CHAPTER 2 Giancoli: Physics



CHAPTER 2,3 Giancoli: Physics Study Guide Dr. Lee

Summary for FREE FALL:

Key Concepts and equations:

Take the upward direction to be positive. g means 9.8 m/s2.

Use the constant acceleration equation with a = - g.

v = vo - g t (1)

The height covered, y, is given by:

y = vo t - ½ g t2 (2)

These two equations is the key. They can be used to solve all problems

A third equation is helpful but not necessary:

v2 = vo2 - 2 g y

Solution of some problems.

(Note: capital X means multiplication)

35. We use a coordinate system with the origin at the top of the building and up positive.

(a) To find the time of fall, we have eqn (2) with y = - 380 and vo = 0.

- 380 = - ½ (9.80 m/s2)t2 , which gives t = 8.81 s.

(b) We find the velocity just before landing from

v = - g t

= - (9.80 m/s2)(8.81 s) = - 86.3 m/s.

(The negative sign means that velocity is downward.)

37. We use a coordinate system with the origin at the ground and up positive.

We use equations (1) and (2):

v = vo - g t (1)

y = vo t - ½ g t2 (2)

Let T be the time it took for the kangaroo to reach the top. From (2) we have:

2.7 = vo T - ½ 9.8 T2 (3)

We cannot find T yet because we do not know vo . To get that we use (1) knowing

that at time T, v = 0. From (1) we have

0 = vo - 9.8 T

vo = 9.8 T

Putting this into eqn (3) gives :

2.7 = 9.8 T2 - ½ 9.8 T2 = ½ 9.8 T2

Therefore T = 0.74 s

It takes the same time to go up and to go down. So the kangaroo is in the air for 1.5 s.

38. We use a coordinate system with the origin at the ground and up positive.

Again we use eqn (1) and (2)

v = vo - g t (1)

y = vo t - ½ g t2 (2)

When the ball reaches the top, its speed is 0 and t = 3.3 / 2 = 1.65 s.

Putting these into (1):

0 = vo - 9.8 X 1.65

Therefore, vo = 9.8 X 1.65 = 16.2 m/s

To find the height reached, use eqn (2):

Height reached = y = 16.2 X 1.65 - - ½ 9.8 X (1.65)2

= 12.9 m

[pic]

41. We use a coordinate system with the origin at 105 m and up positive.

When the parcel is dropped, what is its speed ?

It is NOT 0, it has the speed of the helicopter, + 5.5 m/s. This is the vo .

Again we only need eqn (2). At the time T when it reaches ground,

y = - 105

Therefore: - 105 = +5.5 T - ½ 9.8 T2

This is a quadratic equation for T2 .

We write it in the standard form: 4.9 T2 - 5.5 T – 105 = 0

Using the quadratic formula gives:

T = ( 5.5 +/- SQRT(5.52 + 4 X 4.9 X 105) /9.8

= - 4.1 s and 5.22 s

We take the positive answer. The time to reach ground is 5.22 s.

CHAPTER 3

Key Concepts and equations for section 1 to 4.

Scalers have magnitude only. Examples are mass, temperature.

Vectors have magnitude and direction. Examples are velocity, displacement, force.

Resolving a vector into its components.

If vector A makes an angle θ with the x-axis, we can describe the vector by its

x component of A = Ax = A cos θ

y component of B = Ay = A sin θ

Vector Addition ( A + B = C )

Graphical Method:

When vector A is added to vector B, we can visualize it as going along A first.

When we have reached the end of A, we follow along B. The resultant is a vector

that runs from the beginning of A to the end of B.

Algebraic Method:

Resolve A and B into their components.

The Resultant C has x component = Cx = Ax + Bx

The Resultant C has y component = Cy = Ay + By

Solution of some problems.

[pic]

5. Draw the three vectors in scale as above

The resultant is 31 m, 44° N of E.

11.

[pic]

(a) For the components we have

Rx = Ax + Bx + Cx

= 66.0 cos 28.0° – 40.0 cos 56.0° + 0 = 35.9;

Ry = Ay + By + Cy

= 66.0 sin 28.0° + 40.0 sin 56.0° – 46.8 = 17.3.

(b) We find the resultant from

R = (Rx2 + Ry2)1/2 = [(35.9)2 + (17.3)2]1/2 = 39.9;

tan θ = Ry/Rx = (17.3)/(35.9) = 0.483, which gives

θ = 25.8° above + x-axis.

12. For the components we have

[pic]

Rx = Ax – Cx

= 66.0 cos 28.0° – 0 = 58.3;

Ry = Ay –Cy

= 66.0 sin 28.0° – (– 46.8) = 77.8.

We find the resultant from

R = (Rx2 + Ry2)1/2 = [(58.3)2 + (77.8)2]1/2 = 97.2;

tan θ = Ry/Rx = (77.8)/(58.3) = 1.33, which gives

θ = 53.1° above + x-axis.

15. (a) For the components we have

[pic]

Rx = Cx – Ax – Bx

= 0 – 66.0 cos 28.0° – (– 40.0 cos 56.0°)

= – 35.9;

Ry = Cy – Ay – By

= – 46.8 – 66.0 sin 28.0° – 40.0 sin 56.0°

= – 111.0.

We find the resultant from

R = (Rx2 + Ry2)1/2 = [(– 35.9)2 + (– 111.0)2]1/2

= 117;

tan θ = Ry/Rx = (111.0)/(35.9) = 3.09,

which gives

θ = 72.1° below – x-axis.

(b) For the components we have

Rx = 2Ax – 3Bx + 2Cx

= 2(66.0 cos 28.0°) –

3(– 40.0 cos 56.0°) + 2(0)

= 183.8;

Ry = 2Ay – 3By + 2Cy

= 2(66.0 sin 28.0° ) –

3(40.0 sin 56.0°) + 2(– 46.8)

= – 131.2.

We find the resultant from

R = (Rx2 + Ry2)1/2 = [(183.8)2 + (– 131.2)2]1/2

= 226;

tan θ = Ry/Rx = (131.2)/(183.8) = 0.714,

which gives

θ = 35.5° below + x-axis.

20. We choose a coordinate system with the origin at the takeoff point, with x horizontal and y vertical, with the positive direction up. We find the height of the cliff from the vertical displacement:

y = v0yt + ½ ayt2;

y = 0 + ½ (-9.80 m/s2)(3.0 s)2 = - 44 m.

- 44 m means that it its 44 m beneath the takeoff point.

The horizontal motion will have constant velocity.

We find the distance from the base of the cliff from

x = v0xt;

x = (1.6 m/s)(3.0 s) = 4.8 m.

[pic]

24. We choose a coordinate system with the origin at the release point,

with x horizontal and y vertical, with the positive direction up.

We find the time of fall from the vertical displacement:

y = ½ ayt2;

-56 m  = ½ (-9.80 m/s2)t2,

which gives t = 3.38 s.

The horizontal motion will have constant velocity.

We find the initial speed from

x = v0xt;

45 m = v0(3.38 s),

which gives v0 = 13 m/s.

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