Asst 20 Query

course Phy 201

??w????????assignment #020

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020. `query 20

Physics I

03-27-2007

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22:12:45

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

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RESPONSE -->

Add the x components and add the y components.

Then use this value in the equation hypotenuse = sqrt (x^2 + y^2)

confidence assessment: 2

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22:12:53

** 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 -->

self critique assessment: 3

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assignment #019

019. `query 19

Physics I

03-27-2007

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assignment #020

020. `query 20

Physics I

03-27-2007

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22:21:13

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

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RESPONSE -->

Add the x components. Add the y components. Use the equation: hypotenuse = sqrt (x^2 + y^2)

confidence assessment: 2

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22:21:19

** 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 -->

self critique assessment: 3

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22:23:22

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

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RESPONSE -->

x = hypotenuse * cos (theta)

y = hypotenuse * sin (theta)

confidence assessment: 2

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21:28:23

** 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 -->

self critique assessment: 3

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21:52:33

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 -->

Fnet = `dp / `dt

-25 N =`dp / 20s

`dp = 500

p=mv

500 = 65 kg * v

v = 7.7 m/s/s

confidence assessment: 0

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21:54:06

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 -->

I orginially had the negative signs in place but removed them thinking I did something wrong. Also, I am wondering why we we only consider the frictional force and not the force in the direction the skier is moving.

self critique assessment: 2

If the skier is on a level surface then (provided we choose to neglect air resistance, as we do here) the only forces acting are the gravitational force, the normal force and the frictional force.

The gravitational force is vertical, with no component in the horizontal direction. The normal force exerted by a horizontal surface is vertical (the normal force exerted by any surface is perpendicular to that surface), so it has no component in the horizontal direction.

Since motion is in the horizontal direction, the only force with a component in that direction is the frictional force. So when analyzing motion in the horizontal direction, this is the only force considered.

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22:08:38

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 -->

m1v1 + m2v2 = m3v3 + m4v4

(.023 kg * 230 m/s) + (2.0 kg * 0) = (.023 kg * 170 m/s) + (2.0 kg * v)

5.3 kg m/s + 0 = 3.9 kg m/s + 2.0 g v

subtract 3.9 kg m/s from both sides

divide both sides by 2.0 g

v = .7 m/s

confidence assessment: 1

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22:08:52

**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 -->

self critique assessment: 3

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Asst 20 Query

Good work. See my answer to your question about the skier. Let me know if you have additional questions.