Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

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The document Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev is a part of the JEE Course I. E. Irodov Solutions for Physics Class 11 & Class 12.
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Q. 16. Two particles, 1 and 2, move with constant velocities v1 and v2  along two mutually perpendicular straight lines toward the intersection point 0. At the moment t = 0 the particles were located at the distances l1 a nd l2 from the point 0. How soon will the distance between the particles become the smallest? What is it equal to? 

Ans. Let the particle 1 and 2 be at points B and A at t = 0 at the distances l1 and l2 from intersection point O .
Let us fix the inertial frame with the particle 2. Now the particle 1 moves in relative to this reference frame with a relative velocity  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev  and its trajectory is the straight line BP. Obviously, the minimum distance between the particles is equal to the length of the perpendicular AP dropped from point A on to the straight line BP (Fig.).

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

From Fig. (b),  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev (1) 

The shortest distaiice

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

The sought time can be obtained directly from the condition that  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev is minimum. This gives  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 17. From point A located on a highway (Fig. 1.2) one has to get by car as soon as possible to point B located in the field at a distance 1 from the highway. It is known that the car moves in the field ri times slower than on the highway. At what distance from point D one must turn off the highway? 

Ans. Let the car turn off the highway at a distance x from the point D.
So, CD = x, and if the speed of the car in the field is v, then the time taken by the car to cover the distance AC = AD - x on the highway

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev  (1) 

and the time taken to travel the distance CB in the field

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev     (2)

So, the total time elapsed to move the car from point A to B

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

For t to be minimum

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 18. A point travels along the x axis with a velocity whose projection vx  is presented as a function of time by the plot in Fig. 1.3. 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Assuming the coordinate of the point x = 0 at the moment t = 0, draw the approximate time dependence plots for the acceleration wx, the x coordinate, and the distance covered s.

Ans. To plot x (f), s (t) and wx (t) let us partion the given plot vx (t) into five segments (for detailed analysis) as shown in the figure.
For the part oa : wx =1 and vx = t = v

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

On the basis of these obtained expressions wx (r), x (;t) and s (t) plots can be easily plotted as shown in the figure of answersheet


Q. 19. A point traversed half a circle of radius R = 160 cm during time interval τ = 10.0 s. Calculate the following quantities averaged over that time:

(a) the mean velocity (v);
(b) the modulus of the mean velocity vector |(v)|;
(c) the modulus of the mean vector of the total acceleration |(w)|
if the point moved with constant tangent acceleration. 

Ans. (a) Mean velocity

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(b) Modulus of mean velocity vector
Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(c) Let the point moves from i to /along the half circle (Fig.) and v0 and v be the spe at the points respectively.

We have Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev (as wt is constant, according to the problem)

So,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So, from (1) and (3) Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Now the modulus of the mean vector of total acceleration 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Using (4) in (5), we get :  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q.20. A radius vector of a particle varies with time t as r = at (1 — αt), where a is a constant vector and α is a positive factor.
Find:
(a) the velocity v and the acceleration w of the particle as functions of time;
(b) the time interval Δt taken by the particle to return to the initial points, and the distance s covered during that time. 

Ans. (a) we have  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
and  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(b)  From the equation 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So, the sought time  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

As  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Hence, the sought distance

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Simplifying,   Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q.21. At the moment t = 0 a particle leaves the origin and moves in the positive direction of the x axis. Its velocity varies with time as v = v0, (1 — t/τ), where v0 is the initial velocity vector whose modulus equals v0  = 10.0 cm/s; τ = 5.0 s.
Find:
(a) the x coordinate of the particle at the moments of time 6.0, 10, and 20 s;
(b) the moments of time when the particle is at the distance 10.0 cm from the origin;
(c) the distance s covered by the particle during the first 4.0 and 8.0 s; draw the approximate plot s (t). 

Ans. (a) As the particle leaves the origin at t = 0 

So,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev   (1)

As  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

where  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev s directed towards the +ve x-axis

So,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev    (2)

From (1) and (2), 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev     (3)

Hence x coordinate of the particle at t = 6 s. 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Similarly at 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

and at 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(b) At the moments the particle is at a distance of 10 cm from the origin, x = ± 10 cm.

Putting    x = +10 in Eq. (3)

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Now putting   x = -10in Eqn (3)

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

On solving,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

As t cannot be negative, so, 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Hence the particle is at a distance of 10 cm from the origin at three moments of time :

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(c) We have  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So, Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

and 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev    (A)

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

And for t = 8 s 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

On integrating and simplifying, we get s = 34 cm. 

On the basis of Eqs. (3) and (4), x (t) and s (t) plots can be drawn as shown in the answer sheet.


Q.22. The velocity of a particle moving in the positive direction of the x axis varies a  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev here α is a positive constant. Assuming that at the moment t = 0 the particle was located at the point x = 0, find:
(a) the time dependence of the velocity and the acceleration of the particle;
(b) the mean velocity of the particle averaged over the time that the particle takes to cover the first s metres of the path. 

Ans. As particle is in unidirectional motion it is directed along the jc-axis all the time. As at t = 0, x = 0

So,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Therefore,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
or, Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev  (1)

As, Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

On integrating, Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev   (2)

(b ) Lets be the time to cover first s m of the path. From the Eq. 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev    (3)

The mean velocity of particle 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Q.23. A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev where a is a positive constant. At the initial moment the velocity of the point is equal to v0. 'What distance will it traverse before it stops? What time will it take to cover that distance? 

Ans. According to the problem

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

On integrating w e get  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Again according to the problem 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Thus  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q.24. A radius vector of a point A relative to the origin varies with time t as r = ati — bt2j, where a and b are positive constants, and i and j are the unit vectors of the x and y axes.
Find:
(a) the equation of the point's trajectory y (x); plot this function;
(b) the time dependence of the velocity v and acceleration w vectors, as well as of the moduli of these quantities;
(c) the time dependence of the angle α between the vectors w and v;
(d) the mean velocity vector averaged over the first t seconds of motion, and the modulus of this vector

Ans. (a) As   Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

and therefore   Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

which is Eq. of a parabola, whose graph is shown in the Fig.

(b) As  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev    (1)

So, Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Diff. Eq. (1) w.r.t. time, we get 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(c)  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

so,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(d) The mean velocity vector 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Hence,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q.25. A point moves in the plane xy according to the law x = at, y = at (1. — αt), where a and a are positive constants, and t is time.
Find:
(a) the equation of the point's trajectory y (x); plot this function;
(b) the velocity v and the acceleration w of the point as functions of time;
(c) the moment t0 at which the velocity vector forms an angle π/4 with the acceleration vector. 

Ans. (a) We have

x = at and y = αt (1 - αt)    (1)

Hence, y (x) becomes,

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(b) Diiferentiating Eq. (1) we get

vx = a and vy = a (1 - 2αt)      (2)

So,   Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Diff. Eq. (2) with respect to time

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(c) From Eqs. (2) and (3)

We have

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

On simplifying. Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

As,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q.26.  A point moves in the plane xy according to the law x = a sin cot, y = a (1 — cos ωt), where a and co are positive constants.
Find:
(a) the distance s traversed by the point during the time T;
(b) the angle between the point's velocity and acceleration vectors.

Ans. Differentiating motion law : x = a sin ωt , y = a (1 - cos ωt ), with respect to time, vx = a ω cos ωt, vy = a ω sin ωt

So,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev   (1)

and  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev    (2)

Differentiating Eq. (1) with respect to time

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev   (3)

(a) The distance s traversed by the point during the time τ is given by 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(b) Taking inner product of  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

We get,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So, Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Thus,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev the angle between velocity vector and acceleration vector equals Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q.27. A particle moves in the plane xy with constant acceleration w directed along the negative y axis. The equation of motion of the particle has the form y = ax — bx2, where a and b are positive constants. Find the velocity of the particle at the origin of coordinates. 

Ans. Accordiing to the problem

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

So,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev    (1)

Differentiating Eq. of trajectory, y - ax - bx2, with respect to time

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev  (2)

So,

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Again differentiating with respect to time

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Using (3) in (2)  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Hence, the velocity of the particle at the origin 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Hence,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q.28. A small body is thrown at an angle to the horizontal with the initial velocity v0. Neglecting the air drag,
find:
(a) the displacement of the body as a function of time r (t);
(b) the mean velocity vector (v) averaged over the first t seconds and over the total time of motion. 

Ans. As the body is under gravity of constant accelration  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev it’s ve lo c ity vector and displacemen vectors are:

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev  (1)

and Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev   (2)

So,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev over the first t seconds

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Hence from   Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev over the first t seconds

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev    (4)

For evaluating ty take

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

But we have v = v0 at t = 0 and

Also at t - x (Fig.) (also from energy conservation)

Hence using this propety in Eq. (5)

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

As Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev
Putting this value of τ in Eq. (4), the average velocity over the time of flight 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q.29.  A body is thrown from the surface of the Earth at an angle α to the horizontal with the initial velocity v0. Assuming the air drag to be negligible, find:
(a) the time of motion;
(b) the maximum height of ascent and the horizontal range; at what value of the angle α they will be equal to each other;
(c) the equation of trajectory y (x), where y and x are displacements of the body along the vertical and the horizontal respectively;
(d) the curvature radii of trajectory at its initial point and at its peak.

Ans. The body thrown in air with velocity v0 at an angle a from the horizontal lands at point P on the Earth's surface at same horizontal level (Fig.). The point of projection is taken as origin, so, Δx = x and Δy = y

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(b) At the maximum height of ascent, vy = 0 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Hence maximum height Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

During the time of motion the net horizontal displacement or horizontal range, will be obtained by the equation 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

when  R = H 

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

(c) For the body, x (t) and y (t) are

x = v0 cos α t   (1)

and  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev   (2)

Hence putting the value of t from (1) into (2) we get,

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Which is the sought equation of trajectory i.e. y (x)

(d) As the body thrown in air follows a curve, it has some normal acceleration at all the moments of time during it’s motion in air.
At the initial point (x = 0, y = 0), from the equation :

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

At the peak point   Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev and the angential acceleration is zero.

Now from the Eq.  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Note : We may use the formula of curvature radius of a trajectory y (x), to solve part (d)y

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev


Q. 30.  Using the conditions of the foregoing problem, draw the approximate time dependence of moduli of the normal wn  and tangent wτ, acceleration vectors, as well as of the projection of the total acceleration vector wv on the velocity vector direction. 

Ans. We have, vx = v0 cos α, vy = v0 sin α - gt

As  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev all the moments of time. 

Thus  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Now,  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Hence  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

or  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

As  Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev during time o f motion

Kinematics (Part - 2) - Mechanics, Irodov JEE Notes | EduRev

On the basis of obtained expressions or facts the sought plots can be drawn as shown in the figure of answer sheet

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