# NCERT Solution , Motion in a Plane

Q: 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.

Sol: Scalars: Volume, mass, speed, density, number of moles, angular frequency.

Vectors: Acceleration, velocity, displacement, angular velocity.

Q: 2. Pick out the two scalar quantities in the following lists: force angular momentum, work, current, linear momentum, electric field average velocity, magnetic moment, reaction as per Newton’s third law, relative velocity.

Sol:Work and current are the scalar quantities in the given list.

Q:3.Pick out the only vector quantity in the following lists: temperature, pressure, impulse, time, power, total path-length, energy, gravitational potential, coefficient of friction, charge.

Sol:Since, Impulse = change in momentum = force × time. As momentum and force are vector quantities, hence impulse is a vector quality.

Q:4. State with reasons, whether the following algebraic operations with scalars and vectors 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 (f) adding a component of vector to the same vector.

Sol: (a) No, scalar of same dimensions can be added. e.g . Work can not be added with distance

(b) No , because scalar can not be added to a vector

(c) Yes , e.g. $\large \vec{F} = m \vec{a}$ ; Where F = force , m = mass  , a = acceleration

(d) Yes , e.g . Work = Power × time

(e) No, Vectors of same dimension can be added

(f) Yes , bacause both are vectors of same dimensions .

Q: 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 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 plane can never add up to given a null vector.

Sol: (a) True, because magnitude is pure number

(b) False, because each component is a vector.

(c) True only, if the particle moves along a straight line in the same direction, otherwise false.

(d) True, because the total path length is either greater than or equal to the magnitude of the displacement vector.

(e) True, as they cannot represent the three sides of a triangle taken in the same order.

Q:6.Establish the following inequalities geometrically or otherwise:

(a) $|\vec{A} + \vec{B}| \le |\vec{A}| + |\vec{B}|$

(b) $|\vec{A} + \vec{B}| \ge ||\vec{A}| – |\vec{B}||$

(c) $|\vec{A} – \vec{B}| \le |\vec{A}| + |\vec{B}|$

(d) $|\vec{A} – \vec{B}| \ge | |\vec{A}| – |\vec{B}||$  |

When does the equality sign apply ?

Sol: Consider two vectors  $\vec{A}$  and $\vec{B}$  be represented by the sides $\vec{OP}$ and $\vec{OQ}$ of a parallelogram OPSQ . According to parallelogram law of vector addition , $(\vec{A} + \vec{B})$ will be represented by $\vec{OS}$ as shown in the adjoining figure. (a) To prove $|\vec{A} + \vec{B}| \le |\vec{A}| + |\vec{B}|$

We know that the length of one side of triangle is always less than the sum of the lengths of the other two sides.

Hence from ΔOPS, we have

OS < OP + PS  ; or, OS < OP + OQ

$|\vec{A} + \vec{B}| \le |\vec{A}| + |\vec{B}|$  …(i)

If the two vectors $\vec{A}$   and $\vec{B}$   are acting along the same straight line and in the same direction then,

$|\vec{A} + \vec{B}| = |\vec{A}| + |\vec{B}|$  …(ii)

From (i) & (ii)

$|\vec{A} + \vec{B}| \le |\vec{A}| + |\vec{B}|$

(b) To Prove $|\vec{A} + \vec{B}| \ge ||\vec{A}| – |\vec{B}||$

From ΔOPS, we have, OS + PS  > OP ; or, OS  > |OP – PS|

or, OS >|OP- OQ|     …(iii)  (since PS = OQ )

The modulus of (OP – PS) has been taken because the LHS is always positive but the RHS may be negative if OP < PS.
Thus from (iii) we have,

$|\vec{A} + \vec{B}| > ||\vec{A}| – |\vec{B}||$    ….(iv)

If the two vectors $\vec{A}$    and $\vec{B}$    are acting along a straight line in opposite directions then

$|\vec{A} + \vec{B}| = ||\vec{A}| – |\vec{B}||$    …(v)

From (iv) & (v)

$|\vec{A} + \vec{B}| \ge ||\vec{A}| – |\vec{B}||$

(c) To Prove $|\vec{A} – \vec{B}| \le |\vec{A}| + |\vec{B}|$

From ΔOPR we note that OR < OP + PR

$|\vec{A} – \vec{B}| < |\vec{A}| + |-\vec{B}|$

$|\vec{A} – \vec{B}| < |\vec{A}| + |\vec{B}|$ …(vi)

If the two vectors are acting along the straight line but in opposite direction, then,

$|\vec{A} – \vec{B}| = |\vec{A}| + |\vec{B}|$   …(vii)

From (i) & (ii)

$|\vec{A} – \vec{B}| \le |\vec{A}| + |\vec{B}|$

(d) To Prove $|\vec{A} – \vec{B}| \ge | |\vec{A}| – |\vec{B}||$  |

In fig, from ΔOPR we have,

OR + PR > OP or, OR > |OP-PR|

of, OR > |OP – OT| …(viii)  (since OT = PR)

The modulus of (OP – OT ) has been taken because LHS is positive and RHS may be negative if OP < OT.

From (viii),  $|\vec{A} – \vec{B}| > | |\vec{A}| – |\vec{B}||$  |  …(ix)

If the two vectors $\vec{A}$  and $\vec{B}$  are along the same straight line in the same direction then.

$|\vec{A} – \vec{B}| = | |\vec{A}| – |\vec{B}||$  |  …(x)

From (ix) & (x) we get,

$|\vec{A} – \vec{B}| \ge | |\vec{A}| – |\vec{B}||$  |

Q:7. Given $\vec{A} + \vec{B} + \vec{C} + \vec{D} = 0$ ;  which of the following statements are correct ?

(a) $\vec{A}$ , $\vec{B}$ , $\vec{C}$  and $\vec{D}$ must be a null vector

(b)The magnitude of $(\vec{A} + \vec{C})$ equals the magnitude of $(\vec{B} + \vec{D})$

(c) The magnitude of  $\vec{A}$ can never be greater than the sum of the magnitude of $\vec{B}$ , $\vec{C}$ and $\vec{D}$

(d)$\vec{B} + \vec{C}$ must lie in the plane of $\vec{A} + \vec{D}$ , if $\vec{A}$ and $\vec{D}$ are not collinear and in the line of $\vec{A}$ and $\vec{D}$ , if they are collinear.

Sol: (a)False, because $\vec{A} + \vec{B} + \vec{C} + \vec{D}$ can be zero in many ways other than $\vec{A}$ , $\vec{B}$ , $\vec{C}$  and $\vec{D}$ must each be null vector.

(b)True, since $\vec{A} + \vec{B} + \vec{C} + \vec{D} = 0$ ;

Therefore , $\vec{A} + \vec{C} = -( \vec{B}+ \vec{D} )$  ;or, $|\vec{A} + \vec{C}| = |\vec{B}+ \vec{D} |$

(c) True, since $\vec{A} + \vec{B} + \vec{C} + \vec{D} = 0$ ;

Therefore , $\vec{A} = – ( \vec{B} + \vec{C} + \vec{D} )$

It means the magnitude of $\vec{A}$  is equal to the magnitude of vector $( \vec{B} + \vec{C} + \vec{D} )$ . Since the sum of the magnitude of  $\vec{B}$ , $\vec{C}$ and $\vec{D}$  may be equal or greater than the magnitude of $\vec{A}$ , hence the magnitude of $\vec{A}$ can never be greater than the sum of the magnitude of $\vec{B}$ , $\vec{C}$ and $\vec{D}$

(d) True, since $\vec{A} + \vec{B} + \vec{C} + \vec{D} = 0$ ;

Therefore, $\vec{A} + (\vec{B} + \vec{C} ) + \vec{D} = 0$

The resultant sum of the three vectors $\vec{A} , (\vec{B} + \vec{C} ) , \vec{D}$ can be zero only if $(\vec{B} + \vec{C} )$ lies in the plane of $\vec{A}$ and $\vec{D}$ and these three vectors are represented by the three sides of the triangle taken in one order. If $\vec{A}$ and $\vec{D}$  are collinear , then $(\vec{B} + \vec{D} )$ must be in the line of $\vec{A}$ and $\vec{D}$  , only then vector sum of all the vectors will be zero.

Q: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 shown in figure. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of path skated ? Sol:Displacement for each girl = PQ = Diameter of circular ground

Magnitude of Displacement = 2 × 200 = 400 m

For girl B , the magnitude of displacement is equal to the actual length of path skated .

Q: 9 . A cyclist starts from 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 . If the round trip takes 10 minutes , what is the (a) net displacement (b) average velocity (c) average speed of the cyclist . Sol: (a) Here , net displacement = Zero

(b) Average velocity = net displacement / time = 0

(c) Average Speed = distance covered / time taken

$= \frac{1 km + \frac{2\pi r}{4} + 1 km}{10/60}$

$= \frac{1 km + \frac{2\times (22/7) \times 1}{4} + 1 km}{1/6}$

= 21.4 km/h

Q: 10. 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 eight turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

Sol: In this question, the path is a regular hexagon ABCDEF of side length 500m. In figure Let the motorist start from A.
Third Turn

The motorcyclist will take the 3rd turn at D. Displacement vector at D = AD

Magnitude of this displacement = 500 + 500 = 1000 m

Total path length from A to D = AB + BC + CD = 500 + 500 + 500 = 1500 m

Sixth Turn :
The motorcyclist will take the 6th turn at A.
Displacement vector is null vector.

Total path length = AB + BC + CD + DE + EF

= 500 + 500 + 500 + 500 + 500 + 500

= 3000 m

Eight Turn:
The motorcyclist take the 8th turn at C.

Displacement vector = AC, which is represented by the diagonal of the parallelogram ABCG

$\displaystyle = \sqrt{500^2 + 500^2 + 2 \times 500 \times 500 cos 60 }$

$\displaystyle = \sqrt{500^2 + 500^2 + 500^2 }$

= 866.03 m

$\displaystyle tan\beta = \frac{500 sin60}{500 + 500 cos60}$

$\displaystyle tan\beta = \frac{1}{\sqrt{3}}$

β = 30°

Q:11. A passenger arriving in new town whishes to go from the station to a hotel located 10km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23km long and reaches the hotel in 28minutes. What is (a) the average speed of the taxi, (b) the magnitude of the average velocity? Are the two equal?

Ans.
Here, actual path length travelled, s = 23 km ,

Displacement = 10 km,

Time taken, t = 28 min = 28/60h.

(a) Average speed of the taxi = actual path length/time taken

= 23/28/60 = 49.3 km/h

(b) Magnitude of average velocity = displacement/ time taken

= 10/(28/60) = 21.4 km/h

The average speed is not equal to the magnitude of average velocity. The two are equal for the motion of taxi along a straight path in one direction.

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