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

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RESPONSE -->

Fx = magnitude*(angle).

Fy = magnitude*(angle).

confidence assessment: 2

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10:43:25

** 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). **

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RESPONSE -->

Oh I forgot to put in that for the x component we use cos(angle) not (angle) and sin(angle) for the y component.

self critique assessment: 2

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10:43:35

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?

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RESPONSE -->

OK

confidence assessment: 3

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10:43:41

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.

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RESPONSE -->

OK

self critique assessment: 3

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10:43:46

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

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RESPONSE -->

OK

confidence assessment: 3

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10:43:53

**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.**

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RESPONSE -->

OK

self critique assessment: 3

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11:11:11

**** 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?

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RESPONSE -->

Force of bullet is: .015.m/.00025m *.750N = 45N. F=m*a. a=-45N/1kg = -45m/s/s. vf^2 = v0^2 + 2 a *ds. v0^2 = vf^2-2ads = 0-2*(-45m/s/s)(.015m) =(a)3.4m/s/s.

(b) F=m*a, a=45N/.008kg = 5625m/s/s. vf=0m/s. ds=.015m. v0^2=vf^2-2ads = 0-2*(5625m/s/s)(.015m)=169m/s.

confidence assessment: 2

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11:15:08

** 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. **

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RESPONSE -->

k=.75N/.25cm = 3N/cm.

PE = .5kx^2 = .5*3N/cm*(15cm)^2 = 337.5Ncm. = 3.38J. v=sqrt(2*3.38J/1kg) = 2.6m/s.

2.6m/s * .992kg = 2.6kgm/s.

2.6kgm/s / .008kg = 330m/s.

self critique assessment: 2

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