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NCERT Class 11 Physics Chapter 3 Motion in A Plane
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Motion in A Plane
Chapter: 3
| Part – I |
EXERCISE
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.
Ans:
| Scalars | Vectors |
| Volume | Acceleration |
| Mass | Velocity |
| Speed | Displacement |
| Density | Angular velocity |
| Number of moles | |
| Angular frequency |
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.
Ans: Work and current.
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.
Ans: Impulse.
4. State with reasons, whether the following algebraic operations with scalar and vector physical quantities are meaningful:
(a) Adding any two scalars.
Ans: Meaningful: Multiplying a vector by a scalar scales the vector’s magnitude. The two scalar quantities are meaningful if they both represent the same physical quantity.
(b) Adding a scalar to a vector of the same dimensions.
Ans: Not Meaningful: vectors can be added only to another vector.
(c) Multiplying any vector by any scalar.
Ans: Meaningful: This operation scales the vector’s magnitude without changing its direction. For example, force is multiplied with time to give impulse.
(d) Multiplying any two scalars.
Ans: Meaningful: This operation results in another scalar, representing the product of the two numerical values.
(e) Adding any two vectors.
Ans: May or may not be meaningful: Adding vectors depends on whether they represent the same physical quantity and have the same direction.
(f) Adding a component of a vector to the same vector.
Ans: Meaningful: It is possible to add a component of a vector to the same vector because both have the same dimensions.
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.
Ans: True. Magnitude is a scalar quantity representing the size of a vector.
(b) Each component of a vector is always a scalar.
Ans: False. Each component of a vector is a vector by itself.
(c) The total path length is always equal to the magnitude of the displacement vector of a particle.
Ans: False.The total path-length can also be more than 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.
Ans: True. Average speed is always greater than or equal to the magnitude of average velocity due to the nature of distance and displacement.
(e) Three vectors not lying in a plane can never add up to give a null vector.
Ans: True. In order to give a null vector, the third vector should have the same magnitude and opposite direction to the resultant of two vectors.
6. Establish the following vector inequalities geometrically or otherwise: When does the equality sign above apply?
Ans:
(a)
Ans: To prove:
We know that the length of one side of a triangle is less than the sum of length of the other two sides. Hence from 𝛥OPS, we have
If the two vectors
Are acting along the same straight line and in the same direction then:
Combining the conditions mentioned in (i) and (ii) we have
(b)
Ans: To prove:
From 𝜟OPS, we have
(PS=OQ)
The modulus of (OP-PS) has been taken because the L.H.S. is always positive but the R.H.S. may be negative if OP < PS. Thus from (iii) we have:
If the two vectors
Are acting along a straight line in opposite directions, then
(c)
Ans: To prove:
If the two vectors are acting along a straight line but in the opposite directions, then
Combining the conditions mentioned in (vi) and (vii) we get:
(d)
Ans: To prove:
The modulus of (OP-OT) has been taken because L.H.S. is positive and R.H.S. may be negative of OP<OT. From (viii)
If two vectors
Are acting along the same straight line in the same directions then,
Combining the conditions mentioned in (ix) and (x) we get:
We observe that the equality sign holds in above.
Cases when
Are acting along the same straight line and in the same direction of in opposite direction i.e., they are collinear.
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.
Ans: Incorrect: The equation only states that the vectors sum to zero; they do not need to be null vectors. They can be non-zero vectors that balance each other out.
(b) The magnitude of (a + c) equals the magnitude of ( b + d).
Ans: Correct as,
(c) The magnitude of a can never be greater than the sum of the magnitudes of b, c, and d.
Ans: Correct as,
(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?
Ans: Correct as,
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 the path skate ?
Ans: Diameter= 2 × Radius = 400m
Thus, the magnitude of the displacement vector for each girl is 400 m.
Therefore the magnitude of the displacement for each girl is 400m. Which is equal to the path skated by girl B.
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?
Ans: (a) The initial and the final positions of the cyclist are the same, therefore the net displacement of the cyclist is zero.
(b) Average velocity = Net displacement/ Total time taken
= 0 km/10 min
= 0 km/ min.
The net displacement of the cyclist is zero therefore his average velocity will also be zero.
(c) Total speed covered.
= OP + PQ + QO = R + 2𝜋r/4 + R
Average speed
= Total distance covered/ total time taken
= 25/7 km/ 10 min
= 25/7 km/ ⅙ h.
10. On an open ground, a motorist follows a track that turns to his left by an angle of 600 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.
Ans: (i) 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 third turn at the point C.
= 500√4 = 1000 m = 1 km
Total path length
= 500m + 500m + 500m = 1500m = 1.5km.
The ratio of the magnitude of displacement to the total path-length = 1/1.5 = 1/3 = 0.67
(ii) The motorist will take the sixth turn at O. Displacement is zero.
Path-length is = 3000 m or 3 km.
Ratio of magnitude of displacement and path- length is zero.
(iii) The motorist will take the 8th turn at B. Magnitude of displacement:
= 2 × 500 cos 30° = 500 √3 m = √3/2 km.
Path – length = 8 × 500 = 4 km.
Ratio of magnitude of displacement and path length is:
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?
Ans: Given data:
Distance travelled by taxi: s = 23km
Time taken to reach the hotel: t = 28 min
Straight-line distance from the station to hotel: d = 10km.
As we know that speed is usually expressed in km/h, so we need to convert the time from minutes to hours:
t = 28 min = 28/60 h.
= 7/15 h.
(i) Average speed = Total distance/total time =
= Average speed = 23 km/ 7 km/h
= 49.29 km/h.
(b) Distance between the hotel and the station = 10 km
= Displacement of the car, Average velocity = 10/ (28/60)
= 21. 43 km/h.
Clearly the average speed and magnitude of average velocity are not equal.
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-1 can go without hitting the ceiling of the hall?
Ans: Here H = 25m, u = 40 ms-1
As H = u sin2 θ/ 2g = 25m
= (40)2 sin2 θ / 2 × 9.8 = 25m
= sin2 θ = 490/ (40)2
= sin θ = 0.5534 = θ = 33.6
R = u2 sin (33.6)/ 9.8
= (40)2 × sin 67.2/9.8
= (40)2 × 0.9219/ 9.8
= 150.5m.
13. A cricketer can throw a ball to a maximum horizontal distance of 100 m. How high above the ground can the cricketer throw the same ball?
Ans: R = 100m
The horizontal range for a projection velocity v is given by the relation:
R = v2sin2θ/g
100 = v2/g sin 90
v2/g = 100
The final velocity v is zero at the maximum height H.
Acceleration a = -g
V2 – u2 = 2gH
= (0)2 – (√100g)2 = 2 (-g) H
H = 100g/ 2g
= 50m.
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?
Ans: The angular velocity is:
ω = 2𝜋 × number of revolutions/ time
ω = 2𝜋 × 14 / 25
ω = 28 𝜋 / 25
3.52 rad/s
The centripetal acceleration ac is given by:
= ac = ω2r
= ac = (3.52)2 × 0.8
= ac = 12.38 × 0.8
= 9.9m/s2
15. An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Ans: Here given that r 1 km = 1000m
Speed of the aircraft (v) = 900 km/h = 900 × 5/18 = 250 m/s
= Centripetal acceleration, ac = v2/r
= (250)2 / 1000 = 62.5m/s2
Acceleration due to gravity, g = 9.8 m/s2
Ration ac /g
= 62.5/ 9.8
= ac = 6.38g.
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.
Ans: False. In uniform circular motion, the net acceleration, which is the centripetal acceleration, is indeed directed towards the centre of the circle along the radius.
(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. The velocity is always tangential, if the motion is uniform or not.
(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null vector.
Ans: True. The average of these vectors over one cycle is a null vector.
17. The position of a particle is given by:
r = 3.0t i – 2.0t2 j + 4. 0 km
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:
(b) What is the magnitude and direction of velocity of the particle at t = 2.0 s?
Ans:
18. A particle starts from the origin at t = 0 s with a velocity of 10.0 ĵ m/s and moves in the x-y plane with a constant acceleration of (8.0î + 2.0 ĵ ) + 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?
Ans: Initial conditions:
Initial position: r0 = 0i + 0j
Initial velocity: v0 = 0i + 10.0j m/s
Constant acceleration: a = 8. 0i + 2. 0 j m/s2
Position as a function of time:
r(t) = r0 + v0t + ½ at2
r(t) = 0 + ( 0i + 10.0j)t + ½ (8.0i + 2.0j)t2
= r(t) = (4.0t2)i + (10.0t + 1. 0t2) j
(a) Finding the time when the x coordinate is 16m:
x(t) = 4.0t2
This equal to 16m:
4.0 t2 = 16
Solving for t:
t2 = 16/4.0 = 4
t = 2s
plugging in t = 2
y(2) = 10.0(2) + 1.0(22)
= 20.0 + 4.0
= 24.0m.
(b) What is the speed of the particle at the time?
Ans: = √(16)2 + (14)2
= √256 + 196 = √452
= 21.26 ms-1
19. i and j are unit vectors along the x- and y- axis respectively. What is the magnitude and direction of the vectors i + j , and i – j? What are the components of a vector A= 2i + 3j along the directions of i + j and i – j? [You may use a graphical method].
Ans: (i)
= θ = 45°
So, the vector
Makes and an angle of 45° with x – axis.
(ii)
= √ (1)2 + (1)2 – 2 × 1 × 1 × cos 90° = √2
= 1.414 units.
The vector
Makes an angle of 45° with x – axis.
(iii) Let us now determine the components of
(iv)
20. For any arbitrary motion in space, which of the following relations are true:
(a) v average = (1/2) (v (t1 ) + v (t2)
(b) v average = [r(t2 ) – r(t1 ) ] /(t 2 – t1)
(c) v (t) = v (0) + a t
(d) r (t) = r (0) + v (0) t + (1/2) a t2
(e) a average =[ v (t2 ) – v (t1 )] /( t 2 – t1)
(The ‘average’ stands for average of the quantity over the time interval t1 to t2)
Ans: Since the motion is arbitrary, the accelerations may not be uniform. Therefore relations (c) and (d) cannot be correct. For an arbitrary motion motion the average velocity cannot be defined as an equation therefore (a) is also not correct. Any Arbitrary motion in space can be represented by the given relation. Thus, the given relation is true for arbitrary motion. Hence only relation (b) and (e) are correct.
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.
Ans: False. Scalar quantities are quantities that are described only by a magnitude. They do not have a direction of action.
(b) Can never take negative values.
Ans: False. A scalar quantity is the one that can never take negative values. A scalar quantity is the one that does not vary from one point to another in space.
(c) Must be dimensionless.
Ans: False. A scalar quantity is one that must be dimensionless.
(d) Does not vary from one point to another in space.
Ans: False. A scalar quantity is the one that does not 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.
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?
Ans: tan 15° = AC/OC
= AC = OC tan 15°
= 3400 × tan 15°
𝚫ACO is similar to 𝚫CBO
Therefore AC = CB
2AC = 2 × 3400 tan 15°
= 6800 × 0.268 = 1822.4m
Therefore speed of the aircraft = 1822.4m/10
= 182.24m/s.
