Important Questions for Class 11
Physics
Chapter 4 – Motion in A Plane
Very Short Answer Questions
1 Mark
1. What is trajectory of a projectile?
Ans: Trajectory of a projectile can be defined as the path followed by a projectile.
Trajectory may be of various forms, for example, parabola.
2. A projectile is fired at an angle of 30 with the horizontal with velocity
10m / s . At what angle with the vertical should it be fired to get maximum
range?
Ans: The maximum range that can be achieved is at an angle of 45 .
3. What is the value of angular speed for 1 revolution?
Ans: For one complete revolution,   2 in time period t  T , the angular speed
2
can be given as  
.
T
4. Give an example of a body moving with uniform speed but having a
variable velocity and an acceleration which remains constant in magnitude
but changes in direction
Ans: An example of a body moving with uniform speed having velocity that is
not constant can be given as a body moving in a circular path.
5. What is the direction of centripetal force when particle is following a
circular path?
Ans: The direction of the centripetal force in a circular path is towards the centre
of the circle.
6. Two vectors A and B are perpendicular to each other. What is the value
of A  B ?
Ans: We know,   90
A  B  ABcos 
 AB  0
7. What will be the effect on horizontal range of a projectile when its initial
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velocity is doubled, keeping the angle of projection same?
Ans: The horizontal range will be four times the initial horizontal range.
8. What will be the effect on maximum height of a projectile when its angle
of projection is changed from 30 to 60 , keeping the same initial velocity of
projection?
Ans: The maximum height will be three times the initial vertical height.
9. What is the angular velocity of the hour hand of a clock?

Ans: The angular velocity is radian per hour .
6
10. A body is moving on a curved path with a constant speed. What is the
nature of its acceleration?
Ans: Acceleration should be perpendicular to the direction of motion and the
nature is called as centripetal acceleration.
11. State, for each of the following physical quantities, if it is a scalar or a
vector: volume, mass, speed, acceleration, density, number of moles, velocity,
angular frequency, displacement, angular velocity.
Ans: Scalars: Volume, mass, density, speed, number of moles, angular frequency
Vectors: Acceleration, displacement, velocity, angular velocity
A scalar quantity is expressed by its magnitude only. It does not have any
direction related with it. Volume, mass, speed, density, number of moles, and
angular frequency are some of the scalar physical quantities.
On the other hand, a vector quantity is specified by its magnitude as well as the
direction associated with it.
Acceleration, velocity, displacement, and angular velocity come under this
category.
12. Pick out the two scalar quantities in the following list:
force, angular momentum, work, current, linear momentum, electric field,
average velocity, magnetic moment, relative velocity.
Ans: Work and current are scalar quantities.
Work done is expressed by the dot product of force and displacement. Because
the dot product of two quantities is always a scalar, work is a scalar physical
quantity.
Current is expressed only by its magnitude. Its direction is not taken into
consideration. Therefore, it is a scalar quantity.
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13. 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.
Ans: Impulse is a vector quantity.
Impulse is expressed as the product of force and time. Because the force is a
vector quantity, its product with time (a scalar quantity) gives a vector quantity.
Short Answer Questions
2 Marks
1. What is the angle between two forces of 2N and 3N having resultant as
4N ?
Ans: Using the equation R  A 2  B2  2ABcos 
We can write,
 4  22  32  2  2  3cos 
 16  4  9  12cos 
3
 cos    cos1 0.25
12
   7532' , which is the required angle.
2. What is the angle of projection at which horizontal range and maximum
height are equal?
Ans: We can write,
u 2 sin 2 u 2 sin 2 

g
2g
2
u sin 2
 R  Horizontal Range 
g
Therefore, we can write,
1
sin 2  sin 2 
2
2
2
u sin 
 H max  Maximum Height 
2g
1
 2sin  cos   sin 2 
2
 tan   4
   75.96 , which is the required angle of projection.
3. Prove that for elevations which exceed or fall short of 45 by equal
amounts the ranges are equal.
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Ans: We know that,
u 2 sin 2
R
g
And,
1  45  
2  45  
2
u 2 sin 21 u sin 2  45   
R1 

g
g
2
u sin  90  2 
 R1 
g
And,
2
u 2 sin 22 u sin 2  45   
R2 

g
g
u 2 sin  90  2 
 R2 
g
Therefore, we can write, R 1  R 2 .
4. At what range will a radar set show a fighter plane flying at 3km above
its centre and at distance of 4km from it?
Ans: Here, the straight distance of the object from the radar can be given as  OB
OB  42  32
 OB  16  9
 OB  25
 OB  5
 Range  5km
5. Two forces 5kg wt and 10kg wt are acting with an inclination of 120
between them. What is the angle which the resultant makes with 10kg wt ?
Ans: We can write,
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F1  5kg wt
F2  10kg wt
 F  F12  F2 2  2F1F2 cos 
F2 sin 
tan  
F1  F2 cos 
5sin120
 tan  
10  5cos120
3
5
2
 tan  
1
10  5 
2
1
 tan  
3
 1 
   tan 1 
  30 , which is the required angle here.
 3
6. A stone is thrown vertically upwards and then it returns to the thrower.
Is it a projectile? Explain.
Ans: A stone thrown vertically upwards cannot be called as a projectile since a
projectile must have two perpendicular components of velocities but in the given
case a stone has velocity in one direction while going up or coming downwards.
7. Which is greater: the angular velocity of the hour hand of a watch or
angular velocity of earth around its own axis?
1
Ans: In hour hand of a watch time period can be given as T  h
2
2
H 
12
For rotation of earth can be given as T  24h
2
E 
24

24
 H
2
E 12
 H  2E
Clearly, H  E .
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8. Why does the direction of motion of a projectile become horizontal at the
highest point of its trajectory?
Ans: At the highest point of trajectory, the vertical component of velocity
becomes zero. Thus, direction of motion of projectile becomes horizontal.
9. A vector A has magnitude 2 and another vector B have magnitude 3
and is perpendicular to each other. By vector diagram, find the magnitude
of 2A  B and show its direction in the diagram.
Ans: We can write,
PQ  2A  4cm;QS  B  3cm
PS  PQ2  QS2  PS  42  32  PS  5cm
10.Find a unit vector parallel to the resultant of the vectors A  2iˆ  3jˆ  4kˆ
and B  3iˆ  5jˆ  kˆ .
Ans: We know that R̂ 

R
R
 
R  A  B  2iˆ  3jˆ  4kˆ  3iˆ  5jˆ  kˆ

 R  5iˆ  2ˆj  5kˆ
 R  52   2   52
2
 R  25  4  25
 R  54
 R̂ 
5iˆ  2jˆ  5kˆ
, which is the required unit vector.
54
11. A stone tied at the end of string is whirled in a circle. If the string breaks,
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the stone flies away tangentially. Why?
Ans: When a stone, that is tied at the end point of a string, moving around a
circular path, its velocity acts tangent to the circle.
When the string breaks, the centripetal force will not act. Because of inertia, the
stone continues to move along the tangent to circular path, and flies off
tangentially to the circular path.
12. What are the two angles of projection of a projectile projected with
velocity 30m / s , so that the horizontal range is 45m . Take, g  10m / s2 .
Ans: We can write,
u 2 sin 2 302  sin 2
R

 45
g
10
450
 sin 2  2
30
1
 sin 2 
2
2  30 or150
   15 or 75 , which are the two required angles of projection.
13. The blades of an aeroplane propeller are rotating at the rate of 600
revolutions per minute. Calculate its angular velocity.
Ans: We can write,
  600rev / min
600

rev / sec
60
600
  2  2 
60
   20rad / sec , which is the required angular velocity.
14. What is a uniform circular motion? Explain the terms time period,
frequency and angular velocity. Establish relation between them.
Ans: When an object moves in a circular path with constant speed then the motion
is said to be a uniform circular motion
Time period can be defined as the time taken by the object to complete one
revolution
Frequency can be given as the total number of revolutions in one second.
Angular velocity can be stated as the time rate of change of angular displacement.
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2
 2
T
1
   , which is the required relation.
T

15. A body of mass m is thrown with velocity v at angle of 30 to the
horizontal and another body B of the same mass is thrown with velocity at
an angle of 60 to the horizontal. Find the ratio of the horizontal range and
maximum height of A and B?
Ans: Case 1
When   30
u2
R A   sin 2  30 
g
u2
3
 RA  
g
2
When   60
u2
u2
3
R B   sin 2  60   R B  
 R A : R B  1:1
g
g
2
Case 2:
When   30
u2
u2  1 
2
H A   sin 30  H A   
g
g 4
When   60
u2
u2  3 
2
H B   sin 60  H B     H A : H B  1: 3 , which is the required ratio.
g
g 4
16. Read each statement below carefully and state with reasons, if it is true
or false:
(a) The magnitude of a vector is always a scalar.
Ans: True.
The magnitude of a vector is a number. Therefore, it is a scalar.
(b) each component of a vector is always a scalar.
Ans: False.
Each component of a vector is also a vector.
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(c) the total path length is always equal to the magnitude of the
displacement vector of a particle.
Ans: False.
Total path length is a scalar quantity, whereas displacement is a vector
quantity. Therefore, the total path length is always greater than the magnitude
of displacement. It becomes equal to the magnitude of displacement only when
a particle is moving in a straight line.
(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 average velocity of the particle over the same interval of
time.
Ans: True.
It is because of the fact that the total path length is always greater than or equal
to the magnitude of displacement of a particle.
(e) Three vectors not lying in a plane can never add up to give a null
vector.
Ans: True.
Three vectors, which do not lie in a plane, cannot be represented by the sides
of a triangle taken in the same order.
17. State with reasons, whether the following algebraic operations with
scalar and vector physical quantities are meaningful:
(a) adding any two scalars
Ans: Meaningful.
The addition of two scalar quantities is meaningful only if they both represent
the same physical quantity.
(b) adding a scalar to a vector of the same dimensions.
Ans: Not meaningful.
The addition of a vector quantity with a scalar quantity is not meaningful.
(c) multiplying any vector by any scalar.
Ans: Meaningful.
A scalar can be multiplied with a vector. For example, force is multiplied with
time to give impulse.
(d) multiplying any two scalars.
Ans: Meaningful.
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A scalar, irrespective of the physical quantity it represents, can be multiplied
with another scalar having the same or different dimensions.
(e) adding any two vectors.
Ans: Meaningful.
The addition of two vector quantities is meaningful only if they both represent
the same physical quantity.
(f) adding a component of a vector to the same vector.
Ans: Meaningful.
A component of a vector can be added to the same vector as they both have the
same dimensions.
18. Three girls skating on a circular ice ground of radius 200m 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 Fig. 4.20. What is the magnitude
of the displacement vector for each? For which girl is this equal to the actual
length of the path skated?
Ans: Displacement can be given by the minimum distance between the initial and
final positions of a particle. In the given case, all the girls start from point P and
reach point Q. The magnitudes of their displacements will be equal to the
diameter of the ground.
Radius of the ground can be given as  200m
Diameter of the ground  2  200  400m
Therefore, the magnitude of the displacement for each girl is 400m . This is equal
to the actual length of the path skated by girl B.
19. A man can swim with a speed of 4.0km / h in still water. How long does
he take to cross a river 1km wide if the river flows steadily at 3.0km / 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?
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Ans: Speed of the man is given as vm  4.0km / h
Width of the river is given as 1km
Width of the river
Time taken tocross the river 
Speed of the river
1
1
 Time taken tocross the river  h   60  15min
4
4
Speed of the river, v r  3km / h
Distance covered with flow of the river  v r  t
1 3
3
 3     1000  750m , which is the required distance.
4 4
4
20. A stone tied to the end of a string 80cm long is whirled in a horizontal
circle with a constant speed. If the stone makes 14 revolutions in 25s , what
is the magnitude and direction of acceleration of the stone?
Ans: Length of the string is given as, l  80cm  0.8m
Number of revolutions  14
Time taken  25s
No.of revolutions 14
Frequency,  
 Hz
Time taken
25
Angular frequency can be given as,   2
22 14 88
   2    rad / s
7 25 25
Centripetal acceleration can be given as,
2
 88 
2
a c   r  a c     0.8  a c  9.91m / s 2
 25 
The direction of centripetal acceleration is directed along the string, toward the
centre, at all points.
21. An aircraft executes a horizontal loop of radius 1.0km with a steady
speed of 900km / h . Compare its centripetal acceleration with the
acceleration due to gravity.
Ans: Given that,
Radius of the loop is given as, r  1.0km  1000m
Speed of the aircraft is given as, v  900km / h
Centripetal acceleration,
2
250 
v2

ac   ac 
 62.5m / s 2
r
1000
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Acceleration due to gravity, g  9.8m / s 2
a
62.5
 c
 6.38  a c  6.38g , which is the required relation.
g
9.8
22. Read each 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
Ans: False.
The net acceleration of a particle in circular motion is not always directed along
the radius of the circle toward the centre. It happens only in the case of uniform
circular motion.
(b) The velocity vector of a particle at a point is always along the tangent
to the path of the particle at that point
Ans: True.
At a point on a circular path, a particle appears to move tangentially to the
circular path on which it is moving.
Therefore, the velocity vector of the particle is always along the tangent at a
point.
(c) The acceleration vector of a particle in uniform circular motion
averaged over one cycle is a null vector
Ans: True.
In uniform circular motion (UCM), the direction of the acceleration vector
points toward the centre of the circle. But it constantly changes with time. The
average of these vectors over one cycle is a null vector.
23. A cricketer can throw a ball to a maximum horizontal distance of 100m
. How much high above the ground can the cricketer throw the same ball?
Ans: Maximum horizontal distance can be given as, R  100m
It will be possible for the cricketer to throw the ball to the maximum horizontal
distance when the angle of projection is 45 , i.e.,   45 .
The horizontal range for a projection velocity v , can be expressed by the relation:
u 2 sin 90
u2
u 2 sin 2
 100 

 100
R
g
g
g
The ball will achieve the maximum height when it is thrown vertically upward.
For such motion, the final velocity v is zero at the maximum height H.
Acceleration is given as, a  g
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Using the third equation of motion:
v2  u 2  2gH
1 u2 1
 H     100  50m , which is the required height.
2 g 2


24. The position of a particle is given by r  3tiˆ  2t 2ˆj  4kˆ 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.
Ans: The position of the particle is given by: r  3tiˆ  2t 2ˆj  4kˆ
Velocity v , of the particle is given as:
dr d ˆ
v

3ti  2t 2ˆj  4kˆ  v  3iˆ  4tjˆ
dt dt
Acceleration a , of the particle is given as:
dv d ˆ
a

3i  4tjˆ  a  4ˆj
dt dt




(b) What is the magnitude and direction of velocity of the particle at
t  2.0s ?
Ans: We have velocity vector as, v  3iˆ  4tjˆ
At t  2.0s ,
v  3iˆ  8jˆ
The magnitude of velocity is given by:
v  32   8   73  8.54m / s
2
Direction can be given as,
v 
 8 
  tan 1  y     tan 1    tan 1  2.667   69.45
 3 
 vx 
The negative sign indicates that the direction of velocity is below the x-axis.
25. For any arbitrary motion in space, which of the following relations are
true:
(The ‘average’ stands for average of the quantity over the time interval t 1 to
t2 )
1
(a) vaverage     v  t1   v  t 2  
 2
Ans: False.
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It is given that the motion of the particle is arbitrary. Therefore, the average
velocity of the particle cannot be given by this equation.
r  t 2   r  t1  
(b) vaverage  
 t 2  t1 
Ans: True.
The arbitrary motion of the particle can be represented by this equation.
(c) v  t   v  0   at
Ans: False.
The motion of the particle is arbitrary. The acceleration of the particle may also
be non-uniform. Therefore, this equation cannot represent the motion of the
particle in space.
1
(d) r  t   r  0   v  0  t  at 2
2
Ans: False.
The motion of the particle is arbitrary; acceleration of the particle may also be
non-uniform. Therefore, this equation cannot represent the motion of particle
in space.
 v  t 2   v  t1  
(e) aaverage  
 t 2  t1 
Ans: True.
The arbitrary motion of the particle can be represented by this equation.
26. 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
Ans: False.
Nevertheless, being a scalar quantity, energy is not conserved in inelastic
collisions.
(b) can never take negative values
Ans: False.
Despite being a scalar quantity, temperature can take negative values.
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(c) must be dimensionless
Ans: False.
Total path length is a scalar quantity. Yet it has the dimension of length.
(d) does not vary from one point to another in space
Ans: False.
A scalar quantity such as gravitational potential can vary from one point to
another in space.
(e) has the same value for observers with different orientations of axes
Ans: True.
The value of a scalar does not vary for observers with different orientations of
axes.
27.
(a) A vector has magnitude and direction. Does it have a location in space?
Ans: No.
Generally, a vector has no definite locations in space. This is due to a vector
remains invariant when displaced in such a way that its magnitude and
direction remain the same. However, a position vector has a definite location
in space.
(b) Can it vary with time?
Ans: Yes.
A vector can vary with time. Example can be given as, the displacement vector
of a particle moving with a certain velocity varies with time.
(c) Will two equal vectors a and b at different locations in space
necessarily have identical physical effects? Give examples in support of
your answer.
Ans: No.
Two equal vectors located at different locations in space need not produce the
same physical effect. For example, two equal forces that are acting on an object
at different points can cause the body to rotate, but their combination cannot
produce an equal turning effect.
28.
(a) A vector has both magnitude and direction. Does it mean that anything
that has magnitude and direction is necessarily a vector?
Ans: No.
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A physical quantity having both magnitude and direction need not be always
considered a vector.
For example, despite having magnitude and direction, current is a scalar
quantity. The essential requirement for a physical quantity to be considered a
vector is that it should follow the law of vector addition.
(b) 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?
Ans: No.
Generally, the rotation of a body about an axis is not a vector quantity as it
does not follow the law of vector addition. But a rotation by a certain small
angle follows the law of vector addition and is therefore considered a vector.
29. Can you associate vectors with
(a) the length of a wire bent into a loop?
Ans: No.
One cannot relate a vector with the length of a wire bent into a loop.
(b) a plane area?
Ans: Yes.
One can relate an area vector with a plane area. The direction of this vector is
normal, inward, or outward to the plane area.
(c) a sphere?
Ans: No.
One cannot relate a vector with the volume of a sphere. However, an area
vector can be related with the area of a sphere.
Long Answer Questions
3 Marks
1. Derive expressions for velocity and acceleration for uniform circular
motion.
Ans: We can write,
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If PQ  l ,
l
V
t
And angular velocity can be given as  

l
l
l
Using      …… (1)
r
r
l  Vt and   t
Substituting in (1)
t
t  V
r
 V  r , which is the required expression for velocity.
Also,
dV
d
dr
V
V2
a
r
   V   V 
dt
dt
dt
r
r
2
V
a
, which is the required expression for acceleration.
r
2.Derive an equation for the path of a projectile fired parallel to horizontal.
Ans: Let a projectile having initial uniform horizontal velocity u be under the
influence of gravity, then at any instant t at position P the horizontal and vertical.
1
s  ut  at 2
2
s  x, u  u , t  t , a  0
x  ut
x
 t  …… (1)
u
For vertical motion,
1
s  ut  at 2
2
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s   y , u  0 , t  t , a  g
1
 y   gt 2
2
1
 y  gt 2 …… (2)
2
Using equation (1) and (2)
2
1  x  1 x2
 y  g    g 2 , which is the required equation of the path.
2 u 2 u
3.
(a) Define time of flight and horizontal range.
Ans: The time taken by the projectile to complete its trajectory is defined as
time of flight.
The maximum horizontal distance covered by the projectile from the foot of
the tower to the point where projectile hits the ground is defined as horizontal
range.
(b) From a certain height above the ground a stone A is dropped gently.
Simultaneously another stone B is fired horizontally. Which of the two
stones will arrive on the ground earlier?
Ans: Both the stones will reach the ground simultaneously since the initial
vertical velocity in both cases is zero and both are falling with same
acceleration equal to acceleration due to gravity.
4. At what point of projectile motion
(a) Potential energy is the maximum?
Ans: P.E. will be maximum at the highest point(H).
 P.E.Highest Point  mgH
  P.E.H
 u 2 sin 2  
1
2
2
 mg 
   P.E.H  mu sin 
2
 2g 
(b) Kinetic energy is the minimum?
Ans: K.E will be minimum at the highest point. i.e.,
1
2
 K.E.H  m  vH 
2
Here, vertical component of velocity is zero.
1
  K.E.H  mv2 cos2 
2
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(c) Total mechanical energy is the maximum?
Ans: Total energy at any point of projectile motion will be the same.
Mathematically, at the highest point;
1
1
 K.E.H   P.E.H  mv2 cos2   mu 2 sin 2 
2
2
1
  K.E.H   P.E.H  mv2  cos2   sin 2  
2
1
  K.E.H   P.E.H  mv2
2
5. A cyclist starts from the centre O of a circular park of radius 1km ,
reaches the edge P of the park, then cycles along the circumference, and
returns to the centre along QO as shown in Fig. 4.21. If the round trip takes
10 min, what is the
(a) net displacement?
Ans: Displacement is given by the shortest distance between the initial and
final positions of a body which is travelling. In the provided case, the cyclist
comes to the starting point after cycling for 10 minutes. Therefore, his net
displacement is zero.
(b) average velocity?
Ans: Average velocity is given by the relation:
Net displacement
Average velocity 
Total time
Because the net displacement of the cyclist is zero, his average velocity will
also be zero.
(c) average speed of the cyclist?
Ans: Average speed of the cyclist is given by the formula:
Totalpath length
Averagespeed 
Total time
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Total path length is given as  OP  PQ  QO
1
 1   2 1  1
4
1
 2    3.570km
2
10 1
Time taken can be given as 10min 
 h
60 6
3.570
Average speed 
 21.42km / h
1
6
6. A passenger arriving in a new town wishes to go from the station to a hotel
located 10 km away on a straight road from the station. A dishonest cabman
takes him along a circuitous path 23km long and reaches the hotel in 28min
. What is
(a) the average speed of the taxi?
Ans: Given that,
Total distance travelled can be given as  23km
28
Total time taken can be given as  28min  h
60
Totaldistance travelled
Averagespeed of the taxi 
Total time taken
23
 Averagespeed of the taxi 
 49.29km / h
 28 
 
 60 
(b) the magnitude of average velocity? Are they two equal?
Ans: Given that,
Distance between the hotel and the station  10km  Displacement of the car.
10
Average velocity 
 21.43km / h
 28 
 
 60 
Therefore, the two physical quantities (average speed and average velocity) are
unequal.
7. Rain is falling vertically with a speed of 30m / s . A woman rides a bicycle
with a speed of 10m / s in the north to south direction. What is the direction
in which she should hold her umbrella?
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Ans: The described situation is depicted in the given figure.
Here,
v c  Velocity of the cyclist
v r  Velocity of falling rain
To protect herself from the rain, the woman should hold her umbrella in the
direction of the relative velocity (v) of the rain with respect to the woman.
v  v t   vc   v  30   10   20m / s
Now,
v 10
tan   c 
v r 30
   tan 1  0.333  18
Therefore, the woman must hold the umbrella toward the south, at an angle of
nearly 18 with the vertical.
8. The ceiling of a long hall is 25m high. What is the maximum horizontal
distance that a ball thrown with a speed of 40m / s can go without hitting the
ceiling of the hall?
Ans: Given that,
Speed of the ball is given as, u  40m / s
Maximum height is given as, h = 25 m
In projectile motion, the maximum height attained by a body projected at an angle
θ, is given by the formula:
u 2 sin 2 
h
2g
402
 25 
2  9.8
2
 sin   0.30625
 sin   0.5534
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   sin 1  0.5534   33.60
u 2 sin 2
Horizontal Range, R 
g
402  sin 2  33.60 
R 
9.8
1600  sin 67.2
R
9.8
1600  0.922
R
 150.53m
9.8
9. An aircraft is flying at a height of 3400m above the ground. If the angle
subtended at a ground observation point by the aircraft positions 10s apart
is 30 , what is the speed of the aircraft?
Ans: The positions of the observer and the aircraft are depicted in the following
figure.
Height of the aircraft from ground is given as, OR  3400m
Angle subtended between the positions is given as, POQ  30
Time  10s
In PRO we can write:
PR
tan15 
OR
 PR  OR tan15
 PR  3400  tan15
PRO is similar to RQO .
PR  RQ
 PQ  PR  RQ
 PQ  2PR  2  3400tan15
 PQ  6800  0.268  1822.4m
1822.4
Therefore, speed of the aircraft 
 182.24m / s .
10
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10. A bullet fired at an angle of 30 with the horizontal hits the ground 3km
away. By adjusting its angle of projection, can one hope to hit a target 5km
away? Assume the muzzle speed to the fixed, and neglect air resistance.
Ans: Range is given as, R  3km
Angle of projection is given  30
Acceleration due to gravity, g  9.8m / s 2
Horizontal range for the projection velocity u 0 , is given by the formula:
u 2 sin 2
R
g
u 02
3
sin 60
g
u 02
3
……  3  3  3 
 3 3 

g
2
u 02
 2 3 …… (1)
g
The maximum range, R max , is achieved by the bullet when it is fired at an angle
of 45 with the horizontal, that is,
u 02
R max 
…… (2)
g
On comparing equations (1) and (2), we get:
 R max  2 3  2  1.732  3.46km
Therefore, the bullet will not hit a target 5km away.
Long Answer Questions
4 Marks
1. Given a  b  c  d  0 , which of the following statements are correct:
(a) a , b , c , and d must each be a null vector.
Ans: Incorrect.
To make a  b  c  d  0 , it is not important to have all the four given vectors
to be null vectors. There are many other combinations which can give the
addition of vectors as zero.
(b) The magnitude of a  c equals the magnitude of  b  d  .
Ans: Correct.
abcd0
a  c  b  d
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Taking modulus on both the sides, we obtain:
a  c  b  d  b  d
Therefore, the magnitude of a  c is the same as the magnitude of  b  d  .
(c) The magnitude of a can never be greater than the sum of the
magnitudes of b , c , and d .
Ans: Correct.
abcd0
a bcd
Taking modulus both sides, we obtain:
a  bcd
a  b  c  d ...... (1)
Equation (1) shows that the magnitude of a is equal to or less than the sum of
the magnitude of b , c , and d .
Therefore, the magnitude of vector a can never be greater than the sum of the
magnitudes of b , c , and d .
(d) b  c must lie in the plane of a and d if a and d are not collinear, and
in the line.
Ans: Correct.
For a  b  c  d  0
a   b  c  d  0
The resultant sum of the three vectors a ,  b  c  , and d is equal to zero only if
 b  c  lie in a plane containing a and d , assuming that these three vectors are
represented by the three sides of a triangle.
If a and d are collinear, then it implies that the vector  b  c  is in the line of
a and d. This statement holds only when the vector sum of all the vectors will
be zero.
2. In a harbour, wind is blowing at the speed of 72km / h and the flag on the
mast of a boat anchored in the harbour flutters along the N-E direction. If
the boat starts moving at a speed of 51km / h to the north, what is the
direction of the flag on the mast of the boat?
Ans: Velocity of the boat is given as v b  51km / h
Velocity of the wind is given as v w  72km / h
The flag is fluttering in the north-east direction. It means that the wind is blowing
toward the north-east direction. When the ship starts sailing toward the north, the
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flag will move along the direction of the relative velocity ( v wb ) of the wind with
respect to the boat.
The angle between v w and   v b   90  45 .
tan  
51sin  90  45
72  51cos  90  45
51
1
2
51sin  45 

72  51  cos 45  72  51 1
2
51
51
51
 tan  


72 2  51 72  1.414  51 50.800
   tan 1 1.0038   45.11
Angle with respect to the east direction  45.11  45  0.11 .
Therefore, the flag will flutter almost due east.
 tan  
3. A fighter plane flying horizontally at an altitude of 1.5km with speed
720km / h passes directly overhead an anti-aircraft gun. At what angle from
the vertical should the gun be fired for the shell with muzzle speed to hit the
plane? At what minimum altitude should the pilot fly the plane to avoid
being hit? (Take g  10m / s2 ).
Ans: Height of the fighter plane is given as 1.5km  1500m .
Speed of the fighter plane is given as, v = 720 km/h = 200 m/s
Let  be the angle with the vertical so that the shell hits the plane. The situation
is depicted in the given figure.
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Muzzle velocity of the gun, u  600m / s
Time taken by the shell to hit the plane  t
Horizontal distance travelled by the shell  u x t
Distance travelled by the plane  vt
The shell hits the plane. Therefore, these two distances should be equal.
u x t  vt
usin   v
v
sin  
u
200 1
 sin  
  0.33
600 3
  sin 1  0.33
   19.5
In order to avoid being hit by the shell, the pilot must fly the plane at an altitude
(H) higher than the maximum height achieved by the shell.
u 2 sin 2  90   
H
2g
 600 
H
cos 2 
2g
360000  cos2 19.5
H
2  10
2
 H  18000   0.943
 H  16006.42m
 H  16km , which is the required height.
2
Long Answer Questions
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5 Marks
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1.
(a) What is the angle between A and B . If A and B denote the adjacent
sides of a parallelogram drawn from a point and the area of the
1
parallelogram is AB ?
2
Ans: Area of a parallelogram can be given as  A  B
Area of parallelogram can be given as  ABsin  (Applying cross product)
Given data is,
1
Area of parallelogram  AB
2
1
So,  AB  ABsin 
2
1
  sin 
2
1
   sin 1  
2
   30
Clearly, area of a parallelogram can be given as  A  B
(b) State and prove triangular law of vector addition.
Ans: Triangular law of vector addition states that if two vectors can be
expressed both in magnitude and direction by the sides of a triangle taken in
order then their resultant is given by the third side of the triangle taken in
opposite order.
In ADC ,
2
2
2
 AC   AD    DC
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  AC    AB  BD    DC 
2
2
2
  AC   AB2  BD2  2  AB BD    DC 
2
2
  AC   P2   Qcos    2P  Qcos     Qsin  
2
2
2
  AC  P2  Q2  sin 2   cos2   2PQcos 
2
 BD
 CD


 cos   and 
 sin  

 BC
 BC


 R 2  P2  Q2  2PQcos 
 sin 2   cos2   1
 R  P2  Q2  2PQcos 
Hence proved.
2. Establish the following vector inequalities geometrically or otherwise:
When does the equality sign above apply?
(a) a  b  a  b
Ans: Let two vectors a and b be represented by the adjacent sides of a
parallelogram
OMNP, as shown in the given figure.
Here, we can express:
OM  a …… (1)
MN  OP  b …… (2)
ON     …… (3)
We know that, in a triangle, each side is smaller than the sum of the other two
sides.
Therefore, in OMN , we have:
ON  OM  MN
a  b  a  b …… (4)
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If the two vectors and act along a straight line in the same direction, then it can
be written as:
a  b  a  b …… (5)
Combining equations (4) and (5), we obtain:
ab  a  b
(b) a  b  a  b
Ans: Let two vectors a and b be represented by the adjacent sides of a
parallelogram OMNP, as depicted in the given figure.
Here, we can express:
OM  a
MN  OP  b
ON  a  b
We know that, in a triangle, each side is smaller than the sum of the other two
sides.
Therefore, in OMN , we can write:
ON  MN  OM
ON  OM  MN
ON  OM  OP

OP  MN 
a  b  a  b …... (4)
If the two vectors a and b act along a straight line in the same direction, then
we can express:
a  b  a  b …... (5)
Combining equations (4) and (5), we get:
ab  a  b
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(c) a  b  a  b
Ans: Let two vectors a and b be represented by the adjacent sides of a
parallelogram PORS, as depicted in the given figure.
Here we can write:
OR  PS  b …… (1)
OP  a …… (2)
We know that, in a triangle, each side is smaller than the sum of the other two
sides. Therefore, in OSP we have:
OS  OP  PS
a  b  a  b
a  b  a  b …… (3)
If the two vectors act in a straight line but in opposite directions, then we can
express:
a  b  a  b …… (4)
Combining equations (3) and (4) we get:
ab  a  b
(d) a  b  a  b
Ans: Let two vectors and be represented by the adjacent sides of a
parallelogram PORS, as shown in the given figure.
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The following relations can be written for the given parallelogram.
OS  PS  OP …… (1)
OS  OP  PS …… (2)
a  b  a  b …… (3)
The quantity on the LHS is always positive and that on the RHS can be positive
or negative.
To make both quantities positive, we take modulus on both sides as:
a  b  a  b
a  b  a  b …… (4)
If the two vectors act in a straight line but in the opposite directions, then we
can write:
a  b  a  b …… (5)
Combining equations (4) and (5), we get:
ab  a  b
3. On an open ground, a motorist follows a track that turns to his left by an
angle of 60 after every 500m . Starting from a given turn, specify the
displacement of the motorist at the third, sixth and eighth turn. Compare the
magnitude of the displacement with the total path length covered by the
motorist in each case.
Ans: The path persuaded by the motorist is a regular hexagon with side 500m ,
as shown in the following figure.
Let the motorist start from the point P .
The motorist takes the third turn at S .
∴Magnitude of displacement  PS  PV  VS  500  500  1000m
Total path length  PQ  QR  RS  500  500  500  1500m
The motorist takes the sixth turn at point P , which is the starting point.
∴Magnitude of displacement can be given  0
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Total path length  PQ  QR  RS  ST  TU  UP
 500  500  500  500  500  500  3000m
The motorist takes the eight turns at point R
∴Magnitude of displacement can be given  PR
 PR  PQ2  QR 2  2  PQ   QR  cos60
 PR  5002  5002   2  500  500  cos60 
 PR  250000  250000  500000 
 PR  866.03m
 500sin 60 
  tan 1 

 500  500cos60 
1
2
Therefore, the magnitude of displacement is 866.03m at an angle of 30 with
PR.
Total path length  Circumferenceof thehexagon  PQ  QR
 6  500  500  500  4000m
The magnitude of displacement and the total path length corresponding to the
required turns is depicted in the following table
Turn
Third
Sixth
Eighth
Magnitude
displacement(m)
1000
0
866.03, 30
of Total path length(m)
1500
3000
4000
ˆ / s and
4. A particle starts from the origin at t  0s with a velocity of 10jm
moves in the x-y plane with a constant acceleration of 8iˆ  2jˆ m / s 2 .


(a) At what time is the x-coordinate of the particle 16m ? What is the ycoordinate of the particle at that time?
Ans: Velocity of the particle is given as v  10.0ˆjm / s
dv ˆ ˆ
Acceleration of the particle is given as a 
 8i  2 j
dt
Also,
a  8iˆ  2ˆj
But,
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dv ˆ ˆ
 8i  2 j
dt
dv  8iˆ  2jˆ dt
a


Integrating both sides:
v  t   8tiˆ  2tjˆ  u
Where,
u  Velocity vector of the particle at t = 0
v  Velocity vector of the particle at time t
However,
dr
v
dt
dr  vdt  8tiˆ  2tjˆ  u dt


Integrating the equations with the conditions: at t  0 ; r  0 and at t  t ; r  r
1
1
r  ut   8t 2ˆi   2t 2ˆj
2
2
2ˆ
 r  ut  4t i  t 2ˆj
 r  10jˆ t  4t 2ˆi  t 2ˆj
 
 xiˆ  yjˆ  4t 2ˆi  10t  t 2  ˆj
Since the motion of the particle is confined to the x-y plane, on equating the
coefficients of î and ˆj , we get:
x  4t 2
1
2
x
t  
4
And y  10t  t 2
When x  16m :
1
2
 16 
t     2s , which is the required time.
 4
y  10  2  22  24m , which is the required y coordinate.
(b) What is the speed of the particle at the time?
Ans: Velocity of the particle can be given by:
v  t   8tiˆ  2tjˆ  u
At t  2s
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v  t   8  2iˆ  2  2jˆ  10jˆ
v  t   16iˆ  14jˆ
v  162  142
 v  256  196  452
 v  21.26m / s , which is the required speed.
5. î and ĵ are unit vectors along x- and y-axis respectively. What is the
magnitude and direction of the vectors ˆi  ˆj , and ˆi  ˆj ? What are the
components of a vector along the directions of ˆi  ˆj and ˆi  ˆj ? [You may use
graphical method]
Ans: Let us consider a vector P , given as:
P  ˆi  ˆj
P ˆi  P ˆj  ˆi  ˆj
x
y
On comparing the components on both sides, we obtain:
Px  Py  1
P  Px 2  Py 2  12  12  2
Therefore, the magnitude of the vector is 2 .
Let  be the angle made by the vector P , with the x-axis, as shown in the given
figure.
P 
tan    y 
 Px 
1
   tan 1    45
1
Therefore, the vector ˆi  ˆj makes an angle of 45 with the x-axis.
Let Q  ˆi  ˆj
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Qx ˆi  Q yˆj  ˆi  ˆj
Qx  Q y  1
Q  Q x 2  Q y 2  12  12  2
Therefore, the magnitude of the vector ˆi  ˆj is 2 .
Let  be the angle made by the vector ˆi  ˆj , with the x- axis, as shown in the
following figure.
Q 
tan    y 
 Qx 
1
   tan 1    45
1
Therefore, the vector makes an angle of 45 .
It is given that:
A  2iˆ  3jˆ
A ˆi  A ˆj  2iˆ  3jˆ
x
y
On comparing the coefficients of î and ˆj , we have:
Ax  2
Ay  3
A  Q x 2  Q y 2  22  32  13
Let A x make an angle with the x-axis as depicted in the given figure.
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A 
tan    y 
 Ax 
3
   tan 1    56.31
2
Angle between the vectors 2iˆ  3jˆ and ˆi  ˆj
  56.31  45  11.31
Component of vector A , along the direction of P , making an angle  :
 i  j
  Acos   P   Acos11.31 
2
0.9806 ˆ ˆ
25
5
 13 
i  j  2.5 ˆi  ˆj   2 
10
2
2
Now, let  be the angle between the vectors 2iˆ  3jˆ and ˆi  ˆj .
  45  56.31  101.31
Component of vector A , along the direction of Q , making an angle 
ˆi  ˆj
  A cos   Q   A cos  
2
ˆi  ˆj
 13  cos  901.31 
2
13

sin11.30 ˆi  ˆj
2
 2.550  0.1961 ˆi  ˆj
 
 
 
 
 
 
 
1
5
 0.5 ˆi  ˆj    2  
10
2
6. A cyclist is riding with a speed of 27km / h . As he approaches a circular
turn on the road of radius 80m , he applies brakes and reduces his speed at
the constant rate of 0.5m / s every second. What is the magnitude and
direction of the net acceleration of the cyclist on the circular turn?
Ans: Speed of the cyclist is given as, v  27km / h  7.5m / s
Radius of the circular turn, r  80m
Centripetal acceleration can be given as:
v2
ac 
r
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 ac
 7.5

2
 0.7m / s
80
The situation is depicted in the provided figure:
Let us suppose the cyclist begins cycling from point P and moves toward point
Q. At point Q, he applies the brakes and decelerates the speed of the bicycle by
0.5m / s2 .
This acceleration is along the tangent at Q and in opposite direction to the
direction of motion of the cyclist.
Because the angle between a c and a t is 90 , the resultant acceleration a is given
by:
a  a c2  a t 2
 a  0.7 2  0.52
 a  0.74  0.86m / s2
a
tan   c
at
Where  is the angle of the resultant with the direction of velocity.
0.7
tan  
 1.4
0.5
   tan 1 1.4 
   54.46 , which is the required angle of direction here.
7.
(a) Show that for a projectile the angle between the velocity and the x-axis
 v  gt 
as a function of time is given by   t   tan1  0y

 v0x 
Ans: Let v 0x and v0 y respectively be the initial components of the velocity of
the projectile along horizontal (x) and vertical (y) directions.
Let and respectively be the horizontal and vertical components of velocity at a
point P.
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Time taken by the projectile to reach point P  t
Applying the first equation of motion along the vertical
v y  v0y  gt
And, v x  v0x
v
v  gt
tan   y  0y
vx
v0x
 v  gt 
   tan 1  0y

 v0x 
Hence proved.
(b) Show that the projection angle for a projectile launched from the
 4h 
origin is given by   t   tan 1  m  , where the symbols have their
 R 
usual meaning.
u 0 2 sin 2 
Ans: Maximum vertical height can be given as, h m 
…… (1)
2g
u 0 2 sin 2
R

Horizontal range can be given as,
…… (2)
g
Solving equations (1) and (2), we obtain:
h m sin 2 

R sin 2
h
sin  sin 
 m
R 2  2sin  cos 
h
1 sin  1
 m 
 tan 
R 4 cos  4
Now,
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tan  
4h m
R
 4h 
   tan 1  m 
 R 
Hence proved.
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