SOLVED NUMERICALS IN PHYSICS: XI C OVERING S INDH T EXTBOOK B OARD
ONLY
BY
D R . R A M C HAND P H D( P HYSICS )
Principal/Head of Physics Department Shaheed Benazir Bhutto Govt Girls Degree College, Jhudo, District Mirpurkhas 033325133 033325 13375, 75, ram_r ra m_r25@ 25@hot hotma mail. il.com com http://www.facebook.com/ramcraguel
2017
RAM’S OUTLINE SERIES
Copyright
c 2017, Department of Physics, Shaheed Benazir Bhutto Government Girls
Degree College, Jhudo
C OMPOSED BY D R . R AM C HAND This manuscript is written in LATEX. The diagrams and images are created in open-source
applications IPE, LatexDraw, Freeplane, VUE and Blender 3D.
The author is a visiting scientist to Aspen Center for Physicist, USA, the University of Malaya, Kuala Lumpur, Malaysia, the International Center for Theoretical Physics (ICTP), Italy and the Chinese Academy of Sciences, Beijing, China. The author is also a member of American Association of Physics Teachers Teachers (AAPS), USA. The author author’s ’s research profile
can be found at his LinkedIn page & Google Scholar page.
[email protected],
[email protected] http://www.facebook.com/ramcraguel @RamCRaguel www.linkedin.com/in/ram-chand .linkedin.com/in/ram-chand LinkedIn page: www https://sites.google.co google.com/site/theco m/site/thecomphys/researc mphys/research-1/Soft– h-1/Soft–CondensedCondensed-Matter-Th Matter-Theory eory Research page: https://sites. Firstt printing, January 2017 Firs Revised Edition, August 2017
Contents
1
SCOPE OF PHYSICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
2
SCALARS SCALAR S AND VECTORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
3 MOTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
27 27
43 MOTION IN TWO DIMENSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
5 STATICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 GRAVITATION
55 55
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 65
7
71 WORK, POWER & ENERGY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
8
77 WAVE WA VE MOTION & SOUND . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
9
83 NATURE OF LIGHT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
10 GEOMETRIC GEOMETRICAL AL OPTICS
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 89
Ram’s Outline Series
1. SCOPE OF PHYSICS
1.1 1. 1
Prob Pr oble lems ms Problem 1.1. Find the area of a rectangular plate having length (21.3 ± 0.2)cm and width (9.80 ± 0.10)cm. Data ta:: L = 21.3cm, and w = 9.80cm Solution: Da where ∆ L = 0.2cm and ∆w = 0.10cm are least counts of the instruments. Area of rectangular plate = length × width A = L × w = 21.3 × 9.80 = 208.74 The least number of significant figures in numerical is 3, so we can round off the answer to 3
significant figures: A = 209cm2
Error calculation (least count) can be calculated as: ∆ A ∆ L ∆w = + A L w ∆ A 0.2 0.10 = + =⇒ ∆ A = 4.09 = 4( round off to 1 significant figure ) 209 21.3 9.8 2 ∴ A = 209 ± 4cm Ans.
Problem 1.2. Calculate (a) the circumference of a circle of radius 3.5cm and (b) are area a of a circle 4 . 65 of radius cm. Data ta:: r = = 3.5cm, A = 4.65cm. Solution: Da (a) Circumference of a circle is given by the formula: = 2π r = 2 × 3.142 × 3.5 = 21.994 S = r = Ram’s Outline Series
Chapter 1. SCOPE OF PHYSICS
6
The least number of significant figures is 2, so we should round off the answer to 2: = 22cm Ans. S =
(b) Area of circle is given by: A = π r 2 = 3.142 × (4.65)2 = 3.142 × 21.6225 = 67.9564
The least number of significant figures is 3, so we should round off the answer to 3: A = 67.9cm2 Ans. + 12 at 2 is dimensionally correct. Problem 1.3. Show that the expression x = vot + Here re x is the distance in meter, vo is velocity in m/s, a is acceler acceleration ation in m/s2 and t is i s Solution: He time in second. The dimensions of these physical quantities are: 1 + at 2 x = vot + 2 − 1 + LT −2 × T 2 (∵ 1/2 is dimensionless quantity) L = LT × T + L = LT 0 + LT 0 L = L + L (∵ T 0 = 1) =⇒ L = 2 L =⇒ L = L
Since both sides of the equation have the same dimensions, therefore, the given equation is
dimensionally correct. = Ct 3 . Problem 1.4. Find the dimension of C in the equation S =
Solution: found as:
The dime dimensio nsion n of of S is L and the dimension of t can be S is t is T , then the dimension of C C can
S L = t 3 T 3 −3 Ans. ∴ Dimenions of C are: LT
= C =
Problem 1.5. Estimate the number of liters of gasoline used by cars in Pakistan Pakistan each year (consider cars in Pakistan = 500000, average distance travelled by each = 16000km and gasoline consumed per car = 6km/lit) Solution: No. of cars in Pakistan = 500000 = 5 × 105 Average distance/year = 16000km Gasoline consumption = 6km/litres No. of liters of gasoline used/year = ? total distance covered distance covered per liter 16000 = = 2666.67 lit 6 ∴ gasoline consumed each year by all cars = number of cars × consumption by each car. Gasoline consumed each year by each car =
= 5
× 105 × 2666.67 =
1.33 × 109 liter literss Ans. Ram’s Outline Series
2. SCALARS AND VECTORS
2.1 2. 1
Prob Pr oble lems ms Problem 2.1. State which of the following are scalars and which are vectors. Solution:
Physi siccal Quantity Weight Calorie Specific Heat Momentum Density Energy Volume Distance Speed Magneti Mag neticc fiel fieldd inte intensi nsity ty
Type Vector Scalar Scalar Vector Scalar Scalar Scalar Scalar Scalar Vect ector or
Physical Quantity Entropy Work Centripetal force Temperature Gravitational force Charge Shearing stress Frequency Kinetic energy Electr Ele ctric ic fiel fieldd int intensi ensity ty
Problem 2.2. Find the resultant resultant of the following displacement:
= 20 km 30o south of east B = 50 km due to west, A = = 30 km 60o south west. = 40 km north-east, C D
Solution: Here is what these vectors look like: Ram’s Outline Series
Type Scalar Scalar Vector Scalar Vector Scalar Vector Scalar Scalar Vect ector or
Chapter 2. SCALARS AND VECTORS
8
The resultant vector is found by head to tail rule as shown in right figure. The magnitude of
resultant vector is 20.9km with direction of 21.65o south of west.
Problem 2.3. An aeroplane flies 400 km due to west from city A to city B, then 300 km north east to city C, and finally 100 km north to city D. How far is it from city A to D? In what direction must the aeroplane had to return directly to city D from city A? Solution: The distance from city city A to D is found graphically in following diagram:
The resultant vector from A to D is drawn in dotted line (left-side figure) which is equal to 364km. If aeroplane returns directly from city D to A then the direction would be 31 0 east of south as
shown in right-side figure.
−→ −→ −→ − − −
−→
Problem 2.4. Show graphically that − A − B = − A + B . Solution:
First we cons construc tructt
B graphica A B, graphically lly as shown in figure. Choose negative negative of
then apply head to tail rule. The resultant resultant is A
is shown in dotted line. Make a negative of B + B by taking negativ A. the resultant as shown in figure. Now construct − A negativee of
Now compare both resultant vectors (as shown in 3rd figure); we see that both have same
directions and same magnitudes. Hence, the equation is proved. Ram’s Outline Series
2.1 Problems
9
D as shown in figure below. Construct: Problem 2.5. Given vector A, B, C and
−→ → −
−→ −→
(a) 4 A − 3 B − 2 C + 2 D 1 → −C + 1 A−→ + B−→ + 2 D −→ (b) 2 3
− 3 B + 2 D − 2C Solution: (a) 4 A
by 4, First multiply all vectors by scalar values, for example, multiply A by 2 and B by 3, C by D by 2.
Then add all vectors according to head to tail rule as shown in figure:
1 1 + A + B + 2 D (b) C 2 3
In this case, we also first multiply all vectors by scalar values, like: multiply by C by
1 , D by 2 and 2
+ B + 2 D by 1 . Now add all vectors then finally resultant of A vectors according according to head to tail rule as
shown in figure:
3
Ram’s Outline Series
Chapter 2. SCALARS AND VECTORS
10
Problem 2.6. The following forces act on a particles P: ˆ, ˆ 1 = 2iˆ + 3 j ˆ − 5k 2 = −5iˆ + ˆ F F j + 3k ˆ, ˆ 3 = ˆi − 2 j ˆ + 4k 4 = 4iˆ − 3 j ˆ − 2k F F measured measur ed in newton, find: (a) The resultant force F (b) Magnitude of the resultant force. F is Solution: (a (a)) Le Lett is the resultant of the all forces, then: 2 + F 3 + F 4 = F = F 1 + F ˆ ) + ( 5iˆ+ ˆ j + 3k ˆ ) + (iˆ 2 j ˆ+ 4k ˆ ) + (4iˆ 3 j ˆ 2k ˆ) = ( 2iˆ + 3 j ˆ 5k F = ˆ + 3k ˆ + 4k ˆ 2k ˆ = 2iˆ 5iˆ + iˆ + 4iˆ + 3 j ˆ + ˆ F = j 2 j ˆ 3 j ˆ 5k
−
− ˆ = 2iˆ − ˆ F = j + 0k = 2 i − ˆ F = j
−
− − −
−
−
− −
F is (b) The magnitude of resultant force is given by Pythagoras Theorem:
| F | = F x2 + F y2 + F z2 | F | = (2)2 + (−1)2 + (0)2 = √ 4 + 1 √ | F | = 5 Ans. ˆ ˆ Find: = ˆi + 2 j ˆ − k. A = 3iˆ− ˆ j − 4k, B = −2iˆ + 4 j ˆ − 3kˆ and C = Problem 2.7. If − B + B | − 2 B | + 3C + C + 4C (a) 2 A (b) | A (c) |3 A − 2 B + 4C (d ) a unit vector parallel to: 3 A Solution:
− B = ˆ ) − (−2iˆ + 4 j ˆ − 3k ˆ ) + 3(iˆ + 2 j ˆ− k ˆ) + 3C = 2(3iˆ − ˆ j − 4k (a) 2 A ˆ + 2iˆ − 4 j ˆ + 3k ˆ + 3iˆ + 6 j ˆ − 3k ˆ = 6iˆ − 2 j ˆ − 8k ˆ + 3k ˆ − 3k ˆ = 6iˆ + 2iˆ + 3iˆ− 2 j ˆ− 4 j ˆ + 6 j ˆ − 8k − B = + 3C = 11iˆ− 8kˆ Ans. 2 A
Ram’s Outline Series
2.1 Problems
11
+ B | = |(3iˆ − ˆ ˆ ) + (−2iˆ + 4 j ˆ− 3k ˆ ) + (iˆ + 2 j ˆ − k ˆ )| + C (b) | A j − 4k ˆ − 2iˆ + 4 j ˆ − 3k ˆ + iˆ + 2 j ˆ − k ˆ| = |3iˆ − ˆ j − 4k ˆ − 3k ˆ − k ˆ| = |3iˆ − 2iˆ + iˆ − ˆ j + 4 j ˆ + 2 j ˆ − 4k
− | (2)2 + (5)2 + (−8)2 = √ 4 + 25 + 64 √ | A + B + C | = 93 Ans.
ˆ = = 2iˆ + 5 j ˆ 8k
|
− 2 B | = |3(3iˆ − ˆ ˆ ) − 2(−2iˆ + 4 j ˆ − 3k ˆ ) + 4(iˆ + 2 j ˆ − k ˆ )| + 4C j − 4k (c) |3 A ˆ + 4iˆ − 8 j ˆ + 6k ˆ + 4iˆ + 8 j ˆ − 4k ˆ| = |9iˆ − 3 j ˆ − 12k ˆ + 6k ˆ − 4k ˆ| = |9iˆ + 4iˆ + 4iˆ − 3 j ˆ − 8 j ˆ + 8 j ˆ− 12k
ˆ = = 17iˆ 3 j ˆ 10k
| − − | (17)2 + (−3)2 + (−10)2 √ = 289 + 9 + 100 √ Ans.. |3 A − 2 B + 4C | = 398 Ans − 2 B then + 4C = 3 A then unit vector parallel to this vector is: X = (d) Let − 2 B 3 A + 4C X − 2 B = ˆ + 4C √ = = 17iˆ − 3 j ˆ − 10k But 3 A xˆ = | 398 | X
ˆ 17iˆ − 3 j ˆ − 10k √ 398 Ans. = x
Problem 2.8. Two tugboats are towing a ship. Each exerts a force of 6000N, and the angle between the two ropes is 60o . Calculate the resultant force on the ship.
Figure 2.1: Problem 2.8 Ram’s Outline Series
Chapter 2. SCALARS AND VECTORS
12 Two for force cess whic which h Solution: exerting on ship are equal: 1 = F F 2 = 6000 N ; Angle between ropes: θ = 60o = ?. Resultant force on ship: F =
tugboa tugb oats ts are are
Putting the values (∵ cos60o = 0.5):
| F | =
Following figure the total force can be calculated
as:
× × × ×
2(6000)2 + 2(6000)2 × 0.5
=
2 36 106 + 36000000
=
72 106 + 36 × 106
=
2 = F = F 1 + F
| F | = | F | =
∵
1 = F F 2
| | | |
108 106
√
108000000 = 10392.3 =
1 2 + |F 2 |2 + 2|F 1 ||F 2 | cos60o F
| F | = 10392.3 N
1 2 + 2|F 1 |2 cos60o 2 F
Ans.
ˆ , ˆ. r 1 = 2iˆ+ 3 j ˆ− k r 2 = 4iˆ− 3 j ˆ+ 2k Problem 2.9. The position vectors of points P and Q are given by − → Determine PQ in terms of rectangular unit vector ˆ i,iˆ, jˆ and ˆk and find its magnitude.
Solution: ˆ r 1 = 2iˆ + 3 j ˆ k
−
ˆ r 2 = 4iˆ 3 j ˆ + 2k
−
−PQ → = ∆ r = =? These two vectors are position vectors.
−→ −→ r 2 − r 1 r = = PQ = ∆ −PQ → = 4iˆ− 3 j ˆ+ 2k ˆ − 2iˆ+ 3 j ˆ− k ˆ ˆ − 2iˆ − 3 j ˆ + ˆk = 4iˆ − 3 j ˆ + 2k −PQ → = 2iˆ− 6 j ˆ+ 3kˆ Ans. Length PQ can be calculated as:
Magnitude can be found as:
→| = (2iˆ)2 + (−6 j ˆ)2 + (3k ˆ )2 |−PQ →| = √ 4 + 36 + 9 = √ 49 = 7 |−PQ →| = 7 Ans. |−PQ
ˆ , ˆ and ˆ can = 4iˆ − 2 j ˆ − 6k vectors A = 3iˆ + ˆ j − 2k B = −iˆ + 3 j ˆ + 4k and C = can Problem 2.10. Prove that the vectors form the sides of a triangle. Find the length of the medians of the triangle.
To find:
Data:
Magnitude of length of medians: = 3iˆ+ ˆ ˆ A j − 2k ˆ = −iˆ + 3 j ˆ + 4k B = ˆ = 4iˆ − 2 j ˆ− 6k C
|−a→f | =? →| =? |−bd |−→ ce | =? Ram’s Outline Series
2.1 Problems
13
A, B and C form Let thre three e vec vectors tors form a C triangle as shown in figure. Two vectors B and are added in such a way so that their resultant is
Solution:
according to head-to-tail method: A = A B + C
If above equation is correct then these vectors can and form a triangle. Lets calculate and see if it B + C
is equal to A:
= ˆ + 4iˆ − 2 j ˆ − 6k ˆ + C = −iˆ + 3 j ˆ+ 4k B = ˆ = + C = 3iˆ + ˆ = A B j − 2k = = we see that B + C A, hence it is proved that
these three vector can form a triangle. Now we will calculate the medians and then we will find their magnitudes. Median is straight line that divides divides a side of a tria triangle ngle into two equal parts. parts. Thes These e are also vectors vectors as sho shown wn in figure figure.. Lower case alphabet are written to mark the positions of these vectors. Following vectors addition by head-to-tail rule we can calculate median vectors easily. And then applying Pythagoras Theorem
we can find their magnitude. In a f c :
−a→f = −→ −→ ac + c f −→ 1 → But − ac = B and c f = C 2 → −ac = B + 1 C 2 → −ac = −iˆ+ 3 j ˆ+ 4k ˆ + 1 4iˆ− 2 j ˆ− 6k ˆ → −ac = −iˆ+ 3 j ˆ+ 4k ˆ + 22iˆ− ˆ j − 3k ˆ → −ac = ˆi + 2 j ˆ+ ˆk |−→ ac| = (1)2 + (2)2 + (1)2 √ √ |−→ ac| = 1 + 4 + 1 = 6 √ |−→ ac| = 6 Ans. In abd : :
According to head to tail rule:
−ad → = −→ −→ = ab + bd −→ and −ad → = 1 = B But ab = A
2 Now we will find third median: In ceb : : According According to head to tail rule:
Therefore: → 1 − B = bd + A 2 −bd → = 1 = B − A 2 −bd → = 1 ˆ ˆ ˆ = −i + 3 j + 4k − 3iˆ+ ˆ j − 2k ˆ 2 −bd → = 1 3 ˆ − 3iˆ − ˆ j + 2k ˆ = − iˆ+ j ˆ + 2k 2 2 −bd → = 7 1 ˆ = − iˆ+ j ˆ + 4k 2 2
=
(7/2)2 + (1/2)2 + (4)2
49 1 + + 16 4 4 →| = 49 + 1 + 64 = |−bd 4 √ 1 = 114 2 →| = 1 √ 114 Ans. |−bd 2 =
→ −cb = −→ −→ ce + eb −→ C and −→ 1 But cb = and eb = A 2
Ram’s Outline Series
114 4
Chapter 2. SCALARS AND VECTORS
14
1 1 → → = ˆ − 1 3iˆ + ˆ j − 2k ˆ = − =⇒ − = 4iˆ − 2 j ˆ− 6k C ce + A ce = C − A 2 2 2 5 5 → −ce = 4iˆ− 2 j ˆ− 6k ˆ − 3 iˆ− 1 j ˆ + ˆk = ˆ = iˆ − j ˆ − 5k 2 2 2 2 5 2 5 2 → − | ce| = 2 + − 2 + (−5)2 = 254 + 254 + 25
25 + 25 + 100 = |−→ ce | = 4 1 √ 150 Ans. |−→ ce | =
150 1 √ = 150 4 2
2
, 15 unit long when it forms an angle Problem 2.11. Find the rectangular rectangular components of a vector A o o o with respect to +ve x-axis of (i) 50 , (ii) 130 , (iii) 230 , (iv) 310o .
Data: | A | = 15 unit = 50o (i) θ = = 130o (ii) θ = = 230o (iii) θ = = 310o (iv) θ = To find: Rectangular components of A A: A x = ? and A y =?. Solution: (i) A x = A cos θ A x = 15cos50o A x = 15 × 0.6428 A x = 9.64 unit A y = A sin θ A y = 15sin50o A y = 15 × 0.766 A y = 11.49 unit A x = 9.64 unit and A y = 11.49 unit
(ii) A x = A cos θ A x = 15cos130o A x = 15 × (−0.6428) A x = −9.64 unit A y = A sin θ A y = 15sin130o A y = 15 × 0.766 A y = 11.49 unit A x = −9.64 unit and A y = 11.49 unit Ram’s Outline Series
2.1 Problems
15
(iii) A x = A cos θ A x = 15cos230o = 15 × (−0.6428) A x = −9.64 unit A y = A sin θ A y = 15sin230o = 15 × (−0.766) A y = −11.49 unit
A x = −9.64 unit and A y = −11.49 unit
(iv) A x = A cos θ A x = 15cos310o = 15 × 0.6428 A x = 9.64 unit A y = A sin θ A y = 15sin310o = 15 × (−0.766) A y = −11.49 unit
A x = 9.64 unit and A y = −11.49 unit
Problem 2.12. Two vectors 10 cm and 8 cm long form an angle of (a) 60o , (b) 90o and (c) 120o . Find the magnitude of differe difference nce and angle with respec respectt to the largest vector. vector.
Data: | A | = 10cm | B | = 8cm = 60o , (b) θ = 90o , ( (a) θ = (c) c) θ = 120o To find: − B | =? (i) Magnitude of difference: | A
−
B and = ? (ii) Angle b/w A A : θ =
Solution: (a) θ = 60o : (i) We can apply Pythagoras theorem and dot product to find the magnitude of vectors:
| A − B | =
− − − − | | − | || | | |
| A − B | =
B A
2
=
B B 2 2 A . B . A = A + B 2 A
2 2 A B cos θ + 2 + B A
Ram’s Outline Series
(1)
Chapter 2. SCALARS AND VECTORS
16
| A − B | = (10)2 − 2(10)(8) cos60o + (8)2 = √ 100 + 160 × 0.5 + 64 √ | A − B | = √ 164 − 80 = 84 = 9.165cm | A − B | = 9.165cm Ans. − B ) can be found by: (ii) Angle b/w A and ( A .( A − B || A − B . A − A . B || A − B ) = | A | cos θ =⇒ A = | A | cos θ A | A |2 − | A || B | cos θ = cos−1 θ = | A || A − B |
(2)
2
− (10)(8) cos60o = cos−1 100 − 80 × 0.5 10 × 9.165 91.65 − 100 40 60 = cos−1 = cos−1 = cos−1 0.6547 = 49.1o θ = ( 10) = cos−1 θ =
91.65 = 49.1 Ans. θ =
91.65
o
= 90o : (b) θ = (i) Recall equation (1):
| − | | | − | || | − B = A
2 2 A B cos θ + |2 + | B A
= (10)2
2(10)(8) cos90o + (8)2
√ − × + 100 160 0 64 √ √ = 164 − 0 + 64 = 164 =
= 12.806cm
| A − B | = 12.806cm Ans. − B ) can be found by equation (2): A and ( A (ii) Angle b/w 2 1 | A| − | A|| B| cos θ − = = θ cos | A || A − B | ( 10)2 − (10)(8) cos90o 100 − 80 × 0 100 − 0 = cos−1 = cos−1 = cos−1 θ = 10 × 12.806 128.06 128.06 100 = = cos−1 (0.7809) = 38.66o . θ = 38.66o Ans. θ = 128.06
(c) θ = 120o : (i) Recall equation (1):
| − | | | − | || | − B = A
2 2 A B cos θ + |2 + | B A
= (10)2
2(10)(8) cos120o + (8)2 =
| A − B | = 15.62cm Ans.
√
√
100 − 160 × (−0.5) + 64 = 164 + 80 + 64 = 244
Ram’s Outline Series
2.1 Problems
17
− B ) can be found by equation (2): (ii) Angle b/w A and ( A | A |2 − | A || B | cos θ = cos−1 θ = | A || A − B | ( 10)2 − (10)(8) cos120o 100 − 80 × (−0.5) 100 + 40 θ = = cos−1 = cos−1 = cos−1 10 × 15.62 156.2 128.06 140 = = cos−1 (0.8962) = 26.33o θ = 156.2 θ = = 26.33o Ans.
| = | B | = 1 , calculate A and B is 60o . Given that | A Problem 2.13. The angle between the vectors | , , (b) | B |. − A + A (a) | B
Data: = 60o | A | = | B | = 1, and θ = To find: | =?, and (b) | B | =? − A + A (a) | B
Solution: Apply Pythagoras Theorem and dot product of vectors:
| = − A (a) | B =
(1)2
− − A B
2
2 = B
. B 2 = + A 2 A
− 2(1)(1)(0.5) + (1)2 = √ 1 − 1 + 1 =
| | − | || |
√
B 2 2 A cos θ + |2 + | B B
| = 1 Ans. − A 1 = 1. | B
Ram’s Outline Series
Chapter 2. SCALARS AND VECTORS
18
= + A (b) B
|
=
|
+ A B
2
. B 2 = 2 + 2 A + A = B
(1)2 + 2(1)(1)(0.5) + (1)2 =
| B + A | =
| |
|| B 2 + 2| A | cos θ + |2 + | B B
√ 1 + 1 + 1 = √ 3
√
3 Ans.
weighing hing 10,000 10,000 N on a hill which which makes an angle of 20 o with horizontal. Problem 2.14. A car weig Find the components of car’s weight parallel and perpendicular to the road.
Data: Weight of car: W = 10, 000 N Angle of inclination: θ = 20o To find: Parallel to road component: W =? Perpendicular to road component: W ⊥ = ? Solution:
As car car is moving moving on incl incline ined d plane plane as shown in figure, the parallel and perpendicular component of weight (W ) can be found by
trigonometry as: sin θ W = W sin Putting the values: W = 10000 × sin20o
∵
sin200 = 0.342
= 10000
× 0.342
W = 3420.2 N
Perpendicular component of W W : : W ⊥ = W cos cos θ Putting the values: W ⊥ = 10000 × cos20o
∵
cos200 = 0.9397
= 10000
× 0.9397
W ⊥ = 9397.2 N
Ram’s Outline Series
2.1 Problems
19
ˆ B = 6iˆ − 3 j ˆ + 2k. Problem 2.15. Find the angle between A = 2iˆ + 2 j ˆ− kˆ and Data: = 2iˆ + 2 j ˆ − k ˆ and ˆ and A B = 6iˆ − 3 j ˆ + 2k and Angle le between between vecto vectors rs A B can be Solution: Ang found by dot product: · B || B = | A | cos θ A can be calculated as: Where A · B ˆ · 6iˆ − 3 j ˆ + 2k ˆ = 2iˆ + 2 j ˆ − k
(1)
· · − 2 · 3( j ˆ· ˆ j) − 1 · 2(k ˆ · k ˆ ) ˆ = = 1 Since: ˆi · iˆ = jˆ · ˆ j = ˆk · k · B = 12 − 6 − 2 = 4 A | and | B | are: Magnitudes | A | A | = (2)2 + (2)2 + (−1)2 √ | A | = √ 4 + 4 + 1 = 9 = 3 | B | = (6)2 + (−3)2 + (2)2 = 2 6(iˆ iˆ)
∴
| B | = √ 36 + 9 + 4 =
√
49 = 7 Putting these values into equ(1): · B 4 4 A = = cos θ = | A || B | 3 × 7 21 = cos−1 (0.1905) = 79o θ = θ = 79o Ans.
onto the direction of vector Problem 2.16. Find the projection of the vector A = ˆi − 2 j ˆ + ˆk onto B = ˆ 4iˆ − 4 j ˆ + 7k. Projection can be found by dot product. Projection product. First we have have to find the direction of vector vector B. ˆ We know that direction of any vector can be found by unit vector. Unit vector of B and b is defined
Solution: as:
B bˆ = | B | =
ˆ ˆ ˆ 4 iˆ − 4 j ˆ + 7k ˆ 4iˆ − 4 j ˆ + 7k 4iˆ − 4 j ˆ + 7k 4 iˆ − 4 j ˆ + 7k √ = √ = = 9 16 + 16 + 49 81 (4)2 + (−4)+ (7)2
B can be calculated as: Now projection of A onto the direction of
ˆ 4iˆ 4 j ˆ + 7k 9
− · − − · − ×
· bˆ = iˆ 2 j ˆ+ ˆk A
1 ˆ ˆ ˆ ˆ = 1 4 1(iˆ · iˆ) + 2 × 4( j ˆ · ˆ j) + 1 × 7(k ˆ · k ˆ) = i 2 j + k 4iˆ 4 j ˆ + 7k 9 9 ˆ = = 1 Since: ˆi · iˆ = jˆ · ˆ j = ˆk · k · bˆ = 1 × (4 + 8 + 7) = 19 ∴ A 9 9 B = 19 Ans. Hence projection of A on the the direction of 9
ˆ makes A = 3iˆ − 6 j ˆ+ 2k γ which the vector Problem 2.17. Find the angles α , , β and γ makes with positive x, y and z axis respectiv respectively. ely. Ram’s Outline Series
Chapter 2. SCALARS AND VECTORS
20
can be writAccordin ording g to figure figure the the A Solution: Acc ten in unit vector form as: aˆ = 3iˆ − 6 j ˆ + 2k ˆ A
x A A A ˆ iˆ + y jˆ + z k | A | | A | | A | z A y A x A = = = where cos α = , cos β = and cos γ = | A | | A | | A |
where aˆ =
= And A
|| | | = √ + A
(3)2 + ( 6)2 + (2)2
− √
9 36 + 4 = 49 = 7 z = 2. No A, three components are: A x = 3, A y = −6 and A In a given vector Now w we can find find all angles. A x 3 = = 0.4286 =⇒ θ = = cos−1 (0.4286) = 64.62o | A| 7 A −6 For β angle: angle: cos β = y = = −0.8571 =⇒ θ = = cos−1 (−0.8571) = 149o 7 | A| z 2 A = = = 0.2857 =⇒ θ = = cos−1 (0.2857) = 73.4o For γ angle: angle: cos γ = | A | 7 = For α angle: angle: cos α =
Angles are found: α = 64.62o , β = 149o and γ = = 73.4o Ans.
ˆ if = 3iˆ + 2 j ˆ − 5k if the Problem 2.18. Find the work-done in moving an object along a vector r = ˆ ˆ ˆ = 2i − j − k. applied force is F =
Data: ˆ. Applied force on an object is: = 2iˆ − ˆ F = j − k ˆ = 3iˆ + 2 j ˆ − 5k The displacement an object covered when force is applied: r = To find: The work-done by the force is: W =? Solution: covered:
Work-d ork-done one by force on any object object is the dot product of force and displace displacement ment it
ˆ = 2iˆ − ˆ W = F · r = j − k = (6
ˆ = 2 × 3(iˆ · iˆ) − 1 × 2( j ˆ · ˆ j) + 1 × 5(k ˆ · k ˆ) 3iˆ + 2 j ˆ − 5k
· ·
− 2 + 5) = 9
∵
ˆ = ˆi iˆ = jˆ · ˆ j = ˆk · k = 1
= 9 J Ans. Work-done: W =
Problem 2.19. Find the work-done by a force of 30,000 N in moving an object through a distance of 45 m when (a) the force is in the direction of motion, and (b) the force makes angle of 40 o to the direction of motion. Find the rate at which the force is working at a time when the velocity is
2m/sec.
Data: F | = 30000 N Force applied on an object: | | = 45m Distance covered by an object due to the force: |d v | =? Velocity of the object: | Ram’s Outline Series
2.1 Problems
21
Angle between the direction of force and displacement: θ = 40o To find: Work-done by the force on the object: W =? Rate of doing work (Power): P =? ork-done done is the scalar or dot product of force and displacem displacement ent (in the first case, the Solution: Workforce and the displacement are in the same direction so the angle between them is zero):
= | cos θ = 30000 × 45cos0o = 1350000 × 1 = | W = F · d F ||||d = 1.35 × 106 J Ans. W =
Rate of doing work is equal to power: P = o
F d
·
t
· d t = F · v = | F |||| v | cos θ
= F
P = 30000 × 2cos0 = 60000 × 1 P = 6 × 104W Ans.
In the second case, the force and the displacement make an angle of 40 o : = | cos θ = 30000 × 45cos40o = 1350000 × 0.7660 = 1034160 = | W = F · d F ||||d = 1.034 × 106 J Ans. W =
v | cos θ = 30000 × 2cos40o Rate of doing work OR power: P = F · v = | F |||| P = 60000 × 0.7660 = 45962.667
P = 4.5963 × 104W Ans.
| = 3 , | B | = 4 , and = −5 , find: Problem 2.20. Two vectors A and B are such that | A A. B B (a) the angle between A and + B − B | and | A | (b) the length of | A + B − B and A . (c) the angle between A
Data: = −5. | A | = 3, | B | = 4 and A · B
A and B can be found by dot product as: Solution: (a) angle between · B || B = | A | cos θ A Putting the values from data:
−5 = −0.4167 = − 5 = 3 × 4 × cos θ =⇒ cos θ = 12 − 1 o = cos (−0.4167) = 114.62 θ = θ = = 114.62o Ans.
+ B − B | and | A | can be found by Pythagoras theorem for vectors product as (b) the length of | A follows: (also known as Law of Cosine): + B = A
|
+ B A
2
2 + 2 A B + B 2 = = A
| √ √ = 9 − 10 + 16 = 15 √ | A + B | = 15 Ans.
·
(3)2 + 2
Ram’s Outline Series
× (−5) + (42)
Chapter 2. SCALARS AND VECTORS
22
Figure 2.2: Problem 2.20 Similarly: B = A
− − B A
2
2 = A
| − | √ √ = 9 + 10 + 16 = 35 √ | A − B | = 35 Ans.
B B 2 = 2 A
· −
(3)2
− 2 × (−5) + (42)
−
+ B B and A , then according to dot product: (c) Consider θ is the angle between A
· − + B A
B + B − B = | A || A | cos θ A
· A − A · B − B + B − B + B · A · B = | A || A | cos θ A 2 − + B − B + − B 2 = | A || A | cos θ · · A A B A B 2 − B 2 = A | A + B || A − B | cos θ =
√ √ = cos−1 15 × 35 = (3)2 − (4)2 = 9 − 16 = −7 =⇒ θ = −7 = cos−1 (−0.3055) = 107.79o = cos−1 θ =
−7 3.873 × 5.916
21.913 = 107.79o Ans. θ =
ˆ ˆ Find (a) A = 2iˆ − 3 j ˆ − k, B = ˆi + 4 j ˆ − 2k. A × B, Problem 2.21. If + B − B × A . (c) A
Data: ˆ A = 2iˆ − 3 j ˆ− k ˆ = ˆi + 4 j ˆ − 2k B
and (b) B × A,
To find: , (b) (a) A × B B × A + B − B × A . (c) A
Ram’s Outline Series
2.1 Problems
23
can be determined as: Solution: (a) The cross product A × B
We can use cross arrow method to find the cross product: ˆ )( ˆ )( × B = ( iˆ)(−3)(−2) + ( j ˆ)(−1)(1) + (k )(2)(4) − ( j ˆ)(2)(−2) − (iˆ)(−1)(4) − (k )(−3)(1) A ˆ + 4 j ˆ + 4iˆ + 3k ˆ = ˆ = 6iˆ − ˆ = 10iˆ + 3 j ˆ + 11k j + 8k × B ˆ Ans. = 10iˆ + 3 j ˆ + 11k A
can be determined as: B × A (b) The cross product
We can use cross arrow method to find the cross product: = ( iˆ)(4)(−1) + ( j ˆ)(−2)(2) + (k ˆ )( ˆ )( × A )(1)(−3) − ( j ˆ)(1)(−1) − (iˆ)(−2)(−3) − (k )(4)(2) B ˆ + ˆ j − 6iˆ − 8k ˆ = ˆ = ˆ = −4iˆ − 4 j ˆ− 3k = −10iˆ − 3 j ˆ − 11k = −10iˆ − 3 j ˆ− 11k = −10iˆ − 3 j ˆ− 11k ˆ Ans. × A B
× − × − − − − × − − − − × − − × − − − − × − − × − × −
× B . = − B × A Note that A
+ B (c) The cross product A
B can be determined as: A
+ B A
B ˆ + iˆ + 4 j ˆ 2k ˆ = 2iˆ 3 j ˆ k A
ˆ (iˆ + 4 j ˆ 2k ˆ) 2iˆ 3 j ˆ k
+ B A
B ˆ = 3iˆ + ˆ A j 3k
ˆ iˆ 4 j ˆ + 2k ˆ 2iˆ 3 j ˆ k
+ B A
B ˆ = 3iˆ + ˆ A j 3k
iˆ 7 j ˆ + ˆk
+ B Now we can use cross arrow method to find the cross product of A
Ram’s Outline Series
B : A
Chapter 2. SCALARS AND VECTORS
24
× − × −− + B A
B = A
ˆ )( (iˆ)(1)(1) + ( j ˆ)( 3)(1) + (k )(3)( 7) + B A
B A
− − ( j)(3)(1) − (i)(−3)(−7) − (k )()(1)(1) ˆ − 3 j ˆ − 21iˆ − k ˆ = ˆ = ˆi − 3 j ˆ − 21k = 20iˆ − 6 j ˆ − 22k B ˆ Ans. = 20iˆ − 6 j ˆ − 22k A
× − + B A
ˆ and A = 2iˆ − 6 j ˆ − 3k Problem 2.22. Determine the unit vector perpendicular to the plane of ˆ = 4iˆ + 3 j ˆ− k. B Data: ˆ A = 2iˆ − 6 j ˆ − 3k ˆ = 4iˆ + 3 j ˆ − k B To find: Unit vector ⊥ to A × B
Solution: Let unit vec vector tor is uˆ and defined as: × B A | A × B | −−−− > (1) has to be determined. Where A × B
uˆ =
We can use cross arrow method to find the cross product: × B ˆ )( ˆ )( = (iˆ)(−6)(−1) + ( j ˆ)(−3)(4) + (k )(2)(3) − ( j ˆ)(2)(−1) − (iˆ)(−3)(3) − (k )(−6)(4) A × B ˆ + 2 j ˆ+ 9iˆ + 24k ˆ = ˆ = 6iˆ − 12 j ˆ + 6k = 15iˆ − 10 j ˆ + 30k A Magnitude can be found as: B = A
| × |
(15)2 + ( 10)2 + (30)2 =
−
√
√
225 + 100 + 900 = 1225 = 35
Now putting these values in equation (1), we can determine its unit vector: × B ˆ 15 10 30 3 2 5 15 iˆ − 10 j ˆ + 30k A ˆ = ˆ = = iˆ − j ˆ + k = iˆ − j ˆ + k uˆ = 35 35 35 35 7 7 7 | A × B| 3 2 5 ˆ Ans. uˆ = iˆ − j ˆ + k 7 7 7
Problem 2.23. Using the definition of vector product, prove the law of sines for plane triangles of sides a, b and c. a, b and c added Plane triangle is closed closed geometry shape shape constructed from three vectors according to head to tail rule. Angles are marked as A, B and C . Sum of vectors of any closed shape
Solution:
is always zero, hence: Ram’s Outline Series
2.1 Problems
a + b + c = 0
25
−−−− > (1)
Taking cross product of above eqaution with a :
× a + b + c = a × a + a × b + a × c = 0 a × b + a × c = 0 ( a × a = 0) a × b = − a × c = c × a ( − a × c = c × a) a × b = c × a a
∵
∵
b | sin C and From the figure we see that and c || a × b = | a|| c × a = | a| sin B. Comparaing these equations with equation (1), we get: = | c || | a|| b | sin C = a| sin B | | sin C sin B a| sin C a| sin B = =⇒ | c | | c | = | b | −−−− > (2) | b | Again taking cross product of equation (1) with b b
×
a + b + c = b a + b b + b c = 0
× × × × −× × ×
b a + b c = 0
× × b × c = − b × a = a
∵
b b = 0
b
b a = a b
∵
b || b | sin C . From the figure we see that b × c | sin A and c = | a × b = | a|| b | sin A | b | sin C = | b || c | sin A = | a|| b | sin C =⇒ | | a| | c |
sin A sin C | a| = | c |
Hence:
But
sin C sin B | c | = | b |
sin A sin B sin C | a| = | b | = | c | Law of Sines
r 1 and r 2 are the position vectors (both lie in xy plane) making angles θ 1 and θ θ 2 Problem 2.24. If with the position x-axis measured counter clockwise, clockwise, find their vector product when:
(i) r 1 = 4cm with θ 1 = 30o r 2 = 3cm with θ 1 = 90o (iii) r 1 = 10cm with θ 1 = 20o r 2 = 9cm with θ 1 = 110o
(ii) r 1 = 6cm with θ 1 = 220o r 2 = 3cm with θ 1 = 40o
Ram’s Outline Series
Chapter 2. SCALARS AND VECTORS
26
vector product product of two two vectors vectors can Solution: The vector be found as:
r 1 × r 2 = | r 1 || r 2 | sin ∆θ (i) r 1 × r 2 = | r 1 || r 2 | sin(θ 2 − θ 1 ) r 1 × r 2 = 4 × 3sin(900 − 300 ) = 12sin600 √ 3 √ = 6 3 r 1 × r 2 = 12 × 2 √ r 1 × r 2 = 6 3cm2 Ans. r 1 × r 2 = | r 1 || r 2 | sin ∆θ (ii) r 1 × r 2 = | r 1 || r 2 | sin(θ 1 − θ 2 ) r 1 × r 2 = 6 × 3sin(2200 − 400 ) = 18sin1800 r 1 × r 2 = 18 × 0 r 1
× r 2 = 0cm2
Ans.
r 1 × r 2 = | r 1 || r 2 | sin ∆θ (iii) r 1 × r 2 = | r 1 || r 2 | sin(θ 2 − θ 1 ) r 1 × r 2 = 10 × 9sin(110 − 200 ) = 90sin900 r 1 × r 2 = 90 × 1 r 1
× r 2 = 90cm2
Ans.
Ram’s Outline Series
3. MOTION
3.1 3. 1
Prob Pr oble lems ms Problem 3.1. In an electron gun of a television set, an electron with an initial speed of 103 m/s enterss a region enter region wher wheree it is electr electrical ically ly accelerated. accelerated. It emerges emerges out of this region region after 1 micr micro o
4 × 105 m/s. What is the maximum length of the electr section with speed of 4 electron on gun? Calculate the acceleration.
Given Data: Time = t = 1µ s = 1 × 10−6 s Initial speed of the electron = V i = 103 m/s Final speed of the electron = V f = 4 × 105 m/s
To find: Acceleration of electron = a= ? Length of electron gun = t = ?
Solution: Using the first equation of motion and putting values from the given data: data:
Figure 3.1: Electron gun of TV set Ram’s Outline Series
Chapter 3. MOTION
28
= 4 × 105 = 103 + a × 10−6 V f = V i + at =
10−6 a = 4 × 105 − 103 = 103 4 × 102 − 1 a =
103 4 × 102 − 1 10−6
= 103+6 (400
− 1) = (399) 109
a = 399 × 109 m/s2 Ans.
The length of the electron gun can be found by using 3rd equation of motion: 2as = V f 2
−
V i2
= 2
2
2
× × − − × − −
× 399
109t = =
4 105
103
= 16 × 1010 106 798 × 109t =
6 4 16 × 1010 − 106 10 16 10 1 ( 160000 1) 106−9 = = = t = 798 × 109 798 × 109 798 − 3 159999 × 10 = = 200.5 × 10−3 798 = 0.2005m Ans. t =
Problem 3.2. A car is waiting at a traffic signal and when it turns green, the car starts ahead with 2m/s2 . At the same time a bus traveling with a constant speed of 10 10m/s a constant acceleration of 2 overtakes and passes the car. (a) How far beyond its starting point will the car overtake the bus? (b) How fast will the car be moving? Given Data: For Car: Initial velocity of the car = vi = 0m/s To find: Acceleration of the car = a = 2m/s2 Distance S = ? For Bus: Final velocity of the car = v f =? Speed of bus = v =10m/s (uniform) Time = t Solution: Fo Forr Car Car:: We can find the distance by using 2nd equation of motion: 1 1 1 2 = vit + + at 2 = 0 × t + + × 2t 2 = 0 + 2t S = 2 2 2 = t 2 S =
(1)
For Bus: Since bus is traveling with uniform velocity, so a = 0. In this case distance covered by bus can be
found by using the equation: = vt = = 10t S = = 10t S =
(2)
Distance is common for both car and bus, so comparing equation (1) and (2), we get: t 2
= 10t =
⇒
2 t = 10 t
= 10s t = Ram’s Outline Series
3.1 Problems
29
Figure 3.2: Problem 3.2 Now putting the value of t in equation (1), we can find the distance travelled by car: t in = t 2 = (10)2 = 100m S = = 100m Ans. S =
The velocity which the car overtakes the bus can be found by first equation of motion as: = 0 + 2 × 10 = 20 v f = vi + at =
v f = 20m/s Ans.
12m/s. At a height of 80 80m above the ground, a Problem 3.3. A helicopter ascending ascending at a rate of 12 package is dropped. dropped. How long does the package take to reach the ground? ground?
Solution: Consider upward upward motion from B to C: Initial velocity of the package at B: vi = 12m/s Final velocity of package at C: v f = 0m/s Time from B to C is t 1 = ? And acceleration due to gravity for upward motion is -ve: g = −9.8m/s2 Using the first equation of motion: v f = vi + at 0 = 12 + (−9.8)t 1 = 12 − 9.8t 1 =⇒ 9.8t 1 = 12 12 t 1 = 9 .8 t 1 = 1.2s Ans. Ram’s Outline Series
Chapter 3. MOTION
30
To find the distance h1 from B to C, we use 2nd equation of motion: 1 = vit + + gt 2 S = 2 1 h1 = 12 × 1.2 + (−9.8)(1.2)2 = 14.4 − 4.9 × 1.44 = 14.4 − 7.06 2 h1 = 7.34m
(1)
After reaching the maximum height, package will moves downward with +ve g. The downward
motion from C to A is: h2 = 80 + h1 = 80 + 7.34 h2 = 87.34m vi = 0, g = 9.8m/s2 and t 2 =? Using the equation of motion 1 h2 = vit 2 + gt 22 2 1 87.34 = 0 × t 2 + 9.8t 22 = 4.9t 22 2 . 87 34 = 17.82 t 22 = 4 .9 t 2 = 4.22s Ans.
Total time the package took to reach the ground is: = t 1 + t 2 = 1.2 + 4.22 = 5.42 t = Total time = 5.42s Ans.
Problem 3.4. A boy throws a ball upward upward from the top of a cliff with a speed of 14 14.7m/s. On the way down it just misses the thrower and falls the ground 49 metres below. below. Find (a) How long the ball rises? (b) How high it goes? (c) How long it is in air and (d) with what velocity it strikes the ground. Ram’s Outline Series
3.1 Problems
31
Solution: Initia Initiall velocity velocity of of ball is is vi = 14.7m/s Height of the cliff is h1 = 49m Final velocity at maximum height is v f = 0 (i) Time taken to reach the maximum height is t 1 = ? (ii) Maximum height is h2 =? = ? (iii) Total time taken is t = (iv) Final velocity of ball at ground is v f =? (i) For upward motion:
Time t 1 for upward motion can be calculated as (take g = −9.8m/s2 ) v f = vi − gt 1 0 = 14.7 − 9.8t 1 14 .7 = 1.5 t 1 = 9 .8 t 1 = 1.5s Ans.
(ii) Maximum height reached: According to 3rd equation of motion: 2gh2 = v f 2 − v2i 2(−9.8) × h2 = 02 − (14.7)2 = −216.09 .09 − 19.6h2 = −216.09 =⇒ h2 = 216 19.6 h2 = 11.025m Ans. (iii) Let h3 is the total height from maximum to the ground, then: h3 = h1 + h2 = 49 + 11.025 = 60.025m Ram’s Outline Series
Chapter 3. MOTION
32
Ball will take time t 2 to cover this distance according to: 1 + gt 2 h3 = vit + 2 where vi is the initial velocity of ball at maximum height which is zero. 1 60.025 = 0 + × 9.8t 22 = 4.9t 22 2 √ 60 . 025 = 12.25 =⇒ t 2 = 12.25 = 3.5s t 22 = 4 .9 Total time taken is: is : = t 1 + t 2 = 1.5 + 3.5 = 5s t = = 5s Ans. t =
(iv) Final velocity just before hitting the ground: v f = vi + gt 2 = 0 + 9.8 × 3.5 = 34.5 v f = 34.5m/s Ans.
Problem 3.5. A helicopter weighs 3920 Newton. Calculate the force on it if it is ascending up at a rate of 2 2m/s2 . What will be force on helicopter if it is moving up with the constant speed of 4 m/s? Solution: Weight of helicopter W = 3920 N (i) Acceleration a = 2m/s2 , F 1 = ? (ii) F 2 = ? if the velocity is constant v = 4m/s. When helicopter is ascending up with acceleration then there is net force on it according to Newton’s
2nd law is: = ma unbalanced force= ∑ F = 3920 Where m is the mass of helicopter which can be found as m = W g = 9.8 = 400kg. The direction of is downward while external force which helicopter is applying is upward, therefore net sum of W is
forces on it is F 1 − W or: = 400 × 2 + 3920 = 3920 + 800 = 4720 F 1 = ma + W =
F 1 = 4720 N Ans.
Force (F 2 ) on helicopter when it is ascending with constant velocity can be found as F − W = ma. In this case acceleration is zero: = m × 0 = 0 F 2 − W = F 2 = W = 3920
F 2 = 3920 N Ans. penetrates Problem 3.6. A bullet having a mass of 0.005 kg is moving with a speed of 100 m/s. It penetrates into a bag of sand and is brought to rest after moving 25cm into the bag. Find the deceleration
force on the bullet. Also calculate the time in which it is brought to rest. rest. Given Data: Mass of bullet mb = 0.005kg To find: Initial velocity of bullet vi = 100m/s Decelerating force on bullet F = = ? Distance covere covered d S = 25cm = 25 × 10−2 m = Time in which bullet is brought to rest t = = ? 0.25m Final velocity of the bullet = v f = 0 Ram’s Outline Series
3.1 Problems
33
Solution: According to Newton’s Newton’s 2nd law law of motion, decelerating force on bullet bullet is given by: = mb a F = Where a is the acceleration and can be found by 3rd equation of motion 2aS = = v f 2 − v2i 2 × a × 0.25 = (0)2 − (100)2 = −10000 10000 0.5a = −10000 =⇒ a = − 0.5 2 a = −20000m/s Putting this value in equation (1), we get: = 0.005 × (−20000) = −100 F =
(1)
Decelerating force=100 N Ans.
Now time can be calculated by using first equation of motion: v f = vi + at 0 = 100 + (−20000)t = = 100 − 20000t =⇒ 20000t = = 100 100 = = 0.005s t = 20000 = 0.005s Ans. t =
Problem 3.7. A car weighing 9800 N is moving with a speed of 40 km/h. On the application of the brakes it comes to rest after traveling a distance of 50 metres. Calculate the average retarding force force Solution: Weight of the car W = 9800 N ×1000 = 11.11m/s Initial velocity vi = 40km/h = 403600 Final velocity v f = 0m/s = 50m Distance covered S = = ? Average retarding force F = = ma, where a is accelRetarding force on car can be calculated using Newton’s 2nd law: F = eration of the car and m is the mass of car. Acceleration can be found by using 3rd equation of
motion: = v f 2 − v2i 2aS = 2 × a × 50 = (0)2 − (11.11)2 = −123.457 123.457 = −1.235m/s2 100a = −123.457 =⇒ a = − 100 W 9800 Mass can be calculated as: m = = = 1000kg 9 .8 g Using these values: = ma = 1000 × (−1.23457) = −1234.57 N F = = −1234.57 N Ans. F =
Problem 3.8. An electron in a vacuum tube starting from rest is uniformly accelerated by an electric field so that it has a speed 6 6 × 106 m/s after covering a distance of 1.8cm. Find the forc forcee − 31 acting on the electron. Take Take the mass of electron as 6.1 × 10 kg. Solution: Initia Initiall velo velocity city vi = 0m/s Final velocity v f = 6 × 106 m/s Ram’s Outline Series
Chapter 3. MOTION
34
= 1.8cm = 0.018m Distance covered S = Mass of the electron m = 9.1 × 10−31 kg = ? Force on electron F = Force on electron can be found by using Newton’s 2nd law of motion F = ma, where a is acceleration accelera tion and m is the mass of electron. Acceleration can be found by using 3rd equation of
motion: = v f 2 − v2i 2aS = 2 × a × 0.018 = (6 × 106 )2 − (0)2 = 36 × 1012 36 × 1012 0.036a = 36 × 1012 =⇒ a = = 1015 0.036 Now force on electron can be calculated: = ma = 9.1 × 10−31 × (1015 ) = 9.1 × 10−16 N F = = 9.1 × 10−16 N Ans. F =
Problem 3.9. Two bodies A and B are attached to the ends of a string which passes over a pulley, so that the two bodies hang vertically. If the mass of the body A is 4.8 kg. Find the mass of body B 0.2m/s2 . The value of g can be taken as 9.8m/s2 . which moves down with an acceleration of 0 Solution: Mass of body body A is m2 = 4.8kg Acceleration of body B is a = 0.2m/s2 Mass of body B is m2 =? We know that when two bodies hang vertically vertically,, then acceleration is: ( m1 m2 ) g (m1 + m2 ) ( m 4.8) 0.2 = 1 9 .8 (m1 + 4.8) 0.2 ( m1 4.8) = = 9.8 (m1 + 4.8)
a =
−
−
−
⇒
1 ( m1 − 4.8) = 49 (m1 + 4.8)
By cross multiplication: (m1 − 4.8) (m1 + 4.8) = 49 49m1 − 49 × 4.8 = 49m1 − 235.2 =
m1 − 49m1 = −235.2 − 4.8 =⇒ −240 = 5 m1 = −48 m1 = 5kg Ans.
−48m1 = −240
Problem 3.10. Two bodies of masses 10.2 kg and 4.5 kg are attached to the ends of a string which passes over a pulley in such a way that the body of mass 10.2 kg lies on a smooth surface and the
other body hangs verticall vertically. y. Find the acceleration of the bodies and tension of the string and also the force, which the surface exerts, on the body of mass 10.2 kg.
Given Data: Mass of body A: m1 = 4.5kg Mass of body B: m2 = 10.2kg Acceleration due to gravity: g = 9.8m/s2
To find: (i) Acceleration of bodies: a =? (ii) Tension in string: T =? = ? (iii) Force of surface: F =
Ram’s Outline Series
3.1 Problems
35
Solution: (i) For such system, acceleration of bodies bodies can be found as: a =
4.5 × 9 .8 m1 g = = 3 m1 + m2 4.5 + 10.2
a = 3m/s2 Ans.
(ii) The tension in the string can be found by: T =
(iii) Force on body due to surface is:
m1 m2 g 4 .5 × 10.2 × 9.8 449 .82 = = 4.5 + 10.2 14.7 m1 + m2 T = 30.6 N Ans.
= W 2 = m2 g = 10.2 × 9.8 = 99.96 N F = = 99.96 N Ans. F =
Problem 3.11. A 100 grams bullet is fired from a 10 kg gun with a speed of 1000 m/s. What is the speed of recoil of the gun? Given Data: 100 kg = 0.1kg Mass of bullet: m b = 100g = 1000 Mass of gun: m g = 10kg Velocity of bullet before firing: u b = 0m/s Velocity of gun before firing: u g = 0m/s Velocity of bullet after firing: v b = 1000m/s To find: Velocity of gun after firing: v g =? Solution: According to the law of conservation conservation of linear momentum: momentum: mb ub + mg ug = mb vb + mg vb 0.1 × 0 + 10 × 0 = 0.4 × 1000 + 10vg = 100 + 10vg 0 = 100 + 10vg =⇒ −10vg = 100 −100 = −10 vg = 10 veoclity of gun: v g = −10m/s Ans. The negative sign shows that the direction of velocity of gun is opposite the direction of velocity of
the bullet. It is also known as recoil velocity.
Problem 3.12. A 50 grams bullet is fired into a 10 kg block that is suspended by a long cord so that it can swing as a pendulum. If the block is displaced so that its centre of gravity rises by 10cm, what was the speed of the bullet? Given Data: 50 kg = 0.05kg Mass of bullet: m b = 50g = 1000 Mass of block: m B = 10kg Velocity of block before impact: u B = 0m/s Velocity of block after impact: v B Velocity of bullet after impact: v b 10 m = 0.1m height: h = 10cm = 100 To find: Velocity of bullet before impact: vb =? Ram’s Outline Series
Chapter 3. MOTION
36
Solution:
This is type of inelastic inelastic collision collision because because two bodies bodies after collision collision (impact) (impact) stick stick together. In inelastic collision, the total momentum conserves but total KE doesn’t conserve. After
impact the velocity of both bodies is same (common): v = vb = v B According to law of conservation of linear momentum: mb ub + m B u B = mb vb + m B v B 0.05ub + 10 × 0 = 0.05v + 10v ∵ (v = vb = v B ) 10 .05 0.05ub = 10.05v =⇒ ub = v 0.05 ub = 201v
(1)
Now according to law of conservation og energy (since KE doesn’t remains same but converts into
PE): loss of KE=gain of PE 1 ( m + m ) v2 = ( m m B ) gh b + 2 b B 1 2 v = gh =⇒ v2 = 2gh =⇒ v = 2gh 2 √ √ v = 2 × 9.8 × 0.1 = 1.96 = 1.4m/s Putting this value of v in equ(1): ub = 201v = 201 × 1.4 = 281.4
velocity of bullet before impact: u b = 281.4m/s Ans.
Problem 3.13. A machine gun fires 10 bullets per second into a target. Each bullet weighs 20 gm and had a speed of 1500 m/s. Find the force necessary to hold the gun in position. 20 = 0.02kg Solution: Mass of bullet bullet : m = 20g = 20 × 10−3 = 1000 = 10, so the mass of 10 bullets is M = = N × m = 10 × 0.02 = 0.2kg Number of bullets: N = = 1s and initial velocity of bullets is zero (vi = 0) Time: t = Final velocity of bullets is v f = 1500m/s. = ? Force necessary to hold the gun in position is: F =
Force necessary to hold the gun=change in monetum of bullets per second Mvv f − Mvi 0 .2 × 1500 − 0.2 × 0 ∆P M = = = = 300 − 0 F = 1 t t = 300 N Ans. F = going up a slope of 30 pedaling, 30o with a speed of 3.5 m/s. If he stops pedaling, Problem 3.14. A cyclist is going how much distance will he move before coming to rest? (Assume the friction to be negligible).
Solution: Initia Initiall speed of cyclist: cyclist: v i = 3.5m/s Final speed (as cycle stops): v f = 0m/s Angle of inclination: θ = 30o =? Distance covered: S = Ram’s Outline Series
3.1 Problems
37
The total distance covered by cyclist before coming to rest is given by 3rd equation of motion: = v f 2 − v2i 2aS = Where a is acceleration of cycle on slope: = −9.8sin30o = −9.8 × 0.5 = −4.9m/s2 a = −g sin θ = The negative sign shows that velocity of cycle is reducing. Putting the value of "a" into above equation: = ( 0)2 − (3.5)2 = −12.25 2 × (−4.9)S = 12 .25 = = = 1.25 − 9.8S = − 12.25 =⇒ S = 9 .8 = 1.25m Ans. S =
Problem 3.15. The engine of a motorcar moving up 45o slop with a speed of 63 km/h stops working suddenly. How far will the car move before coming to rest? (Assume the friction to be negligible).
×1000 = 17.5m/s Initiall velocity velocity of of car: v i = 63km/h = 633600 Solution: Initia Final velocity of car (as car is going to stop): v f = 0m/s Angle of inclination: θ = 45o =?. Distance covered before coming to rest: S = Using the 3rd equation of motion, we can find distance covered by car before coming to rest: = v f 2 − v2i 2aS = Where a is acceleration of car on inclination: = −9.8sin45o = −9.8 × 0.707 = −6.928m/s2 a = −g sin θ = The negative sign shows that car is going to stop. Putting the value of "a" into above equation: = ( 0)2 − (17.5)2 = −306.25 2 × (−6.928)S = 306 .25 = = = 22.102 − 13.856S = − 306.25 =⇒ S = 12.856 = 22.102m Ans. S =
Problem 3.16. In question 3.15, find the distance that the car moves, if it weighs 19.600N and the frictional force is 2000 N.
×1000 = 17.5m/s Solution: Initia Initiall velocity velocity of of car: v i = 63km/h = 633600 Final velocity of car (as car is going to stop): v f = 0m/s Angle of inclination: θ = 45o =?. Distance covered before coming to rest: S = Weight of car: W = 19600 N Frictional force: f = 2000 N Ram’s Outline Series
Chapter 3. MOTION
38
Using the 3rd equation of motion, we can find distance covered by car before coming to rest: = v f 2 − v2i 2aS = Where a is acceleration of car on inclination: Acceleration of car on inclination with friction can be found by: + f ) − ma = (mg sin θ + The negative sign shows that car is going to stop. W 19600 = 2000kg Here m is the mass of the car: m = = 9 .8 g o ∴ −(2000)a = ( 2000 × 9.8 × sin45 + 2000) = 19600 × 0.707 + 2000 .2 = −7.928m/s2 − 2000a = 13857.2 + 2000 = 15857.2 =⇒ a = − 15857 2000 Putting the value of "a" into equation(1): = ( 0)2 − (17.5)2 = −306.25 2 × (−7.928)S = 306 .25 = = = 19.313m − 15.857S = − 306.25 =⇒ S = 15.857 = 19.313m Ans. S =
(1)
Problem 3.17. In the Figure 3.3 find the acceleration of the masses and the tension in the string.
Figure 3.3: Problem No. 3.17 Solution: Consider the downward motion of block A: Weight of block A: W 1 = 98 N Mass of block A: m1 = W g1 = 998.8 = 10kg Slope of inclined plane: θ = 30o Tension in the string: T =? Acceleration of the system: a =? Resultant force on block A is: F 1 = W 1 − T =⇒ m1 a = W 1 − T (∵ F 1 = m1 a) 10a = 98 − T
Consider the upward motion of block B: Weight of block B: W 2 = 147 N Ram’s Outline Series
(1)
3.1 Problems
39
Mass of block B: m2 = W g2 = 147 9.8 = 15kg Resultant force on block B is: F 2 = T − W 2 sin θ =⇒ m2 a = T − W 2 sin θ (∵ F 2 = m2 a) 15a = T − 147sin30 o = T − 147 × 0.5 = T − 73.5 15a = T − 73.5 Adding equ(1) and (2), we get: 10a = 98 − T 15a = T − 73.5
(2 )
25a = 98 − − 73.5 T + T − 24 .5 25a = 24.5 =⇒ a = = 0.98m/s2 25 a = 0.98m/s2 Ans. Tension can be found by equation (1): 10 × 0.98 = 98 − T =⇒ 9.8 = 98 − T =⇒ T = 98 − 9.8 = 88.2 N = 88.2 N Ans. T =
Problem 3.18.
Two blocks are connected connected as shown in Figure Figure 3.4. If the pulle pulleyy and the planes on which the blocks are resting are frictionless, find the acceleration of the blocks and the tension in
the string.
Figure 3.4: Problem No. 3.18 Solution: Consider the downward motion of block A: Weight of block A: W 1 = 490 N Mass of block A: m1 = W g1 = 490 9.8 = 50kg Tension in the string: T =? Acceleration of the system: a =? Resultant force on block A is: F 1 = W 1 sin θ 1 − T =⇒ m1 a = W 1 sin θ 1 − T (∵ F 1 = m1 a) 50a = 490sin30o − T = 490 × 0.5 − T = = 245 − T 50a = 245 − T
Consider the upward motion of block B: Weight of block B: W 2 = 245 N Ram’s Outline Series
(1)
Chapter 3. MOTION
40
Mass of block B: m2 = W g2 = 245 9.8 = 25kg Resultant force on block B is: F 2 = T − W 2 sin θ 2 =⇒ m2 a = T − W 2 sin θ 2 (∵ F 2 = m2 a) 25a = T − 245sin60 o = T − 245 × 0.866 25a = T − 212.17 Adding equ(1) and (2), we get: 50a = 245 − T 25a = T − 212.17
(2 )
− 212.17 75a = 245 − T + T − 32 .83 75a = 32.83 =⇒ a = = 0.437m/s2 75 a = 0.437m/s2 Ans. Tension can be found by equation (1): 50 × 0.437 = 245 − T =⇒ 21.5 = 245 − T =⇒ T = = 245 − 21.5 = 223.5 N = 223.5 N Ans. T =
Problem 3.19. Two blocks each weighing 196N rest on planes as shown in Figure 3.5. If the planes and pulleys are frictionless, find the acceleration and tension in the cord.
Figure 3.5: Problem No. 3.19
Solution: Weight of block A placed on the inclined inclined surface: W 1 = 196 N W 1 196 Mass of block A: m1 = g = 9.8 = 20kg Weight of block B placed on the flat surface: W 2 = 196 N Mass of block B: m2 = 20kg Angle of inclination: θ = 300 Acceleration of blocks: a =? Tension in the string: T =? Consider the downward motion of block A: Resultant force on block A is: F 1 = W 1 sin θ − T =⇒ m1 a = W 1 sin θ − T (∵ F 1 = m1 a) = 98 − T 20a = 196sin30o − T = 196 × 0.5 − T = 20a = 98 − T Ram’s Outline Series
(1)
3.1 Problems
41
Since block B is movi moving ng in the direction direction of T along x-axis, therefore, unbalanced force acting on T along
block B is T : T = m2 a = 20a Subsituting this value of T in equation (1): 98 20a = 98 − 20a =⇒ 40a = 98 =⇒ a = = 2.45m/s2 40 a = 2.45m/s2 Ans.
Now subsituting the value of "a" in equation (2), we get: T = 20a = 20 × 2.45 = 49 N = 49 N Ans. T =
Ram’s Outline Series
(2)
4. MOTION IN TWO DIMENSION
4.1 4. 1
Prob Pr oble lems ms Problem 4.1. A rescue helicopter drops a package of emergency ration to a stranded party on the ground. groun d. If the helicopter is traveling traveling horizontally horizontally at 40 m/s at a height of 100 m above the ground, (a) where does the package strike the ground relative to the point at which it was released? (b)
What are the horizontal and vertical component of the velocity of the package just before it hits the ground?
Data: When ration package leaves helicopter, then it gets the velocity of helicopter which becomes package’s initial velocity in horizontal di-
rection: vix = 40m/s Initial velocity in vertical direction: viy = 0m/s
Since origin of coordinate system lies at the dropping point of the package so vertical distance
(height) covered would be negative in y-axis: h = −100m In this case g is also along − y so: g = −9.8m/s2 To find: (a) Horizontal distance covered: x =? (b) v f x = ? and v f y =?
Solution: (a) Since helicopter is moving along horizontal direction, therefore there would be horizontal velocity in the package just after the releasing off the helicopter. There is no acceleration along the
horizontal direction so we can find the distance covered by the equation: 1
x = vixt Ram’s Outline Series
Chapter 4. MOTION IN TWO DIMENSION
44
where time t can can be calculated using 2nd equation of motion: 1 + gt 2 h = viyt + 2 − 100 = 0 × t − 12 × 9.8 × t 2 = −4.9 × t 2 √ 100 = 20.41 =⇒ t = = 20.41 = 4.52s t 2 = 4 .9 Putting the value of time into equation (1), we get: x = 40 × 4.52 = 180.7m x = 180.7m Ans.
(b) There is no acceleration along the horizontal direction, so velocity along this direction remains
uniform throughout motion which is equal to helicopter’s horizontal velocity: v x = v f x = vix = 40m/s v x = 40m/s Ans.
Final velocity along y-axis can be found by using 1st equation of motion: = 0 − 9.8 × 4.52 = −44.3 v f y = viy − gt = v f y = −44.3m/s Ans.
Negative sign shows that the direction of velocity component is along negative y-axis. Problem 4.2. A long-jumper leaves the ground at an angle of 20 20o to the horizontal and at a speed of 11m/s (a) How far does he jump? (b) What is the maximum height reached? Assume the motion of the long jumper is that of projectile. Data: Initial speed of the long-jumper: v o = 11m/s Angle of projection with x-axis: θ = 20o To find: (a) Horizontal distance covered (range): R =? (b) Vertical Vertical distance (maximum height reached): h =? Solution: (a) It is assumed that the motion of the long jumper is projectile. Range or total horizontal distance
of long jumper can be found by: ( 11)2 v 2o 121 = sin2 × 20o = sin40o R = sin2θ = 9 .8 9 .8 g o R = 12.35 × 0.6428 (∵ sin40 = 0.6428) R = 7.94m Ans. (b) Height of the long jumper can be found by: v2 ( v sin θ )2 h = oy = o ∵ voy = v0 sin θ 2g 2g ( 11sin20o )2 ( 11 × 0.3420)2 ( 3.7622)2 14 .154 = = = = 0.722m h = 2 × 9 .8 19.6 19.6 19.6 h = 0.722m Ans.
Ram’s Outline Series
4.1 Problems
45
Problem 4.3. A stone is thrown upward from the top of a building at an angle of 30 o
to the horizontal and with a initial speed of 20 m/s. If the height of building is 45 m. (a) Calculate the totall time the ston tota stonee in fligh flightt (b) What is the speed of stone just before before it stri strikes kes the ground? ground? (c)
Where does the stone strike the ground?
Data: Initial speed of the stone: vo = 20m/s Angle of projection with x-axis: θ = 30o Height of the building: h = − y = −45m To find: (a) Total time of flight: t = = ? (b) Final speed of stone just before it strikes the
ground: v =? (c) Horizontal distance covered: R =?
Solution: First we will find the horizontal and vertical components of the initial velocity vo : = 20cos30o = 20 × 0.866 v0 x = vo cos θ = vox = 17.32m/s Similarly: voy = vo sin θ = 20sin30o = 20 × 0.5 voy = 10m/s
(a) Total time of flight can be calculated using the 2nd equation of motion: 1 + gt 2 h = y = voyt + 2 1 + × (−9.8) × t 2 =⇒ −45 = 10t − 4.9t 2 − 45 = 10 × t + 2 2 4.9t − 10 − 45 = 0 This is quadratic equation and can be solved by: √ − b ± b2 − 4ac = Quadratic Formula t = 2a
Here a = 4.9, b = −10 and c = −45, putting these values in quadratic formula, we get: √ √ (−10) ± (−10)2 − 4 × 4.9 × (−45) 10 ± 100 + 882 10 ± 982 − = = = t = 2 × 4 .9 9 .8 9 .8 10 ± 31.34 10 − 31.34 −21.34 = = = = −2.175s t = t = 9 .8 9 .8 9 .8 10 + 31.34 41 .34 = = = 4.22s OR t = 9 .8 9.8 Since time can not be negative so total time of flight is:
= 4.22s t =
Ram’s Outline Series
Chapter 4. MOTION IN TWO DIMENSION
46
(b) Final velocity of the stone at the ground is: v =
v x2 + v y2 where v x is horizontal and v y are vertical components.
In projectile motion the horizontal component of velocity remains constant since there is no acceleration: v x = vox = 17.32
v x = 17.32m/s
Vertical component can be found by using first equation of motion: = 10 + (−9.8) × 4.22 = 10 − 41.34 = −31.34m/s v y = voy + gt = v y = −31.34m/s
(-ve sign shows that the direction of velocity is in the -y-axis )
The resultant velocity can be found by: v =
v x2 + v y2 =
(17.32)2 + ( 31.34)2 =
−
√ 300 + 982 = √ 1282 = 35.8
m/s
v = 35.8m/s Ans.
(c) Horizontal distance covered (range) is given by: = 17.32 × 4.22 = 73m R = v xt = R = 73m Ans.
Problem 4.4. A ball is thrown in horizontal direction from a height of 10 m with a velocity of 21 m/s (a) How far will it hit the ground from its initial position on the ground? and with what
velocity?
Data: Initial horizontal velocity of the ball: v ox = 10m/s Initial vertical velocity of the ball: v oy = 0m/s Vertical distance covered (height) h = − y = −10m To find: (a) Horizontal distance covered: R =? (b) Final velocity of the ball just before it strikes the ground: v =? Solution: (a) Horizontal distance (range) is given by the equation: R = v x × t −−−− > (1) where time "t" has to be calculated first using the motion equation: 1 + gt 2 h = y = voyt + 2 1 + (−9.8)t 2 = 0 − 4.9t 2 =⇒ 4.9t 2 = 10 − 10 = 0 × t + 2 √ 10 = 2.041 = 1.429s t 2 = = 2.041 =⇒ t = 4.9 Putting this value into equation (1), we get: R = 21 × 1.429 = 30m Horizontal distance = 30m Ans.
Ram’s Outline Series
4.1 Problems
47
(b) Net velocity of the ball at the ground is given by: v =
v x2 + v y2 − −− > (2) where v x is horizontal and v y are vertical components.
The horizontal component of velocity remains constant v x = vox = 21
∵
a = 0
v x = 21m/s
Vertical component can be found by using first equation of motion: = 0 + (−9.8) × 1.429 = 0 − 14 = −14m/s v y = voy + gt = v y = −14m/s
(-ve sign shows that the direction of velocity is in the -y-axis )
Putting these values into equation (2), we get net velocity: v =
(21)2 + ( 14)2 =
−
√ 441 + 196 = √ 637 = 25.24
ms/s
v = 25.24m/s Ans.
Problem 4.5. A rocket is launched at an angle of 53 53o to the horizontal with an initial speed of 100 m/s. It moves along its initial line of motion with an acceleration of 30 30m/s2 for 3s. At this time the engine fails and the rocket proceeds to move as a free body. Find (a) the maximum altitude reached by the rocket (b) its total time of flight, and (c) its horizontal range.
For first 3 seconds, the motion of rocke rockett is not projectile motion as engine is providing force and
when the engine fails then it follows projectile motion: Data: Launch angle of the rocket: θ = 53o Initial speed: vo = 100m/s with acceleration of: a = 30m/s2 Time taken before the engine failed: t 1 = 3s To find: (a) Maximum height reached: h =? Ram’s Outline Series
Chapter 4. MOTION IN TWO DIMENSION
48
= ? (b) Total time of flight: t = (c) Horizontal range: R =?
Solution:
When rocket fails, fails, the final velocity which which rocket achieved achieved during 3 seconds of flight will become the initial velocity of the rocket for following projectile motion. Therefore first we
will calculate the speed the rocket reaches after the engines have failed: = 100 + 30 × 3 = 100 + 90 = 190m/s. v f = vo + at = This velocity v f = 190m/s would be initial velocity for projectile motion: v o = v f = 190m/s. (a) The maximum altitude of the rocket is h = h1 + h2 , where h1 is the height reached when the engines
are running and h2 is the height reached when rocket follows projectile motion.
h1 can be calculated as: 1 + at 2 = 100sin53o × 3 + 0.5 × 30 × (3)2 = 300 × 0.798 + 15 × 9 h1 = vo sin θ t t + 2 h1 = 239.4 + 135 = 374.4m And h2 is given by: v f 2 − v2oy = 2gh2 ∵ v f = 0, v oy = vo sin θ 0 − (v0 sin θ )2 = 2gh2 =⇒ −(190sin53 0 )2 = 2(−9.8)h2 (190 × 0.7986)2 ( 151.74)2 23025.25 − = = = 1174.76m h2 = 19.6 19.6 −19.6 ∴ h = h1 + h2 = 374.4 + 1174.76 = 1549.2m h = 1549.2m Ans. (b) The total time is the falling time plus the time the engines are running plus the time to reach
maximum height: = t 1 + t 2 + t 3 = 3 + t 2 + t 3 −−−− > (1) t = v sin θ 190sin530 151 .74 = = = 15.48s where t 2 = 0 9 .8 9.8 g And t 3 can be worked out as: v f − v0 v f − 0 v f = = = −−−− > (2) v f = vo + gt =⇒ t = g g g 2 where final velocity can be found as: v f − v2o = 2gh
v f 2 = 2gh − 0 = 2gh =⇒ v f = 2gh Putting this value into eqation (2), we get: √ 2gh √ 2 × 9.8 × 1549.2 √ 30364.32 174 .25 = = = = 17.78s t 3 = 9 .8 9.8 9 .8 g Putting the values of t t 2 and t 3 into equation (1):
= 3 + 15.48 + 17.78 = 36.3s t = = 36.3s Ans. t = (c) Horizontal range R = R1 + R2 , where R1 is the range when engines are running while R2 is the
Ram’s Outline Series
4.1 Problems
49
range when engines are failed. First we calculate R1 and R2 as follow: R1 can be worked out using the equation of motion: 1 1 1 + a xt 2 = vo cos θ + + a cos θ t S x = v xt + t 2 = 100cos530 + × 30cos53o × (3)2 2 2 2 R1 = S x = 100 × 0.602 × 3 + 15 × 0.602 × 9 = 180.54 + 81.25 = 261.8m And R3 = v x (t 2 + t 3 ) = vo cos θ × 33.3 = 190cos53o × 33.3 = 3807.7m ∴ R = R1 + R2 = 261.8 + 3807.7 = 4070m R = 4070m Ans.
Problem 4.6. A diver leaps from a tower with an initial horizontal velocity component of 7 m/s and upward velocity component of 3 m/s. find the component of her position and velocity after 1
second.
Data: Initial horizontal velocity of the diver diver:: vox = 7m/s
Initial vertical velocity of the diver: v oy = 3m/s To find: (i) Horizontal distance covered: x =? (ii) Vertical distance covered: y =? (iii) Final horizontal velocity component: v x = ? (iv) Final vertical velocity component: v y =?
Solution: (i) Horizontal distance covered by the diver is given by: = 7 × 1 = 7m x = vox × t =
x = 7m Ans. (ii) Vertical Vertical distance covered by the diver is given by: 1 1 + gt 2 = 3 × 1 + (−9.8) × (1)2 = 3 − 4.9 × 1 = 3 − 4.9 = −1.9m y = voyt + 2 2 y = −1.9m Ans.
(iii) In projectile motion, the horizontal velocity component remains constant, ∴
v x = v0 x = 7m/s v x = 7m/s Ans.
(iv) Vertical Vertical velocity component after one second is given by: = 3 + (−9.8) × 1 = 3 − 9.8 = −6.8m/s v y = voy + gt = v y = −6.8m/s Ans.
Problem 4.7.
A boy standing 10m from a buildi building ng can just barely reach reach the roof 12m above him when he throws a ball at the optimum angle with respect to the ground. Find the initial velocity
components of the ball. Ram’s Outline Series
Chapter 4. MOTION IN TWO DIMENSION
50
Data: Height of the building: y = 12m Horizontal distance: x = 10m At optimum angle θ , the ball will be at its maximum height ( y = 12m) when it barely reaches the roof which is 10m away and 12m above. At the maximum height, the ball’s final vertical velocity v y would be 0, but his
horizontal velocity remains constant. To find: Horizontal velocity component: v ox = v x = ? Vertical velocity component: v oy
Solution:
Choosing up and forward as positive positive and then solving for both both components of velocity:
For vertical motion: voy can be found by: v f 2 − v2i = 2gh, Here in our case: h = y = 12, v f = 0, v i = voy , and for upward motion g is taken -ve: 0 − v2oy = 2 × (−9.8) × 12 = −235.2
√
voy = 235.2 = 15.34m/s voy = 15.34m/s
For horizontal direction: x −−− −− > (1) x = v xt =⇒ v x = − t where time ’t’ can be calculated from data of vertical motion:
v f y − voy 0 − 15.34 −15.34 = −9.8 = −9.8 = 1.565s g Putting tha value of time in equation (1): 10 x = 6.4m/s v x = = t 1.565 v x = 6.4m/s voy = 15.34m/s Ans. = v f y = voy + gt =⇒ t =
Problem 4.8. A mortar shell is fired at a ground level target 500m distance with an initial velocity of 90 m/s. What is its launch angle? Data: Horizontal range of the shell: R = 500m? Initial velocity of the shell: v o = 90m/s To find: The launch angle of the shell: θ =? Solution:
Since Sinc e range of the mortal mortal shell is given given so we can find the launch angle angle by using the
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projectile range equation: v 2 R = o sin2θ g ( 90)2 8100 500 = sin2θ = = sin2θ = = 826.53sin2θ 9 .8 9 .8 500 sin2θ = = = 0.6049 =⇒ 2θ = sin−1 (0.6049) = 37.2o 826.53 37 .2o = = 18.6o θ = 2
In projectile motion the there are two launch angles for the same range according to the relation: θ 1 + θ 2 = 90
We have found one launch angle, the second launch angle can be: θ 2 = 90
− 18.6 = 71.4o
Hence mortal shell could be fired at 18.6o OR at 71.4o for the same range of R R = 500m.
Problem 4.9. What is the take off speed of a locust if its launch angle is 55o and its range is 0.8m? Data: Horizontal range of the locust: R = 0.8m? Launch angle of the locust: θ = 55o To find: Take off speed of the locust: v o =? Solution: off speed:
As range is given, given, so we can use the equation equation for range of the projectile projectile to find the take
v 2 R = o sin2θ g v2 v2 v2 0.8 = 0 si sinn 2 × 55o = o sin110o = o × 0.9397 9 .8 9.8 9 .8 √ . . . × 0 8 9 8 7 84 = = 8.34 =⇒ vo = 8.34 = 2.89m/s v2o = 0.9397 0.9397 vo = 2.89m/s Ans.
Problem 4.10. A car is traveling on a flat circular track of radius 200m at 20 m/s and has a centripetal acceleration a acceleration ac = 4.5m/s2 (a) If the mass of the car is 1000 kg, what frictional force is required to provide the acceleration? (b) if the coefficient of static frictions µ s is 0.8, what is the
maximum speed at which the car can circle the track?
Data: Speed of car: v = 30m/s Centripetal acceleration: a c = 4.5m/s2 Radius of track: r = = 200m Mass of car: m = 1000kg To find: (a) Frictional force: f =? (b) Maximum speed: v max = ? when µ s = 0.8 Ram’s Outline Series
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Solution:
(a) Since car is travelling travelling in circular motion, motion, so centripetal centripetal force is required to keep car on the circu circular lar track. This force is provided by frict frictional ional force between between car and the track: f = F c ,
where F c is centripetal force:
f = F c = mac = 1000 × 4.5 = 4500 N f = 4500 N Ans.
(b) The maximum speed with which car moves is related to maximum frictional force:
where R is normal force and equal to its weight: R = mg = 1000 × 9.8 = 9800 N . ∴
f max max = µ s R,
f max max = 0.8 × 9800 = 7840 N .
This maximum force provides the centripetal force with which the car moves with maximum speed:
00 15698 m (vmax )2 f max max r 7849 × 2 =⇒ v2max = = = = 1569 10 00 10 r m √ vmax = 1569 = 39.6
f max max = mac =
vmax = 39.6m/s Ans.
Problem 4.11. The turntable of a record player rotates initially at a rate of 33 rev/min and takes 20s to come to rest (a) What is the angular acceleration of the turntable, assuming the acceleration is constant? (b) How many rotation does the turntable make before coming to rest? (c) If the radius of the turntable is 0.14m, what is the magnitude of the tangential acceleration of the bug at time t = 0? Data: Initial angular speed: ω i = 33rev/min Final angular speed: ω f f = 0 = 20s Time taken: t = = 0.14m Radius of turntable: r = To find: (a) Angular acceleration: α = ? = ? (b) Number of rotations completed θ = (c) Initial speed of bug: v i = ? = 0: at = = ? (d) Tangential acceleration of bug at t = Solution: First conv convert ert units units into SI: ω i = 33rev/min = 33/60 = 0.55rev/s But 1 revolution = 2 π radians, radians, hence: ω i = 0.55 × 2π rad rad /s = 3.46rad /s (a) Angular acceleration α is is found by: ω f f − ω i 0 − 3.46 = = = −0.173 rad/s2 α = t 20 α = = −0.173 rad/s2 Ans. Ram’s Outline Series
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(b) Numebr of cycles completed is angular distance: θ : = ω avg θ = avgt − −− > (1). Where ω avg avg is avergae angular velocity: ω i + ω f f 0 .55 − 0 = = 0.725rev/s. Put this value in equ(1): ω avg avg = 2 2 = 0.725 × 20 = 5.5 revolutions. θ = Number of revolutions completed in 20 seconds = 5.5 Ans.
ω i = 0.14 × 3.46 = 0.4844m/s (c) Initial linear (tangential speed):vi = r ω
vi = 0.4844m/s Ans.
α = (d) Tangential acceleration of the bug is given by:at = = r α = 0.14 × 0.173 = 0.0242
= 0.0242m/s2 Ans. at =
Problem 4.12.
Tarzan swings swings on a vine of length 4m in a vertical circle circle under the influ influence ence of o gravity. When the vine makes an angle of θ = 20 with the vertical, Tarzan has a speed of 5 m/s. Find (a) his centripetal acceleration at this instant, (b) his tangential acceleration, and (c) the
resultant acceleration.
Data: Tangential speed of Tarzan: v = 5m/s Length of vine (radius): r = = 4m Angle with vertical: θ = 20o To find: (a) Centripetal acceleration: a c =? (b) Tangential acceleration: at = ? (c) Resultant acceleration: a =?
Solution: (a) Centripetal acceleration of tarzan is given by: a c = ( 5)2
v 2 r
25 = 6.25m/s2 ac = 6.25m/s2 Ans. 4 4 = g sin θ = = 9.8sin20o (b) Tangential acceleration of Tarzan is given by: at = ac =
=
= 9.8 × 0.3420 = 3.35. at = = 3.35m/s2 Ans. at =
(c) Since ac and at are are perpendicular to each other, so a is given by: a =
a2c + at 2
=
(6.25)2 + (3.35)2 =
√ 39.06 + 11.24 = √ 50.3 = 7.1
a = 7.1m/s2 Ans.
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5. STATICS
5.1 5. 1
Prob Pr oble lems ms Locatee the centr centree of mass of a system of particles particles each of mass ‘m’, arranged arranged to Problem 5.1. Locat correspond corres pond in position to the six corners of a regular (planar) hexagon.
according to co-ordinate system. Take Take x- and y-axis such that origin Solution: Divide the system according O lies at the centre of the hexagon. The co-ordinates X c and Y c of the center of masses are given by: 6
6 mi x mi y xi yi , Y c = ∑ X c = ∑ mi mi i=1 i=1
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By expanding the equations: m x + m2 x2 + m3 x3 + m4 x4 + m5 x5 + m6 x6 X c = 1 1 m1 + m2 + m3 + m4 + m5 + m6 m (− x) + mx + mx + mx + m(− x) + m(− x ) = m+m+m+m+m+m + + + − − − mx + mx + mx mx mx mx = . X c = 0 6m And Y c : m y + m2 y2 + m3 y3 + m4 y4 + m5 y5 + m6 y6 Y c = 1 1 m1 + m2 + m3 + m4 + m5 + m6 + − − + + 0 + my + my + m(0) + m(− y) + m(− y) + m(0) my my my my + 0 = = 6m m+m+m+m+m+m 0 = . Y = 0 6m c Thus the coordinates of centre of masses are: ( X c , Y c ) = ( 0, 0).
Problem 5.2. Find the position of centre of mass of five equal-mass particles located at the five corners of a square-based right pyramid with sides of length ‘l’ and altitude ‘h’.
right pyramid pyramid is 3D structure, so we need three three coordinates to locate locate Solution: Since square based right the CM as given by:
5
5 5 mi x xi mi y yi mi z zi , Y c = ∑ , Z c = ∑ X c = ∑ mi mi mi i=1 i=1 i=1
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Using the above equations and figure, we can find the coordinates of center of masses as: m 1 x1 + m2 x2 + m3 x3 + m4 x4 + m5 x5 m1 + m2 + m3 + m4 + m5 + − − + m(0) 0 mx + mx − mx − mx + m(0) mx mx mx mx = = = . X = 0 5m 5m c m+m+m+m+m For Y c : m y + m2 y2 + m3 y3 + m4 y4 + m5 y5 Y c = 1 1 m1 + m2 + m3 + m4 + m5 + − − + m(0) 0 my + my − my − my + m(0) my my my my = = = . Y = 0 5m 5m c m+m+m+m+m For Z c : m z + m2 z2 + m3 z3 + m4 z4 + m5 z5 Z c = 1 1 m1 + m2 + m3 + m4 + m5 m (0) + m(0) + m(0) + m(0) + mh mh m h h h = = = = . Z c = 5m 5 5 m+m+m+m+m m 5 X c =
Hence the coordinates coordinates of center of masses are: ( X c , Y c , Z c ) = ( 0, 0, h/5). The center of mass lies at
h/5 along z-axis.
Problem 5.3. The mass of the sun is 329390 times the earth’s mass and the mean distance from the centre of the sun to the centre of the earth is 1.496 × 108 km km.. Treating the earth and sun as particles with each mass concentrated at the respective geometric centre, how far from the centre of the sun is the C.M (centre of mass) of the earth-sun system? Compare this distance with the mean radius of the sun (6.9960 × 105 km) Solution: Let m1 be the mass of the earth and m2 that of the sun. Let the centre of mass of the earth-sun system be located at distance r 1 from the centre of the earth and at distance r 2 from = r 1 + r 2 is the the centre of the sun, so that r = distance between the centres of earth and sun. Taking origin at the centre of mass of the sun, we
can find (whole system lies on x-axis): 2
X c = ∑
i=1
mi r i m 1 r 1 + m2 r 2 = mi m1 + m2
= 1.496 × 108 km = 1.496 × 1011 m; using this data we can find: Here m2 = 329390m1 and r =
X c =
m 1 × 1.496 × 1011 + 329390m1 × 0 1 .496 × 1011 m1 1 .496 × 1011 m 1 = = 329391m1 329391 m1 + 329390m1 m 1
X c = 4.54 × 105 m Ans. Coordinate X c is located at C in in figur figure. e. Center of earth-sun earth-sun system lies inside of the sun. Let mean
radius of the sun is r s = 6.9960 × 105 km = 6.9960 × 108 m, then comparing both values: 4.54 × 105 X c = = 6.49 × 10−4 r s 6.9960 × 108 Ram’s Outline Series
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Problem 5.4. A particle of mass 4 kg moves along the x-axis with a velocity v = 15t m/s, where t = 0 is the instant that the particle is at the origin. (a) at t=2s, what is the angular momentum momentum of the particle about a point P located on the +y-axis, 6m from the origin. (b) what torque about P acts
on the particle?
Data: Mass of particle: m = 4kg Time taken is: t = = 2s and the velocity of particle: v = 15tm /s = 15 × 2m/s = 30m/s Point P located on y-axis: y = 6m To find: (a) Angular momentum L = ? and (b) torque τ = = ?
Solution: (a) Angular momentum can be found as: where sin θ = = L = mvr sin θ where ∴
y 6 = PQ r
6 L = mv r × = 4 × 30 × 6 = 720. L = 720kgm2 /s Ans. r
(b) We know the torque acting on the particle is equal to the rate of change of angular momentum: L 720 = . τ = = 360 N .m Ans. 2 t Problem 5.5. A particle of mass ‘m’ is located at the vector position r and has a linear momentum = τ =
vector P. The vector r and P are non zero. If the particle moves only in the x, y plane, prove that
= 0. L x = L y = 0 and L z
Solution: are:
Particle Par ticle of mass mass m moves only in x-y plane, so its position vector and momentum vectors
= r x iˆ + r y j ˆ; r = P = P x iˆ + P y j ˆ = The angular momentum can be found as: L = r × P.
ˆ iˆ jˆ k r 0 ˆ r 0 ˆ r r = + k x y − L r x r y 0 = ˆi y j x P y 0 P x 0 P x P y x P y 0 P ˆ = ˆi(0) − ˆ j(0) + ˆk (r x P y , −r y P x ) = 0iˆ − 0 j ˆ + (r x P y , −r y P x )k
Comparing the components of L L with the coefficient of ˆ iˆ, jˆ and ˆk , we get: = 0. Hence proved. L x = 0 and L y = 0 and L z = ( r x P y , −r y P x )
Problem 5.6. A light rigid rod 1m in length rotates in the xy-plane about a pivot through the rod’s centre. centr e. Two particles particles of mass 2kg and 3kg are connected connected to its ends. Deter Determined mined the angular angular
momentum of the system about the origin at the instant the speed of each particle is 5m/s. Ram’s Outline Series
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Solution: Mass of particle particle A: m1 = 2kg Mass of particle B: m2 = 3kg Velocity of each particle: v = 5m/s Length of rod: L = 1m = 0.5m. Distance of each particle from the pivot: r = The total angular momentum L of the system can be found as: + m2 vr 2 sin θ = = 2 × 5 × 0.5sin90o + 3 × 5 × 0.5sin90o L = L1 + L2 = m1 vr 1 sin θ + L = 5 × 1 + 7.5 × 1(∵ sin900 = 1) L = 5 + 7.5 = 12.5
L = 12.5 Js Ans. masses m1 and m m2 . If the knife edge of the Problem 5.7. A uniform beam of mass ‘M’ supports two masses m support is under the beam’s centre of gravity and m m1 is at a distance ‘d’ from the centre, determine the position of m2 such that the system is balanced.
= Mkg Solution: Mass of beam: M = Mass of body A: m1 = m1 kg and that of B: m2 = m2 kg Distance of body A from the center: r 1 = d , and the dista distance nce of body B from the center center is r 2 = D. Since the system is balanced, so we can apply 2nd condition of equilibrium: = 0 about C ∑ τ =
− − W 2r 2 + W 3r 3 = m1gd − m2gD − Mg × 0 = 0 = m1 gd − m2 gD = 0 =⇒ m1 d = = m2 D = τ 1 + ( τ 2 ) + τ 3 = W 1 r 1
D =
m 1 d Ans. m2
Problem 5.8. A uniform ladder of length l and weight W = 50 N rests against a smooth vertical wall. If the coefficient of friction between the ladder and the ground is 0.40, find the minimum angle θ min min such that the ladder may not slip.
Data: Length of ladder = L = 50 N Weight of ladder: W = Coefficient of friction: µ = 0.4 To find: The minimum angle made by ladder with horizontal: θ min min =
?
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Solution:
Accord Acco rdin ing g to to figu figure re,, R is reaction force of wall and H and and V are horizontal and vertical com compone ponents nts of react reaction ion of grou ground nd resp respecti ectively vely..
H H =⇒ 0.4 = V V = 0.4V H = = µ =
(1)
Since the ladder is in equilibrium, hence it must
satisfy all conditions of equilibrium. Apply first condition: Along x-axis: ∑ F x = 0 + (− R)0 H + = R H − R = 0, =⇒ H = Along y-axis: + (−W ) = 0 ∑ F y = 0 = V + V − 50 = 0 =⇒ V = 50 N From equation (1): H = 0.4V H = = 0.4 × 50 = 20 N ∴ R = 20 N H =
Now apply 2nd condition: = 0 about A ∑ τ = = 0 R × DC − W × AE = Consider ADC : : DC DC = = sin θ = AC A C L = L sin θ DC = Consider AEB AE AE cos θ = = AB L/2 L = AE = 2 Put these values in equation (2): L 20 × L sin θ − 50 × cos θ = = 0 2 20 L sin θ = = 25 L cos θ sin θ 25 = = 1.25 cos θ 20 = 1.25 tan θ = = tan−1 (1.25) = 51.3o θ = o θ min min = 51.3 Ans.
Problem 5.9. A ladder with a uniform density and a mass ‘m’ rests against a frictionless vertical wall at an angle of 60 60o . The lower end rests on a flat surface where the coefficient of friction (static) is 0.40. A student with a mass (M = 2m) attempts to climb the ladder. ladder. What fraction fraction of the length
‘L’ of the ladder will the student have reached when the ladder begins to slip?
Let the ladder AD of length L of mass m having weight W = mg act acting ing at C (center of mass of the ladder). ladder). The
ladder makes θ = 60o with x-axis. Mass of student: ms = 2m Weight of student: W s = 2mg Coefficient of friction: µ = 0.40 To find: Fraction of length reached by student= LX
According ing to figure, figure, R is reaction force of wall and H and and V are are horizontal and vertical Solution: Accord components of reaction of ground respectively. respectively.
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