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State for each of the following physical quantities, if it is a scalar or vector: Volume, mass, speed, acceleration, density, number of moles, velocity, angular frequency, displacement, angular velocity.

Scalar: Volume, mass, speed, density, number of moles, angular frequency.
Vector: acceleration, velocity, displacement, angular velocity.
[Scalars have only physical magnitude. Vectors have direction in addition.]


Pick out the two scalar quantities in the following list: force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic moment, reaction as per Newton's third law of motion, relative velocity.

Work, Current


Pick out the only vector quantity in the following list:
Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

Impulse (change in momentum).


State with reasons, whether the following algebraic operations with scalar and       vector physical quantities are meaningful.
(a) Adding any two scalars,
(b) Adding a scalar to a vector of the same dimension,
(c) Multiplying any vector by any scalar,
(d) Multiplying any two scalars,
(e) Adding any two vectors, and
(f) Adding a component of a vector to the same vector.

a. Not for any two scalars. But it is meaningful, if both the scalars are represented the same physical equalities.
b. No. A scalar cannot be added to a vector.
c. Yes. (The result is a vector). It is permissible.
d. Yes. It is permissible, e.g., Volume multiplied with velocity.
e. Not for any two vectors. They should represent the same physical dimension to be added together.
f. It is not permissible as the component of a vector is scalar and we cannot add a scalar to a vector.


Read each of the following statements carefully and state with reasons, if they are true or false:
(a) The magnitude of a vector is always a scalar.
(b) Each component of a vector is always a scalar.
(c) The total path length is always equal to the magnitude of displacement     vector of a particle. 
(d) The average speed of a particle (defined as total path length divided by the time taken to cover the path) is either greater or equal to the magnitude of the average velocity of the particle over the same length of time.
(e) Three vectors not lying in a plane can never be added to give a null vector.

(a) True. Magnitude represents the physical value of the vector and is indicated by the length of the line representing the vector.
(b) False. Components represent the equivalent of the vector in those directions.
(c) False. It is true only if the particle travels in a straight line. [Displacement vector is a straight line joining the physical position of the starting and end point.]
(d) True. If the particle travels in a straight line, the average speed will be numerically equal to the average velocity. Any other path should be longer. Dividing a longer distance by the same time will give a higher numerical value for the speed.
(e) True. When forces are represented in a vector polygon the closing line represents the resultant vector. If three vectors add to give a null vector, they should form a closed triangle. All three lines of any closed triangle are necessarily to lie in a plane. Hence the given statement is true.


When does the equality sign above apply?




Vectors and are represented in the above figure. + is represented by the vertical line. For finding
  - , the vector is reversed and is added to . The horizontal line as shown in the figure represents this. Now, the inequalities are proved making use of the facts that (a) the two sides of a triangle are together greater than the third and (b) one side is more than the difference between the other two.

(i) In the above figure,

If |a| and |b| are collinear
Both the above are combined to one inequality
(ii) By the logic (b) above,
If |a| and |b| are in opposite directions, 
Both the above are combined to one inequality 
(iii) By the logic (a) above,

If |a| and |b| are collinear, 
Both the above are combined to one inequality
(iv) By the logic (b) above,

If |a| and |b| are in opposite directions, 
Both the above are combined to one inequality.


Given = 0, which of the following statements are correct:
(a) , , and must each be a null vector. 
(b) The magnitude of ( + ) equals the magnitude of ( + ). (c) The magnitude of can never be greater than the sum of the magnitudes of , and
(d) + must lie in the plane of and , if and are not collinear, and in the line of and , if they are collinear?

Vector sum + + + = 0 can be represented by figure 1. Since + + + = + + + , the same sum can be represented alternatively by figure 2.


(a) Incorrect except in case where = = = = 0. For the sum of the vectors to be equal to zero, the vector polygon should close.
(b) Correct. The figure 2 is equivalent to figure1. The closing lines for + and + are same. Hence their magnitudes should be equal. (The magnitudes are same, but the directions are opposite).
(c) Correct. See figure 1. The vector should be the closing line of the open figure + + . Hence the resultant of sum of + + is exactly match with the magnitude of .
(d) Correct. In the figure 1, the diagonal shown represents resultant of both + and + . So, this diagonal should lie in both the planes i.e. of and as well as of and . However, if and are collinear the resultant of and should necessarily be in the same line to close the triangle.


Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in the figure. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skate?

Radius of the circular path = 200m, hence, PQ = 400m. The path is the straight line joining the starting and ending position. Hence the paths traced by the three girls A, B and C is PAQ, PBQ, and PCQ. All the girls start from the point P and finally reach Q. The magnitude of the displacement for all the girls is 400m. However, magnitude of the displacement is equal to the actual length of path selected only for the girls B.


A cyclist from the centre O of a circular park of radius 1 km, reaches the edge P of the park, then cycles along the circumference, and returns to the centre along QO as shown in figure. If the round trip takes 10 minutes, what is the (a) net displacement, (b) average velocity and average speed of the cyclist ?         

As the cyclist comes back to the centre O via OPQO.
Net displacement is zero since the starting and end points are same.

(a) Average velocity is also zero because the displacement is zero.
(b) Average speed = Total distance traveled / time = OP + PQ + QO
                  = 1km + (2
πx1)/4x km + 1 km =2km + (2 x 22) / (4 x 7) = 3.5 Km.
     Time taken = 10 min = 1/6hrs
     Average speed = 3.57 / (1/6) = 21.42 Km / hr.


On an open ground, a motorist follows a track that turns to his left by an angle of 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth road turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

(a) Displacement at the third turn is represented by the vector . It is 100 m at an angle of 60o to the initial direction of travel.
     Total path length covered = 3 x 500 m = 1.5 km

(b) The sixth turn is the starting point itself. Displacement at this point is zero irrespective of the distance of actual travel.
     Total path length covered = 6 x 500 m = 3.0 km

(c) Displacement at the eighth turn is represented by the vector . It is 866 m at 30o the initial direction of travel.
     Total path length covered = 8 x 500 m = 4.0 km.


A passenger arriving in a new town wishes to go from station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is the (a) average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?

(a) Average speed = total path length / time
                            = 23 / 28 km/min
                            = 49.26 km/h

(b) Average velocity = straight length / time
                              = 10 / 28 km/min
                              = 21.43 km/h

(c) They are not equal because, velocity is calculated from displacement, which is the shortest distance between starting and end points. Speed is calculated from actual path length. They are equal only if the travel is in a straight line.


Rain is falling vertically with a speed of 30 ms-1. A woman rides a bicycle with a speed of 10 ms-1 in the north to south direction. What is the direction in which she should hold the umbrella?


Rain will fall on the person at an angle of the resultant velocities of the rain drops and the cycle. The velocity diagram is given on the sketch above.
Downward velocity of rain Vr = 30 m/s
Southward velocity of the cycle Vc = 10 m/s
®       ®            ®               ®
Relative velocity with respect to woman = Vr - Vc = Vr + (- Vc)
Angle of fall of rain drops on the person = tan-1 (10/30)
                                                        = 18o towards south.
The woman should keep her umbrella at angle of 18o towards the south.


A man can swim with a speed of 4 km/h in still water. How long does he take to cross a river 1 km wide, if the river flows steadily at 3 km/h and he makes his strokes normal to the river current? How far down the river does he go when he reaches the other bank?

Velocity of the river Vr = 3 km/h
Velocity of the swimmer Vs = 4 km/h
Relative velocity = Vs - Vr
The component of the velocity across the river is 4 km/h.
This is the velocity that will carry the swimmer across the river 1 km wide.
Time to swim the distance of 1 km across the river
with a velocity Vs of 4 km/h} = ¼ hr = 15 min.
During the time, he crosses the river the current would have carried him downstream to some distance.
The velocity component along the river will carry the
swimmer during the above 15 minutes} = 3 x ¼ km = 750 m.


In a harbour, wind is blowing at a speed of 72 km/h, and the flag on the mast of a boat anchored in the harbour flutters along N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat? [The velocities on the flag are represented in the sketch.]



Boat speed = b = 51 km/h
Wind speed = w = OB = 72km/h
OA = Boat speed reversed = -b
OAB = 45o (AB is N-E)
When the boat does not move the flag flutters in the direction of the wind, i.e., North-East.
When the boat starts moving, it will flutter in the direction of resultant velocity of the boat and the wind.
Resultant velocity = w - b
i.e. compound with w, the reversed velocity of b. (i.e., -b). 
In the triangle OAB,
OA = Boat speed = 51 km/h
AB = Wind velocity = 72 km/h
OAB = 45o
Solving triangle ABC, in this particular case, OAB works out to 90o.
The flag will flutter in OB direction i.e., East.


The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 ms-1 can go without hitting the ceiling of the hall?

The maximum height attained by the project is given by H = (u2sin2θ)/(2g )
Maximum height H = Height of ceiling = 25m

25 =  

θ =

θ =  
Range of the ball R =   =
                                         = m.


A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball?

Maximum range is reached when the angle of throw is 45o and
the maximum range R = ( u2 × sin 2
α) /( g) 

n the given problem,
Maximum range = u2 sin (2 × 45o)/g = 100 m.
... R= u2 / g =100m (sin 900 = 1)
The projectile can gain maximum height only when it thrown vertically up i.e. when angle of projection
αis 90o
h = u2sin2
α /2g
h = ( u2/2g ) = ( 100/2 )
Maximum height = h = 50 m


A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 seconds what is the magnitude and direction of acceleration of the stone?

Angular velocity = ω = 2πN = 2π × (14/25) m/s = 3.52 m/s (where N is the frequency)
Acceleration = ω 2 r = (3.52)2 × 0.8 ms-2 = 9.91 ms-2
The acceleration is directed towards the centre.


An aircraft executes a horizontal loop of radius 1 km with a steady speed of 900 km / h. Compare its centripetal acceleration with the acceleration due to gravity.

Tangential velocity = r × ω = 900 km/h
                                         = 900 × 1000/3600 m/s
                                         = 250 m/s
Angular velocity ω = 250 / 1000 = 0.25 radians/s
Centripetal acceleration = ω 2 r = (0.25)2 × 1000 ms-2 = 62.5 ms-2
Ratio to g = 62.5 / 9.8 = 6.38.


Read each of the statement below carefully and state, with reasons, if it is true or false:
(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards the centre. 
(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at that point.
(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.
(d) The radial component of acceleration of a particle in circular motion is given by v2/R (in magnitude) even when the speed v is not constant.

(a) The acceleration of a particle is towards the centre of rotation only when the circular motion is with uniform angular velocity. The word net acceleration implies that there are other accelerations also. Hence this statement is not true.
(b) True. 
(c) True. For uniform circular motion the acceleration is directed toward the centre and the magnitude is constant. Averaged over a full circle, this is a null vector.
(d) True. The radial component of acceleration is given by the formula v2 / r at any moment. [The statement does not refer to actual value of acceleration, but only about the validity of the formula v2 / r].


The position of a particle is given by r = m where t is in seconds and the coefficients have the proper units for r to be in metres.
(a) Find the ‘v’ and ‘a’ of the particle? (b) What is the magnitude and direction of velocity of the; particle at t = 2s?

(a) Velocity  m/s and
     Acceleration     ms-2    

(b) At time t = 2s,
     Velocity m/s
     Magnitude, v =   
                                              = 8.54 m/s
     For direction
θ, tan θ = = 2.67 = tan 70°Angle with x-axis, t = 70°


A particle starts from the origin at t = 0s with a velocity of 10.0m/s and moves in the x-y plane with a constant acceleration of ms-2.
(a) At what time is the x-coordinate of the particle 16 m? What is the y-coordinate of the particle at that time?
(b) What is the speed of the particle at the time?

Initial velocity u = 10.0 m/s
Acceleration = m/s2

(a) We know that r = ut + at2
     Substituting the values, we get,
                         r = 10.0 t + () t2
                           = 4.0 t2 + (10.0 t + 1.0 t2)
     Hence x = 4.0 t2 
              y = 10.0 t + 1.0 t2

     Given that x = 16 m
     Therefore 16 = 4.0 t2
                      t2 = 4
                      t = 2 s

     y = 10.0 t + 1.0 t2
     = 10.0
×2 + 1.0 ×2 ×2
        = 24 m

(b) Velocity v = = 8.0t + (10.0 + 2.0t)
     At t = 2s, v = 8.0
×2 + (10.0 + 2.0 ×2)
                      = 16.0 + 14.0
     Therefore magnitude of velocity v =
                                                    = 21.26 m/s.


and are unit vectors along x and y- axis respectively. What is the magnitude and direction of the vectors + , and - ? What are the components of a vector A = 2 + 3along the directions of +, and - ?


The resultant = + = √ + 12) = 2 units
The direction = tan β==1/1=1.Therefore β= 45o

The resultant = + ( - ) = = 12 )+ ( -12 ) = 2units
Direction = -45o with the x axis.


OP represents vector A = 2i+3j
OM represents the component of A on the direction of i + j 
ON represents the component of A on the direction of i - j

[Projection of A on B is given by A.B and is equal to AB cos θ , where θ = angle between A and B]
A.B = A B cos θ
Component of A in the direction of i + j = A cos θ = A.B/B
= (2i+ 3j) . (i+ j) / (i2+ j2)
= 2i.i + 3j.i + 2i.j + 3j.j / 2
= 5 / 2 units
Component of A in the direction of i-j 
= (2i+ 3j) . (i- j) / (i2+ j2)
= 2i.i + 3i.j - 2i.j - 3j.j/ (i2+ j2)
= - 5 / 2 units
[Notice the use of vector and scalar (modulus) values in the above arguments].


For any arbitrary motion in space, which of the following relations are true?
          a. Vaverage = (1/2) (v (t1) + v (t2))
          b. Vaverage = ( r(t2) + r (t1) ) / (t2 - t1)
          c. v (t) = v (0) + at
          d. r (t) = r (0) + v (0) t + (1/2)at2 
          e. aaverage = [v(t2) - v(t1)] / (t2 - t1)
[The 'average' stands for average of the quantity over the time interval t1 to t2].

a. False
b. True
c. False
d. False
e. True


Which of the following quantities are independent of the choice of orientation of the coordinate axes: a + b, 3ax+ 2by, | a + b - c|, angle between b and c, a where ‘a’ is a scalar?

In the case of vector operations, if a new coordinate orientation is chosen they will get resolved into the new system. The resultant on the new coordinate system will be interpreted by scalar addition of the projections on the new axes. Thus all vector results are independent of orientation of coordinate axes. Thus, all the factors above are independent of coordinate systems, except 3ax + 2by, which is a scalar result. Though the result of |a + b - c| is a scalar, the result is obtained by a vector operation and hence will be constant. When a vector is multiplied by a scalar as in λ, the result is a vector.


Read each statement below carefully and state, with reasons and examples, if it is true or false: A scalar quantity is one that
           a. is conserved in a process,
           b. can never take negative values,
           c. must be dimensionless,
           d. does not vary from one point to another in space,
           e. has the same value for observers with different orientation axes.

(a) False. (A scalar multiplied by a vector is a vector).
(b) False. (Distances measured in the negative direction in the coordinate axes, negative temperature).
(c) False. (Speed, density) A scalar quantity may not be dimensionless.
(d) False. (Temperature, volume) A scalar quantity may vary from point to point in space.
(e) True. A scalar quantity is independent of orientation of axes.


An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point is by the aircraft positions 10 s apart is 30o, what is the speed of the aircraft?

Distance traveled by the aircraft in 10 sec = AB
From the figure AB   = 2 x tan 15o x 3400 m
                              = 1822 m
Speed of the aircraft = 182.2 m/s.


 A vector has magnitude and direction. Does it have a location in space? Can it vary with time? Will two equal vectors ‘a’ and ‘b’ at different locations in space necessarily have identical physical effects? Give examples in support of your answer.

Position and displacement vectors have a location in space and they are defined by the coordinate system used. In the case of velocity and acceleration vectors, while the line of action may remain constant the tail and tip have no significance with reference to space. They may also vary with time along the same axes. Two equal vectors may not always give the same results. A force applied on one side of a lever gives a different effect than identical force acting on them.


A vector has both magnitude and direction. Does that mean that anything that has magnitude and direction is necessarily a vector? The rotation of a body can be specified by the direction of the axis of rotation, and the angle of rotation about the axis. Does that make any rotation a vector?

While vectors have direction and magnitude, the converse is not universally true. The vectors should obey the laws of addition for vectors, i.e. the components resolved in the two axes should constitute the components of the resultant. 
The rotation of a body specified by the direction of the axis of rotation, and the angle of rotation about the axis, will not obey this law of vector addition. Such a representation cannot be a vector.


Can you associate vectors with: (a) the length of a wire bent into a loop? (b) a plane area (c) a sphere? Explain.

The values of length, area and volume are all scalar quantities. However any point on the line or surface or in the volume of a sphere can be represented by vectors, if they are related to a coordinate system.


A bullet fired at an angle of 30o with the horizontal hits the ground 3 km away. By adjusting its angle of projection, can one hope to hit a target 5 km away? Assume the muzzle speed to be fixed, and neglect air-resistance? Take g = 10 ms-2


Let α be the angle of projection to the horizontal and u be the velocity of projection. 
Vertical component of velocity = u sin
The bullet goes up against the gravity say for a time t seconds and reaches zero velocity at the top most point. Considering this motion,
0 = u sin
α - g t
-u sin
α = - gt
t = u sin
Time to touch ground = 2 t = 2 u sin
During this time the horizontal component would have taken the bullet to the range = R (say)
Horizontal component of the velocity = u cos
Range = 2 u sin α . u cos α/g = u2 sin 2α/g
Maximum range occurs at sin
θ = 1. i.e., at θ = 45o
R = (u2 sin 2
θ) / (g)
3000 = (u2 sin 60o ) / (10)
u2 = (3000
× 10 × 2) / (3)
     = 34641
Solving we get, u = 186.12 ms-1
For g = 10 ms-2
sin 2
θ = sin 90o = 1
Rmax =       
         = 3464.1 m
Rmax = 3.46 km We cannot hope to hit the target since maximum range is less than 5km.


A machine gun is mounted in the top of a tower 100 m high. At what angle should the gun be inclined to cover a maximum range of firing on the ground below? The muzzle speed of the bullet is 150ms-1. [Take g = 10 ms-2 ].


Let the angle of projection to the horizontal = α
Horizontal component = 150 cos α
Let the time to reach maximum height = t1
0 = (150 sin
α) - gt1 t1 = (150 sin α)/g
    = 15sin
Horizontal displacement during the time t1 = 150 cos α . 15 sin α
on reaching the same level as that of projection the downward angle will be the same as that of projection gets directed downward. Also the speed will be the same as projection speed. 
Similarly after reaching the projection level, the horizontal displacement
= 150 cos
Total horizontal displacement during the time 2t1 + t2
= 150 cos
α [15 sin α + ]
Since it is complicated to maximize this function by calculus, use graph to find the maximum value. Since the projection angle is 45o for level ground, the angle in the present case should be slightly below that. Try out values starting from values 45o .


A fighter plane flying horizontally at an altitude of 1.5 km with a speed of 720 km/h passes directly overhead an anti-aircraft gun. At what angle from the vertical should the gun be fired for the shell with a muzzle speed of 600 ms-1 to hit the plane? At what minimum altitude should the pilot fly the plane to avoid being hit ? [Take g = 10 m/s2].



Speed of the plane = 720 km/hr = = 200m/s
Speed of the shell = u = 600 m/s
Time for the shell to reach the point B = Time t for the plane to travel from A to B
u sin
θ x t = 200 x t ( u sin θ is the horizontal component of the velocity of the shell,  u = 600 m/s) 
θ = = sin (19.5o)        
sin -1
θ = 1/3 = sin -1 θ = (0.33)
600 sin
θ = 200
θ = 19.5o
The shell should be fired at 19.5o to the vertical.
To avoid the shell, the plane should fly above the maximum height attainable by the shell.
Maximum height reached by the projectile H =
Thus the plane should fly at a height = 1800 x (8/3)2 = 16000 m
                                                                         H = 16 km
The lowest altitude for the plane is 16 km to avoid being hit.


A cyclist is riding with a speed of 27 km/h. As he approaches a circular turn on the road of radius 80 m, he applies brakes and reduces speed with a constant rate of 0.5 m /s every second. What is the magnitude and direction of the net acceleration of the cyclist on the circular turn?

Tangential velocity at the start of the curve = v = 27000 / 3600 m/s
Angular velocity at the start of the curve = w = v/r
                                                           = (27000/3600) / 80 rad/s
                                                           = 0.09375 rad/s
Centripetal acceleration = w 2r
                                   = (0.09375)2 x 80 m/s-2
                                   = 0.703 m/s-2

Acceleration vector diagram:

Linear acceleration = OA = 0.5 ms-2
Centripetal acceleration = OB
(Due to movement in the curve)
Resultant acceleration = OR
The resultant acceleration = = 0.5087 m/s
at an angle q as shown in the sketch = tan-1 (0.09375 / 0.5)
                                                     = 10°37'
[The above calculations are worked out for values at the start of the curve].


Show that the projection angle θ0 for a projectile launched from the origin is given by θ0 = where the symbols have usual meaning.



θ =  


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