NCERT Solution Class 11 Physics Chapter 3 Motion in a plane - Param Himalaya
3.1 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.
Solution :
Scalars : Volume, mass , speed , density, number of moles , angular frequency.
Vectors : Acceleration, velocity, displacement, angular velocity.
3.2 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.
Solution :
The dot product of force and displacement is the work done. Work is scalar quantity since the dot product of two quantities is always scalar.
Current is a scalar quantity as it is described only by its magnitude and it is independent of direction.
3.3 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.
Solution :
Impulse is the product of force and time. Since force is vector quantity , its product with time which is a scalar quantity, gives a vector quantity.
3.4 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 dimensions ,
(c) multiplying any vector by any scalar,
(d) multiplying any two scalars,
(e) adding any two vectors,
(f) adding a component of a vector to the same vector.
Solution :
(a) Meaningful
The addition of two scalar quantities is meaningful only both represent the same physical quantity.
we added does temperature (scalar) cannot be added to a length (scalar).
(b) Not meaningful
The addition of a vector quantity with a scalar quantity is not meaningful.
(c) Meaningful
A scalar can be multiplied with a vector.
For example F=ma [i.e. mass (scalar) multiplied by acceleration (vector)] gives an force.
(d) Meaningful
A scalar , irrespective of the physical quantity it represents can be multiplied by another scalar having the same or different dimensions.
For example W= Pt [ power (scalar) multiplied by time (scalar) give rise to work done (W).
(e) Meaningful
The addition of a vector can be added to the same vector as they both have the same dimensions.
(f) Meaningful
A component of a vector can be added to the same vector as they both have the same dimensions
3.5 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,
(b) each component of a vector is always a scalar,
(c) the total path length is always equal to the magnitude of the 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 average velocity of the particle over the same interval of time,
(e)Three vectors not lying in a plane can never add up to give a null vector.
Solution:
(a) True. The magnitude of a vector gives its length, which is always a scalar.
(b) False. Each component of a vector is a vector by itself.
(c) False. The total path length is always greater than or equal to the magnitude of the displacement vector of the particle.
(d) True. Since the path length is always greater than or equal to the magnitude of the displacement vector of the particle, the average speed is always greater than or equal to the magnitude of the average velocity.
(e) True. If the vectors are not coplanar, their addition will always yield some nonzero component in at least one direction.
3.6 Establish the following vector inequalities geometrically or otherwise :
(a) |a+b| < |a| + |b|
(b) |a+b| > ||a| −|b||
(c) |a−b| < |a| + |b|
(d) |a−b| > ||a| − |b||
When does the equality sign above apply?
3.7 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,
(b) The magnitude of (a + c) equals 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,
(d) b + c must lie in the plane of a and d if a and d are
not collinear, and in the line of a and d, if they are
collinear ?
3.8 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 Fig. 3.19. What is the magnitude of the displacement vector for each ? For which girl is this equal to the actual length of path skate ?
Solution:
Given: The radius of the circular ice ground is 200 m. The girls skating on a circular ice ground along the three different paths as shown in the figure below:
Displacement is the shortest distance between the starting and end point. The displacement for each girl is given as,
$S= \overrightarrow{PQ}$
The magnitude of displacement is given as,
PQ=Diameter of the circular ice ground PQ=R×2 =200×2 =400 m
Thus, the displacement for each girl is 400 m.
The actual path length of girl B is same as the shortest distance between the starting and end point. Therefore for girl B, the magnitude of displacement is equal to the actual length of the path stake.
3.9 A cyclist starts 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 Fig. 3.20. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist ?
Solution:
3.10 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 turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
Solution:
(a) The path followed by the motorist will be a closed hexagonal path.
Suppose the motorist starts his journey from the point O . He takes the turn at the point C.
Displacement= $\overrightarrow{OC}$
Here
$$OC = \sqrt{OB^{2}+BC^{2}}$$
$$OC=\sqrt{(OF+FB)^{2}+BC^{2}}$$
$$OC=\sqrt{(500cos30^{\circ}+500cos30^{\circ})^{2}+500^{2}}$$
$$OC=\sqrt{(2 \times 500cos30^{\circ})^{2}+500^{2}}$$
$$OC= \sqrt{\left(2 \times 500 \times \frac{\sqrt{3}}{2}\right)^{2} + (500)^{2}}$$
$$OC= \sqrt{\left( 500 \times \sqrt{3} \right)^{2}+\left( 500 \right)^{2}}$$
$$OC=\sqrt{3 \times \left(500 \right)^{2} + \left(500 \right)^{2}}$$
$$OC= 500 \sqrt{3+1}$$
$$OC= 500 \sqrt{4}$$
$$OC=500 \times 2$$
$$OC= 1000m$$
$$OC =1Km$$
Total path length = 500m + 500 m = 1500m = 1.5 Km
$$\frac{magnitude \ of \ displacement }{Total \ path \ length} = \frac{1}{1.5}$$
$$\frac{magnitude \ of \ displacement }{Total \ $path \ length} = \frac{2}{3}$$
$$\frac{magnitude \ of \ displacement }{Total \ path \ length} = 0.67$$
(ii) After completing six turns , the motorist returns to the starting point, creating a closed hexagonal path.
The displacement is zero
The total path length is 6 × 500 = 3000 m
Ratio of magnitude of displacement and path length is zero
(iii) At the eighth turn , the motorist reaches point B. The displacement is the straight line distance from the starting line distance from the starting point O to point B.
Since $\angle BOA = \angle OBA = 30^{\circ}$, By trigonometry, the horizontal and vertical components of the displacement are equal to OAcos30° and ABcos30° respectively.
The total displacement $$\overrightarrow{OB} =(500cos30^{\circ}+500cos30^{\circ})$$
$$\overrightarrow{OB} =(2 \times 500 \times \frac{ \sqrt{3}}{2}$$
$$\overrightarrow{OB} = 500 \sqrt{3}m$$
$$\overrightarrow{OB} = 0.866km$$
because OA = AB , each being 500 m.Thus
The total Path Length after eight turns is 8 ×500 = 4000m
$$\frac{magnitude \ of \ displacement }{Total \ path \ length} = \frac{500 \sqrt{3}}{4000}$$
$$\frac{magnitude \ of \ displacement }{Total \ path \ length} = \frac{\sqrt{3}}{8}$$
3.11 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 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity ? Are the two equal ?
Solution:
(a) Total distance traveled (S)= 23 km
Total time taken = 28 min
$$T= \frac{28}{60}h$$
So Average speed of the taxi
$$V_{AS}= \frac{S}{T}$$
$$V_{AS} = \frac{23}{\frac{28}{60}}$$
$$V_{AS} = \frac{23 \times 60}{28}$$
$$V_{AS} = \frac{1380}{28}$$
$$V_{AS} = 49.29 m/s$$
(b) Distance between the hotal and the station = Displacement of the car = 10 km
So, Average velocity of the taxi
$$V_{AV} = \frac{D}{T}$$
$$V_{AV}= \frac{10}{\frac{28}{60}}$$
$$V_{AV}=\frac{10 \times 60}{28}$$
$$V_{AV}= \frac{600}{28}$$
$$V_{AV}=21.43km/h$$
Therefore , the two physical quantities (average speed and average velocity) are not equal.
3.12 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 m/s can go without hitting the ceiling of the hall ?
Solution:
Step 1: - Calculate the value of θ (angle made with horizontal). As per question, a ball is thrown in a hall such that it should not hit the ceiling of the hall and given, the speed of the ball u = 40 m/s
As we know that, in projectile motion, the maximum height reached by a body projected at an angle θ, is given by,
$$H_{max} = \frac{u^{2}sin^{2} \theta}{2g}$$
Substituting the values , we get ,
$$25 = \frac{(40)^{2}sin^{2} \theta}{2 \times 9.8}$$
$$Sin^{2} \theta = \frac{25 \times 2 \times 9.8}{1600}$$
$$Sin^{2} \theta = 0.30625$$
$$sin \theta = \sqrt{0.30625}$$
$$sin \theta = 0.5534$$
$$\theta = sin^{-1}(0.5534) = 33.60^\circ$$
Step 2 - Find horizontal range in projectile motion. As the horizontal range is given by
$$R = \frac{u^{2}sin2 \theta}{g}$$
Substituting the value of u, g and θ , we get
$$R = \frac{40^{2} sin ( 2 × 33.60^\circ)}{9.8}$ $
$$R = \frac{40^{2} sin ( 67.20^\circ)}{9.8}$$
$$R = \frac{1600 \times 0.922}{9.8}$$
$$R= 150.53 m $$
Final Answer: The maximum horizontal distance that a ball thrown, can go without hitting the ceiling of the hall is 150.53 m.
3.13 A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball ?
Solution:
Step 1: Find velocity of projection using range of projectile motion.
Given, maximum horizontal distance (𝑅) = 100m
The ball will cover maximum horizontal distance, only when the cricketer throws the ball with angle of projection of 45°
i.e $$\theta = 45^\circ$$
Now, horizontal range is given by R
$$R = \frac{u^{2}sin 2 \theta}{g}$$
Where u = initial velocity
$$100 = \frac{u^{2}sin 90^ \circ}{g}$$
$$u^{2} = 100g$$
$$\frac{u^{2}}{g} = 100$$
Step 2. Find maximum height attained by the ball
The ball can attain maximum height when it is thrown vertically upwards
At maximum height (H) , final velocity of ball becomes zero , v = 0
Acceleration a = -g
using third equation of motion
$$v^{2} = u^{2} + 2as$$
$$0 = u^{2} - 2gH$$
$$u^{2} = 2gH$$
$$H = \frac{u^{2}}{2g}$$
$$H = \frac{1}{2} \times \frac{u^{2}}{g}$$
$$H = \frac{1}{2} \times 100$$
$$H = 50 m$$
The cricketer can throw the same ball 50m high above the ground.
3.14 A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ?
Solution:
Length of the string , l = 80 cm = 0.8 m
Number of revolutions , n = 14
Time taken , t = 25 s
$$Frequency = \frac{number \ of \ revolution}{Time \ taken }$$
$$f = \frac{n}{t}$$
$$f = \frac{14}{25} H_{z}$$
Angular frequency, $$\omega = 2\pi f$$
$$\omega = 2 \times \frac{22}{7} \times \frac{14}{25}$$
$$\omega = \frac{88}{25}rad/s$$
Centripetal acceration
$$a_{c} = \omega^{2}r$$
$$a_{c} = \left ( \frac{88}{25} \right )^{2} \times 0.8$$
$$a_{c} = 0.80 \times \frac{88}{25} \times \frac{88}{25}$$
$$a_{c}= 9.90 m/s^{2}$$
The direction of centripetal acceleration is always directed along the string towards the centre at all points.
3.15 An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Solution:
Radius of horizontal loop r= 1 km = 1000 m
$$v = 900 km/hr$$
$$v = \frac{900 \times 1000}{3600}$$
$$v = \frac{900 \times 5}{18}$$
$$v = 50 \times 5$$
$$v = 250 m/s$$
Therefore centripetal acceleration
$$a = \frac{v^{2}}{r}$$
$$a = \frac{250 \times 250}{1000}$$
$$a = \frac{625}{10}$$
$$a = 62.5 m/s^{2}$$
Ratio of centripetal acceleration with the acceleration due to gravity.
$$\frac{a}{g} = \frac{62.5}{10}$$
$$\frac{a}{g} = 6.25$$
3.16 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
(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
Solution:
(a) False, the net acceleration of a particle in circular motion is along the radius of the circle towards the centre only in uniform circular motion.
(b) True, because while leaving the circular path, the particle moves tangentially to the circular path.
(c) True, the direction of acceleration vector in a uniform circular motion is directed towards the centre of circular path. It is constantly changing with time. The resultant of all these vectors will be a zero vector.
3.17 The position of a particle is given by
ˆ ˆ ˆ 2 r i j k = − + 3.0 2.0 4.0 m t t
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 = 2.0 s ?
3.18 A particle starts from the origin at t = 0 s with a velocity of 10.0 jɵ m/s and moves in
the x-y plane with a constant acceleration of (8.0 2.0 )
ɵ ɵ
i j + m s-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 ?
3.19 ɵ
i
and ɵ
j are unit vectors along x- and y- axis respectively. What is the magnitude
and direction of the vectors ɵ ɵ
i j + , and ɵ ɵ
i j − ? What are the components of a vector
A= 2 ɵ ɵ
i j + 3 along the directions of ɵ ɵ
i j + and ɵ ɵ
i j − ? [You may use graphical method]
3.20 For any arbitrary motion in space, which of the following relations are true :
(a) vaverage
= (1/2) (v (t
1
) + v (t
2
))
(b) v
average
= [r(t
2
) - r(t
1
) ] /(t
2
– t
1
)
(c) v (t) = v (0) + a t
(d) r (t) = r (0) + v (0) t + (1/2) a t
2
(e) a average
=[ v (t
2
) - v (t
1
)] /( t
2
– t
1
)
(The ‘average’ stands for average of the quantity over the time interval t
1
to t
2
)
3.21 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 orientations of axes.
Solution :
Solution A) Step 1: Recall “conservation of energy” in different processes. The statement is false. Reason:- A scalar quantity need not be conserved in all processes. Example:- Kinetic energy is a scalar quantity, but it is not conserved in inelastic collisions.
B) Step 1: Recall “Scalar quantity” properties. The statement is false. Reason: - A scalar quantity need not be always positive. Example: - Temperature is a scalar quantity, but it can be negative.
C) Step 1: Recall some of the scalar quantities which have dimensions. The statement is false. Reason: - A scalar quantity need not be dimensionless. Example: - Density is a scalar quantity but it has dimension.
$$Density \ (\rho) = \frac{mass \ (m)}{volume \ (v)}$$
$$Density \ (\rho) = \frac{ [ M^{1} ] }{[ L^{3}]}$$
$$Density = [ M^{1} L ^{− 3} T^{0}]$$
D) Step 1: Recall path length/distance covered by particle. The Statement is false Reason: - A scalar quantity can also vary in space. Example: - Path-length is scalar-quantity; but it varies from one point to another.
E) Step 1: Scalar quantities are independent of direction. The statement is true. Scalar quantity only has magnitude, so the value of a scalar does not vary for observers with different orientation of axes. Example: - Mass is invariant for observers with different orientation of axes.
3.22 An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s a part is 30°, what is the speed of the aircraft ?
Solution :
O is the observation point at the ground. A and B are the position of aircraft for which
$\angle AOB = 30$.Time taken by aircraft from A to B is 10 s.
In $\bigtriangleup$ AOB
$$tan30 = \frac{AB}{AO}$$
$$tan30 = \frac{AB}{3400}$$
$$AB = 3400 \times tan30$$
$$AB = 3400 \times \frac{1}{\sqrt{3}}$$
So speed of aircraft,
$$V = \frac{AB}{10}$$
$$V = \frac{3400}{10 \sqrt{3}}$$
$$V = \frac{3400}{10 \times 1.73}$$
$$V = \frac{3400}{17.3}$$
$$V = 196.3 m/s$$