# NCERT Solutions For Class 11 Physics Chapter 4

## NCERT Solutions Class 11 Physics Motion in a Plane

NCERT solutions for class 11 physics chapter 4 is one of the important study material for class 11 students because motion in a plane is on of the most scoring chapter in kinematics. Students who are studying in class 11 must try to understand each and every topic in a detailed way so that they can write appropriate answers in class 11th final examination. To score good marks in class 11 examination one must solve the questions provided at the end of each chapter in the NCERT book.

NCERT solutions for class 11 physics chapter 4 motion in a plane is provided here so that students can have look at these for better clarification and understanding of the chapters.

Q-1: State whether the following physical quantities are scalar or vector.

(i) Mass

(ii) Volume

(iii) Speed

(iv) Acceleration

(v) Density

(vi) Number of moles

(vii) Velocity

(viii) Angular frequency

(ix) Displacement

(x) Angular velocity

Ans:

Scalar: Density, mass, speed, volume, angular frequency, number of moles.

Vector: Velocity, acceleration, angular velocity, displacement.

A scalar quantity depends only on the magnitude and it is independent of the direction. Density, mass, speed, volume, angular frequency and number of moles are scalar quantities.

A vector quantity depends on the magnitude as well as the direction. Velocity, acceleration, angular velocity, displacement comes under this.

Q-2: From the following pick any two scalar quantities:

Force, angular momentum, work, current, linear momentum, electric field, average velocity, magnetic momentum, relative velocity.

Ans:

Dot product of force and displacement is the work done. Work is a scalar quantity since dot product of two quantity is always scalar.

Current is a scalar quantity as it is described only by its magnitude and it is independent of direction.

Q-3: From the following identify the vector quantities:

Pressure, temperature, energy, time, gravitational potential, power, total path length, charge, co-efficient of friction, impulse.

Ans:

Impulse is the product of force and time. Since force is a vector quantity, its product with time which is a scalar quantity gives a vector quantity.

Q-4: Explain whether the following algebraic operations meaningful or not:

(a) Addition of any two scalars

(b) Adding a scalar to a vector which has the same dimensions

(c) Multiplying a vector by any scalar

(d) Multiplying any two scalars

(f) Addition of a vector component to the same vector.

Ans:

(a) 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 by another scalar having the same or different dimensions.

(e) 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.

Q-5: State with reasons whether the following are 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.

Ans:

(a) True:

The magnitude of a vector is a number. So, 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.

Q-6: Draw the given vector inequalities geometrically or otherwise:

(a) |a+b||a|+|b|$\left | a + b \right |\leq \left | a \right | + \left | b \right |$

(b)|ab|||a||b||$|a-b|\geq||a|-|b||$

(c) |ab||a|+|b|$\left | a – b \right |\leq \left | a \right | + \left | b \right |$

(d)|ab|||a||b||$| a – b |\geq | | a | – | b | |$

When does the equality sign above apply?

Ans:

(a) Let two vectors a⃗ $\vec{a}$ and b⃗ $\vec{b}$ be represented by the adjacent sides of a parallelogram PQRS, as given in the figure.

Here,

QR=|a⃗ |$\left | \vec{QR} \right | = \left | \vec{a} \right |$ —– (i)

RS=QP=b⃗ $\left | \vec{RS} \right | = \left | \vec{QP} \right | = \left | \vec{b} \right |$ —– (ii)

QS=a⃗ +b⃗ $\left | \vec{QS} \right | = \left | \vec{a} + \vec{b} \right |$ —– (iii)

Each side in a triangle is smaller than the sum of the other two sides.

Therefore, in ΔQRS$\Delta QRS$,

QS < (QR + RS)

a⃗ +b⃗ <|a⃗ |+b⃗ $\left | \vec{a} + \vec{b} \right | < \left | \vec{a} \right | + \left | \vec{b} \right |$ —– (iv)

If the two vectors a⃗ $\vec{a}$ and b⃗ $\vec{b}$  act along a straight line in the same direction, then:

a⃗ +b⃗ =|a⃗ |+b⃗ $\left | \vec{a} + \vec{b} \right | = \left | \vec{a} \right | + \left | \vec{b} \right |$ —– (v)

Combine equation (iv) and (v),

a⃗ +b⃗ |a⃗ |+b⃗ $\left | \vec{a} + \vec{b} \right | \leq \left | \vec{a} \right | + \left | \vec{b} \right |$

(b) Let two vectors a⃗ $\vec{a}$ and b⃗ $\vec{b}$ be represented by the adjacent sides of a parallelogram PQRS, as given in the figure.

Here,

QR=|a⃗ |$\left | \vec{QR} \right | = \left | \vec{a} \right |$ —– (i)

RS=QP=b⃗ $\left | \vec{RS} \right | = \left | \vec{QP} \right | = \left | \vec{b} \right |$ —– (ii)

QS=a⃗ +b⃗ $\left | \vec{QS} \right | = \left | \vec{a} + \vec{b} \right |$ —– (iii)

Each side in a triangle is smaller than the sum of the other two sides.

Therefore, in ΔQRS$\Delta QRS$,

QS + RS > QR

QS + QR > RS

QS>QRQP$\left | \vec{QS} \right | > \left | \vec{QR} – \vec{QP} \right |$ (QP = RS)

|a⃗ +b⃗ |>||a⃗ ||b⃗ ||(iv)$| \vec{a} + \vec{b} |> | | \vec{a} | – | \vec{b} | | —— (iv)$

If the two vectors a⃗ $\vec{a}$ and b⃗ $\vec{b}$  act along a straight line in the same direction, then:

|a⃗ +b⃗ |=||a⃗ ||b⃗ ||(v)$| \vec{a} + \vec{b} | = | | \vec{a} | – | \vec{b} | | —– (v)$

Combine equation (iv) and (v):

|a⃗ +b⃗ |||a⃗ ||b⃗ ||$| \vec{a} + \vec{b} | \geq | | \vec{a} | – | \vec{b} | |$

(c) Let two vectors a⃗ $\vec{a}$ and b⃗ $\vec{b}$ be represented by the adjacent sides of a parallelogram PQRS, as given in the figure.

Here,

PQ=SR=b⃗ $\left | \vec{PQ} \right | = \left | \vec{SR} \right | = \left | \vec{b} \right |$ —– (i)

PS=|a⃗ |$\left | \vec{PS} \right | = \left | \vec{a} \right |$ —– (ii)

Each side in a triangle is smaller than the sum of the other two sides.

Therefore, in ΔPSR$\Delta PSR$,

PR < PS + SR

a⃗ b⃗ <|a⃗ |+b⃗ $\left | \vec{a} – \vec{b} \right | < \left | \vec{a} \right | + \left |- \vec{b} \right |$

a⃗ b⃗ <|a⃗ |+b⃗ $\left | \vec{a} – \vec{b} \right | < \left | \vec{a} \right | + \left | \vec{b} \right |$ —– (iii)

If the two vectors act along a straight line in the opposite direction, then:

a⃗ b⃗ =|a⃗ |+b⃗ $\left | \vec{a} – \vec{b} \right | = \left | \vec{a} \right | + \left | \vec{b} \right |$ —– (iv)

Combine (iii) and (iv),

a⃗ b⃗ |a⃗ |+b⃗ $\left | \vec{a} – \vec{b} \right | \leq \left | \vec{a} \right | + \left | \vec{b} \right |$

(d) Let two vectors a⃗ $\vec{a}$ and b⃗ $\vec{b}$ be represented by the adjacent sides of a parallelogram PQRS, as given in the figure.

Here,

PR + SR > PS —– (i)

PR > PS – SR —– (ii)

a⃗ b⃗ >|a⃗ |b⃗ $\left | \vec{a} – \vec{b} \right | > \left | \vec{a} \right | – \left | \vec{b} \right |$ —– (iii)

The quantity on the left hand side is always positive and that on the right hand side can be positive or negative.

We take modulus on both the sides to make both quantities positive:

a⃗ b⃗ >|a⃗ |b⃗ $\left |\left | \vec{a} – \vec{b} \right | \right | > \left |\left | \vec{a} \right | – \left | \vec{b} \right | \right |$

a⃗ b⃗ >|a⃗ |b⃗ $\left | \vec{a} – \vec{b} \right | > \left |\left | \vec{a} \right | – \left | \vec{b} \right | \right |$ —–  (iv)

If the two vectors act along a straight line in the opposite direction, then:

a⃗ b⃗ =|a⃗ |b⃗ $\left | \vec{a} – \vec{b} \right | = \left |\left | \vec{a} \right | – \left | \vec{b} \right | \right |$ —– (v)

Combine (iv) and (v):

a⃗ b⃗ |a⃗ |b⃗ $\left | \vec{a} – \vec{b} \right | \geq \left |\left | \vec{a} \right | – \left | \vec{b} \right | \right |$

Q-7: Given that l + m + n + o = 0, which of the given statements are true:

(a) l, m, n and o each must be a null vector.

(b) The magnitude of (l + n) equals the magnitude of (m+ o).

(c) The magnitude of l can never be greater than the sum of the magnitudes of m, n and o.

(d) m + n must lie in the plane of l and o if l and o are not collinear, and in the line of l and o, if they are collinear?

Ans:

(a) False

In order to make l + m + n + o = 0, it is not necessary to have all the four given vectors to be null vectors. There are other combinations which can give the sum zero.

(b) True

l + m + n + o = 0

l + n = – (m + o)

Taking mode on both the sides,

|l+n|=|(m+o)|=|m+o|$\left | l + n \right | = \left | -\left ( m + o \right ) \right | = \left | m + o \right |$

Therefore, the magnitude of (l + n) is the same as the magnitude of (m + o).

(c) True

l + m + n + o = 0

l = (m + n + o)

Taking mode on both the sides,

|l|=|m+n+o|$\left | l \right | = \left | m + n + o \right |$

|l||l|+|m|+|n|$\left | l \right | \leq \left | l \right | + \left | m \right | + \left | n \right |$ —– (i)

Equation (i) shows the magnitude of l is equal to or less than the sum of the magnitudes of m, n and o.

(d) True

For,

l + m + n + o = 0

The resultant sum of the three vectors l, (m + n), and o can be zero only if (m + n) lie in a plane containing l and o, assuming that these three vectors are represented by the three sides of a triangle.

If l and o are collinear, then it implies that the vector (m + n) is in the line of l and o. This implication holds only then the vector sum of all the vectors will be zero.

Q-8: On a circular ground whose radius is 200 m, three girls are skating from point P to point Q which is diametrically opposite to each other. Find the magnitude of the displacement vector. Which girl skated the actual path?

Ans:

The distance between the initial and the final position of the particle is called as the displacement. All the three girls reach from point P to Q. The diameter of the ground is the magnitude of displacement.

Diameter = 200 x 2 = 400 m

Hence, the magnitude of displacement is 400 m for each girl. This magnitude is equal to the path skated by girl B.

Q-9: A person starts to cycle from point O to point P of a circular park whose radius is 1 km and he cycles along the circumference to point Q. After which he travels to O, back to where he started. The total time taken for the trip is 20 min, then find

(i) Net displacement

(ii) Average velocity and

(iii) Average speed of the cyclist.

Ans:

(i) The distance between the initial and final position of the body is called as displacement.  The cyclist comes back to the place where he had started in 20 minutes. So, the displacement is zero.

(ii) Average Velocity = netdisplacementtimetaken$\frac{net\;displacement}{time\;taken}$

As the displacement is zero, the average velocity is zero.

(iii) Average speed = TotalpathlengthTotaltime$\frac{Total\;path\;length}{Total\;time}$

Total path length = OP + PQ + QO

= 1 + 14$\frac{1}{4}$(2π$\pi$ x 1) + 1

= 2 + 12$\frac{1}{2}$ π$\pi$

= 3.570 km

Time = 20 minutes = 2060$\frac{20}{60}$ = 13$\frac{1}{3}$ h

Average Speed = 3.57013$\frac{ 3.570}{\frac{1}{3}}$ = 10.71 km/h

Q-10: A motorist on an open ground follows a track that turns to his left by an angle of 60° after every 500 m. starting from the given turn, specify the displacement of the motorist at the 3rd, 6th and 8th turn. Compare the magnitude of the displacement with the total path length covered y the motorist in each case.

Ans.

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

Let the motorist start from point P.

The motorist takes the third turn at S.

Therefore,

Magnitude of the displacement = PS = PV + VS

= 500 + 500 = 1000 m

Total path length = PQ + QR + RS

= 500 + 500 + 500 = 1500 m

The motorist takes the 6th 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.

Magnitude of displacement = PR

= PQ2+QR2+2(PQ)×(QR)cos60$\sqrt{PQ^{2} + QR^{2} + 2\left ( PQ \right )\times \left ( QR \right )\cos 60^{\circ}}$

= 5002+5002+(2(500)×(500)cos60)$\sqrt{500^{2} + 500^{2} + \left (2\left ( 500 \right )\times \left ( 500 \right )\cos 60^{\circ} \right )}$

= 250000+250000+(500000×12)$\sqrt{250000 + 250000 + \left ( 500000 \times \frac{1}{2} \right )}$

= 866.03 m

β=tan1(500sin60500+500cos60)$\beta = \tan ^{-1}\left ( \frac{500 \sin 60^{\circ}}{500 + 500 \cos 60^{\circ}} \right )$

= 30°

Therefore, the magnitude of displacement is 866.03 m at an angle of 30° 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 following table:

 Turn Magnitude of displacement (m) Total path length (m) 3rd 1000 1500 6th 0 3000 8th 866.03;  30∘$30^{\circ}$ 4000

Q-11: A passenger arriving in a new town wants 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.

(a) What is the average speed of the taxi?

(b) What is the magnitude of average velocity? Are the two equal?

Ans.

(a) Total distance travelled = 23 km

Total time taken = 28 min = 2860$\frac{28}{60}$ h

Therefore,

Average speed = TotaldistancetravelledTotaltimetaken$\frac{Total \; distance \; travelled}{Total \; time \; taken }$

= 232860$\frac{23}{\frac{28}{60}}$ = 49.29 km/h

(b) Distance between the hotel and the station = 10 km = Displacement of the car

Therefore,

Average velocity = 102860$\frac{10}{\frac{28}{60}}$ = 21.43 km/h

The two physical quantities are not equal.

Q-12: Rain is falling vertically with a speed of 30 ms1$m s^{-1}$. A woman rides a bicycle with a speed of 10 ms1$m s^{-1}$ in the north to south direction. What is the direction in which she should hold her umbrella?

Ans:
The described situation is shown in the given figure

Here,

vc$v_{c}$ = Velocity of the cyclist

vr$v_{r}$ = Velocity of falling rain

In order to protect herself from the rain, the woman must hold her umbrella in the direction of the relative velocity (v) of the rain with respect to the woman.

v=vr+(vc)$v = v_{r} + \left ( -v_{c} \right )$

= 30 + (-10) = 20 m/s

tanθ=vcvr$\tan \theta = \frac{v_{c}}{v_{r}}$

= 1030$\frac{10}{30}$

θ=tan1(13)$\theta = \tan ^{-1}\left ( \frac{1}{3} \right )$

= tan1(0.333)$\tan ^{-1}\left ( 0.333 \right )$ = 18°C

Hence, the woman must hold the umbrella toward the south, at an angle of nearly 18° with the vertical.

Q-13: A man can swim with a speed of 4 km/h in still water. How long does he take to cross a river 1 km wide if the river flows steadily at 3 km/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?

Ans:

Speed of the man vm$v_{m}$ = 4 km/h

Width of the river = 1 km

Time taken to cross the river = WidthoftheriverSpeedoftheriver$\frac{Width \; of \; the \; river }{Speed \; of \; the \; river}$

= 14$\frac{1}{4}$ h

= 14×60$\frac{1}{4} \times 60$ = 15 min

Speed of the river, vr$v_{r}$ = 3 kmh$\frac{km}{h}$

Distance covered with flow of the river = vr×t$v_{r} \times t$

= 3×14$3 \times \frac{1}{4}$

= 34$\frac{3}{4}$

= 34×1000$\frac{3}{4} \times 1000$ = 750 m

Q-14: In a harbor, wind is blowing at the speed of 72 km/h and the flag on the mast of a boat anchored in the harbor flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?

Ans:

Velocity of the boat = vb$v_{b}$ = 51 km/h

Velocity of the wind = vw$v_{w}$ = 72 km/h

The flag is fluttering in the north – east direction. It shows that the wind is blowing toward the north-east direction. When the ship begins sailing toward the north, the flag will move along the direction of the relative velocity (vwb$v_{wb}$) of the wind with respect to the boat.

The angle between vw$v_{w}$ and (vb)$\left (-v_{b} \right )$ = 90+45$90^{\circ} + 45^{\circ}$

tanβ=51sin(90+45)72+51cos(90+45)$\tan \beta = \frac{51 \sin \left ( 90 + 45 \right )}{72 + 51\cos \left ( 90 + 45 \right )}$

= 51sin4572+51(cos45)$\\\frac{51 \sin 45}{72 + 51\left ( -\cos 45 \right )}$

= 51×127251×12$\\\frac{51 \times \frac{1}{\sqrt{2}} }{72 – 51 \times \frac{1}{\sqrt{2}}}$

= 5172251$\\\frac{51}{72\sqrt{2} – 51}$

= 5172×1.41451$\\\frac{51}{72 \times 1.414 – 51}$

= frac5150.800$\\frac{51}{50.800}$

= tan1(1.0038)$\\\tan ^{-1}\left ( 1.0038 \right )$ = 45.11°

Angle with respect to the east direction = 45.11° – 45° = 0.11°

Hence, the flag will flutter almost due east.

Q-15: The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball is thrown with a speed of 40 ms1$ms^{-1}$ can go without hitting the ceiling of the hall?

Ans:

Speed of the ball, u = 40 ms-1

Maximum height, h = 25 m

In projectile motion, the maximum height achieved by a body projected at an angle θ$\theta$, is given as:

h=u2sin2θ2g$h = \frac{u^{2}\sin ^{2}\theta }{2g}$ 25=(40)2sin2θ2×9.8$25 = \frac{\left (40 \right )^{2}\sin ^{2}\theta }{2 \times 9.8}$ sin2θ=0.30625$\sin ^{2}\theta = 0.30625$ sinθ=0.5534$\sin \theta = 0.5534$

θ=sin1(0.5534)$\theta = \sin ^{-1}\left (0.5534 \right )$ = 33.60°

Horizontal range, R:

= u2sin2θg$\frac{u^{2}\sin 2\theta}{g}$

= (40)2sin2×33.609.8$\frac{\left (40 \right )^{2}\sin 2 \times 33.60}{9.8}$

= 1600×sin67.29.8$\frac{1600 \times \sin 67.2}{9.8}$

= 1600×0.9229.8$\frac{1600 \times 0.922}{9.8}$ = 150.53 m

Q-16: 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?

Ans:

Maximum horizontal distance, R = 100 m

The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is 45° i.e., θ$\theta$ = 33.60°

The horizontal range for a projection velocity v, is given as:

R = u2sin2θg$\frac{u^{2}\sin 2\theta}{g}$

100=u2gsin90$100 = \frac{u^{2}}{g}\sin 90^{\circ}$

u2g=100$\frac{u^{2}}{g} = 100$ —— (i)

The ball will achieve the max height when it is thrown vertically upward. For such motion, the final velocity v is 0 at the max height H.

Acceleration, a = – g

Using the 3rd equation of motion:

v2u2=2gH$v^{2} – u^{2} = -2gH$ H=12×u2g$H = \frac{1}{2} \times \frac{u^{2}}{g}$

= H=12×100$H = \frac{1}{2} \times 100$ = 50 m

Q-17: 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 direction and magnitude of the acceleration of the stone?

Ans:

Length of the string, l = 80 cm = 0.8 m

No. of revolutions = 14

Time taken = 25 s

Frequency, v = No.ofrevolutionTimetaken$\frac{No. \; of \; revolution}{Time \; taken}$ = 1425$\frac{14}{25}$ Hz

Angular frequency ω$\omega$,

= 2πv$2\pi v$

= 2×227×1425$2 \times \frac{22}{7} \times \frac{14}{25}$

= 8825rads1$\frac{88}{25} \; rad \; s^{-1}$

Centripetal acceleration:

ac=ω2r$a_{c} = \omega ^{2}r$ = (8825)2×0.8$\left (\frac{88}{25} \right )^{2} \times 0.8$ = 9.91 ms-2

The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.

Q-18: An aircraft executes a horizontal loop of radius 1 km with a steady speed of 900 km h-1.Compare its centripetal acceleration with the acceleration due to gravity.

Ans:

Radius of the loop, r = 1 km = 1000 m

Speed, v = 900 km h-1 = 900×518$900 \times \frac{5}{18}$ = 250 ms-1

Centripetal acceleration: ac=v2r$a_{c} = \frac{v^{2}}{r}$

= (250)21000$\frac{\left (250 \right )^{2}}{1000}$ = 62.5 ms-2

Acceleration due to gravity, g = 9.8 ms-2

acg=62.59.8$\frac{a_{c}}{g} = \frac{62.5}{9.8}$ = 6.38

ac=6.38g$a_{c}= 6.38 \; g$

Q-19: True or False. Explain the reason.

(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.

Ans:

(a) False

The net acceleration of a particle in circular motion is always directed along the radius of the circle toward the centre only in the case of uniform circular motion.

(b) True

At a point on a circular path, a particle appears to move tangentially to the circular path.

(c) True

In uniform circular motion (UCM), the direction of the acceleration vector points toward the centre of the circle. However, it constantly changes with time. The average of these vectors over one cycle is a null vector.

Q-20: The position of a particle is given by

r=3.0ti^2.0t2j^+4.0k^m$r = 3.0t \;\hat{i} – 2.0t^{2} \;\hat{j} + 4.0 \hat{k} \; \;m$

Where t is in seconds and the coefficients have the proper units for r to be in meters.

(a) Find the ‘v’ and ‘a’ of the particle?

(b) What is the magnitude and direction of the velocity of the particle at t = 2.0 s?

Ans:

(a) The position of the particle is given by:

r⃗ =3.0ti^2.0t2j^+4.0k^$\vec{r} = 3.0t \;\hat{i} – 2.0t^{2} \;\hat{j} + 4.0 \hat{k}$

Velocity v⃗ $\vec{v}$, of the particle is given as:

v⃗ =drdt=ddt(3.0ti^2.0t2j^+4.0k^)v⃗ =3.0i^4.0tj^$\\\vec{v} = \frac{\vec{dr}}{dt} = \frac{d}{dt}\left ( 3.0t \;\hat{i} – 2.0t^{2} \;\hat{j} + 4.0 \hat{k} \right )\\ \\ \vec{v} = 3.0\; \hat{i} – 4.0t \; \hat{j}$

Acceleration a⃗ $\vec{a}$, of the particle is given as:

a⃗ =dv⃗ dt=ddt(3.0i^4.0tj^)a⃗ =4.0j^$\\\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}\left ( 3.0\; \hat{i} – 4.0t \; \hat{j} \right ) \\ \\ \vec{a} = -4.0 \hat{j}$

8.54 m/s, 69.45$69.45^{\circ}$ below the x – axis

(b) we have velocity vector, v⃗ =3.0i^4.0tj^$\vec{v} = 3.0\; \hat{i} – 4.0t \; \hat{j}$

At t = 2.0 s:

v⃗ =3.0i^8.0j^$\vec{v} = 3.0\; \hat{i} – 8.0 \; \hat{j}$

The magnitude of velocity is given by:

|v⃗ |=3.02+8.02=73=8.54m/s$\left |\vec{v} \right | = \sqrt{3.0^{2} + – 8.0^{2}} = \sqrt{73} = 8.54 \;m/s$

Direction, θ=tan1(vyvx)=tan1(83)=tan1(2.667)=69.45$\theta = \tan ^{-1} \left ( \frac{v_{y}}{v_{x}} \right )\\ \\ = \tan ^{-1} \left ( \frac{-8}{3} \right ) = – \tan ^{-1} \left ( 2.667 \right )\\ \\ = 69.45^{\circ}$

The negative sign indicates that the direction of velocity is below the x – axis.

Q-21: A particle starts from the origin at t = 0 s with the velocity of 10j^ms1$10 \; \hat{j} \; m \; s^{-1}$ and moves in the x –y plane with a constant acceleration of (8.0i^+2.0j^)ms2$\left ( 8.0 \; \hat{i} + 2.0 \; \hat{j}\right ) \; m \; s^{-2}$.

(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?

Ans:

(a) Velocity of the particle = 10j^ms1$10 \; \hat{j} \; m \; s^{-1}$

Acceleration of the particle = (8.0i^+2.0j^)ms2$\left (8.0 \; \hat{i} + 2.0 \; \hat{j}\right) \; m \; s^{-2}$

But, a⃗ =dvdt=8.0i^+2.0j^$\vec{a} = \frac{\vec{dv}}{dt} = 8.0 \; \hat{i} + 2.0 \; \hat{j}$

dv=(8.0i^+2.0j^) ;dt$\vec{dv} = \left (8.0 \; \hat{i} + 2.0 \; \hat{j} \right )\ ;dt$

Integrating both the sides:

v⃗ (t)=8.0ti^+2.0tj^+u⃗ $\vec{v}\left ( t \right ) = 8.0 t \; \hat{i} + 2.0 t \; \hat{j} + \vec{u}$

Where, u⃗ $\vec{u}$ = velocity vector of the particle at t = 0

v⃗ $\vec{v}$ = velocity vector of the particle at time t

But, v⃗ =drdt$\vec{v} = \frac{\vec{dr}}{dt}$

dr=v⃗ dt$\vec{dr} = \vec{v}\; dt$

= (8.0ti^+2.0tj^+u⃗ )dt$\left (8.0 t \; \hat{i} + 2.0 t \; \hat{j} + \vec{u} \right ) \; dt$

Integrating the equations with the conditions:

At t = 0; r = 0 and at t = t; r = r.

r⃗ =u⃗ t+128.0t2i^+12×2.0t2j^$\vec{r} = \vec{u}t + \frac{1}{2}8.0t^{2} \; \hat{i} + \frac{1}{2} \times 2.0t^{2} \; \hat{j}$ r⃗ =u⃗ t+4.0t2i^+t2j^$\vec{r} = \vec{u}t + 4.0t^{2} \; \hat{i} + t^{2} \; \hat{j}$ r⃗ =(10.0j^)t+4.0t2i^+t2j^$\vec{r} = \left (10.0 \; \hat{j} \right)t + 4.0t^{2} \; \hat{i} + t^{2} \; \hat{j}$ xi^+yj^=4.0t2i^+(10.0t+t2)j^$x \;\hat{i} + y \;\hat{j} = 4.0t^{2} \; \hat{i} + \left ( 10.0 \; t + t^{2} \right )\; \hat{j}$

Since the motion of the particle is confined to the x-y plane, on equating the coefficients of i^andj^$\hat{i} \; and \; \hat{j}$, we get:

x=4t2$x = 4t^{2}$

t = (x4)12$\left (\frac{x}{4} \right)^{\frac{1}{2}}$

y = 10t+t2$10 t + t^{2}$

When x = 16 m:

t=(164)12$t = \left (\frac{16}{4} \right )^{\frac{1}{2}}$ = 2 s

Therefore, y = 10 × 2 + (2)2$\left (2 \right)^{2}$ = 24 m

(b) Velocity of the particle:

v⃗ (t)=8.0ti^+2.0tj^+u⃗ $\vec{v} ( t ) = 8.0 t \; \hat{i} + 2.0 t \; \hat{j} + \vec{u}$

At t = 2 s:

v⃗ (t)=8.0×2i^+2.0×2j^+20j^$\vec{v} ( t ) = 8.0 \times 2 \; \hat{i} + 2.0 \times 2 \; \hat{j} + 20 \; \hat{j}$ v⃗ (t)=16i^+14j^$\vec{v} ( t ) = 16 \; \hat{i} + 14\; \hat{j}$

Therefore, Speed of the particle:

|v⃗ |=(16)2+(14)2$\left |\vec{v} \right | = \sqrt{\left ( 16 \right )^{2} + \left ( 14 \right )^{2}}$ |v⃗ |=256+196$\left |\vec{v} \right | = \sqrt{256 + 196}$ |v⃗ |=452$\left |\vec{v} \right | = \sqrt{452}$ |v⃗ |=21.26ms1$\left |\vec{v} \right | = 21.26 \; m \; s^{-1}$

Q-22: i⃗ $\vec{i}$ and j⃗ $\vec{j}$ are unit vectors along x- and y-axis respectively. What is the magnitude and direction of the vectors i⃗ $\vec{i}$ + j⃗ $\vec{j}$ and i⃗ $\vec{i}$j⃗ $\vec{j}$  ? What are the components of a vector A⃗ $\vec{A}$ = 2 i⃗ $\vec{i}$ and 3j⃗ $\vec{j}$  along the directions of i⃗ $\vec{i}$ + j⃗ $\vec{j}$  and i⃗ $\vec{i}$  – j⃗ $\vec{j}$? [You may use graphical method]

Ans:

Consider a vector P⃗ $\vec{P}$, given as below,

P⃗ =i^+j^$\vec{P} = \hat{i} + \hat{j}$ Pxi^+Pyj^=i^+j^$P_{x} \;\hat{i} + P_{y} \;\hat{j} = \hat{i} + \hat{j}$

Comparing the components on both the sides, we get:

Px=Py=1$P_{x} = P_{y} = 1$ P⃗ =P2x+P2y$\left | \vec{P} \right | = \sqrt{P_{x}^{2} + P_{y}^{2}}$ P⃗ =12+12$\left | \vec{P} \right | = \sqrt{1^{2} + 1^{2}}$

P⃗ =2$\left | \vec{P} \right | = \sqrt{2}$ ——- (i)

Therefore, the magnitude of the vector i⃗ +j⃗ $\vec{i} + \vec{j}$ is 2$\sqrt{2}$.

Let θ$\theta$ be the angle made by the vector P⃗ $\vec{P}$, with the x – axis as given in the figure below.

tanθ=(PxPy)$\tan \theta = \left ( \frac{P_{x}}{P_{y}} \right )$ θ=tan1(11)$\theta = \tan ^{-1} \left ( \frac{1}{1} \right )$

θ=45$\theta = 45^{\circ}$ —– (ii)

Therefore, the vector i⃗ j⃗ $\vec{i} – \vec{j}$ makes an angle of 45$45^{\circ}$ with the x axis.

Let Q⃗ =i^j^$\vec{Q} = \hat{i} – \hat{j}$

Qxi^Qyj^=i^j^$Q_{x}\; \hat{i} – Q_{y}\; \hat{j}= \hat{i} – \hat{j}$ Qx=Qy=1$Q_{x}\; = Q_{y}\; = 1$ Q⃗ =Q2x+Q2y$\left | \vec{Q} \right | = \sqrt{Q_{x}^{2} + Q_{y}^{2}}$

Q⃗ =2$\left | \vec{Q} \right | = \sqrt{2}$ —— (iii)

Therefore, the magnitude of the vector 1^j^$\hat{1} – \hat{j}$ is 2$\sqrt{2}$.

Let θ$\theta$ be the angle by the vector Q⃗ $\vec{Q}$, with the x – axis as given in the figure below,

tanθ=(QxQy)$\tan \theta = \left ( \frac{Q_{x}}{Q_{y}} \right )$ θ=tan1(11)$\theta = -\tan ^{-1} \left ( -\frac{1}{1} \right )$

θ=45$\theta = -45^{\circ}$ —— (iv)

Therefore, the vector i⃗ j⃗ $\vec{i} – \vec{j}$ makes an angle of 45$-45^{\circ}$ with the x axis.

It is given that,

A⃗ =2i^+3j^$\vec{A} = 2 \hat{i} + 3 \hat{j}$ Axi^+Ayj^=2i^+3j^$A_{x} \;\hat{i} + A_{y}\; \hat{j} = 2 \hat{i} + 3 \hat{j}$

Comparing the components on both the sides, we get:

Ax=2andAy=3$A_{x} = 2\; and \;A_{y}\; = 3$ A⃗ =22+32$\left | \vec{A} \right | = \sqrt{2^{2} + 3^{2}}$ A⃗ =13$\left | \vec{A} \right | = \sqrt{13}$

Let Ax$\vec{A_{x}}$ make an angle θ$\theta$ with the x- axis, as it is shown in the figure,

tanθ=(AxAy)$\tan \theta = \left ( \frac{A_{x}}{A_{y}} \right)$ θ=tan1(32)$\theta = \tan ^{-1} \left ( \frac{3}{2} \right )$ θ=tan1(1.5)$\theta = \tan ^{-1} \left ( 1.5 \right )$ θ=56.31$\theta = 56.31^{\circ}$

Angle between the vectors (2i^+3j^)$\left ( 2 \hat{i} + 3 \hat{j}\right )$ and (i^+j^)$\left ( \hat{i} + \hat{j}\right )$,

θ=56.3145=11.31$\theta’ = 56.31 – 45 = 11.31^{\circ}$

Components of vector A⃗ $\vec{A}$, along the direction of P⃗ $\vec{P}$, making an angle θ$\theta$.

= (Acosθ)P^$\left ( A\cos \theta’ \right )\hat{P}$

= (Acos11.31)(i^+j^)2$\left ( A\cos 11.31 \right )\frac{\left ( \hat{i} + \hat{j} \right )}{\sqrt{2}}$

= 13×0.98062(i^+j^)$\sqrt{13} \times \frac{0.9806}{\sqrt{2}}\left ( \hat{i} + \hat{j}\right )$

= 2.5(i^+j^)$2.5 \left (\hat{i} + \hat{j} \right )$

= 2510×2$\frac{25}{10} \times \sqrt{2}$

= 52$\frac{5}{\sqrt{2}}$ —– (v)

Let θ$\theta’$ be the angle between the vectors (2i^+3j^)$\left (2 \hat{i} + 3 \hat{j} \right )$ and (i^j^)$\left ( \hat{i} – \hat{j} \right )$

θ=45+56.31=101.31$\theta” = 45 + 56.31 = 101.31^{\circ}$

Component of vector A⃗ $\vec{A}$, along the direction of Q⃗ $\vec{Q}$, making an angle θ$\theta$.

= (Acosθ)Q⃗ $\left ( A \cos \theta ” \right )\vec{Q}$

= (Acosθ)i^j^2$\left ( A \cos \theta ” \right )\frac{\hat{i} – \hat{j}}{\sqrt{2}}$

= 13cos(901.31)