NCERT Exemplar Class 11 Physics Solutions Chapter 4 Motion in a Plane

NCERT Exemplar Class 11 Physics Solutions Chapter 4: MCQI

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), ${90}^o$
Explanation :
We know from the formula that $A.B=\Vert A\Vert \Vert B\Vert \cos \theta$
Given that $\overrightarrow{A}=\hat{\iota }+\hat{J}$ and $\overrightarrow{B}=\hat{\iota }-\hat{J}$
$\cos \theta =\frac{(\hat{\iota }+\hat{J})(\hat{\iota }-\hat{J})}{\sqrt{1^2+1^2}(\sqrt{1^2+(-1^2)})}$
$=0=\cos 90$
Hence the angle is ${90}^o$

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.
Explanation:
Unlike vector quantities, a scalar quantity has no dependence on the direction, and hence its value remains the same for all the 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.
Explanation:
Now, $\overrightarrow{u}=a\hat{\iota }+b\hat{J}$ and $\overrightarrow{v}=p\hat{\iota }+q\hat{J}$
For vector u, both the components a and b are in the positive direction of the x-axis. Hence, they are positive.
For vector v, p is in the positive direction, but q is in the opposite direction. Hence, p is positive, and 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-axis
Explanation: if a vector B forms an angle $\theta$ along the x-axis. Then its components along the x-axis can be written as B Cos θ.
The maximum value of $\cos \theta =1$, which is when $\theta =0$.
Hence, for the vector r, the maximum value will be when it is along the positive x-axis.

Question: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 m
Explanation:
According to the formula:
$R=\frac{u^2\sin 2\theta }{g}$
Given in the question:
$\theta =15,R=50m$
Putting in the formula we get:
$u^2=100g$
For $\theta =45,$
The value of the range is:
$R=\frac{100g\times \sin 90}{g}=100m$

Question: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 area
Explanation:
Impulse is a vector quantity and not a scalar quantity as the impulse is the rate of change of momentum, and it involves direction.
The area is a vector quantity

Question: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.
Explanation: when instantaneous speed is positive and constant since speed is a scalar quantity equal paths will be covered in an equal amount of time.

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 motion
Explanation: In a two-dimensional motion, instantaneous speed is positive, and constant is given. Since velocity is constant, acceleration which is the rate of change in velocity is also constant. Hence, the acceleration will be in the plane of motion.

Question: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) $(A^{\ast }B).C=\left|A\right|\left|B\right|\left|C\right|\to C^2=A^2+B^2$

Answer:

The answer is the option (a) and (c)
Explanation:
Given $\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C}=0$
Option a : $\overrightarrow{B}\times (\overrightarrow{A}+\overrightarrow{B}+\overrightarrow{C})=\overrightarrow{B}\times 0=0$
$0=\overrightarrow{B}\times \overrightarrow{A}+\overrightarrow{B}\times \overrightarrow{B}+\overrightarrow{B}\times \overrightarrow{C}$
$0=\overrightarrow{B}\times \overrightarrow{A}+0+\overrightarrow{B}\times \overrightarrow{C}$
$\overrightarrow{A}\times \overrightarrow{B}=\overrightarrow{B}\times \overrightarrow{C}$(this cannot be zero)
Only if B and C are antiparallel or parallel $\overrightarrow{B}\times \overrightarrow{C}$ will be zero
Hence for the whole quantity to be zero, $\overrightarrow{B}\parallel \overrightarrow{C}$ should be true.
Option C : $\overrightarrow{A}\times \overrightarrow{B}=\overrightarrow{X}$
X is perpendicular to the planes which have vector A and vector B
Vector Y is perpendicular to the planes which have vector A and vector B
Vector Y is perpendicular to the plane containing X and C which is, in turn, the plane containing vectors A, B and C

Question:4.10

It is found that $|A+B|=\left|A\right|.$ This necessarily implies,
(a)$B=0$
(b) A,B are antiparallel
(c) A,B are perpendicular
(d) $A.B\le 0$

Answer:

The answer is the option (a) $B=0$
Explanation :
Squaring both sides and opening the brackets we get,
${\left|\overrightarrow{A}\right|}^2+{\left|\overrightarrow{B}\right|}^2+2\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \theta ={\left|\overrightarrow{A}\right|}^2$
${\left|\overrightarrow{B}\right|}^2+2\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \theta =0$
$cos\theta =\frac{-B}{2A}$
$ifB=0$
$cos\theta =0$

NCERT Exemplar Class 11 Physics Solutions Chapter 4: MCQII

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)
Explanation:
According to formula. Max height of a projectile is
$H=\frac{u^2{\sin }^2\theta }{2g}$
Option a :$H1>H2$
${\sin }^2\theta 1>{\sin }^2\theta 2$
$(\sin \theta 1+\sin \theta 2)(\sin \theta 1-\sin \theta 2)>0$
So, $(\sin \theta 1+\sin \theta 2)>0or(\sin \theta 1-\sin \theta 2)>0$
So, $\theta 1>\theta 2$ and $\theta$ lies etween 0 and 90 degree i.e. acute
Option b :
$\frac{T1}{T2}=\frac{\sin \theta 1}{\sin \theta 2}$
$T1\sin \theta 2=T2\sin \theta 1$
Since $\sin \theta 1>\sin \theta 2$
$T1>T2$

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)
Explanation: total energy at a point in the path remains the same as there is no friction between air and plane on paths A, B and C
As we know through the energy conservation law, total energy at point P will be same as total energy at point A.
Hence, option c is correct.

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) $V_{av}=\frac{1}{2}[V(t_1)+V(t_2)]$
(b) $V_{av}=\frac{r(t_2)-r(t_1)}{t_2-t_1}$
(c) $r=\frac{1}{2}(V(t_2)-V(t_1))(t_2-t_1)$
(d) $a_{av}=\frac{V(t_2)-V(t_1)}{t_2-t_1}$

Answer:

The answer is the option (a) and (c)
Explanation:
Option a:
The given relation is correct when the acceleration is uniform
Option c:
$\overrightarrow{r}=\frac{1}{2}(\overrightarrow{v}(t_2)-\overrightarrow{v}(t_1))÷(t_2-t_1)$
This is the relationship given in the question, but it is not possible as the LHS and RHS dimensions $\left[M^0{L}^1{T}^0\right]$ and $\left[M^0{L}^1{T}^2\right]$do not match and hence the relationship cannot be considered valid.

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)
Explanation:
Option a: speed is constant at all times in the case of uniform circular motion
Option b: in case of velocity, in a circular motion it is measured tangentially to the direction of motion of the particle which is in turn perpendicular to the radius.
Option c: the direction of the acceleration is in the direction of force always. This can concur with the newton’s second law of motion. So as the particle moves in the circular motion, the direction of force keeps on changing and hence that of acceleration also changes.

Question:4.15

For two vectors A and B, $|A+B|=|A-B|$ is always true when
(a) $\left|A\right|=\left|B\right|\ne 0$
(b) $\left|A\right|\perp \left|B\right|$
(c) $\left|A\right|=\left|B\right|\ne 0$ 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)
Explanation:
Given : $|\overrightarrow{A}+\overrightarrow{B}|=|\overrightarrow{A}-\overrightarrow{B}|$
Now we square both sides and open the brackets to get:
${\left|\overrightarrow{A}\right|}^2+{\left|\overrightarrow{B}\right|}^2+2\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \theta ={\left|\overrightarrow{A}\right|}^2+{\left|\overrightarrow{B}\right|}^2-2\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \theta$
$4\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \theta =0$
So, we have, $\theta =90$ degree and $\left|\overrightarrow{A}\right|=\left|\overrightarrow{B}\right|=0$
Hence option b and option d are correct.

NCERT Exemplar Class 11 Physics Solutions Chapter 4: Very Short Answer

Question:4.16

A cyclist starts from centre O of a circular park of radius 1km and moves along the path OPRQO as shown Fig. 4.3. If he maintains constant speed of 10 ms–1, what is his acceleration at point R in magnitude and direction?

Answer:

To calculate acceleration in a circular motion, we have a formula.
$A=\frac{v^2}{R}$
Given, $R=1000m,v=10m/s$
Hence, $A=10\times \frac{10}{1000}=\frac{1}{10}=0.1m/s^2$

Question:4.17

A particle is projected in air at some angle to the horizontal, moves along parabola as shown in the figure, where x and y indicate horizontal and vertical directions respectively. Show in the diagram, direction of velocity and acceleration at points A, B, and C.

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.

Here we have the vertical and horizontal components of B which are $V_y=0$ and $V_x=u\cos \theta$

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:

The ball is thrown from point O. The angle with the horizontal is 45 degree. As the height increases from O to A, the speed decreases. At B, the speed becomes equal to its initial speed. The height decreases further from B to C and becomes maximum at point C.
$V_x=u\cos 45=\frac{u}{\sqrt{2}}m/s$
a) greatest speed of the ball is at c as $V_y$ and $V_x=u\cos 45=\frac{u}{\sqrt{2}}m/s$
b) slowest speed will be at point A. here $V_y=0$, and the horizontal speed is $V_x=\frac{u}{\sqrt{2}}$
c) the greatest acceleration will be the one caused by the gravitational force downward, which will be equal to g.

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.
b) When the ball is thrown upward, no horizontal forces act on it and hence the horizontal component velocity becomes nil. The vertical component of velocity at the highest point is $V_y=0$. Hence, the net velocity when the ball reaches the highest point is also zero.

Question:4.20

A, B, and C are three non-collinear, non co-planar vectors. What can you say about direction of $A(B\times C)$?

Answer:

Given: A, B, and C are non-collinear and no co-planar vectors.
For finding out the direction, we used the right-hand grip rule.
For $(\overrightarrow{B}\times \overrightarrow{C})$, the direction will be in a plane perpendicular to the plane containing $\overrightarrow{B}$ and $\overrightarrow{C}$.
However, for the direction of $\overrightarrow{A}\times (\overrightarrow{B}\times \overrightarrow{C}):$ it will be perpendicular to both $\overrightarrow{A}$ and the plane which has $(\overrightarrow{B}\times \overrightarrow{C}).$

NCERT Exemplar Class 11 Physics Solutions Chapter 4: Short Answer

Question:4.21

A boy travelling in an open car moving on a labelled road with constant speed tosses a ball vertically up in the air and catches it back. Sketch the motion of the ball as observed by a boy standing on the footpath. Give explanation to support your diagram.

Answer:

Let v be the vertical velocity, and u be the horizontal velocity of the car.

In the case of the ball, as it has both horizontal and vertical components of velocity, it will have a parabolic path when a person standing on a footpath sees the trajectory.

When a boy sitting in the same car sees the ball, the path will vertical up and down. However, he will be able to catch up with the ball provided the car moves at a constant velocity.

Question:4.22

A boy throws a ball in air at ${60}^o$ to the horizontal along a road with a speed of 10 m/s. Another boy sitting in a passing by car observes the ball. Sketch the motion of the ball as observed by the bot in the car, if car has a speed of 18 km/h. Give explanation to support your diagram.

Answer:

Given : $u=36km/h=10m/s$

$u_x=u\cos 60=5m/s$
Now, in the direction of the ball, the speed of the car is $18\times \frac{5}{18}=5m/s$
since the horizontal speeds are the same, in the situation of a ball thrown by the boy as soon as the car passes him, they also cover equal horizontal distances.

Now, when vertical components are considered, $u_y=u\cos 30=5\sqrt{3}m/s$ and the motion of the ball is vertically up-down as seen by the boy.

Question:4.23

In dealing with motion of projectile in air, we ignore effect of air resistance on motion. This gives trajectory as a parabola as you have studied. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.

Answer:

When we are dealing with projectile motion generally we neglect the air resistance. But if air resistance is included the horizontal component of velocity will not be constant and obviously trajectory will change.
Due to air resistance, particle energy, as well as horizontal component of velocity, keep on decreasing making the fall steeper than rise as shown in the figure.
When we are neglecting air resistance path was symmetric parabola (OAC). When air resistance is considered path is asymmetric parabola (OAB).

Question:4.23

In dealing with motion of projectile in air, we ignore effect of air resistance on motion. This gives trajectory as a parabola as you have studied. What would the trajectory look like if air resistance is included? Sketch such a trajectory and explain why you have drawn it that way.

Answer:

The horizontal and vertical components of the velocity decrease due to air resistance. So, as a result, the max height also becomes lesser than in the ideal case.
Now, $R=u^2\sin 2\theta /g$ and max height $H=u^2{\sin }^2\theta /2g$
So, $h1<h2andR1<R2$
But, in the second case when $h1<h2$, due to smaller time taken to rise, the overall time of flight for both cases stands equal

Question:4.24

A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km/h. At what angle of sight when the target is seen, should the pilot drop the bomb in order to attack the target?

Answer:

$u=720km/h=200m/s$
we assume that the bomb is dropped by the pilot t seconds vertically above Q before the target T
vertical component velocity of the bomb will be zero and the horizontal component value will be equal to that of the plane. So, the bomb covers distance TQ as a free fall.
Now, $u=0,H=1.5,g=10$
So, $H=ut+\frac{1}{2}g{t}^2=0+\frac{1}{2}10\times t\times t=1500$
So, we get, $t=10\sqrt{3}s$
ut is the distance covered by plane or the bomb equal to PQ
hence, $PQ=ut=200\sqrt{3}m$
now, $\tan \theta =\frac{TQ}{PQ}=\sqrt{\frac{3}{4}}={23}^o{42}^{\prime }$

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 $g=9.8m/s^2?$
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 $g=9.8m/s^2?$
$(Hint:acceleration\frac{V^2}{R}=\frac{4{\pi }^{2}R}{T^2})$

Answer:

The angular acceleration will have its direction towards the centre and the value will be $a=w^{2}R$
Now, $w=\frac{2\pi }{T}$
$R=6.4\times {10}^{6}m$ and $T=24\times 3600=86400s$
Acceleration of the person on the surface of the earth,
$a=w^{2}R={\left(\frac{2\pi }{T}\right)}^2\times R$
$=\frac{4\times 22\times 22\times 6.4\times {10}^6}{7\times 7\times 24\times 24\times 3600\times 3600}=\frac{1210}{81\times 49\times 9}=0.034m/s^2$
Now, At equator latitude = 0
Hence $\frac{a}{g}=\frac{0.034}{9.8}=\frac{1}{288}=0.0034$ which is much smaller than the previous case.
b) acceleration of Earth revolving around the sun
$R=1.5\times {10}^{11}m$ and $w=\frac{2\pi }{T}$
$T=365\times 24\times 3600=3.15\times {10}^{7}seconds$
Now, $a=w^{2}R=(\frac{2\pi }{T}{)}^2\times R$
$=\frac{4\times 3.14\times 3.14\times 1.5\times {10}^{11}}{3.15\times 3.15\times {10}^7\times {10}^7}=6\times {10}^{-3}m/s^2$
Also, $\frac{a}{g}=\frac{0.0006}{9.8}$

Question:4.26

Given below in column I are the relations between vectors a, b, and c and in column II are the orientations of a, b, and c in the XY plane. Match the relation in column I to correct orientations in column II.

Answer:

a) $\overrightarrow{a}+\overrightarrow{b}=\overrightarrow{c}$ matches with (iv)
b) $\overrightarrow{c}+\overrightarrow{b}=\overrightarrow{a}$ matches with (iii)
c) $\overrightarrow{a}+\overrightarrow{c}=\overrightarrow{b}$ matches with (i)
d) $\overrightarrow{a}+\overrightarrow{c}+\overrightarrow{b}=0$ matches with (ii)

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)
$\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \theta =0$
$\cos \theta =0$
$\theta =90$
b) Matches with (i)
$\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \theta =8$
$2\times 4\cos \theta =8$
$\cos \theta =1$
$\theta =0$
c) Matches with (iv)
$\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \theta =4$
$2\times 4\cos \theta =84$
$\cos \theta =\frac{1}{2}$
$\theta =60$
d) Matches with (iii)
$\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\cos \theta =-8$
$2\times 4\cos \theta =-8$
$\cos \theta =-1$
$\theta =180$

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)$|A\times B|=0$ (i) $\theta ={30}^o$
(b) $|A\times B|=8$ (ii) $\theta ={45}^o$
(c) $|A\times B|=4$ (iii) $\theta ={90}^o$
(d) $|A\times B|=4\sqrt{2}$ (iv) $\theta =0^o$

Answer:

Given : $\left|\overrightarrow{A}\right|=2$ and $\left|\overrightarrow{B}\right|=4$
a) Matches with (iv)
$|\overrightarrow{A}\times \overrightarrow{B}|=0$
$\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin \theta =0$
$2\times 4\sin \theta =0$
$\sin \theta =\sin 0$
$\theta =0$
b) matches with (iii)
$|\overrightarrow{A}\times \overrightarrow{B}|=8$
$\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin \theta =8$
$2\times 4\sin \theta =8$
$\sin \theta =1$
$\theta =90$
c) matches with (i)
$|\overrightarrow{A}\times \overrightarrow{B}|=4$
$\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin \theta =4$
$2\times 4\sin \theta =4$
$\sin \theta =\frac{1}{2}$
$\theta =30$
d) matches with (ii)
$|\overrightarrow{A}\times \overrightarrow{B}|=4\sqrt{2}$
$\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\sin \theta =4\sqrt{2}$
$2\times 4\sin \theta =4\sqrt{2}$
$\sin \theta =\frac{1}{\sqrt{2}}$
$\theta =45$

NCERT Exemplar Class 11 Physics Solutions Chapter 4: Long Answer

Question:4.29

A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800 m from the foot of hill and can be moved on the ground at a speed of 2 m/s so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill? Take g = 10 m/s².

Answer:

Packet speed = 125 m/s, height of hill = 500m
In order to cross the hill, the vertical component of the packet should be reduced to make the height of 500m attainable. The distance between canon and hill should also be half of that of the packet’s range.
$V^2-u^2=2gh$
$U_y=\sqrt{2gh}=\sqrt{10000}=100m/s$
Now, $U=u{{}_x}^2+u{{}_y}^2$
$u{{}_x}^2=125\times 125-(100\times 100)$
$u_x=75m/s$
now we consider the packet's vertical motion,
$v_y=u_y+gt$
t= total time of flight = 10 sec
so, $v_y=75\times 10=750$
so the distance between canon and hill is 750m
distance for which the canon needs to move$=800-750=50m$
time taken for the canon to move $50m=\frac{50}{2}=25sec$
so the total time taken by the packet $=25+10+10=45seconds$

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 $\mathrm{\Delta }x$ 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
$h=\mathrm{\Delta }x[1+\frac{\mathrm{\Delta }x}{R}]$

Answer:

The solution to this problem can be given in two ways:
i) the target is present at the horizontal distance of $(R+\mathrm{\Delta }x)$ and is h meters below the projection point. $Y=-h$
ii) the motion of projectile starting from point P till reaching point T. vertical height covered is -h and horizontal range is $\mathrm{\Delta }x$
max range of a projectile is given by $R=\frac{v^2}{g}$
here, $\theta =45$
we assume that the gun is raised to a height h to hit the target T
total range $=(R+\mathrm{\Delta }x)$
the horizontal component of velocity is, $v\cos \theta$
horizontal velocity at $P=V_x=-V\cos \theta$
vertical velocity at $P=V_y=V\sin \theta$
now, $h=ut+\frac{1}{2}a{t}^2$
So, $h=-V\sin \theta t+\frac{1}{2}g{t}^2---------------(1)$
$V\cos \theta .t=(R+\mathrm{\Delta }x)$
$t=\frac{(R+\mathrm{\Delta }x)}{V\cos \theta }$
Substituting the value of t in (1) we get,
$h=-V\sin \theta \left(\frac{(R+\mathrm{\Delta }x)}{V\cos \theta }\right)+\frac{1}{2}g{\left(\frac{(R+\mathrm{\Delta }x)}{V\cos \theta }\right)}^2$
$h=-(R+\mathrm{\Delta }x)+\frac{1}{R}(R^2+\mathrm{\Delta }{x}^2+2R\mathrm{\Delta }x)=-R-\mathrm{\Delta }x+R+\frac{\mathrm{\Delta }{x}^2}{R}+2\mathrm{\Delta }x$
$h=\mathrm{\Delta }x[1+\frac{\mathrm{\Delta }x}{R}]$
hence proved.

Question:4.31

A particle is projected in air at an angle $\beta$ to a surface which itself is inclined at an angle $\alpha$ to the horizontal.
a) find an expression of range on the plane surface
b) time of flight
c) $\beta$ at which range will be maximum

Answer:

a) expression of range on the plane surface
Now, $a_y=-g\cos \alpha$ and $a_x=g\sin \alpha$
$Y=0$ at O and P,
So
$U_y=v\sin \beta$ where t = T
We calculate Time of Flight Part (b) before Part (a)
b) motion of particle along Y axis
$s=ut+\frac{1}{2}g{t}^2$
$s=0,u=v\sin \beta ,g=a=-g\cos \alpha ,t=T$
$0=v\sin \beta .T+\frac{1}{2}(-g\cos \alpha )T^2$
$T[v\sin \beta -T.\frac{g}{2}\cos \alpha ]=0$
$\frac{Tg}{2}.\cos \alpha =v\sin \beta$
So, $T=2v\sin \beta /g.\cos \alpha$
a) Now we continue the part a
$T=2v\sin \beta /g.\cos \alpha$
$s=ut+\frac{1}{2}g{t}^2$
$L=v\cos \beta (T)+\frac{1}{2}(-g\sin \alpha )T^2$
$L=2v^2\sin \beta [\cos \beta \cos \alpha -\sin \beta \sin \alpha ]/g{\cos }^2\alpha$
$L=2v^2\sin \beta \times \cos [\alpha +\beta ]/g{\cos }^2\alpha$
c) on the axis X, L will be maximum when $\sin \beta \cos [\alpha +\beta ]$ will be maximum
let $z=\sin \beta \cos [\alpha +\beta ]$
$=\sin \beta [\cos \beta \cos \alpha -\sin \beta \sin \alpha ]$
$=\frac{1}{2}[\cos \alpha 2\sin \beta +\sin \alpha \cos 2\beta -\sin \alpha ]$
$=\frac{1}{2}[\sin (2\beta +\alpha )-\sin \alpha ]$
In order to make Z maximum, we put $[\sin (2\beta +\alpha )-\sin \alpha ]=1$
Opening the brackets, we get $(2\beta +\alpha )=90$
$2\beta =90-\alpha$
$\beta =45-\frac{\alpha }{2}$radian

Question:4.32

A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle $\theta$ with speed $v_o$ and rebounds elastically (Fig 4.7). Find the distance along the plane where it will hit second time.

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:

Focusing on the motion of the particle from O to A,
$s=ut+\frac{1}{2}g{t}^2$
Here, t=T which is the time of flight
$0=T[v\cos \theta -\frac{1}{2}g\sin \theta T]$
So, either $T=0or[v\cos \theta -\frac{1}{2}g\sin \theta T]=0$
$S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$
$L=[\frac{2v}{g}]v\sin \theta +\frac{1}{2}g\sin \theta [\frac{2v}{g}{]}^2$
$L=\frac{4v^2}{g}.\sin \theta$

Question:4.33

A girl riding a bicycle with a speed of 5 m/s towards north direction, observes rain falling vertically down. If she increases her speed to 10 m/s, rain appears to meet her at ${45}^o$ to the vertical. What is the speed of the rain? In what direction does rain fall as observed by a ground based observer?

Answer:

Let the north direction be i and south direction be j
The velocity of rain is $v=a\hat{i}+b\hat{J}$
Case 1 (v=5i)

The velocity of rain with respect to the girl is:
$v_r-v_g=(a\hat{i}+b\hat{j})-5\hat{i}=(a-5)\hat{i}+b\hat{j}$
Since the horizontal component is zero, $a-5=0ora=5$
Case 2 (v=10i)

$v_r-v_g=(a\hat{i}+b\hat{j})-10\hat{i}=(a-10)\hat{i}+b\hat{j}$
The angle of rain appears to be 45 degrees.
$\tan 45=\frac{b}{a}=\frac{b}{-5}$
$b=-5$
So,
$\left|v_r\right|=\sqrt{5^2+(-5{)}^2}=\sqrt{50}=5\sqrt{2}m/s$

Question: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 velocity

$V_s=4m/s$ due north
$V_r=4m/s$ due east
Now since both the directions are perpendicular,
${\left|V_r\right|}^2=4^2+3^2=5m/s$
$\tan \theta =\frac{V_r}{V_s}=0.75={36}^o{54}^{\prime }$ in the North direction

b) 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, $\theta ={48}^o{29}^{\prime }{30}^{″}$ 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,
$\frac{t1}{t2}=\frac{(\frac{w}{4})}{(\frac{w}{\sqrt{7}})}$
$4t1=\sqrt{7}t2$
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.

$U_x=u+v\cos \theta$
Vertical component,
$U_y=v\sin \theta$
$\tan \theta =\frac{v\sin \theta }{u+v\cos \theta }$
$\theta ={\tan }^{-1}[\frac{v\sin \theta }{u+v\cos \theta }]$
b) Time of flight
$S_y=U_{y}t+\frac{1}{2}{a}_y{t}^2$
Since the ball returns back to the same position, Sy = 0
$U_y=v\sin \theta$
So,$0=v\sin \theta (T)-\frac{1}{2}g{T}^2$
$T[v\sin \theta -\frac{1}{2}gT]=0$
Since T cannot be zero, we have
$T=2v\sin \frac{\theta }{g}$
c) Maximum range
for the max range, the condition is
$\frac{dR}{d\theta }=0$
$d\frac{\{\frac{v}{g}[2u]\sin \theta +\frac{v}{g}\sin 2\theta \}}{d\theta }=0$
$\theta ={\cos }^{-1}\left[\frac{-u\pm \sqrt{u^2+8v_0^2}}{4v_0}\right]$
$\cos \theta =\left[\frac{-u\pm \sqrt{u^2+8v_0^2}}{4v_0}\right]$
e) In the case when $u=v,\cos \theta =$
$\frac{-v_0\pm \sqrt{v_0^2+8v_0^2}}{4v_0}$
$=\frac{-v_0+3v_0}{4v_0}$
$\cos \theta =-1+\frac{(-3)}{4}$
$\cos \theta =\frac{1}{2}$ (as $\theta$ is taken as an acute angle here)
hence, $\theta ={60}^o$
for the case of u << v
$\cos \theta =-u+\frac{(-2\sqrt{2}v)}{4v}$
Since $\theta$ is an acute angle here,
as u << v here, we can neglect the last term
$\cos \theta =\frac{1}{\sqrt{2}}$
$\theta =\frac{\pi }{4}$
For u >> v,
$\cos \theta =\frac{-u+(-u)}{4v}$
$\cos \theta =0=\cos 90$
$\theta =\frac{\pi }{2}$
f) when $u=0$
$\cos \theta =\frac{-u\pm \sqrt{u^2+8v_0^2}}{4v_0}$
$\cos \theta =\frac{2\sqrt{2}v}{4v}=\frac{1}{\sqrt{2}}$
$\cos \theta =45$
$\cos \theta =\frac{\pi }{4}$

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 ${}^{\prime }{\theta }^{\prime }$ 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) $\hat{r}=\cos \theta \hat{i}+\sin \theta \hat{J}-------(1)$
$\hat{\theta }=-\sin \theta \hat{i}+\cos \theta \hat{J}----------(2)$
When we multiply 1 by $\sin \theta$ and 2 by $\cos \theta$, we get :
$\hat{r}\sin \theta +\hat{\theta }\cos \theta =({\sin }^2\theta +{\cos }^2\theta )\hat{J}$
$\hat{r}\sin \theta +\hat{\theta }\cos \theta =\hat{J}---------(3)$
When we multiply 1 by $\cos \theta$ and 2 by $\sin \theta$ we get :
$\hat{r}\cos \theta ={\cos }^2\theta \hat{i}+\sin \theta \cos \theta \hat{J}-----(4)$
$\hat{\theta }\sin \theta =-{\sin }^2\theta \hat{i}+\sin \theta \cos \theta \hat{J}-----(5)$
Subtracting equation 5 from 4 and comparing the coefficients we get,
$\hat{r}\cos \theta -\hat{\theta }\sin \theta =\hat{i}$
$\hat{r}\sin \theta -\hat{\theta }\cos \theta =\hat{J}$
b) from equation 1 and 2, through the dot product method we get,
$\hat{r}.\hat{\theta }=(\cos \theta \hat{i}+\sin \theta \hat{J}).(-\sin \theta \hat{i}+\cos \theta \hat{J})$
$\left|\hat{r}\right|\left|\hat{\theta }\right|\cos \theta =0$
Since LHS elements cannot be zero,
$\cos \theta =0$
and $\theta =90$
c) $\hat{r}=\cos \theta \hat{i}+\sin \theta \hat{J}$
$\frac{d\hat{r}}{dt}=\frac{d(\cos \theta \hat{i}+\sin \theta \hat{J})}{dt}$
$=(-\sin \theta \hat{i}+\cos \theta \hat{J})\frac{d\theta }{dt}$
Since, $\omega =\frac{d\theta }{dt}$
$\frac{d\hat{r}}{dt}=\omega \hat{\theta }$
d) $\hat{r}=\left|\hat{a}\right|\left|\hat{\theta }\right|\hat{r}$
now looking at the dimensions of the quantities on the LHS and the RHS,
$[a]=\frac{[\hat{r}]}{[\hat{\theta }][\hat{r}]}$
$=\frac{[M^0{L}^1{T}^0]}{[M^0{L}^0{T}^0][M^0{L}^0{T}^0]}$
$=[M^0{L}^1{T}^0]$
e) $a=1$
$\hat{r}=\hat{\theta }[\cos \theta \hat{i}+\sin \theta \hat{J}]$
$V=\frac{d\hat{r}}{dt}=\frac{d\theta }{dt}\hat{r}+\theta (-\sin \theta \hat{i}+\cos \theta \hat{J})\frac{d\theta }{dt}$
$V=\omega \hat{r}+\theta .\hat{\theta }\omega$
$\overrightarrow{a}=\frac{dv}{dt}$
$=\frac{d(\omega \hat{r}+\theta .\hat{\theta }\omega )}{dt}$
$=d^2\frac{\theta }{d{t}^2}\hat{r}+d\frac{\theta }{dt}d\hat{r}\frac{\hat{r}}{dt}+d^2\frac{\theta }{d{t}^2}(\theta .\hat{\theta })+d\frac{\theta }{dt}(\theta .\hat{\theta })$
$=d^2\frac{\theta }{d{t}^2}\hat{r}+\omega (\frac{-\sin \theta \hat{i}d}{dt}+\frac{\cos \theta \hat{J}d\theta }{dt})+\frac{d^2\theta }{d{t}^2}(\theta .\hat{\theta })+\frac{\omega d\theta }{dt}(\theta .\hat{\theta })$
$=\frac{d^2\theta }{d{t}^2\hat{r}}+{\omega }^2\hat{\theta }+\frac{d^2\theta }{d{t}^2}(\theta .\hat{\theta })+{\omega }^2\hat{\theta }+{\omega }^2(-\hat{r})$
$\overrightarrow{a}=\left(\frac{d^2\theta }{d{t}^2-{\omega }^2}\right)\hat{r}+\left(\frac{2{\omega }^{2+}{d}^2\theta }{d{t}^2\theta }\right)\hat{\theta }$