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This chapter covers projectile motion, which can be seen in real-life examples like a football being kicked into the air or a missile being launched. By breaking down the motion into horizontal and vertical components, students can predict the path and landing point of the projectile. Circular motion—such as a satellite orbiting the Earth or an automobile navigating a curved road—is another significant topic discussed in this chapter.
The NCERT Exemplar Class 11 Science Solutions for Chapter 4 takes the fundamental idea of motion and expands it to describe the movement of objects in two or three dimensions. This chapter helps students understand the concepts of motion in a plane by utilizing vector operations and graphical representation, making it easier to visualize and grasp the complex ideas.
Question:4.1
The angle between $A=\hat{i}+\hat{j}$ and $B=\hat{i}-\hat{j}$ is
(a) $45^{o}$
(b) $90^{o}$
(c) $-45^{o}$
(d)$180^{o}$
Answer:
The answer is the option (b), $90^{o}$Question:4.2
Which one of the following statements is true?
(a) A scalar quantity is the one that is conserved in a process.
(b) A scalar quantity is the one that can never take negative values.
(c) A scalar quantity is the one that does not vary from one point to another in space.
(d) A scalar quantity has the same value for observers with different orientations of the axes.
Answer:
The answer is the option (d), A scalar quantity has the same value for observers with different orientations of the axes.Question:4.3
Figure 4.1 shows the orientation of two vectors u and v in the XY plane.
If $u=a\hat{i}+b\hat{j}$ and
$v=p\hat{i}+q\hat{j}$
which of the following is correct?
(a) a and p are positive while b and q are negative.
(b) a, p and b are positive while q is negative.
(c) a, q and b are positive while p is negative.
(d) a, b, p and q are all positive.
Answer:
The answer is the option (b) a, p and b are positive while q is negative.Question:4.4
The component of a vector r along X-axis will have maximum value if
(a) r is along positive Y-axis
(b) r is along positive X-axis
(c) r makes an angle of 45° with the X-axis
(d) r is along negative Y-axis
Answer:
The answer is the option (b) r is along positive X-axisQuestion:4.5
The horizontal range of a projectile fired at an angle of $15^{o}$ is 50 m. If it is fired with the same speed at an angle of $45^{o}$, its range will be
a) 60 m
b) 71 m
c) 100 m
d) 141 m
Answer:
The answer is the option (c) 100 mQuestion:4.6
Consider the quantities pressure, power, energy, impulse gravitational potential, electric charge, temperature, area. Out of these, the only vector quantities are
a) impulse, pressure, and area
b) impulse and area
c) area and gravitational potential
d) impulse and pressure
Answer:
The answer is the option (b) impulse and areaQuestion:4.7
In a two dimensional motion, instantaneous speed v0 is a positive constant. Then which of the following are necessarily true?
a) the average velocity is not zero at any time
b) average acceleration must always vanish
c) displacements in equal time intervals are equal
d) equal path lengths are traversed in equal intervals
Answer:
The answer is the option (d) equal path lengths are traversed in equal intervals.Question:4.8
In a two dimensional motion, instantaneous speed $v_{0}$ is a positive constant. Then which of the following are necessarily true?
a) the acceleration of the particle is zero
b) the acceleration of the particle is bounded
c) the acceleration of the particle is necessarily in the plane of motion
d) the particle must be undergoing a uniform circular motion
Answer:
The answer is the option (c) the acceleration of the particle is necessarily in the plane of motionQuestion:4.9
Three vectors A,B and C add up to zero. Find which is false.
(a) (A * B) * C is not zero unless B,C are parallel
(b) (A * B).C is not zero unless B,C are parallel
(c) If A,B,C define a plane, (A * B) *C is in that plane
(d) $(A^{*}B).C=\left | A \right |\left | B \right |\left | C \right |\rightarrow C^{2}=A^{2}+B^{2}$
Answer:
The answer is the option (a) and (c)Question:4.10
It is found that $\left | A+B \right |=\left | A \right |.$ This necessarily implies,
(a)$B=0$
(b) A,B are antiparallel
(c) A,B are perpendicular
(d) $A.B\leq 0$
Answer:
The answer is the option (a) $B=0$Question:4.11
Two particles are projected in air with speed $v_{0}$, at angles $\theta _{1}$ and $\theta _{2}$ to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices
a) angle of project: $q_{1} > q_{2}$
b) time of flight: $T_{1} > T_{2}$
c) horizontal range: $R_{1} > R_{2}$
d) total energy: $U_{1} > U_{2}$
Answer:
The answer is the option (a) and (b)Question:4.12
A particle slides down a frictionless parabolic $(y=x^{2})$ track $(A - B - C)$ starting from rest at point A (Fig. 4.2). Point B is at the vertex of parabola and point C is at a height less than that of point A. After C, the particle moves freely in air as a projectile. If the particle reaches highest point at P, then
(a) KE at P = KE at B
(b) height at P = height at A
(c) total energy at P = total energy at A
(d) time of travel from A to B = time of travel from B to P.
Answer:
The answer is the option (c)Question:4.13
Following are four different relations about displacement, velocity, and acceleration for the motion of a particle in general. Choose the incorrect one(s) :
(a) $V_{av}=\frac{1}{2}\left [ V(t_{1})+V(t_{2}) \right ]$
(b) $V_{av}=\frac{r(t_{2})-r(t_{1})}{t_{2}-t_{1}}$
(c) $r=\frac{1}{2}\left ( V(t_{2})-V(t_{1}) \right )\left ( t_{2}-t_{1} \right )$
(d) $a_{av}=\frac{V(t_{2})-V(t_{1})}{t_{2}-t_{1}}$
Answer:
The answer is the option (a) and (c)Question:4.14
For a particle performing uniform circular motion, choose the correct statement(s) from the following:
(a) Magnitude of particle velocity (speed) remains constant.
(b) Particle velocity remains directed perpendicular to radius vector.
(c) Direction of acceleration keeps changing as particle moves.
(d) Angular momentum is constant in magnitude but direction keeps changing.
Answer:
The answer is the option (a), (b), and (c)Question:4.15
For two vectors A and B, $|A + B| = |A - B|$ is always true when
(a) $\left | A \right |=\left | B \right |\neq 0$
(b) $\left | A \right |\perp \left | B \right |$
(c) $\left | A \right |= \left | B \right |\neq 0$ and A and B are parallel or anti parallel
(d) when either $\left | A \right |$ or $\left | B \right |$ is zero.
Answer:
The answer is the option (b) and (d)Question:4.16
Answer:
To calculate acceleration in a circular motion, we have a formula.Question:4.17
Answer:
A particle travelling in a projectile motion will definitely have a path which is parabolic in nature. The velocity of such a particle is always measured tangential to the path of motion. Point B is where the particle reaches maximum height during the projectile motion.Question:4.18
A ball is thrown from a roof top at an angle of $45^{o}$ above the horizontal. It hits the ground a few seconds later. At what point during its motion, does the ball have
a) greatest speed
b) smallest speed
c) greatest acceleration?
Answer:
Question:4.19
A football is kicked into the air vertically upwards. What is its
a) acceleration
b) velocity at the highest point
Answer:
a) When the football is kicked into the air, the only force working on it is the gravitational force. Hence the acceleration will be caused due to this force which will be in the downward direction towards the centre of the Earth.Question:4.20
Answer:
Given: A, B, and C are non-collinear and no co-planar vectors.Question:4.21
Answer:
Let v be the vertical velocity, and u be the horizontal velocity of the car.Question:4.22
Answer:
Given : $u=36\; km/h=10\; m/s$Question:4.23
Answer:
Question:4.23
Answer:
Question:4.24
Answer:
Question:4.25
a) Earth can be thought of as a sphere of radius 6400 km. Any object is performing circular motion around the axis of earth due to earth’s rotation. What is the acceleration of object on the surface of the earth towards its centre? What is it at latitude $\theta$? How does these accelerations compare with $g=9.8\; m/s^{2}?$
b) Earth also moves in circular orbit around sun once every year with an orbital radius of $1.5\times 10^{11}\; m.$ What is the acceleration of earth towards the centre of the sun? How does this acceleration compare with $g=9.8\; m/s^{2}?$
$\left ( Hint: acceleration\frac{V^{2}}{R}=\frac{4\pi ^{2}R}{T^{2}} \right )$
Answer:
The angular acceleration will have its direction towards the centre and the value will be $a=w^{2}R$Question:4.26
Column I | Column II |
a) a+b=c | |
5b) a-c=b | |
c) b - a = c | |
d) a+b+c=0 |
Answer:
a) $\vec{a}+\vec{b}=\vec{c}$ matches with (iv)Question:4.27
If $\left | A \right |=2$ and $\left | B \right |=4$, then match the relations in column I with the angle $\theta$ between A and B in column II.
Column I Column II
(a) $A.B=0$ (i) $\theta =0$
(b) $A.B=+8$ (ii) $\theta =90^{o}$
(c) $A.B=4$ (iii) $\theta =180^{o}$
(d) $A.B=-8$ (iv) $\theta =60^{o}$
Answer:
a) Matches with (ii)Question:4.28
If $\left | A \right |=2$ and $\left | B \right |=4$ then match the relations in column I with the angle $\theta$ between A and B in column II.
Column I Column II
(a)$\left | A\times B \right |=0$ (i) $\theta =30^{o}$
(b) $\left | A\times B \right |=8$ (ii) $\theta =45^{o}$
(c) $\left | A\times B \right |=4$ (iii) $\theta =90^{o}$
(d) $\left | A\times B \right |=4\sqrt{2}$ (iv) $\theta =0^{o}$
Answer:
Given : $\left | \vec{A} \right |=2$ and $\left | \vec{B} \right |=4$Question:4.29
Answer:
Question:4.3
A gun can fire shells with maximum speed $v_{0}$ and the maximum horizontal range that can be achieved is $R=v{_{0}}^{2}/g$
If a target farther away by $\Delta x$ beyond R has to be hit with the same gun as shown in the figure. Show that it could be achieved by rating the gun to a height at least
$h=\Delta x[1+\frac{\Delta x}{R}]$
Answer:
Question:4.31
Answer:
Question:4.32
Answer:
The particle rebounds from P. when it strikes plane inclined at v0 speed. Hence the speed of particle after it rebounds from P will be v0 We assume the new axis X’OX and YOY’ axis at P as origin ‘O’. The components of g and v0 in the new OX and OY axis are:Question:4.33
Answer:
Let the north direction be i and south direction be jQuestion:4.34
A river is flowing due east with a speed 3m/s. A swimmer can swim in still water at a speed of 4 m/s (Fig. 4.8).
(a) If swimmer starts swimming due north, what will be his resultant velocity (magnitude and direction)?
(b) If he wants to start from point A on south bank and reach opposite point B on north bank,
(a) which direction should he swim?
(b) what will be his resultant speed?
(c) From two different cases as mentioned in (a) and (b) above, in which case will he reach opposite bank in shorter time?
Answer:
a) If swimmer starts swimming due north, what will be his resultant velocityb) The swimmer wants to start from point A on the south bank and reaches the opposite point B on the north bank
The swimmer makes an angle $\theta$ with the north.
From the figure we have the relation,
$V^{2}=v{_{s}}^{2}-v{_{r}}^{2}=16-9=7$
Henece$v=\sqrt{7}\; m/s$
Now we calculate the value of θ through the below formula,
$\tan \theta =\frac{v_{r}}{v}=\frac{3\sqrt{7}}{7}=1.13$
So, $\theta =48^{o}29'30''$ in th edirection from North to West
c) we need to find from the above two scenarios that for the swimmer to reach the opposite bank in the shorter time
we know that the velocity component perpendicular to the river is 4m/s
let us assume the width of the river to be ‘w’
Time taken - North
$\frac{w}{4}=t1$
time taken in part b) when $v=\sqrt{7}\; m/s$
$\frac{w}{\sqrt{7}}=t2$
taking ratio,
$\frac{t1}{t2}=\frac{(\frac{w}{4})}{(\frac{w}{\sqrt{7}})}$
$4\; t1=\sqrt{7}\; t2$
Now as, $4>\sqrt{7}$
$t1<t2$
So, the swimmer will take a shorter time in case a)
Question:4.35
A cricket fielder can throw the cricket ball with a speed vo. If he throws the ball while running with speed u at an angle $\theta$ to the horizontal, find
a) the effective angle to the horizontal at which the ball is projected in air as seen by a spectator
b) what will be time of flight?
c) what is the distance from the point of projection at which the ball will land?
d) find $\theta$ at which he should throw the ball that would maximise the horizontal range as found c)
e) how does $\theta$ for maximum range change if $u>v_{o}=v_{o}$ and $u<v_{o}$
f) how does $\theta$ in e) compare with that for u = 0?
Answer:
a) u is the horizontal velocity with which the cricketer runs. The ball is thrown by him while running and hence the speed of ball also contains a component of the cricketer’s speed.Question:4.36
Motion in two dimensions, in a plane can be studied by expressing position, velocity and acceleration as vectors in Cartesian co-ordinates $A=A_{x}\hat{i}+A_{y}\hat{j}$ where $\hat{i}$ and $\hat{j}$ are unit vector long x and y directions, respectively and $A_{x}$ and $A_{y}$ are corresponding components of A (fig.4.9). Motion can also be studied by expressing vectors in circular polar co-ordinates as $A=A_{r}\hat{r}+A_{\theta }\hat{\theta }$ where $\hat{r}=\frac{r}{r}=\cos \theta \hat{i}+\sin \theta \hat{j}$ and $\hat{\theta }=-\sin \theta \hat{i}+\cos \theta \hat{j}$ are unit vectors along direction in which 'r' and $'\theta '$ are increasing.
(a) Express $\hat{i}$ and $\hat{j}$ in terms of $\hat{r}$ and $\hat{\theta }$
(b) Show that both $\hat{r}$ and $\hat{\theta }$ are unit vectors and are perpendicular to each other.
(c) Show that $\frac{d}{dt}(\hat{r})=\omega \hat{\theta }$ where $\omega =\frac{d\theta }{dt}$ and $\frac{d}{dt}(\theta )=-\omega \hat{r}$
(d) For particle moving along a spiral given by $r=a\theta \hat{r},$ where $a=1$ (unit), find dimensions of 'a'.
(e) Find velocity and acceleration in polar vector represention for particle moving along spiral described in (d) above.
Answer:
a) $\hat{r}=\cos\; \theta \hat{i}+\sin\; \theta \hat{J}\; \; \; \; \; \; -------(1)$Question:4.37
Answer:
The principles and terminology relevant to motion in a plane are explained in NCERT Exemplar Class 11 Physics Chapter 4, which focuses on the concepts covered in the previous chapter on motion in a straight line. In order to carry out functions like addition, subtraction, multiplication, and vector resolution, it explores a variety of mathematical approaches, including theoretical and graphical approaches.
This chapter additionally explores the properties of dimensions and how they are used to characterize two- or three-dimensional object motion, including uniform circular motion and constant acceleration.
Furthermore, Chapter 4 of NCERT Exemplar Class 11 Physics addresses the diverse uses of projectile motion in anticipating the future location and trajectory of moving objects. Additionally, it highlights how crucial uniform circular motion is. By focusing on the ideas of relative velocity, Chapter 3 answers offer a broader perspective and improve understanding of motion in general.
In Motion in a Plane, the key difference between scalar and vector quantities is:
Scalar Quantities: These have only magnitude and no direction. Examples include speed, distance, time, and mass.
Vector Quantities: These have both magnitude and direction. Examples include velocity, displacement, acceleration, and force.
Relative velocity in two dimensions is the velocity of one object relative to another, calculated by subtracting their velocity vectors. It involves breaking down velocity into horizontal and vertical components.
Uniform circular motion is accelerated because the direction of velocity constantly changes, resulting in centripetal acceleration towards the center, even though speed remains constant.
Tricky problems in Chapter 4 include:
Projectile motion: Finding range, height, or flight time.
Relative velocity: Solving with objects moving in different directions.
Circular motion: Calculating forces, tension, or angular velocity.
Acceleration in a plane: Resolving forces in 2D/3D motion.
Real-life applications include:
Projectile motion: Sports (e.g., throwing a ball, shooting a basketball).
Relative velocity: Navigation (e.g., airplanes, trains).
Circular motion: Satellites orbiting Earth, car turns on a circular track.
Acceleration: Motion of vehicles, falling objects under gravity.
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