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Physics 1101: Vectors Review Solutions

1. Convert the following:
   (a) 3.5 km² to m²
   (b) 56 g/cm³ to kg/m³
   (c) 70 km/h to m/s
   (d) 15 m/s to km/h
   

   (a) 3.5 km² = 3.5 km × km = 3.5 km × km × (1000 m / 1km) × (1000 m / 1km) = 3.5 × 10⁶ m²

   (b) 56 g/cm³ = 56 g/(cm × cm × cm) = 56 g/(cm × cm × cm) × (1 kg / 1000 g) × (100 cm / 1 m) × (100 cm / 1 m) × (100 cm / 1 m) = 56 × 10³ kg/m³

   (c) 70 km/h = 72 km/ h × (1 kg / 1000 g) × (1 km / 1000 m) × (1 h / 3600 s) = 19.44 m/s

   (d) 15 m/s = 15 m/s × (1 km / 1000 m) × (3600 s / 1 h) = 54 km/h

   

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2. Write vectors equations for each diagram below.
   

   (a)		(b)
   (c)		(d)
   (e)		(f)

   (a)		(b)
   (c)		(d)
   (e)		(f)

   
   

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3. The following diagram shows a variety of displacement vectors. Express each vector in component (ij) notation.
   

   (i)		(ii)
   (iii)		(iv)
   (v)		(vi)
   (vii)		(viii)
   (ix)

   Note that a vector such as (i) may be written as A = i7 + j3 when typed, as it is easier to produce since arrow and hat symbols are not common, or as  in math class.

   

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4. Neatly sketch the x and y components on the graphs of the vectors shown below. Indicate the sign of the x and y components of the vectors shown below. Express the vectors in component (i, j) form.

   

   sign of x comp =	+	sign of x comp =	–	sign of x comp =	–
   sign of y comp =	+	sign of y comp =	+	sign of y comp =	+

   (a)	A	= i[+9.0sin(49°)] + j[+9.0cos(49°)]
   		= i[6.7924] + j[5.9045]
   (b)	B	= i[–4.3sin(61° )] + j[+4.3cos(61° )]
   		= i[–3.7609] + j[2.0847]
   (c)	C	= i[–10cos(34° )] + j[+10sin(34° )]
   		= i[–8.2904] + j[5.5919]

   

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5. Use the parallelogram method to sketch in the resultant vector which has the components shown in the diagrams below. In each case, write the vector in component (i, j) form. In each case write the vector in standard form.



A	= 10i - 8j	B	= 6i + 3j	C	= -12i - 4j
A	= [(10.0)2+(8.0)2]½	B	= [(6.0)2+(3.0)2]½	C	= [(12.0)2+(4.0)2]½
	= 12.8062		= 6.7082		= 12.6491
θ	= arctan(8/10)	θ	= arctan(3/6)	θ	= arctan (4/12)
	= 38.66°		= 26.57°		= 18.43°
A	= (12.81, 38.7° b.h.)	B	= (6.71, 26.6° a.h.)	C	= (12.65, 198.43° a.h.)



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6. Convert the following vectors to component form. Include sketches.

   (a)	A = 7.50 m/s2 at 23° a.h.	(b)	B = 55 m at 47° b.h.
   (c)	C = 15.5 m/s at 155° a.h.	(d)	D = 42 m at 35° S of E

   (a)				(b)
   	A	= i[+7.5cos(23°)] + j[+7.5sin(23°)]			B		= i[+55cos(47°)] + j[–55sin(47°)]
   		= i[6.9038] + j[2.9305]					= i[35.5099] – j[40.2245]
   (c)				(d)
   	C	= i[+15.5cos(155°)] + j[+15.5sin(155°)]			D	= i[+42cos(35°)] + j[–42sin(35°)]
   		= i[–14.0478] + j[6.5506]				= i[34.4044] – j[24.0902]

   

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7. A person walks 57.0 m at 47.0° north of east, turns and walks 72.0 m at 15.0° south of east, and then turns and walks 24.0 m 30.0° west of north.
   (a) How far and at what angle is the person's final position from his/her initial position?
   (b) In what direction would the person have to head to return to his/her initial position?

   (a) For vector problems, we first draw a neat sketch of the vectors and the vector operation of interest. Here we are adding three vectors.

   

   Then to solve the problem numerically, we break the vectors into their components and add them to get the components of vector D. Then we convert to polar coordinate form.

   A = i57cos(47°) + j57sin(47°) = i38.8739 + j41.6872 .

   B = i72cos(15°) − j–72sin(15°) = i69.5467 − j18.6350 .

   C = −i24cos(30°) + j24cos(30° ) = −i12 + j20.7846 .

   The vector relationship is D = A + B + C. The result is

   D = i96.4206 + j43.8368.

   Using Pythagoras’ Theorem, D = [(96.4206)² + (43.8368)²]^½ = 105.92 m. The angle θ = tan^–1(|D_y/D_x|) = tan^–1(43.8368/96.4206) = 24.45° . The components of D are in the NE quadrant, thus the person's displacement is 106 m at 24.4° north of east.

   (b) To return, the person would have to travel in the direction opposite to his/her displacement, i.e. 106 m at 24.4° south of west.

   

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8. Momentum is an important vector quantity which we will encounter later in this course. Momentum is a product of mass and velocity (p = mv). A car of mass 1200 kg and velocity 22.0 m/s at 30.0° south of west is hit by a second car of mass 1450 kg travelling at 18.0 m/s at 55.0° west of north.
   (a) Find each car's momentum in standard form.
   (b) Find the magnitude and direction of the total momentum of the two cars.

   (a) The magnitude of the momenta of the cars are:

   p₁ = m₁v₁ = (1200 kg)(22 m/s) = 2.64 × 10⁴ kg-m/s;

   p₂ = m₂v₂ = (1450 kg)(18 m/s) = 2.61 × 10⁴ kg-m/s.

   Therefore

   p₁ = 2.64 × 10⁴ kg-m/s at 30.0° S of W

   p₂ = 2.61 × 10⁴ kg-m/s at 55.0° W of N

   (b) For vector problems, we first draw a neat sketch of the vectors and the vector operation of interest. Here we are adding two vectors.

   

   Then to solve the problem numerically, we break the vectors into their components and add them to get the components of vector P_net. Then we convert to polar coordinate form.

   p₁ = −i(2.64×10⁴)cos(30°) + −j(2.64×10⁴)sin(30°) = −i(2.2863 × 10⁴) − j(1.3200 × 10⁴) .

   p₂ = −i(2.61×10⁴)sin(55°) + j(2.61×10⁴)cos(55°) = −i(2.1380 × 10⁴) + j(1.4970 × 10⁴)

   The resultant P_net = p₁ + p₂ is

   P_net = −i(4.4243 × 10⁴) + j(0.1770 × 10⁴).

   Using Pythagoras' Theorem, P_net = [(–4.4243× 10⁴)² + (0.1770× 10⁴)²]^½ = 4.428 × 10⁴ kg-m/s. The angle θ = tan^–1(|P_{net y}/P_{net x}|) = tan^–1(0.1770/4.4243) = 2.3° . P_net is in the N-W quandrant. Thus, the net momentum of the two cars 4.428 × 10⁴ kg-m/s at 2.3° north of west.

   

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9. Forces are vector quantities. Two forces F₁ and F₂ act on a body such the total force F₃ has a magnitude of 150 N at 15.0° east of north. If F₁ has magnitude 100 N at 10.0° west of north, what is the magnitude and direction of F₂?

We are looking for F₂ = F₃ – F₁, a vector subtraction. For vector problems, we first draw a neat sketch of the vectors and the vector operation of interest.



Then to solve the problem numerically, we break the vectors into their components and subtract them to get the components of vector F₂. Then we convert to polar coordinate form.

F₃ = i150sin(15°) + j150cos(15°) = i38.823 + j144.889 .

F₁ = −i100sin(10°) + j10cos(10°) = −i17.365 + j98.481

The difference F₂ = F₃ − F₁ is

F₂ = i56.188 + j46.408.

Using Pythagoras’ Theorem, F₂ = [(56.188)² + (46.408)²]^½ = 72.875 N. The angle θ = tan^–1(|F_2y/F_2x|) = tan^–1(46.408/56.188) = 39.55° . Force F₂ in in the N-E quadrant, thus the force is 72.9 N at 39.6° north of east.



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10. A person is located 225 m and 35.0° north of west from her initial position. She walked from her initial position to her final position in two stages. In the first stage of her walk, she walked 150 m at 20.0° south of east. How far and in what direction did she walk in the second part?

We are looking for R₂ = R_f – R₁, a vector subtraction. For vector problems, we first draw a neat sketch of the vectors and the vector operation of interest.



Then to solve the problem numerically, we break the vectors into their components and subtract them to get the components of vector R₂. Then we convert to polar coordinate form.

R_f = −i225cos(35°) + j225sin(35°) = −i184/309 + j129.055 .

R₁ = i150cos(20°) + −j150sin(20°) = i140.954 − j51.303

The difference R₂ = R_f − R₁ is

R₂ = −i325.263 + j180.358.

Using Pythagoras’ Theorem, R₂ = [(–325.263)² + (180.358)²]^½ = 371.921 m. The angle θ = tan^–1(|R_2y/R_2x|) = tan^–1(180.358/325.263) = 29.01° . R₂ is in the N-W quadrant, thus the displacement is 372 m at 29.0° north of west.

Questions? mike.coombes@kpu.ca