IB PHYSICS SL NORDEEN

Solutions to HW 21 Problems Giancoli pp. 64-65 #1-19 Odd 1.

The resultant vector displacement of the car is given by DR Dwest Dsouth- . The westward displacement is west

Dwest DR

Dsouth-

215 85cos 45o 275.1 km and the south displacement is

west

85sin 45o 60.1 km . The resultant displacement has a magnitude of

275.12 60.12 281.6 km

282 km . The direction is tan 1 60.1 275.1 12.3o 12o south of west .

3.

relationship V1 + X = V2 via the tail-to-tip method. Thus X V2 V1 .

5.

The vectors for the problem are drawn approximately to scale. The resultant has a length of 58 m and a direction 48o north of east. If calculations are done, the actual resultant should be 57.4 m at 47.5o north of east.

X

V2

Label the “INCORRECT” vector as vector X . Then Fig. 3-6 (c) illustrates the

V1

V3 VR V1 V2 V3

V2

V1 7.

(a)

See the accompanying diagram

(b) Vx 14.3cos 34.8o 11.7 units (c)

V V V 2 x

2 y

11.7

Vy 14.3sin 34.8o 8.16 units 2

8.16 14.3 units 2

V 34.8o

Vx

tan 1

9.

(a)

8.16 11.7

34.8o above the x axis

vnorth 735 km h cos 41.5o 550 km h

vwest 735 km h sin 41.5o 487 km h

(b) d north vnorth t 550 km h 3.00 h 1650 km

Vy

IB PHYSICS SL NORDEEN d west vwest t 487 km h 3.00 h 1460 km

11.

Ax 44.0 cos 28.0o 38.85 Cx 31.0 cos 270o 0.0

A C

x

AC

13.

C y 31.0sin 270o 31.0

38.85 0.0 38.85

38.85

2

Ay 44.0sin 28.0o 20.66

A C

51.66 64.6

20.66 31.0 51.66

tan 1

2

Ax 44.0 cos 28.0o 38.85

y

51.66

53.1o

38.85

Ay 44.0sin 28.0o 20.66

Bx 26.5 cos 56.0o 14.82

By 26.5sin 56.0 o 21.97

Cx 31.0 cos 270o 0.0

C y 31.0 sin 270 o 31.0

(a)

A B C

A B C

y

x

38.85 14.82 0.0 53.67

20.66 21.97 31.0 32.31

Note that since the x component is positive and the y component is negative, the vector is in the 4th quadrant. ABC

53.67

2

32.31 62.6

tan 1

2

32.31 53.67

31.0o below x axis

(b) A B C x 38.85 14.82 0.0 24.03

A B C

y

ABC

(c)

20.66 21.97 31.0 73.63

24.03

C A B

C A B

y

x

2

73.63 77.5 2

tan 1

73.63 24.03

71.9o

0.0 38.85 14.82 24.03

31.0 20.66 21.97 73.63

Note that since both components are negative, the vector is in the 3rd quadrant. CAB

24.03

2

73.63 77.5 2

tan 1

73.63 24.03

71.9o below x axis

IB PHYSICS SL NORDEEN Note that the answer to (c) is the exact opposite of the answer to (b). 15. The x component is negative and the y component is positive, since the summit is to the west of north. The angle measured counterclockwise from the positive x axis would be 122.4o. Thus the components are found to be x 4580 sin 32.4o 2454 m

y 4580 cos 32.4 o 3867 m

r 2450 m,3870 m,2450 m

r

2454

2

z 2450 m

4580 2450 5190 m 2

2

17. Choose downward to be the positive y direction. The origin will be at the point where the tiger leaps from the rock. In the horizontal direction, vx 0 3.5 m s and ax 0 . In the vertical direction, v y 0 0 , a y 9.80 m s 2 , y0 0 , and the final location y 6.5 m . The time for the tiger to reach the ground is found from applying Eq. 2-11b to the vertical motion.

y y0 v y 0t 12 a y t 2

6.5m 0 0 12 9.8 m s 2 t 2

t

2 6.5m 9.8 m s 2

1.15 sec

The horizontal displacement is calculated from the constant horizontal velocity. x vx t 3.5 m s 1.15 sec 4.0 m

2.5

19.

Apply the range formula from Example 3-8. 2

R

v02 sin 2 0 g

sin 2 0

Rg 2 0

v

1.5

2.0 m 9.8 m 2 6.8 m s

2 0 sin 1 0.4239

s2

0.4239

0 13o , 77 o

1

0.5

0 0

0.5

1

1.5

2

-0.5

There are two angles because each angle gives the same range. If one angle is 45o , then

45o is also a solution. The two paths are shown in the graph.

Solutions to HW 21 Problems Giancoli pp. 64-65 #1-19 Odd 1.

The resultant vector displacement of the car is given by DR Dwest Dsouth- . The westward displacement is west

Dwest DR

Dsouth-

215 85cos 45o 275.1 km and the south displacement is

west

85sin 45o 60.1 km . The resultant displacement has a magnitude of

275.12 60.12 281.6 km

282 km . The direction is tan 1 60.1 275.1 12.3o 12o south of west .

3.

relationship V1 + X = V2 via the tail-to-tip method. Thus X V2 V1 .

5.

The vectors for the problem are drawn approximately to scale. The resultant has a length of 58 m and a direction 48o north of east. If calculations are done, the actual resultant should be 57.4 m at 47.5o north of east.

X

V2

Label the “INCORRECT” vector as vector X . Then Fig. 3-6 (c) illustrates the

V1

V3 VR V1 V2 V3

V2

V1 7.

(a)

See the accompanying diagram

(b) Vx 14.3cos 34.8o 11.7 units (c)

V V V 2 x

2 y

11.7

Vy 14.3sin 34.8o 8.16 units 2

8.16 14.3 units 2

V 34.8o

Vx

tan 1

9.

(a)

8.16 11.7

34.8o above the x axis

vnorth 735 km h cos 41.5o 550 km h

vwest 735 km h sin 41.5o 487 km h

(b) d north vnorth t 550 km h 3.00 h 1650 km

Vy

IB PHYSICS SL NORDEEN d west vwest t 487 km h 3.00 h 1460 km

11.

Ax 44.0 cos 28.0o 38.85 Cx 31.0 cos 270o 0.0

A C

x

AC

13.

C y 31.0sin 270o 31.0

38.85 0.0 38.85

38.85

2

Ay 44.0sin 28.0o 20.66

A C

51.66 64.6

20.66 31.0 51.66

tan 1

2

Ax 44.0 cos 28.0o 38.85

y

51.66

53.1o

38.85

Ay 44.0sin 28.0o 20.66

Bx 26.5 cos 56.0o 14.82

By 26.5sin 56.0 o 21.97

Cx 31.0 cos 270o 0.0

C y 31.0 sin 270 o 31.0

(a)

A B C

A B C

y

x

38.85 14.82 0.0 53.67

20.66 21.97 31.0 32.31

Note that since the x component is positive and the y component is negative, the vector is in the 4th quadrant. ABC

53.67

2

32.31 62.6

tan 1

2

32.31 53.67

31.0o below x axis

(b) A B C x 38.85 14.82 0.0 24.03

A B C

y

ABC

(c)

20.66 21.97 31.0 73.63

24.03

C A B

C A B

y

x

2

73.63 77.5 2

tan 1

73.63 24.03

71.9o

0.0 38.85 14.82 24.03

31.0 20.66 21.97 73.63

Note that since both components are negative, the vector is in the 3rd quadrant. CAB

24.03

2

73.63 77.5 2

tan 1

73.63 24.03

71.9o below x axis

IB PHYSICS SL NORDEEN Note that the answer to (c) is the exact opposite of the answer to (b). 15. The x component is negative and the y component is positive, since the summit is to the west of north. The angle measured counterclockwise from the positive x axis would be 122.4o. Thus the components are found to be x 4580 sin 32.4o 2454 m

y 4580 cos 32.4 o 3867 m

r 2450 m,3870 m,2450 m

r

2454

2

z 2450 m

4580 2450 5190 m 2

2

17. Choose downward to be the positive y direction. The origin will be at the point where the tiger leaps from the rock. In the horizontal direction, vx 0 3.5 m s and ax 0 . In the vertical direction, v y 0 0 , a y 9.80 m s 2 , y0 0 , and the final location y 6.5 m . The time for the tiger to reach the ground is found from applying Eq. 2-11b to the vertical motion.

y y0 v y 0t 12 a y t 2

6.5m 0 0 12 9.8 m s 2 t 2

t

2 6.5m 9.8 m s 2

1.15 sec

The horizontal displacement is calculated from the constant horizontal velocity. x vx t 3.5 m s 1.15 sec 4.0 m

2.5

19.

Apply the range formula from Example 3-8. 2

R

v02 sin 2 0 g

sin 2 0

Rg 2 0

v

1.5

2.0 m 9.8 m 2 6.8 m s

2 0 sin 1 0.4239

s2

0.4239

0 13o , 77 o

1

0.5

0 0

0.5

1

1.5

2

-0.5

There are two angles because each angle gives the same range. If one angle is 45o , then

45o is also a solution. The two paths are shown in the graph.