Motion in a Plane

Class 11 NCERT Physics

NCERT

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 :

$\textbf{Scalar}:$ Volume, mass, speed, density, number of moles, angular frequency$\\$ $\textbf{Vector}:$ Acceleration, velocity, displacement, angular velocity$\\$ A scalar quantity is specified by its magnitude only. It does not have any direction associated with it.$\\$ Volume, mass, speed, density, number of moles, and angular frequency are some of the scalar physical quantities.$\\$ A vector quantity is specified by its magnitude as well as the direction associated with it.$\\$ Acceleration, velocity, displacement, and angular velocity belong to this category.

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 :

$\textbf{Work}$ and $\textbf{Current}$ are scalar quantities.$\\$ $\text{Work}$ done is given by the dot product of force and displacement. Since the dot product of two quantities is always a scalar, work is a scalar physical quantity.$\\$ $\text{Current}$ is described only by its magnitude. Its direction is not taken into account. Hence, it is a scalar quantity.

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 :

$\text{Impulse}$$\\$ $\text{Impulse}$ is given by the product of force and time. Since force is a vector quantity, its product with time (a scalar quantity) gives a vector quantity.

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 dimension s,$\\$ (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) Not meaningful. The addition of two scalar quantities is meaningful only if they both represent the same physical quantity.$\\$ (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, force is multiplied with time to give impulse.$\\$ (d) Meaningful. A scalar, irrespective of the physical quantity it represents, can be multiplied with another scalar having the same or different dimensions.$\\$ (e) Not meaningful. The addition of two vector quantities is meaningful only if they both represent the same physical quantity.$\\$ (f) Meaningful A component of a vector can be added to the same vector as they 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,$\\$ (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 is a number. Hence, it is a scalar.$\\$ (b) False. Each component of a vector is also a vector.$\\$ (c) False. Total path length is a scalar quantity, whereas displacement is a vector quantity. Hence, 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) 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) True. Three vectors, which do not lie in a plane, cannot be represented by the sides of a triangle taken in the same order.

6   Establish the following vector inequalities geometrically or otherwise:$\\$ $(a) |a+b| \leq |a|+|b|\\ (b) |a+b| \geq ||a|-|b||\\ (c)|a-b| \leq |a|+|b|\\ (d)|a-b| \geq ||a|-|b||$$\\$When does the equality sign above apply?

Solution :

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 write $|\bar{OM}|=|\bar{a } \quad ...(i)\\ |\bar{MN}|=|\bar{OP}|=|\bar{b}|\quad ...(ii)\\ |\bar{ON}|=|\bar{a}+\bar{b}| \quad ...(iii)$$\\$ In a triangle, each side is smaller than the sum of the other two sides. Therefore, in $\Delta OMN$ , we have:$\\$ $ON < ( OM + MN )\\ |\bar{a}+\bar{b}| < |\bar{a}|+|\bar{b}| \quad ....(iv)$$\\$ If the two vectors $\bar{ a}$ and $\bar{b} $ act along a straight line in the same direction, then we can write:$\\$ $|\bar{a}+\bar{b}|=|\bar{a}|+|\bar{b}| \quad ...(v)$$\\$ Combining equations (iv) and (v), we get:$\\$ $|\bar{a}+\bar{b}| \leq |\bar{a}|+|\bar{b}|$$\\$ Let two vectors $\bar{a}$ and $\bar{b}$ be represented by the adjacent sides of a parallelogram $OMNP$, as shown in the given figure.$\\$ Here,we have: $\\$ $|\bar{OM}|=|\bar{a } \quad ...(i)\\ |\bar{MN}|=|\bar{OP}|=|\bar{b}|\quad ...(ii)\\ |\bar{ON}|=|\bar{a}+\bar{b}| \quad ...(iii)$$\\$ In a triangle, each side is smaller than the sum of the other two sides. Therefore, in $\Delta OMN$ , we have:$\\$ $ON+MN > OM\\ ON+ OM > MN\\ |ON| > |\bar{OM}-\bar{OM}| \quad (\therefore OP=MN)\\ |\bar{a}+\bar{b}| > ||\bar{a}|-|\bar{b}|| \quad (iv)$$\\$ If the two vectors $\bar{a}$ and $\bar{b}$ act along a straight line in the same direction, then we can write: $\\$ $|\bar{a}+\bar{b}|=||\bar{a}|-|\bar{b}||\quad ...(v)$$\\$ Combining equations (iv) and (v), we get: $\\$ $|\bar{a}+\bar{b}| \geq ||\bar{a}|-|\bar{b}||$$\\$ Let two vectors a and b be represented by the adjacent sides of a parallelogram $PORS,$ as shown in the given figure.$\\$ $\\$

Here we have:$\\$ $|\bar{OR}|=|\bar{PS}|=|\bar{-b}| \quad (i)\\ |\bar{OP}|=|\bar{a}| \quad (ii)$$\\$ In a triangle, each side is smaller than the sum of the other two sides. Therefore, in $\Delta OPS$, we have:$\\$ $OS < OP + PS \\ |\bar{a}-\bar{b}| < | \bar{a}| +| \bar{-b}|\\ |\bar{a-b}| < |\bar{a}|+| \bar{b}| \quad (iii)$$\\$ If the two vectors act in a straight line but in opposite directions, then we can write:$\\$ $|\bar{a}-\bar{b}|=|\bar{a}|+|\bar{b}| \quad (iv)$$\\$ Combining equations (iii) and (iv), we get:$\\$ $|\bar{a}-\bar{b}| \geq | \bar{a}| + |\bar{b}|$$\\ $ Let two vectors $\bar{a}$ and b be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.$\\$ The following relations can be written for the given parallelogram$\\$ $OS+PS > OP \quad ...(i)\\ OS > OP -PS \quad ....(ii)\\ |\bar{a}-\bar{b}| > |\bar{a}|-|\bar{b}| \quad ....(iii)$$\\$ The quantity on the LHS is always positive and that on the RHS can be positive or negative. To make both quantities positive, we take modulus on both sides a s:$\\$ $||\bar{a}-\bar{b}|| > ||\bar{a}| -|\bar{b}||\\ |\bar{a}-\bar{b}| > ||\bar{a}| -|\bar{b}|| \quad ....(iv)$$\\$ If the two vectors act in a straight line but in the same directions, then we can write:$\\$ $|\bar{a}-\bar{b}|=||\bar{a}|-|\bar{b}|| \quad ....(v)$$\\$ Combining equations (iv) an d (v), we get:$\\$ $|\bar{a}-\bar{b}| \leq ||\bar{a}|-|\bar{b}||$

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?$\\$

Solution :

(a) Incorrect$\\$ In order to make, $\vec{a }+ \vec{b} +\vec{ c} + \vec{d} = 0$ it is not necessary to have all the four given vectors to be null vectors. There are many other combinations which can give the sum zero.$\\$ (b) Correct$\\$ $\vec{a }+ \vec{b} + \vec{c }+ \vec{d} = 0\\ \vec{a} +\vec{ c} =- (\vec{ b} + \vec{d} )$$\\$ Taking modulus on both the sides, we get:$\\$ $| \vec{a} + \vec{c} | = | - ( \vec{b} + \vec{d }) | = | \vec{b} + \vec{d} |$$\\$ Hence, the magnitude of $( \vec{a }+ \vec{c} )$ is the same as the magnitude of $(\vec{b} + \vec{d} )$$\\$

(c) Correct$\\$ $\vec{a} + \vec{b} +\vec{ c} + \vec{d} = 0\\ \vec{a} =- ( \vec{b} + \vec{c} + \vec{d} )$$\\$ Taking modulus both sides, we get the magnitude of a to be equal to the magnitude of $( \vec{b} + \vec{c} + \vec{d} ) :\\ \vec{a} = | ( \vec{b} + \vec{c} + \vec{d} ) |\\ | \vec{a} | \leq | \vec{b} | + | \vec{c} | + | \vec{d} | ................( i )\\ $$\\$ Now, $(\vec{b}+ \vec{c} +\vec{ d})$ is the sum of vectors $\vec{b} , \vec{c}$ and $\vec{d}$ . Therefore, the magnitude of $(\vec{b} + \vec{c} + \vec{d} )$ is less than, or equal to the sum of the magnitudes of $\vec{b} , \vec{c}$ and $\vec{d} $. $\\$Hence, the magnitude of a can never be greater than the sum of the magnitudes of $\vec{b} , \vec{c}$ and $\vec{d} $. Equation (i) shows that the magnitude of a is equal to or less than the sum of the magnitudes of $\vec{b} , \vec{c}$ and $\vec{d} .$$\\$

(d) Correct$\\$ For, $\vec{a} +\vec{ b }+ \vec{c }d = 0$$\\$ $\vec{a }=- (\vec{ b} +\vec{ c} \vec{ d}) = 0$$\\$ The resultant sum of the three vectors $\vec{a }, (\vec{b}+\vec{c})$ and $\vec{d}$ can be zero only if $(\vec{b}+\vec{c})$ lie in the same plane as $\vec{a} $ and $\vec{b}$$\\$ If $\vec{a} $ and $ \vec{b} $ are collinear,then it implies that the vector $(\vec{b}+\vec{c})$ lie in the same plane as $ \vec{a } $ and $\vec{b}$$\\$ If $\vec{a}$ and $\vec{b}$ are collinear, then it implies that the vector$(\vec{b}+\vec{c})$ is in the line of $\vec{a} $ and $\vec{b}$. this implication holds true in this scneario and the vector sum of all the vectors will be zero.

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. $4.20$. What is the magnitude of the displacement vector for each? For which girl is this equal to the actual length of the path skated?

Solution :

Displacement is 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 = $200 m$$\\$ Diameter of the ground = $2 × 200 = 400 m$$\\$ Hence, the magnitude of the displacement for each girl is $400 m.$ This is equal to the actual length of 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 the following figure. If the round trip takes 10 min, what is the (a) net displacement, (b) average velocity, and (c) average speed of the cyclist?

Solution :

(a) Displacement is given by the minimum distance between the initial and final positions of a body. In the given case, the cyclist comes to the starting point after cycling for $10$ minutes. Hence, his net displacement is zero.$\\$ (b) Average velocity is given by the relation:$\\$ Average velocity = $\dfrac{\text{Net displacement}}{\text{Total time}}$$\\$ Since the net displacement of the cyclist is zero, his average velocity will also be zero.$\\$ (c) Average speed of the cyclist is given by the relation:$\\$ Averagespeed =$\dfrac{\text{Total path length}}{\text{Total time}}$$\\$ Total path length $=OP+PQ+QO$$\\$ $=1+\dfrac{1}{4}(2\pi * 1)+1\\ =2+\dfrac{1}{2}\pi 3.570 km$$\\$ Time taken =$10 min =\dfrac{10}{60}=\dfrac{1}{6}h$$\\$ $\therefore $ Average speed =$\dfrac{3.570}{\dfrac{1}{6}}=21.42 km/h$$\\$

10   On an open ground, a motorist follows a track that turns to his left by an angle of $60^o$ 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 :

The path followed by the motorist is a regular hexagon with side $500 m,$ as shown in the given figure

Let the motorist start from point P.$\\$ The motorist takes the third turn at S.$\\$ $\therefore $Magnitude of displacement = $PS = PV + VS = 500 + 500 = 1000 m$$\\$ Total path length = $PQ + QR + RS = 500 + 500 +500 = 1500 m$$\\$ The motorist takes the sixth turn at point P, which is the starting point. $\therefore $ Magnitude of displacement = 0$\\$ Total path length = $PQ + QR + RS + ST + TU + UP$$\\$ $= 500 + 500 + 500 + 500 + 500 + 500 = 3000 m$$\\$ The motorist takes the eight turn at point R$\\$ $\therefore $ Magnitude of displacement = $PR$$\\$ $=\sqrt{PQ^2 + QR ^2 + 2 (PQ ). ( QR ) \cos 60 ^o}\\ =\sqrt{500 ^2 + 500^ 2 +( 2 * 500 * 500 * \cos 60^o)}\\ =\sqrt{250000 + 250000 +(500000*\dfrac{1}{2})}\\ =866.03m $$\\$ $\beta =\tan^{-1}(\dfrac{500\sin60^o}{500+500 \cos60^o})=30^o$$\\$ Therefore, the magnitude of displacement is $866.03 m$ at an angle of $30^o$ with PR.$\\$ Total path length = Circumference of the hexagon + PQ + QR $= 6 × 500 + 500 + 500 = 4000 m$$\\$ The magnitude of displacement and the total path length corresponding to the required turns is shown in the given table$\\$ $\begin{array}{|c|c|} \hline \textbf{Turn}& \textbf{Magnitude of displacement(m)}& \textbf{Total path length(m)} \\ \hline \text{Third } & 1000 & 1500 \\ \hline \text{Sixth} & 0 & 3000\\ \hline \text{Eighth} & 866.03;30^o & 4000 \\ \hline \end{array}$