1 Physics Formulae
1 Physics Formulae
1 Physics Formulae
BY
D R . R AM C HAND R AGUEL
P H D(P HYSICS )
Principal & Head of the Physics Department
Government Girls Degree College, Jhudo
District Mirpurkhas
0233878056, ggdcjhudo@gmail.com
http://www.facebook.com/ggdcjhudo
2017
RAM’S OUTLINE SERIES
Copyright
c 2017, Department of Physics, Government Girls Degree College, Jhudo
This manuscript is written in LATEX. The diagrams and images are created in open-source
applications IPE, LatexDraw, 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 (AAPS), USA. The author’s research profile
can be found at his Google Scholar page.
ram_r25@hotmail.com, raguelmoon@gmail.com
http://www.facebook.com/ramcraguel
@RamCRaguel
1 SCOPE OF PHYSICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
3 MOTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
5 STATICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
6 GRAVITATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
9 NATURE OF LIGHT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
10 GEOMETRICAL OPTICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93
1.1 Problems
Problem 1.1. Find the area of a rectangular plate having length (21.3 ± 0.2)cm and width
(9.80 ± 0.10)cm.
The least number of significant figures in numerical is 3, so we can round off the answer to
3 significant figures:
A = 209cm2
Problem 1.2. Calculate (a) the circumference of a circle of radius 3.5cm and (b) area of
a circle of radius 4.65cm.
The least number of significant figures is 2, so we should round off the answer to 2:
S = 22cm Ans.
The least number of significant figures is 3, so we should round off the answer to 3:
A = 67.9cm2 Ans.
L = LT 0 + LT 0
L = L + L (∵ T 0 = 1) =⇒ L = 2L =⇒ L = L
Since both sides of the equation have the same dimensions, therefore, the given equation is
dimensionally correct.
Problem 1.5. Estimate the number of liters of gasoline used by cars in 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
2.1 Problems
Problem 2.1. State which of the following are scalars and which are vectors.
Solution:
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?
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 310 east
of south as shown in right-side figure.
→
− → − →
− → −
Problem 2.4. Show graphically that − A − B = − A + B .
Solution: First we construct − ~A − ~B graphically as shown in figure. Choose negative
of ~B, then apply head to tail rule. The resultant is ~A − ~B is shown in dotted line. Make a
negative of the resultant as shown in figure. Now construct −~A + ~B by taking negative of
~A.
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.
Problem 2.5. Given vector ~A, ~B, C
~ and ~D as shown in figure below. Construct:
→
− →
− →
− →
−
(a) 4 A − 3 B − 2 C + 2 D
1 →
− 1 → − → − →
−
(b) C + A + B +2D
2 3
Solution: (a) 4~A − 3~B − 2C
~ + 2~D
First multiply all vectors by scalar values, for example, multiply ~A by 4, ~B by 3, C
~ by 2
and ~D by 2. Then add all vectors according to head to tail rule as shown in figure:
1 ~ 1 ~ ~
(b) C + A + B + 2~D
2 3
~ by 1 , ~D by
In this case, we also first multiply all vectors by scalar values, like: multiply C
2
~ ~ ~
1
2 and then finally resultant of A + B + 2D by . Now add all vectors according to head
3
to tail rule as shown in figure:
~1 = 2î + 3 jˆ − 5k̂,
F ~2 = −5î + jˆ + 3k̂
F
~3 = î − 2 jˆ + 4k̂,
F ~4 = 4î − 3 jˆ − 2k̂
F
measured in newton, find: (a) The resultant force ~F (b) Magnitude of the resultant force.
~F = F
~1 + F~2 + F ~3 + F ~4
~F = (2î + 3 jˆ − 5k̂) + (−5î + jˆ + 3k̂) + (î − 2 jˆ + 4k̂) + (4î − 3 jˆ − 2k̂)
~F = 2î − 5î + î + 4î + 3 jˆ + jˆ − 2 jˆ − 3 jˆ − 5k̂ + 3k̂ + 4k̂ − 2k̂
~F = 2î − jˆ + 0k̂
~F = 2~i − jˆ
(a) 2~A − ~B + 3C
~ (b) |~A + ~B + C|
~ (c) |3~A − 2~B + 4C|
~
(d) a unit vector parallel to: 3~A − 2~B + 4C
~
Solution:
(a) 2~A − ~B + 3C
~ = 2(3î − jˆ − 4k̂) − (−2î + 4 jˆ − 3k̂) + 3(î + 2 jˆ − k̂)
= 6î − 2 jˆ − 8k̂ + 2î − 4 jˆ + 3k̂ + 3î + 6 jˆ − 3k̂
= 6î + 2î + 3î − 2 jˆ − 4 jˆ + 6 jˆ − 8k̂ + 3k̂ − 3k̂
2~A − ~B + 3C
~ = 11î − 8k̂ Ans.
(b) |~A + ~B + C|
~ = |(3î − jˆ − 4k̂) + (−2î + 4 jˆ − 3k̂) + (î + 2 jˆ − k̂)|
= |3î − jˆ − 4k̂ − 2î + 4 jˆ − 3k̂ + î + 2 jˆ − k̂|
= |3î − 2î + î − jˆ + 4 jˆ + 2 jˆ − 4k̂ − 3k̂ − k̂|
q √
ˆ
= |2î + 5 j − 8k̂| = (2)2 + (5)2 + (−8)2 = 4 + 25 + 64
√
|~A + ~B + C|
~ = 93 Ans.
(d) Let ~X = 3~A − 2~B + 4C ~ then unit vector parallel to this vector is:
~X 3~A − 2~B + 4C ~
x̂ = = √ But 3~A − 2~B + 4C
~ = 17î − 3 jˆ − 10k̂
|~X| 398
17î − 3 jˆ − 10k̂
~x = √ Ans.
398
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.
Solution: Two forces which tugboats are Putting the values (∵ cos 60o = 0.5):
exerting on ship are equal:
~1 = F~2 = 6000N;
q
F |~F| = 2(6000)2 + 2(6000)2 × 0.5
Angle between ropes: θ = 60o p
Resultant force on ship: ~F =?. = 2 × 36 × 106 + 36000000
Following figure the total force can be p
= 72 × 106 + 36 × 106
calculated as: p
= 108 × 106
√
~F = F ~1 + F
~2 ∵ F ~1 = F ~2
= 108000000
q = 10392.3
~
|F| = |F ~1 |2 + |F~2 |2 + 2|F~1 ||F
~2 | cos 60o
q |~F| = 10392.3N Ans.
|~F| = 2|F ~1 |2 + 2|F~1 |2 cos 60o
Problem 2.9. The position vectors of points P and Q are given by ~r1 = 2î + 3 jˆ − k̂,
−→
~r2 = 4î − 3 jˆ + 2k̂. Determine PQ in terms of rectangular unit vector î, jˆ and k̂ and find its
magnitude.
Solution:
~r1 = 2î + 3 jˆ − k̂
~r2 = 4î − 3 jˆ + 2k̂
−→
PQ = ∆~r =?
These two vectors are position vectors.
−→
Length PQ can be calculated as:
~ =−
∆r
→
PQ = ~r2 − ~r1
−→ ˆ
ˆ
PQ = 4î − 3 j + 2k̂ − 2î + 3 j − k̂ Magnitude can be found as:
−→
q
= 4î − 3 jˆ + 2k̂ − 2î − 3 jˆ + k̂ |PQ| = (2î)2 + (−6 jˆ)2 + (3k̂)2
−→ −→ √ √
PQ = 2î − 6 jˆ + 3k̂ Ans. |PQ| = 4 + 36 + 9 = 49 = 7
−→
|PQ| = 7 Ans.
Problem 2.10. Prove that the vectors ~A = 3î + jˆ − 2k̂, ~B = −î + 3 jˆ + 4k̂ and C~ = 4î −
2 jˆ − 6k̂ can form the sides of a triangle. Find the length of the medians of the triangle.
Data:
~A = 3î + jˆ − 2k̂
~B = −î + 3 jˆ + 4k̂
~ = 4î − 2 jˆ − 6k̂
C
To find:
~ = ~A, hence it is proved that these three vector can form a triangle.
we see that ~B + C
Now we will calculate the medians and then we will find their magnitudes. Median is
straight line that divides a side of a triangle into two equal parts. These are also vectors as
shown in 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 4a f c : Therefore:
−
→ − − →
af = →
ac + c f 1~ − →
B = bd + ~A
−
→ 1~ 2
But →
−
ac = ~B and c f = C −
→ 1~ ~
2 bd = B − A
→
− 1 2
ac = ~B + C ~ −
→ 1
2 bd = −î + 3 jˆ + 4k̂ − 3î + jˆ − 2k̂
→
− 1 2
ˆ
ac = −î + 3 j + 4k̂ + ˆ
4î − 2 j − 6k̂ −
→ 1 3
2 bd = − î + jˆ + 2k̂ − 3î − jˆ + 2k̂
→
−
ac = −î + 3 jˆ + 4k̂ + 2î − jˆ − 3k̂ 2 2
−
→ 7 1ˆ
→
−
ac = î + 2 jˆ + k̂ bd = − î + j + 4k̂
q q 2 2
|→
−
ac| = (1)2 + (2)2 + (1)2 = (7/2)2 + (1/2)2 + (4)2
√ √
|→
−
r
ac| = 1 + 4 + 1 = 6 49 1
√ = + + 16
4 4
|→
−
ac| = 6 Ans. r r
−
→ 49 + 1 + 64 114
In 4abd : |bd| = =
4 4
According to head to tail rule: 1√
−
→ → − − → = 114
ad = ab + bd 2
→
− −
→ 1 −
→ 1√
But ab = ~A and ad = ~B |bd| = 114 Ans.
2 2
————————————————-
Problem 2.11. Find the rectangular components of a vector ~A, 15 unit long when it forms
an angle with respect to +ve x-axis of (i) 50o , (ii) 130o , (iii) 230o , (iv) 310o .
Data:
|~A| = 15 unit
(i) θ = 50o
(ii) θ = 130o
(iii) θ = 230o
(iv) θ = 310o
To find:
Rectangular components of A: Ax =? and
Ay =?.
Solution:
(i) Ax = A cos θ
Ax = 15 cos 50o
Ax = 15 × 0.6428
Ax = 9.64 unit
Ay = A sin θ
Ay = 15 sin 50o
Ay = 15 × 0.766
Ay = 11.49 unit
Ax = 9.64 unit and Ay = 11.49 unit
(ii) Ax = A cos θ
Ax = 15 cos 130o
Ax = 15 × (−0.6428)
Ax = −9.64 unit
Ay = A sin θ
Ay = 15 sin 130o
Ay = 15 × 0.766
Ay = 11.49 unit
Ax = −9.64 unit and Ay = 11.49 unit
(iii) Ax = A cos θ
Ax = 15 cos 230o = 15 × (−0.6428)
Ax = −9.64 unit
Ay = A sin θ
Ay = 15 sin 230o = 15 × (−0.766)
Ay = −11.49 unit
Ax = −9.64 unit and Ay = −11.49 unit
(iv) Ax = A cos θ
Ax = 15 cos 310o = 15 × 0.6428
Ax = 9.64 unit
Ay = A sin θ
Ay = 15 sin 310o = 15 × (−0.766)
Ay = −11.49 unit
Ax = 9.64 unit and Ay = −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 difference and angle with respect to the largest vector.
Data:
|~A| = 10cm
|~B| = 8cm
(a) θ = 60o , (b) θ = 90o , (c) θ = 120o
To find:
(i) Magnitude of difference: |~A − ~B| =?
(ii) Angle b/w ~A − ~B and ~A : θ =?
(ii) Angle b/w ~A and (~A − ~B) can be found by equation (2):
|~A|2 − |~A||~B| cos θ
θ = cos−1
|~A||~A − ~B|
(10)2 − (10)(8) cos 90o 100 − 80 × 0 100 − 0
θ = cos−1 = cos−1 = cos−1
10 × 12.806 128.06 128.06
100
θ= = cos−1 (0.7809) = 38.66o . θ = 38.66o Ans.
128.06
(c) θ = 120o :
(i) Recall equation (1):
q
|A − B| = |~A|2 − 2|~A||~B| cos θ + |~B|2
~ ~
q
= (10)2 − 2(10)(8) cos 120o + (8)2
p
= 100 − 160 × (−0.5) + 64
√ √
= 164 + 80 + 64 = 244
|~A − ~B| = 15.62cm Ans.
(ii) Angle b/w ~A and (~A − ~B) can be found by equation (2):
|~A|2 − |~A||~B| cos θ
θ = cos−1
|~A||~A − ~B|
(10)2 − (10)(8) cos 120o 100 − 80 × (−0.5) 100 + 40
θ = cos−1 = cos−1 = cos−1
10 × 15.62 156.2 128.06
140
θ= = cos−1 (0.8962) = 26.33o
156.2
θ = 26.33o Ans.
Problem 2.13. The angle between the vectors ~A and ~B is 60o . Given that |~A| = |~B| = 1,
calculate (a) |~B − ~A|, (b) |~B + ~A|.
Data:
|~A| = |~B| = 1, and θ = 60o
To find:
(a) |~B − ~A| =?, and (b) |~B + ~A| =?
r 2 q q
~ ~
(b) |B + A| = B + A = B + 2A.B + A = |~B|2 + 2|~A||~B| cos θ + |~B|2
~ ~ ~ 2 ~ ~ ~ 2
q √ √
= (1)2 + 2(1)(1)(0.5) + (1)2 = 1 + 1 + 1 = 3
√
|~B + ~A| = 3 Ans.
Problem 2.14. A car weighing 10,000 N on a hill which makes an angle of 20o with
horizontal. Find the components of car’s weight parallel and perpendicular to the road.
Problem 2.15. Find the angle between ~A = 2î + 2 jˆ − k̂ and ~B = 6î − 3 jˆ + 2k̂.
Data:
~A = 2î + 2 jˆ − k̂ and ~B = 6î − 3 jˆ + 2k̂
Problem 2.16. Find the projection of the vector ~A = î − 2 jˆ + k̂ onto the direction of vector
~B = 4î − 4 jˆ + 7k̂.
Solution: Projection can be found by dot product. First we have to find the direction of
vector ~B. We know that direction of any vector can be found by unit vector. Unit vector of
~B and b̂ is defined as:
Problem 2.17. Find the angles α, β and γ which the vector ~A = 3î − 6 jˆ + 2k̂ makes with
positive x, y and z axis respectively.
~x
A 3
For α angle: cos α = = = 0.4286 =⇒ θ = cos−1 (0.4286) = 64.62o
|~A| 7
A~y −6
For β angle: cos β = = = −0.8571 =⇒ θ = cos−1 (−0.8571) = 149o
|~A| 7
A~z 2
For γ angle: cos γ = = = 0.2857 =⇒ θ = cos−1 (0.2857) = 73.4o
|~A| 7
Angles are found: α = 64.62o , β = 149o and γ = 73.4o Ans.
Problem 2.18. Find the work-done in moving an object along a vector~r = 3î + 2 jˆ − 5k̂ if
the applied force is ~F = 2î − jˆ − k̂.
Data:
Applied force on an object is: ~F = 2î − jˆ − k̂.
The displacement an object covered when force is applied: ~r = 3î + 2 jˆ − 5k̂
To find:
The work-done by the force is: W =?
Solution: Work-done by force on any object is the dot product of force and displacement
it covered:
W =~F ·~r = 2î − jˆ − k̂ · 3î + 2 jˆ − 5k̂ = 2 × 3(î · î) − 1 × 2( jˆ · jˆ) + 1 × 5(k̂ · k̂)
h i
= (6 − 2 + 5) = 9 ∵ î · î = jˆ · jˆ = k̂ · k̂ = 1
Work-done: W = 9J Ans.
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 40o to the direction of motion. Find the rate at which the force is working at a
time when the velocity is 2m/sec.
Data:
Force applied on an object: |~F| = 30000N
~ = 45m
Distance covered by an object due to the force: |d|
Velocity of the object: |~v| =?
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 =?
Solution: Work-done is the scalar or dot product of force and displacement (in the first
case, the force and the displacement are in the same direction so the angle between them is
zero):
W = ~F · d~ = |~F||d|
~ cos θ = 30000 × 45 cos 0o = 1350000 × 1
In the second case, the force and the displacement make an angle of 40o :
W = ~F · d~ = |~F||d|
~ cos θ = 30000 × 45 cos 40o = 1350000 × 0.7660 = 1034160
Problem 2.20. Two vectors ~A and ~B are such that |~A| = 3, |~B| = 4, and ~A.~B = −5, find:
(a) the angle between ~A and ~B
(b) the length of |~A + ~B| ~A − ~B|
and |
~ ~
(c) the angle between A + B and A − B . ~ ~
Data:
|~A| = 3, |~B| = 4 and ~A · ~B = −5.
Solution: (a) angle between ~A and ~B can be found by dot product as:
~A · ~B = |~A||~B| cos θ
Putting the values from data:
−5
− 5 = 3 × 4 × cos θ =⇒ cos θ = = −0.4167
12
θ = cos−1 (−0.4167) = 114.62o
θ = 114.62o Ans.
(b) the length of |~A + ~B| and |~A − ~B| can be found by Pythagoras theorem for vectors
product as follows: (also known as Law of Cosine):
Similarly:
r 2 q q
~ ~
|A − B| = ~ ~ ~ 2 ~ ~ ~ 2
A − B = A − 2A · B − B = (3)2 − 2 × (−5) + (42 )
√ √
= 9 + 10 + 16 = 35
√
|~A − ~B| = 35 Ans.
~ ~ ~ ~
(c) Consider θ is the angle between A + B and A − B , then according to dot product:
~A + ~B · ~A − ~B = |~A + ~B||~A − ~B| cos θ
~A · ~A − ~A · ~B + ~B · ~A − ~B · ~B = |~A + ~B||~A − ~B| cos θ
~A2 − ~A ·
~B +
~A ·
~B − ~B2 = |~A + ~B||~A − ~B| cos θ
√ √
2 2 −1 −7
15 × 35 = (3) − (4) = 9 − 16 = −7 =⇒ θ = cos
3.873 × 5.916
−7
θ = cos−1 = cos−1 (−0.3055) = 107.79o
21.913
θ = 107.79o Ans.
Data: To find:
~A = 2î − 3 jˆ − k̂ (a) ~
A × ~B, (b)~B × ~A
~B = î + 4 jˆ − 2k̂ (c) ~A + ~B × ~A − ~B .
~A + ~B × ~A − ~B =
Problem 2.22. Determine the unit vector perpendicular to the plane of ~A = 2î − 6 jˆ − 3k̂
and ~B = 4î + 3 jˆ − k̂.
Data:
~A = 2î − 6 jˆ − 3k̂
~B = 4î + 3 jˆ − k̂
To find:
Unit vector ⊥ to ~A × ~B
~A × ~B
û = − − − − > (1)
|~A × ~B|
Where ~A × ~B has to be determined.
Problem 2.23. Using the definition of vector product, prove the law of sines for plane
triangles of sides a, b and c.
Solution: Plane triangle is closed geometry shape constructed from three vectors ~a, ~b and
~c added according to head to tail rule. Angles are marked as A, B and C. Sum of vectors of
any closed shape is always zero, hence:
From the figure we see that ~a ×~b = |~a||~b| sinC and ~c ×~a = |~c||~a| sin B.
Comparaing these equations with equation (1), we get:
|~a||~b| sinC = |~c||~a| sin B
|~a| sinC |~a| sin B sinC sin B
= =⇒ = − − − − > (2)
|~c| ~
|b| |~c| |~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 ×~b = 0
~b ×~c = −~b ×~a = ~a ×~b ∵ −~b ×~a = ~a ×~b
From the figure we see that ~b ×~c = |~b||~c| sin A and ~a ×~b = |~a||~b| sinC.
|~b|
sin A |~b|
sinC
|~b||~c| sin A = |~a||~b| sinC =⇒ =
|~a| |~c|
!
sin A sinC sinC sin B
= But =
|~a| |~c| |~c| |~b|
sin A sin B sinC
Hence: = = Law of Sines
|~a| |~b| |~c|
Problem 2.24. If ~r1 and ~r2 are the position vectors (both lie in xy plane) making angles
θ1 and θ2 with the position x-axis measured counter clockwise, find their vector product
when:
(i) ~r1 = 4cm with θ1 = 30o (ii) ~r1 = 6cm with θ1 = 220o
~r2 = 3cm with θ1 = 90o ~r2 = 3cm with θ1 = 40o
(iii) ~r1 = 10cm with θ1 = 20o
~r2 = 9cm with θ1 = 110o
3.1 Problems
Problem 3.1. In an electron gun of a television set, an electron with an initial speed of
103 m/s enters a region where it is electrically accelerated. It emerges out of this region
after 1 micro section with speed of 4 × 105 m/s. What is the maximum length of the electron
gun? Calculate the acceleration.
Given Data:
To find:
Time = t = 1µs = 1 × 10−6 s
Acceleration of electron = a= ?
Initial speed of the electron = Vi = 103 m/s
Length of electron gun = t = ?
Final speed of the electron = V f = 4×105 m/s
Solution: Using the first equation of motion and putting values from the given data:
The length of the electron gun can be found by using 3rd equation of motion:
2 2
2as = V f2 −Vi2 = 2 × 399 × 109t = 4 × 105 − 103
798 × 109t = 16 × 1010 − 106
10 6 106 16 × 104 − 1
16 × 10 − 10 (160000 − 1) 106−9
t= = =
798 × 109 798 × 109 798
159999 × 10 −3
= = 200.5 × 10−3
798
t = 0.2005m Ans.
Problem 3.2. A car is waiting at a traffic signal and when it turns green, the car starts
ahead with a constant acceleration of 2m/s2 . At the same time a bus traveling with a
constant speed of 10m/s 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
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:
S = vt = 10t
S = 10t (2)
Distance is common for both car and bus, so comparing equation (1) and (2), we get:
t 2
t 2 = 10t =⇒ = 10
t
t = 10s
Now putting the value of t in equation (1), we can find the distance travelled by car:
S = t 2 = (10)2 = 100m
S = 100m Ans.
The velocity which the car overtakes the bus can be found by first equation of motion as:
v f = vi + at = 0 + 2 × 10 = 20
v f = 20m/s Ans.
Problem 3.3. A helicopter ascending at a rate of 12m/s. At a height of 80m above the
ground, a package is dropped. How long does the package take to reach the ground?
v f = vi + at
0 = 12 + (−9.8)t1 = 12 − 9.8t1 =⇒ 9.8t1 = 12
12
t1 =
9.8
t1 = 1.2s Ans.
1
S = vit + gt 2
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
h2 = 80 + h1 = 80 + 7.34
h2 = 87.34m
vi = 0, g = 9.8m/s2 and t2 =?
Using the equation of motion
1
h2 = vit2 + gt22
2
1
87.34 = 0 × t2 + 9.8t22 = 4.9t22
2
87.34
t22 = = 17.82
4.9
t2 = 4.22s Ans.
Total time the package took to reach the ground is:
t = t1 + t2 = 1.2 + 4.22 = 5.42
Total time = 5.42s Ans.
Problem 3.4. A boy throws a ball upward from the top of a cliff with a speed of 14.7m/s.
On the way down it just misses the thrower and falls the ground 49 metres 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.
v f = vi − gt1
0 = 14.7 − 9.8t1
14.7
t1 = = 1.5
9.8
t1 = 1.5s Ans.
(iii) Let h3 is the total height from maximum to the ground, then:
h3 = h1 + h2 = 49 + 11.025 = 60.025m
Ball will take time t2 to cover this distance according to:
1
h3 = vit + gt 2
2
where vi is the initial velocity of ball at maximum height which is zero.
1
60.025 = 0 + × 9.8t22 = 4.9t22
2
60.025 √
t22 = = 12.25 =⇒ t2 = 12.25 = 3.5s
4.9
Total time taken is:
t = t1 + t2 = 1.5 + 3.5 = 5s
t = 5s Ans.
Problem 3.5. A helicopter weighs 3920 Newton. Calculate the force on it if it is ascending
up at a rate of 2m/s2 . What will be force on helicopter if it is moving up with the constant
speed of 4 m/s?
unbalanced force= ∑ F = ma
Where m is the mass of helicopter which can be found as m = Wg = 39209.8 = 400kg. The
direction of W is downward while external force which helicopter is applying is upward,
therefore net sum of forces on it is F1 −W or:
Force (F2 ) on helicopter when it is ascending with constant velocity can be found as
F −W = ma. In this case acceleration is zero:
F2 −W = m × 0 = 0
F2 = W = 3920
F2 = 3920N Ans.
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.
Given Data:
Mass of bullet mb = 0.005kg
To find:
Initial velocity of bullet vi = 100m/s
Decelerating force on bullet F =?
Distance covered 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
Solution: According to Newton’s 2nd law of motion, decelerating force on bullet is given
by:
F = mb a (1)
Where a is the acceleration and can be found by 3rd equation of motion
2aS = v2f − 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:
F = 0.005 × (−20000) = −100
Decelerating force=100N 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
t= = 0.005s
20000
t = 0.005s Ans.
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
Solution: Weight of the car W = 9800N
Initial velocity vi = 40km/h = 40×1000
3600 = 11.11m/s
Final velocity v f = 0m/s
Distance covered S = 50m
Average retarding force F =?
Retarding force on car can be calculated using Newton’s 2nd law: F = ma, where a is
acceleration of the car and m is the mass of car. Acceleration can be found by using 3rd
equation of motion:
2aS = v2f − v2i
2 × a × 50 = (0)2 − (11.11)2 = −123.457
123.457
100a = −123.457 =⇒ a = − = −1.235m/s2
100
W 9800
Mass can be calculated as: m = = = 1000kg
g 9.8
Using these values:
F = ma = 1000 × (−1.23457) = −1234.57N
F = −1234.57N Ans.
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 × 106 m/s after covering a distance of 1.8cm. Find
the force acting on the electron. Take the mass of electron as 6.1 × 10−31 kg.
Solution: (i) For such system, acceleration of bodies can be found as:
m1 g 4.5 × 9.8
a= = =3
m1 + m2 4.5 + 10.2
a = 3m/s2 Ans.
(ii) The tension in the string can be found by: (iii) Force on body due to surface is:
m1 m2 g 4.5 × 10.2 × 9.8 449.82
T= = = F = W2 = m2 g = 10.2 × 9.8 = 99.96N
m1 + m2 4.5 + 10.2 14.7
T = 30.6N Ans. F = 99.96N Ans.
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
Mass of bullet: mb = 100g = 1000 kg = 0.1kg
Mass of gun: mg = 10kg
Velocity of bullet before firing: ub = 0m/s
Velocity of gun before firing: ug = 0m/s
Velocity of bullet after firing: vb = 1000m/s
To find:
Velocity of gun after firing: vg =?
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
vg = = −10
10
veoclity of gun: vg = −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?
Solution: This is type of inelastic collision because two bodies after collision (impact)
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 = vB
According to law of conservation of linear momentum:
mb ub + mB uB = mb vb + mB vB
0.05ub + 10 × 0 = 0.05v + 10v ∵ (v = vb = vB )
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 2
( mb+mB ) v = (
mb+m B ) gh
2
1 2 p
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: ub = 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.
Problem 3.14. A cyclist is going up a slope of 30o with a speed of 3.5 m/s. If he stops
pedaling, how much distance will he move before coming to rest? (Assume the friction to
be negligible).
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).
rest:
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.
Problem 3.17. In the Figure 3.3 find the acceleration of the masses and the tension in the
string.
F1 = W1 − T =⇒ m1 a = W1 − T (∵ F1 = m1 a)
10a = 98 − T (1)
25a = 98 − T + T − 73.5
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 = 88.2N
T = 88.2N Ans.
Problem 3.18. Two blocks are connected as shown in Figure 3.4. If the pulley and the
planes on which the blocks are resting are frictionless, find the acceleration of the blocks
and the tension in the string.
F1 = W1 sin θ1 − T =⇒ m1 a = W1 sin θ1 − T (∵ F1 = m1 a)
50a = 490 sin 30o − T = 490 × 0.5 − T = 245 − T
50a = 245 − T (1)
Consider the upward motion of block B:
Weight of block B: W2 = 245N
Mass of block B: m2 = Wg2 = 2459.8 = 25kg
Resultant force on block B is:
F2 = T −W2 sin θ2 =⇒ m2 a = T −W2 sin θ2 (∵ F2 = m2 a)
25a = T − 245 sin 60o = T − 245 × 0.866
25a = T − 212.17 (2)
Adding equ(1) and (2), we get:
50a = 245 − T
25a = T − 212.17
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.
F1 = W1 sin θ − T =⇒ m1 a = W1 sin θ − T (∵ F1 = m1 a)
20a = 196 sin 30o − T = 196 × 0.5 − T = 98 − T
20a = 98 − T (1)
Since block B is moving in the direction of T along x-axis, therefore, unbalanced force
acting on block B is T :
T = m2 a = 20a (2)
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 = 49N
T = 49N Ans.
4.1 Problems
Problem 4.1. A rescue helicopter drops a package of emergency ration to a stranded party
on the ground. If the helicopter is traveling 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?
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 horizon-
tal 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:
x = vixt 1
where time t can be calculated using 2nd equation of motion:
1
h = viyt + gt 2
2
1
− 100 = 0 × t − × 9.8 × t 2 = −4.9 × t 2
2
100 √
t2 = = 20.41 =⇒ t = 20.41 = 4.52s
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:
vx = v f x = vix = 40m/s
vx = 40m/s Ans.
Final velocity along y-axis can be found by using 1st equation of motion:
v f y = viy − gt = 0 − 9.8 × 4.52 = −44.3
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 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: vo = 11m/s
Angle of projection with x-axis: θ = 20o
To find:
(a) Horizontal distance covered (range): R =?
(b) 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:
v2o (11)2 121
R= sin 2θ = sin 2 × 20o = sin 40o
g 9.8 9.8
R = 12.35 × 0.6428 (∵ sin 40o = 0.6428)
R = 7.94m Ans.
Problem 4.3. A stone is thrown upward from the top of a building at an angle of 30o to
the horizontal and with a initial speed of 20 m/s. If the height of building is 45 m. (a)
Calculate the total time the stone in flight (b) What is the speed of stone just before it
strikes the 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 :
Here a = 4.9, b = −10 and c = −45, putting these values in quadratic formula, we get:
p √ √
−(−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
t= t= = = −2.175s
9.8 9.8 9.8
10 + 31.34 41.34
OR t = = = 4.22s
9.8 9.8
Since time can not be negative so total time of flight is:
t = 4.22s
v = 35.8m/s Ans.
(c) Horizontal distance covered (range) is given by:
R = vxt = 17.32 × 4.22 = 73m
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: vox = 10m/s
Initial vertical velocity of the ball: voy = 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:
Problem 4.5. A rocket is launched at an angle of 53o to the horizontal with an initial
speed of 100 m/s. It moves along its initial line of motion with an acceleration of 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 rocket 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: t1 = 3s
To find:
(a) Maximum height reached: h =?
Solution: When rocket fails, the final velocity which rocket 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:
(b) The total time is the falling time plus the time the engines are running plus the time to
t = t1 + t2 + t3 = 3 + t2 + t3 − − − − > (1)
v0 sin θ 190 sin 530 151.74
where t2 = = = = 15.48s
g 9.8 9.8
And t3 can be worked out as:
v f − v0 v f − 0 v f
v f = vo + gt =⇒ t = = = − − − − > (2)
g g g
where final velocity can be found as: v f − v2o = 2gh
2
p
v2f = 2gh − 0 = 2gh =⇒ v f = 2gh
Putting this value into eqation (2), we get:
√ √ √
2gh 2 × 9.8 × 1549.2 30364.32 174.25
t3 = = = = = 17.78s
g 9.8 9.8 9.8
Putting the values of t2 and t3 into equation (1):
t = 3 + 15.48 + 17.78 = 36.3s t = 36.3s Ans.
(c) Horizontal range R = R1 + R2 , where R1 is the range when engines are running while
R2 is the range when engines are failed. First we calculate R1 and R2 as follow:
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: vox =
7m/s
Initial vertical velocity of the diver: voy =
3m/s
To find:
(i) Horizontal distance covered: x =?
(ii) Vertical distance covered: y =?
(iii) Final horizontal velocity component:
vx =?
(iv) Final vertical velocity component: vy =?
Solution:
Problem 4.7. A boy standing 10m from a building can just barely 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.
Data:
Height of the building: y = 12m
Horizontal distance: x = 10m
At optimum angle θ , the ball will be at its maxi-
mum 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
vy would be 0, but his horizontal velocity remains
constant.
To find:
Horizontal velocity component: vox = vx =?
Vertical velocity component: voy
Solution: Choosing up and forward as positive and then solving for both components of
velocity:
For vertical motion: voy can be found by: v2f − v2i = 2gh,
Here in our case: h = y = 12, v f = 0, vi = 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
x = vxt =⇒ vx = − −− > (1)
t
where time ’t’ can be calculated from data of vertical motion:
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: vo = 90m/s
To find:
The launch angle of the shell: θ =?
Solution: Since range of the mortal shell is given so we can find the launch angle by
using the projectile range equation:
v2o
R= sin 2θ
g
(90)2 8100
500 = sin 2θ = sin 2θ = 826.53 sin 2θ
9.8 9.8
500
sin 2θ = = 0.6049 =⇒ 2θ = sin−1 (0.6049) = 37.2o
826.53
37.2o
θ= = 18.6o
2
In projectile motion the there are two launch angles for the same range according to the
relation:
θ1 + θ2 = 90
We have found one launch angle, the second launch angle can be:
θ2 = 90 − 18.6 = 71.4o
Hence mortal shell could be fired at 18.6o OR at 71.4o for the same range of 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: vo =?
Solution: As range is given, so we can use the equation for range of the projectile to find
the take off speed:
v2o
R= sin 2θ
g
v2 v2 v2
0.8 = 0 sin 2 × 55o = o sin 110o = o × 0.9397
9.8 9.8 9.8
0.8 × 9.8 7.84 √
v2o = = = 8.34 =⇒ vo = 8.34 = 2.89m/s
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 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 µs 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: ac = 4.5m/s2
Radius of track: r = 200m
Mass of car: m = 1000kg
To find:
(a) Frictional force: f =?
(b) Maximum speed: vmax =? when µs = 0.8
Solution: (a) Since car is travelling in circular motion, so centripetal force is required to
keep car on the circular track. This force is provided by frictional force between car and
the track: f = Fc , where Fc is centripetal force:
(b) The maximum speed with which car moves is related to maximum frictional force:
fmax = µs R, where R is normal force and equal to its weight: R = mg = 1000 × 9.8 =
9800N.
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: ωi = 33rev/min
Final angular speed: ω f = 0
Time taken: t = 20s
Radius of turntable: r = 0.14m
To find:
(a) Angular acceleration: α =?
(b) Number of rotations completedθ =?
(c) Initial speed of bug: vi =?
(d) Tangential acceleration of bug at t = 0:
at =?
Problem 4.12. Tarzan swings on a vine of length 4m in a vertical circle under the influence
of gravity. When the vine makes an angle of θ = 20o 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: ac =?
(b) Tangential acceleration: at =?
(c) Resultant acceleration: a =?
Solution:
v2
(a) Centripetal acceleration of tarzan is given by: ac =
r
(5)2 25
ac = = = 6.25m/s2 ac = 6.25m/s2 Ans.
4 4
(b) Tangential acceleration of Tarzan is given by: at = g sin θ = 9.8 sin 20o
at = 9.8 × 0.3420 = 3.35. at = 3.35m/s2 Ans.
a = 7.1m/s2 Ans.
5.1 Problems
Problem 5.1. Locate the centre of mass of a system of particles each of mass ‘m’, arranged
to correspond in position to the six corners of a regular (planar) hexagon.
Solution: Divide the system according to co-ordinate system. Take x- and y-axis such
that origin O lies at the centre of the hexagon. The co-ordinates Xc and Yc of the center of
masses are given by:
6 6
mi xi mi yi
Xc = ∑ , Yc = ∑
i=1 mi i=1 mi
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’.
Solution: Since square based right pyramid is 3D structure, so we need three coordinates
to locate the CM as given by:
5 5 5
mi xi mi yi mi zi
Xc = ∑ , Yc = ∑ , Zc = ∑
i=1 mi i=1 mi i=1 mi
Using the above equations and figure, we can find the coordinates of center of masses as:
m1 x1 + m2 x2 + m3 x3 + m4 x4 + m5 x5
Xc =
m1 + m2 + m3 + m4 + m5
mx + mx − mx − mx + m(0) +
mx mx− −
mx + m(0)
mx 0
= = = . Xc = 0
m+m+m+m+m 5m 5m
For Yc :
m1 y1 + m2 y2 + m3 y3 + m4 y4 + m5 y5
Yc =
m1 + m2 + m3 + m4 + m5
my + my − my − my + m(0) +
my my−
my− + m(0)
my 0
= = = . Yc = 0
m+m+m+m+m 5m 5m
For Zc :
m1 z1 + m2 z2 + m3 z3 + m4 z4 + m5 z5
Zc =
m1 + m2 + m3 + m4 + m5
m(0) + m(0) + m(0) + m(0) + mh mh mh h h
= = = = . Zc =
m+m+m+m+m 5m 5 m 5 5
Hence the coordinates of center of masses are: (Xc ,Yc , Zc ) = (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. 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)
2
mi ri m1 r1 + m2 r2
Xc = ∑ =
i=1 mi m1 + m2
Here m2 = 329390m1 and r = 1.496 × 108 km = 1.496 × 1011 m; using this data we can
find:
Coordinate Xc is located at C in figure. Center of earth-sun system lies inside of the sun.
Let mean radius of the sun is rs = 6.9960 × 105 km = 6.9960 × 108 m, then comparing both
values:
Xc 4.54 × 105
= = 6.49 × 10−4
rs 6.9960 × 108
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 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 par-
ticle: 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:
y 6
L = mvr sin θ where sin θ = =
PQ r
6
∴ L = mvr × = 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= = . τ = 360N.m Ans.
t 2
Problem 5.5. A particle of mass ‘m’ is located at the vector position r and has a linear
momentum vector P. The vector r and P are non zero. If the particle moves only in the x, y
plane, prove that Lx = Ly = 0 and Lz 6= 0.
Solution: Particle of mass m moves only in x-y plane, so its position vector and momentum
vectors are:
~r = rx î + ry jˆ; ~P = Px î + Py jˆ
The angular momentum can be found as: ~L =~r × ~P.
î jˆ k̂
~r 0 ~r 0 ~r ~r
~L = ~rx ~ry 0 = î y − jˆ x + k̂ x y
~ ~ P~y 0 P~x 0 P~x P
~y
Px Py 0
Comparing the components of L with the coefficient of î, jˆ and k̂, we get:
Lx = 0 and Ly = 0 and Lz = (rx Py , −ry Px ) 6= 0. Hence proved.
Problem 5.6. A light rigid rod 1m in length rotates in the xy-plane about a pivot through
the rod’s centre. Two particles of mass 2kg and 3kg are connected to its ends. Determined
the angular momentum of the system about the origin at the instant the speed of each
particle is 5m/s.
L = L1 + L2 = m1 vr1 sin θ + m2 vr2 sin θ = 2 × 5 × 0.5 sin 90o + 3 × 5 × 0.5 sin 90o
L = 5 × 1 + 7.5 × 1(∵ sin 900 = 1)
L = 5 + 7.5 = 12.5
L = 12.5Js Ans.
Problem 5.7. A uniform beam of mass ‘M’ supports two masses m1 and m2 . If the knife
edge of the support is under the beam’s centre of gravity and m1 is at a distance ‘d’ from
the centre, determine the position of m2 such that the system is balanced.
∑ τ = 0 about C
= τ1 + (−τ2 ) + τ3 = W1 r1 −W2 r2 +W3 r3 = m1 gd − m2 gD − Mg × 0 = 0
= m1 gd − m2 gD = 0 =⇒ m1 d = m2 D
m1
D= 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 θmin such that the ladder may not slip.
Data:
Length of ladder = L
Weight of ladder: W = 50N
Coefficient of friction: µ = 0.4
To find:
The minimum angle made by ladder with horizontal:
θmin =?
Problem 5.9. A ladder with a uniform density and a mass ‘m’ rests against a frictionless
vertical wall at an angle of 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.
What fraction of the length ‘L’ of the ladder will the student have reached when the ladder
begins to slip?
∑ Fy = 0 = V + (−Ws) + (−mg) = 0 AF AF
cos 60o = =
V − 2mg − mg = V − 3mg = 0 =⇒ V = 3mg AC L/2
From equation (1):H = 0.4V AF
0.5 × 2 =
L
H = 0.4 × 3mg = 1.2mg ∴ H = R
AF = 0.25L
————————————————-
Put these values in equation (2):
1.2mg × 0.866L − 2mg × 0.5X − mg × 0.25L = 0
1.0392mgL − mgX − 0.25mgL = 0
0.789 =
mgL =⇒ 0.789L = x
mgX
X
= 0.789 Ans.
L
Problem 5.10. A particle of mass 0.3 kg moves in the xy-plane. At the instant its coor-
dinates are (2, 4)m, its velocity is (3î + 4 jˆ)m/s. At this instant determine the angular
momentum of the particle relative to the origin.
~L = −1.2k̂J.s Ans.
Problem 5.11. A uniform horizontal beam of length 8m and weighing 200N is pivoted
at the wall with its far end supported by a cable that makes an angle of 53o with the
horizontal. If a person weighing 600N stands 2m from the wall, find the tension and the
reaction force at the pivot.
Data: To find:
Length of beam: L = 8m Tension in cable: T =?
Weight of beam: W1 = 200N Horizontal reaction: H =?
Weight of person: W2 = 600N Vertical reaction: V =?
Angle of cable with beam: θ = 53o Since system is in equilibrium so we can ap-
Centre of mass of beam= 4m ply all conditions of equilibrium.
∑ Fx = 0 =⇒ H + (−T x) = 0
H − T cos 30o = H − 0.602T = 0
H = 0.602T (1)
Along y-axis:
∑ Fy = 0
V + Ty + (−W1 ) + (−W2 ) = 0
V + T sin 530 − 200 − 600 = 0
V + T × 0.799 − 800 = 0
V + 0.799 − 800 (2)
6.1 Problems
Problem 6.1. A 10 kg mass is at a distance of 1 m from a 100 kg mass. Find the
gravitational force of attraction when (i) 10 kg mass exerts force on the 100 kg mass (ii)
100 kg mass exerts force on the 10 kg mass.
Solution:
Putting the values:
Given Data: 10 × 100
Mass of body one: m1 = 10kg F12 = 6.67408 × 10−11
12
Mass of body 2nd: m2 = 100kg −11
F12 = 6.67408 × 10 × 103
Distance between them: r = 1m
To find: F12 = 6.67408 × 10−8 N Ans.
Gravitational force: F =?
(ii) Force by body two on body one:
m1 m2
(i) Force by body one on body two. F21 = G = 6.67408 × 10−8 N
By Newton’s Law of Gravitation: r2
F12 = G m1r2m2 Force will same
F21 = 6.67408 × 10−8 N Ans.
This shows that both the bodies will attract each other with equal and opposite force.
Problem 6.2. Compute gravitational acceleration at the surface of the planet Jupiter
which has a diameter as 11 times as compared with that of the earth and a mass equal to
318 times that of earth.
Solution:
Problem 6.3. The mass of the planet Jupiter 1.9 × 1027 kg and that of the sun is 2.0 ×
1030 kg. If the average distance between them is 7.8 × 1011 m, find the gravitational force
of the sun on Jupiter.
Problem 6.4. The radius of the moon is 27% of the earth’s radius and its mass is 1.2% of
the earth’s mass. Find the acceleration due to gravity on the surface of the moon. How
much will a 424N body weight there?
Problem 6.5. What is the value of the gravitational acceleration at a distance of (i) earth’s
radius above the earth’s surface? (ii) Twice earth’s radius above the earth’s surface?
Solution:
(i) when h1 = Re , then equation (1) becomes:
Data:
Distance from earth’s surface: h1 = Re GMe GMe 1 GMe
Distance from earth’s surface: h2 = 2Re gh1 = 2
= 2
=
(Re + Re ) (2Re ) 4 R2e
To find: GMe
(i) Acceleration due to gravity for h1 : gh1 =? from equation (2): g = 2
Re
(ii)Acceleration due to gravity for h2 : gh2 =?
————————————– 1
gh1 = g Ans.
To calculate the acceleration due to gravity at 4
height "h" from the earth’s surface:
(ii) when h2 = 2Re , then equation (1) be-
GMe comes:
gh = (1)
(Re + h)2 GMe GMe 1 GMe
gh2 = = =
(Re + 2Re )2 (3Re )2 9 R2e
GMe
And gravitation acceleration on surface of from equation (2): g = 2
Re
earth:
1
GMe gh2 = g Ans.
g= 2 (2) 9
Re
Problem 6.6. At what distance from the center of the earth does the gravitational acceler-
ation have one half the value that it has on the earth’s surface?
Solution: GMe
Data: from equation (1): g0 =
R2
Gravitational acceleration: g0 = 21 g GMe 1 GMe
To find: ∴ =
R2 2 R2e
Distance from the center of the earth above
GM
e 1GM
e
the surface: R =? =
R 2 2 Re2
Distance from the center of the earth below p
the surface: d =? R2 = 2R2e =⇒ R = 2Re = 1.41Re
————————————– R = 1.41Re Ans.
Value of the acceleration due to gravity at any
distance from the earth’s center is given by: The value of g0 becomes one-half below the
GMe surface (at d):
g0 = 2 (1)
R
0 d
g = 1− g
Re
1 d
And: g ∵ g0 = 1/2g
g = 1−
2 Re
GMe −1 d
g= 2 =−
Re 2 Re
1 1 GM e
But g0 = g = 1
2 2 R2e d = Re Ans.
2
————————————————————————————-
These are two positions at which the value of g becomes one-half of its value at the surface
of earth.
Problem 6.7. Compute the gravitational attraction between two college students of mass
80 & 50 kg respectively, 2m apart from each other. Is this force worth worrying about?
Solution:
Mass of first student: m1 = 80kg
Mass of second student: m2 = 50kg
Distance between two students: r = 2m
According to Newton’s law of gravitational force, the force between these two students
can be calculated as:
m1 m2 80 × 50
F = G 2 = 6.67 × 10−11
r (2)2
4000
F = 6.67 × 10−11 = 6.67 × 10−11 × 1000 = 6.67 × 10−11 × 103
4
F = 6.67 × 10−11+3 = 6.67 × 10−8 N
F = 6.67 × 10−8 N Ans.
This force is very small as compared to the Earth’s gravitational force on us, therefore, it is
not worth worrying about.
Problem 6.8. Determine the gravitation between the proton and the electron in a hydrogen
atom, assuming that the electron describes a circular orbit with a radius of 0.53 × 10−10 m
(mass of proton = 1.67 × 10−27 kg, mass of electron = 9.1 × 10−31 kg).
Solution:
Mass of electron: me = 9.1 × 10−31 kg
Mass of proton: m p = 1.67 × 10−27 kg
Radius between electron and proton:: r = 0.53 × 10−10 m
According to Newton’s law of gravitational force, the force between proton and electron
can be calculated as:
−31 × 1.67 × 10−27
me m p −11 9.1 × 10
F =G = 6.67 × 10
r2 (0.53 × 10−10 )2
15.197 × 10−31−27 101.36399 × 10−58−11+20
F = 6.67 × 10−11 =
0.2809 × 10−20 0.2809
−49 −47
F = 360.85 × 10 = 3.6 × 10
F = 3.6 × 10−47 N Ans.
This force is very very small as compared to electrostatic force between them and it is
often ignored.
Solution:
Mass of woman: m = 45kg
Acceleration of elevator: a = 1.2m/s2
Apparent weight Fw of woman in elevator when it is moving up is given as:
Fw = ma + mg
= 45 × 1.2 + 45 × 9.8
= 54 + 441 = 495N
Apparent weight of woman Fw = 495N Ans.
7.1 Problems
Problem 7.1. Calculate the work done by a force ~F specified by ~F = 3î + 4 jˆ + 5k̂ in
displacing a body from position B to position A along a straight path. The position vectors
A & B are respectively given as~rA = 2î + 5 jˆ − 2k̂ &~rB = 7î + 3 jˆ − 5k̂.
Given Data:
~F = 3î + 4 jˆ + 5k̂ ~rA = 2î + 5 jˆ − 2k̂ and ~rB = 7î + 3 jˆ − 5k̂.
To find: Work done: W =?
Problem 7.2. A 2000 kg car traveling at 20 m/s comes to rest on a level ground in a
distance of 100 m. How large is the average frictional force tending to stop it?
Given Data:
Mass of car: m = 2000kg Initial velocity of car: vi = 20m/s
Final velocity of car: v f = 0m/s Distance covered by car: S = 100m.
To find: Frictional force:: f =?
1 1
W = mv2f − mv2i (But W = f .S)
2 2
1 1
f × 100 = × 2000 × 0 − × 2000 × (20)2 = 0 − 1000 × 400 = 400000
2 2
400000
f =− = −4000N
100
f = −4000N Ans.
Problem 7.3. A 100-kg man is in a car traveling at 20 m/s. (a) Find his kinetic energy. (b)
The car strikes a concrete wall and comes to rest after the front of the car has collapsed 1
m. The man is wearing a seat bell and harness. What is the average force exerted by the
belt and harness during the crash?
Data:
Mass of the man: m = 100kg (b) Since car is suddenly stopping, so KE
Speed of the man: v = 20m/s possessed by man is doing work against the
collapsed distance: S = 1m belt which stops him at the end of crash:
To find:
(a) Kinetic energy of the man: KE =? Work-done against belt=lost in KE
(b) Average force exerted by the belt: Favg =? 1
W = mv2
2
Solution: (a): KE of man can be found by: 1
Favg × S = mv2
1 2
KE = mv2 1
2 Favg × 1 = × 100 × (20)2
1 2
KE = × 100 × (20)2 Favg = 50 × 400
2
KE = 50 × 400 Favg = 20000N
KE = 20000J Ans. Favg = 20000N Ans.
Problem 7.4. When an object is thrown upward, it rises to a height ‘h’. How high is the
object, in terms of h, when it has lost one-third of its original kinetic energy?
Solution: Object rises to a height h; consider an object lost one third of its original KE at
the height h0 , then according to law of conservation of energy:
loss of KE=gain in PE
1 1 1 2
KE = PE =⇒ × mv = m gh0
3 3 2
1 2 0
v = gh − − − − > (1)
6
v can be found by using the equation of motion:
2gh = v2f − v2i
here g = −9.8, v f = 0, vi = v
− 2 × 9.8h = (0)2 − v2 =⇒ −2gh = −v2
v2 = 2gh − −− > (2)
Problem 7.5. A pump is needed to lift water through a height of 2.5 m at the rate of 500
g/min. What must the minimum horse power of the pump be?
Data:
m
Height: h = 2.5 and rate of lifting water: t = 500g/min
To find: Power of pump: P =?
m 500×10−3 0.5
Solution: Rate of lifting water in SI is: t = 60 = 60 = 0.00833kg/s
The power of pump is given by:
work done by pump mgh m
P= = = × (gh)
t t t
P = 0.00833 × 9.8 × 2.3 = 0.2042J/s = 0.02042W
1HP
∵ 1HP = 746W =⇒ 1W =
746W
1HP
∴ P = 0.2042W = 0.000274HP = 2.74 × 10−4 HP
746
W
P = 2.74 × 10−4 HP Ans.
Problem 7.6. A horse pulls a cart horizontally with a force of 40 lb at an angle 30o above
the horizontal and moves along at a speed of 6.0 miles/hr. (a) How much work does the
horse do in 10 minutes? (b) What is the power out put of the horse?
Data:
Force that horse applied on cart: F = 40lb
Angle: θ = 30o
Time taken: t = 10min = 600s
Velocity of cart: v = 6miles/hour
To find:
(a) Work done by horse: W =?
(b) Power of horse: P =?.
Solution: Because force is given in British system so we will convert all units into British:
6 × 1760 × 3
v = 6miles/hour = = 8.8 f t/s [∵ 1mile = 1760yard = 1760 × 3 f t]
60 × 60
Problem 7.7. A body of mass ‘m’ accelerates uniformly from rest to a speed V f in time t f .
Show that the work done on the body as a function of time ‘t’, in terms of V f and t f , is
2
1 Vf 2
m t
2 t 2f
W = FS
According to Newton’s sencond law of motion:F = ma
W = ma × S (1)
We can find ’a’ in term of v f and t f from the equation of motion:
v f = vi + at = 0 + at f [∵ vi = 0]
vf
a=
tf
Also we find S from the equation:
1 1 1
S = vit + at 2 = 0 × t + at 2 = at 2
2 2 2
Putting these values into equation (1):
vf 1 1 vf vf 2
W = m × × at 2 = m t
tf 2 2 tf tf
2
1 vf 2
Work − done = m 2 t Ans.
2 tf
Problem 7.8. A rocket of mass 0.200 kg is launched from rest. It reaches a point p lying at
a height 30.0 m above the surface of the earth from the starting point. In the process + 425
J of work is done on the rocket by the burning chemical propellant. Ignoring air-resistance
and the amount of mass lost due to the burning propellant, find the speed V f of the rocket
at the point P.
Data:
mass of rocket: m = 0.2kg
height reached by rocket: h = 30.0m
work-done by the rocket: W = +425J
To find: final velocity of the rocket at point P: v f =?
1 2 1
mv + mgh = 425 =⇒ × 0.2v2f + 0.2 × 9.8 × 30 = 425
2 f 2
0.1v2f + 58.8 = 425 =⇒ 0.1v2f = 425 − 58.8 = 366.2
366.2 √
v2f = = 3662 =⇒ v f = 3662 = 60.5m/s
0.1
v f = 60.5m/s Ans.
8.1 Problems
Problem 8.1. An object is connected to one end of a horizontal spring whose other end is
fixed. The object is pulled to the right (in the positive x-direction) by an externally applied
force of magnitude 20 N causing the spring to stretch through a displacement of 1 cm (a)
Determine the value of force constant if, the mass of the object is 4 kg (b) Determine the
period of oscillation when the applied force is suddenly removed.
Data:
Force applied on the object: F = 20N
Mass of the object: m = 4kg
Spring stretch distance: x = 1cm = 1 ×
10−2 = 0.01m
To find:
(q) Spring constant: k =?
(b) Period of oscillation: T =?
Solution:
Problem 8.2. A body hanging from a spring is set into motion and the period of oscillation
is found to be 0.50 s. After the body has come to rest, it is removed. How much shorter
will the spring be when it comes to rest?
Data:
Time period of oscillation T = 0.5s
Mass of the object: m = 4kg
To find:
Extension in spring: x =?
Solution:
r
m
The time period of oscillation is given by:T = 2π − −− > (1)
k
And extension in spring is given by Hooke’s Law: F = kx,
Where F is the weight of object: mg, ∴ Hooke’s Law becomes: mg = kx
m x
Re-arranging: =
k g
r
x
Putting this value in equation (1), we can find: T = 2π
g
x
Squaring both sides:T 2 = 4π 2 , and re-arranging order:
g
2
T g 2
(0.5) 9.8 0.25 × 9.8
x= 2 = 2
= = 0.062m = 6.2cm
4π 4(3.142) 4 × 9.872
x = 6.2cm Ans.
Problem 8.3. A pipe has a length of 2.46 m. (a) Determine the frequencies of the funda-
mental mode and the first two overtones if the pipe is open at both ends. Take v = 344 m/s
as the speed of sound in air. (b) What are the frequencies determined in (a) if the pipe is
closed at one end? (c) For the case of open pipe, how many harmonics are present in the
normal human being hearing range (20 to 20000 Hz)?
Data:
Length of pipe L = 2.46m
Speed of sound of air: v = 344m/s
To find:
(a) Pipe is opened at both ends: f1 =?, f2 =? and f3 =?
(b) Pipe is closed at both ends: ( f1 )closed =?, ( f2 )closed =? and ( f3 )closed =?
(c) Number of harmonics: N =?
Solution:
Problem 8.4. A standing wave is established in a 120 cm long string fixed at both ends.
The string vibrates in four segments when driven at 120 Hz (a) Determine the wavelength
(b) What is the fundamental frequency?
Data:
To find:
Length of string L = 120cm = 1.2m
Wavelength of standing wave: λ1 =?
Frequency of standing wave: f4 = 120Hz
Fundamental frequency: f1 =?
Number of loops: n = 4
Solution:
The string vibrates in four loops, so length could be divided into 4 segments:
λ1 λ1 L
L = n = 4 = 2λ1 =⇒ λ1 =
2 2 2
1.2
λ1 = = 0.6m. λ1 = 0.60m Ans.
2
fn
Fundamental frequancy can be given the relationship: fn = n f1 =⇒ f1 =
n
120
f1 = = 30Hz. f1 = 30Hz. ans.
4
Problem 8.5. Calculate the speed of sound in air at atmospheric pressure P = 1.01 ×
105 N/m2 , taking γ = 1.40 and ρ = 1.2kg/m3 .
Data:
Air pressure: P = 1.01 × 105 N/m2 , γ = 1.4, and the density of air: ρ = 1.2kg/m3
To find: velocity of sound in air: v =?
Solution: According to the relationship of Laplace’s correction:
s r
γP 1.4 × 1.01 × 105 p
v= = = 11.78 × 104 = 3.43 × 102 = 343m/s
ρ 1.2
v = 343m/s Ans.
Problem 8.6. A sound wave propagating in air has a frequency of 4000 Hz. Calculate the
percent change in wave length when the wave front, initially in a region where T = 27oC,
enters a region where the air temperature decreases to 10oC.
Data:
Frequency of sound wave: f = 4000Hz
Initial temperature of air: T1 = 27o + 273 = 300K
Final temperature of air: T2 = 10o + 273 = 283K
∆λ
To find: percentage fraction change in wave length: λ1 % =?
∆λ
Solution: Fraction change in wave length is given by: λ1 :
λ1 − λ2 λ1 λ2 λ2
= − = 1−
λ1 λ1 λ1 λ1
Where λ1 is wavalength at initial temp and λ2 at final temp:
v1 v2
But λ1 = and λ2 =
f f
v2 f v2
∴ 1 − = 1 − − −− > (1)
v2 f v2
Speed of sound at any temperature is given by:
r
T
v = 332
273
Therefore, equation (1), becomes:
q
T2
332 √
r r
∆λ 273 T2 283
= 1− q = 1− = 1− = 1 − 0.943 = 1 − 0.97 = 0.029
λ1 332
T1 T1 300
273
The percentage fraction change in wave length is given by:
∆λ ∆λ
% = 0.029 × 100 = 2.9%. % = 2.9% Ans.
λ1 λ1
Problem 8.7. The frequency of the second harmonic of an open pipe (open at both ends)
is equal to the frequency of the second harmonic of a closed pipe (open at one end). (a)
Find the ratio of the length of the closed pipe to the length of the open pipe.(b) If the
fundamental frequency of the open pipe is 256 Hz, what is the length of pipe? (Use v =
340 m/s).
Problem 8.9. An ambulance travels down a highway at a speed of 75 mi/h. Its siren
emits sound at a frequency heard by a person in a car traveling at 55 mi/h in the opposite
direction as the car approaches the ambulance and as the car moves away from the
ambulance.
Data:
Speed of ambulance (source): vs = 75mil/h
Speed of car (observer): vo = 55mil/h, and frequency of sound: f = 4000Hz
To find: Frequency of sound heard by the listener: f 0 =?
Solution:
When both source and listener are moving toward each other, then:
0 v + vo
f = f , where v is speed of sound = 750mil/h
v − vs
0 750 + 55 805
f = × 400 = × 400 = 477Hz
750 − 75 675
f 0 = 477Hz Ans.
When both source and listener are moving away from each other, then:
0 v − vo
f = f
v + vs
0 750 − 55 695
f = × 400 = × 400 = 337Hz
750 + 75 825
f 0 = 337Hz Ans.
Problem 8.10. A car has siren sounding a 2 kHz tone. What frequency will be detected as
stationary observer as the car approaches him at 80 km/h? Speed of sound = 1200 km/h.
Data:
Speed of car (source): vs = 80km/h, Speed of sound: v = 1200km/h
Frequency of sound: f = 2kHz = 2000Hz
To find: Frequency of sound heard by the listener: f 0 =?
Solution:
In this case listener is at rest and source is moving away from him. So we can use:
0 v 1200 1200
f = f= × 2000 = × 2000 = 2143Hz
v − vs 1200 − 80 1120
f 0 = 2.143kHz Ans.
9.1 Problems
Problem 9.1. How many fringes will pass a reference point if the mirror of a Michelson’s
interferometer is moved by 0.08 mm. The wavelength of light used is 5800Å.
Data:
Displacement of mirror: x = 0.08mm = 0.08 × 10−3 m
Wavelength of light used: λ = 5800Å= 5800 × 10−10 m
To find: Number of fringes: m =?
Solution:
Problem 9.2. In a double slit experiment the separation of the slits is 1.9 mm and the
fringe spacing is 0.31mm at a distance of 1 metre from the slits. Find the wavelength of
light?
Data:
Separation of slits: d = 1.9mm = 1.9 × 10−3 m
Fringe spacing: ∆Y = 0.31mm = 0.31 × 10−3 m and distance from slits: L = 1m
To find: Wavelength of light used: λ =?
Solution:
Lλ ∆Y d
Fringe spacing in YDSE is given by: ∆Y = =⇒ λ =
d L
0.31 × 10−3 × 1.9 × 10−3
λ= = 0.589 × 10−3−3 = 0.589 × 10−6 = 5.89 × 10−7
1
λ = 5.89 × 10−7 m Ans.
Problem 9.3. Interference fringes were produced by two slits 0.25 mm apart on a screen
150 mm from the slits. If eight fringes occupy 2.62 mm. What is the wavelength of the light
producing the fringes?
Data:
Separation of slits: d = 0.25mm = 0.25 × 10−3 m
8 fringes occupy: Y = 2.621mm = 2.62 × 10−3 m
Distance from slits: L = 150mm = 150 × 10−3
To find: Wavelength of light used: λ =?
Solution:
∆Y d Y
Wavelength used in YDSE is given by: λ = , where ∆Y =
L 8
2.62 × 10−3
∆Y = = 0.3275 × 10−3 m
8
0.3275 × 10−3 × 0.25 × 10−3
∴λ = = 0.0005458 × 10−6+3 = 5458 × 10−10
150 × 10−3
λ = 5458Å Ans.
Problem 9.4. Green light of a wavelength 5400Å is diffracted by grating having 2000
line/cm. (a) Compute the angular deviation of the third order image. (b) Is a 10th order
image possible?
Data:
Length of grating: L = 1cm = 1 × 10−2 m
Wavelength of light used: λ = 5400Å= 5400 × 10−10 m
No. of lines on grating: N = 2000 lines, No. of order of image: m3 = 3 and m10 = 10.
To find:
(a) Angular deviation: θ =?
(b) is 10th order image possible?
Solution:
Data:
Wavelength of light used: λ = 6 × 10−7 m
No. of lines on grating per mm: N = 400
Grating element: d = 4001
mm = 25 × 10−7 m
To find:
Angular deviation: θ =? with order of images: m = 1, 2, 3.
Solution:
Problem 9.6. If a diffraction grating produced a 1st order spectrum of light of wavelength
6 × 10−7 m at an angle of 20o from the normal. What is its spacing and also calculate the
number of lines per mm?
Data:
Length of grating: L = 1mm = 1 × 10−3 m
Wavelength of light used: λ = 6 × 10−7 m
Angular deviation: θ = 20o No. of order of image: m = 1
To find:
(a) Grating element: d =? (b) No. of lines/mm=?
Solution:
Problem 9.7. Newton’s rings are formed between a lens and a flat glass surface of
wavelength 5.88 × 10−7 m. If the light passes through the gab at 30o to the vertical and
the fifth dark ring is of diameter 9 mm. What is the radius of the curvature of the lens?
Data:
Light passes at angle: θ = 30o
Wavelength of light used: λ = 5.88 × 10−7 m
No. of dark ring: m = 5, diameter of 5th dark ring d5 = 9mm = 9 × 10−3
To find: Radius of curvature of lens used: R =?
Solution:
Problem 9.8. How far apart are the diffracting planes in a NaCl crystal for which X-rays
of wavelength 1.54Å make a glancing angle of 15o − 540 in the 1st order?
Data:
Glancing angle: θ = 15o 540 = 15 + 60
54
= 15.9o
Wavelength of X-rays used: λ = 1.54Å=1.54 × 10−10 m
No. of order: m = 1
To find: Distance between diffraction plane: d =?
Solution:
Problem 9.9. A parallel beam of X-rays is diffracted by rocksalt crystal the 1st order
maximum being obtained when the glancing angle of incidence is 6 degree and 5 minutes.
The distance between the atomic planes of the crystal is 2.8 × 10−10 m. Calculate the
wavelength of the radiation.
Data:
Glancing angle: θ = 6o 50 = 15 + 60
5
= 6.0833o
Separation of planes: d = 2.8Å= 2.8 × 10−10 m
No. of order: m = 1
To find: Wavelength of X-rays used: λ =?
Solution:
10.1 Problems
Problem 10.1. An object 4 cm tall is placed near the axis of a thin converging lens. If the
focal length of the lens is 25 cm, where will the image be formed and what will be the size
of the image? Sketch the principal ray diagram.
Data:
Focal length: f = 25cm
Size of object: ho = 4cm
Object distance: p = 33.33cm
To find:
Image distance: q =?
Size of image: hi =?
Solution:
1 1 1 1 1 1 1 1
Using thin lens formula: = + =⇒ = − = −
f p q q f p 25 33.33
1 33.33 − 25 8.33 1
= = = 0.01 =⇒ q = = 100cm. q = 100cm Ans.
q 25 × 33.33 833.25 0.01
hi q
Image height can be found by magnification formula as: =
ho p
q 100
hi = ho = × 4 = 12cm. hi = 12cm Ans.
p 33.33
Problem 10.2. A convex lens has a focal length of 10 cm. Determine the image distances
when an object is placed at the following distances from the lens.
(50 cm, 20 cm, 15 cm, 10 cm and 5 cm).
Data:
Focal length: f = 10cm
Object distances: p = 50cm, 20cm, 15cm, 10cm, 5cm
To find: Image distance: q =?
Solution: 1 1 1
(iii) For p = 15cm : = −
q 10 15
Using thins lens formula: 1 15 − 10 5
= =
1 1 1 q 150 150
= −
q f p 150
q= = 30cm. q = 30cm Ans.
1 1 1 5
(i) For p = 50cm : = −
q 10 50 1 1 1
(iv) For p = 10cm : = −
1 50 − 10 40 q 10 10
= =
q 500 500 1 10 − 10 0
= =
500 q 100 100
q= = 12.5cm. q = 12.5cm Ans. 100
40 q= = ∞. q = ∞ Ans.
1 1 1 0
(ii) For p = 20cm : = − 1 1 1
q 10 20 (v) For p = 5cm : = −
1 20 − 10 10 q 10 5
= =
q 200 200 1 5 − 10 −5
= =
200 q 50 50
q= = 20cm. q = 20cm Ans.
10 50
q= = −10cm. q = −10cm Ans.
−5
Problem 10.3. Two converging lenses of focal lengths 40 cm and 50 cm are placed in
contact. What is the focal length of this lens combination? What is the power of the
combination in diopters?
Data: To find:
Focal length of first lens: f1 = 40cm Focal length of combination: f =?
Focal length of 2nd lens: f2 = 50cm Power of combination: P =?
Solution:
1 1 1 1 1 5+4
Using the combination of lens formula: = + = + =
f f1 f2 40 50 200
1 9 200
= =⇒ f = = 22.2cm. f = 22.2cm Ans.
f 200 9
1 1 100
Power of lens is given by: P = = −2
= = 4.5 diopters. Ans.
f 22.2 × 10 m 22.2
Data:
To find:
Focal length of first lens: f1 = 20cm
Distance between two lenses: L =?
Focal length of 2nd lens: f2 = 4cm
Solution: Since rays are parallel therefore the object p1 and image q2 are at infinity.
Problem 10.5. A parallel light beam is diverged by a concave lens of focal length −12.5cm
and then made parallel once more by a convex lens of focal length 50cm. How far are the
two lenses apart.
Data:
To find:
Focal length of first lens: f1 = −12.5cm
Distance between two lenses: L =?
Focal length of 2nd lens: f2 = 50cm
Solution: When a beam of parallel rays are diverged by concave lens and again made
parallel by a convex lens then principal focus of both lenses must be at same point. In this
case the distance between lenses is given by:
Problem 10.6. Two lenses are in contact, a converging one of focal length 30cm and
a diverging one of focal length −10cm. What is the focal length and power of the
combination?
Data: To find:
Focal length of first lens: f1 = 30cm Focal length of combination: f =?
Focal length of 2nd lens: f2 = −10cm Power of combination: P =?
Solution:
1 1 1 1 1 1−3
Using the combination of lens formula: = + = + =
f f1 f2 30 −10 30
1 2 30
=− =⇒ f = − = −15cm. f = −15cm Ans.
f 30 2
1 1 100
Power of lens is given by: P = = −2
= = −6.67 diopters. Ans.
f −15 × 10 m −15
The negative sign shows that the combination behaves as a concave lens.
Problem 10.7. Moon light passes through a converging lens of focal length 19 cm, which
is 20.5 cm from a second converging lens of focal length 2 cm. Where is the image of the
moon produced by the lens combination?
Data: Distance between two lenses: L = 20.5cm
Focal length of first lens: f1 = 19cm To find:
Focal length of 2nd lens: f2 = −2cm Image distance: q =?
Solution: Since rays from moon are parallel therefore the object p1 is at infinity.
Problem 10.8. Find the distance at which an object should be placed in front of a convex
lens of focal length 10 cm to obtain an image of double its size?
Solution:
q
Magnification is given by: M =
p
q
2= =⇒ q = 2p, but object distance is given by:
p
1 1 1
= +
f p q
1 1 1 2+1 3
For real image: = + = = . By cross multiplication:
10 p 2p 2p 2p
2p = 30 =⇒ p = 15cm. p = 15cm Ans.
1 1 1 2−1 1
For virtual image: q = −2p : = + = = .
10 p −2p 2p 2p
By cross multiplication: 2p = 10 =⇒ p = 5cm. p = 5cm Ans.
Solution:
Magnification M of compound microscope is given by: M = Mo × Me − −− > (1)
Where Mo and Me are objective and eye piece magnifications.
qo
Magnification of objective is given by: Mo = − −− > (2)
po
Where qo and po have to be determined:
For eye piece lens: q = −225mm and pe is given by using lens formula:
1 1 1 1 1 3+1 4
= − = − = =
pe fe qe 75 −225 225 225
pe = 56.25mm. But qo = L − pe = 200 − 56.25 = 143.75mm
For objective:
1 1 1 1 1 143.55 − 12 131.75
= − = − = = =⇒ po = 13.09mm.
po fo qo 12 143.75 12 × 143.75 1725
143.75
Now putthing these values into equation (2): Mo = = 10.98
13.09
d 225
And magnification for eye piece is given by: Me = 1 + = 1 + = 1+3 = 4
fe 75
Magnification of compound microscope is given by equation(1):
M = 10.98 × 4 = 44. M = 44 Ans.
Problem 10.10. An astronomical telescope of angular magnification 1000 has an objective
of 15 m focal length. What is the focal length of the eye piece?
Data:
Angular magnification of telescope: M = 1000 To find:
Focal length of objective lens: fo = 15m Focal length of eye piece lens: fe =?
Solution:
fo
The magnification of astronomical telescope is given by: M =
fe
fo 15
fe = = = 0.015m. fe = 0.015m Ans.
M 1000
Problem 10.11. A Galilean telescope has an objective of 120 mm focal length and an eye
piece of 50 mm focal length. If the image seen by the eye is 300 mm from the eye piece,
what is angular magnification?
Solution:
Data:
Focal length of objective lens: fo = 120mm Angular magnification is given as:
Focal length of eye lens: fe = 50mm
fo fe
Distance of final image from eye piece: M= 1+
fe d
d = 300mm 120
50
To find: M= 1+ = 2.4(1.1666)
50 300
Magnification: M =?
M = 2.8 Ans.
Problem 10.14. A convex lens forms image of an object on a fixed screen 20 cm from
the lens. On moving the lens 60 cm towards the object, the image is again formed on the
screen. What is the focal length of the lens?
Data: In second case when object is moved:
In first case when image is formed at 20cm: Object distance: p2 = x − 60
Objective distance: p1 = x Image distance: q2 = 80cm
Image distance: q1 = 20cm To find: Focal length: f =?
Solution:
In first case when object is placed at x distance:
1 1 1 1 1 20 + x
= + = + = − − − − > (1)
f p1 q1 x 20 20x
When object is moved to new position:
1 1 1 1 1 80 + x − 60 20 + x
= + = + = = − −− > (2)
f p2 q2 x − 60 80 80(x − 60) 80(x − 60)
Comparing eq.(1) and eq.(2)
20+
x 20+
x 1 1
= = = By cross multp:
20x 80(x − 60) 20x 80(x − 60)
20x = 80(x − 60) =⇒ x = 4(x − 60) = 4x − 240 =⇒ 4x − x = 240
240
3x = 240 =⇒ x = = 80cm
30
20x 20 × 80 1600
Focal length is given by equation (1): f = = = = 16cm
20 + x 20 + 80 100
f = 16cm Ans.
Problem 10.15. Two converging lenses are 25 cm part. Focal length of each is 10 cm. An
object is placed in front of one lens at 50 cm. Find the distance between the object and the
final image?
Data: Object distance from 1st lens: p1 = 50cm
Focal length of each lens: To find:
f1 = f2 = f = 10cm Distance between object and final image:
Distance between lenses: L = 25cm d = p1 + q1 + p2 + q2 =?
Solution:
Distance between object and final image is given by: d = p1 + q1 + p2 + q2 − − > (1)
Therefore, first we need to find these quantities using lens formula:
1 1 1 1 1 5−1 50
For first lens: = − = − = =⇒ q1 = = 12.5cm
q1 f1 p1 10 50 50 4
p2 can be calculated as: p2 = L − q1 = 25 − 12.5 = 12.5cm
1 1 1 1 1 12.5 − 10 2.5 125
For second lens: = − = − = = =⇒ q2 = = 50cm
q2 f2 p2 10 12.5 125 125 2.5
Putting these values in equation (1), we get: d = 50 + 12.5 + 12.5 + 56 = 125cm.
d = 125cm Ans.