Query 19

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course PHY 121

7/23 1

019. `query 19

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Question: `qQuery class notes #20

Explain how we calculate the components of a vector given its magnitude and its angle

with the positive x axis.

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Your solution:

x = magnitude (cos) (degrees)

y = magnitude (sin) (degrees)

confidence rating #$&*:

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Given Solution:

`a** STUDENT RESPONSE:

x component of the vector = magnitude * cos of the angle

y component of the vector = magnitude * sin of the angle

To get the magnitude and angle from components:

angle = arctan( y component / x component ); if the x component is less than 0 than we

add 180 deg to the solution

To get the magnitude we take the `sqrt of ( x component^2 + y component^2) **

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Self-critique (if necessary): OK

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Self-critique rating: 3

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Question: `qExplain what we mean when we say that the effect of a force is completely

equivalent to the effect of two forces equal to its components.

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Your solution:

If you have 2 forces that are working on the same object, you add the x components

together and you add the y components together. Those sums are the x and y components

for the other force.

confidence rating #$&*:

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Given Solution:

`a** If one person pulls with the given force F in the given direction the effect is

identical to what would happen if two people pulled, one in the x direction with force

Fx and the other in the y direction with force Fy. **

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Self-critique (if necessary): OK

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Self-critique rating: 2

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Question: `qExplain how we can calculate the magnitude and direction of the velocity of

a projectile at a given point.

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Your solution:

If you have a point, then you have the x and y components. Using the Pythagorean

Theorum, you can calculate the magnitude with x^2 + y^2 = magnitude^2. Take the square

root of the answer to find the magnitude.

Using the same x and y, you can find the angle with tan-1 (y/x). If the x is negative,

you have to add 180 degrees to the angle you get for the actual angle. If the y

component is negative, you have to add 360 degrees to the angle you calculate for the

actual angle.

confidence rating #$&*:

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Given Solution:

`a** Using initial conditions and the equations of motion we can determine the x and y

velocities vx and vy at a given point, using the usual procedures for projectiles.

The magnitude of the velocity is sqrt(vx^2 + vy^2) and the angle with the pos x axis is

arctan(vy / vx), plus 180 deg if x is negative. **

STUDENT QUESTION

this says that there are the magnitude and the angle with respect to the positvie x

aixs, I am not quite clear on this ar ethey added

together?

INSTRUCTOR RESPONSE

If an object is thrown straight up in the air, its initial velocity is all in the

vertical direction. Its angle as measured from the horizontal x axis is 90 degrees. It

has no horizontal velocity; the horizontal component of its velocity is zero. In this

case our calculations would verify the obvious:

cos(90 deg) = 0, so the x component of the velocity is v_x = v cos(90 deg) = v * 0 = 0.

sin(90 deg) = 1, so the y component of the velocity is v_y = v sin(90 deg) = v * 1 = v.

If an object is thrown in the horizontal direction, its angle with the horizontal is 0

degrees. Its velocity is wholly in the horizontal direction. The vertical component of

its velocity is zero. Our calculations again verify this:

cos(0 deg) = 0, so the x component of this velocity is v_x = v cos(0 deg) = v * 0 = 0.

sin(0 deg) = 1, so the y component of this velocity is v_y = v sin(0 deg) = v * 1 = v.

Now if an object is thrown at some nonzero angle with horizontal, as it typically the

case, the magnitudes of its velocity components are less than the magnitude of its

velocity.

For example an object thrown at angle 45 degrees, halfway between the direction of the x

axis and that of the y axis, has equal x and y components. Our calculation verifies

this

cos(45 deg) = .71, approx., so the x component of this velocity is v_x = v cos(45 deg) =

v * .71 = .71 v.

sin(45 deg) = .71, so the y component of this velocity is v_y = v sin(45 deg) = v * .71

= .71 v.

An object thrown at 30 degrees, closer to the direction of the x axis that to that of

the y axis, has a velocity component in the x direction which is greater than that in

the y direction. Our calculation will verify this:

cos(30 deg) = .87, approx., so the x component of this velocity is v_x = v cos(30 deg) =

v * .87 = .87 v.

sin(30 deg) = .50, so the y component of this velocity is v_y = v sin(30 deg) = v * .50

= .50 v.

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Self-critique (if necessary): OK

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Self-critique rating: 3

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Question: `qExplain how we can calculate the initial velocities of a projectile in the

horizontal and vertical directions given the magnitude and direction of the initial

velocity.

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Your solution:

The horizontal would be found by solving for x: magnitude (cos) (degrees of angle)

The vertical would be found by solving for y: magnitude (sin) (degrees of angle)

confidence rating #$&*:

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Given Solution:

`a** Initial vel in the x direction is v cos(theta), where v and theta are the magnitude

and the angle with respect to the positive x axis.

Initial vel in the y direction is v sin(theta). **

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Self-critique (if necessary): OK

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Self-critique rating: 3

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Question: `qUniv. 8.63 (11th edition 8.58) (8.56 10th edition). 40 g, dropped from

2.00 m, rebounds to 1.60 m, .200 ms contact. Impulse? Ave. force?

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Your solution:

confidence rating #$&*:

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Given Solution:

`a** You have to find the momentum of the ball immediately before and immediately after

the encounter with the floor. This allows you to find change in momentum.

Using downward as positive direction throughout:

Dropped from 2 m the ball will attain velocity of about 6.3 m/s by the time it hits the

floor (v0=0, a = 9.8 m/s^2, `ds = 2 m, etc.).

It rebounds with a velocity v0 such that `ds = -1.6 m, a = 9.8 m/s^2, vf = 0. This

gives rebound velocity v0 = -5.6 m/s approx.

Change in velocity is -5.6 m/s - 6.3 m/s = -11.9 m/s, approx. So change in momentum is

about .04 kg * -11.9 m/s = -.48 kg m/s.

In .2 millliseconds of contact we have F `dt = `dp or F = `dp / `dt = -.48 kg m/s /

(.0002 s) = -2400 Newtons, approx. **

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