1.
IIT- JEE Syllabus
Displacement, velocity, acceleration, kinematics in one and two dimensions, projectiles, circular motion, concept of relative motion.
2.
Linear Kinematics
2.1 Rest and motion
Motion means change of position with time. When a body is at rest, its position does not change with time. But how do we describe the position of a body? We describe it relative to another body for reference.
Thus, rest and motion are relative. A man sitting in a car moving at 55 mph on a highway is at rest relative to a co-passenger, while he is in motion relative to a person standing on the highway.
In order to describe rest and motion, we select a frame of reference and then describe rest or motion relative to this frame of reference.
Motion can be of two types - translational & rotational.
When a body moves such that it always remains parallel to itself throughout the motion it undergoes translation. When a body moves so that each point in the body maintains a constant distance relative to a fixed axis in space, the motion is rotation.
2.2 Position & frames of reference Position:
If a particle is moves along a given straight line (assumed along x-axis), its position is represented by the x-coordinate relative to a fixed origin.
If the particle moves in a plane (let x-y plane) its position is completely known when the x-and y coordinates of its position are known with respect to the given coordinate axis ox and oy.
Similarly for a particle moving in space, three coordinates (x, y, z) are required.
In vector notation, the position vector OP r in the three cases mentioned above are represented as
iˆ x r , r xiˆyjˆ, r xiˆyjˆzkˆ P (x, y) y x O A B
Several types of coordinate frames may be used to describe position.
Frames of reference:
Two of the commonest kinds of coordinate system in use are (a) Rectangular Cartesian Coordinates
(I) 2 dimensions
Cartesian: A point P with coordinates (x, y) in a rectangular cartesian coordinate system (fig.1) is described by its position vector OP , also represented by r. fig. 1 P(x,y)=(r,) L O y x y jˆ x iˆ r
, Where iˆ , ˆj are unit vectors along x, y respectively. . . . (1)
Polar: If the initial ray is chosen to be OX, the point P can be described by the coordinates (r, ) in the polar coordinate system. The position vector OP, is
rˆ r r
, rˆ being the unit vector along OP . . . (2) Equation (1) and (2) relate the two coordinate systems.
From figure 1,
x = r cos r = x2y2
y = r sin tan = y/x . . . (3)
Putting (3) in (1) and (2) and comparing,
iˆcos jˆsin rˆ
(II) 3 - dimensions
The simplest system of coordinate used in 3 dimensions is the rectangular cartesian coordinate system which consists of an origin O and three mutually perpendicular axis x, y & z.
A point P having coordinates (x, y, z) in this system has the position vector
z kˆ y jˆ x iˆ P O r P(x,y,z) L O z y x Illustration 1: A particle is kept at the point of
intersection of the diagonals. Find its position w.r.t. point O. y x O P 5 m 4 m C B A Solution: Here x = (2.5m) y = (2 m) OP = x i y j = (2.5mi +2 mj ) = (2.5i +2j ) m P O 2m 2.5m x y
Exercise 1: In the above illustration, a plane mirror is placed along x – z plane. Find the position of the image of particle w.r.t. point O.
2.3 Displacement
If a particle moves from its initial position A to a final position B, the vector joining A & B and directed along the line AB is known as the displacement vector r. The actual distance covered may have any value which is greater than or equal to the length of the straight line joining the two points A and B.
Suppose that the position vector of A = r1 =
OA= kˆ z jˆ y iˆ x1 1 1
Suppose that the position vector of B = r2 =
OB= kˆ z jˆ y iˆ x2 2 2 Using vectors, OA+ AB = OB AB = r = r2 r1 =
x2x1
iˆ y2 y1
jˆ z2 z1
kˆ = xiˆyjˆzkˆHere x, y and z represent the components of displacement vector r along x, y and z directions.
2 r 1 r A A B O Y x r
Illustration 2 : In the above problem, if the ball shifts from points P to point C, find the displacement of the ball.
y x O P 5 m 4 m C B A Solution: r1= ` OP = (2.5 iˆ +2ˆj) m 2 r = OC` = (4ˆj ) m
Displacement Δr r2 r1 = { 4ˆj – (2.5 iˆ +2ˆj)}m = (– 2.5 iˆ +2ˆj )m.Exercise 2 : In the above illustration, if the ball moves through 2m from P to point Q such that line PQ is perpendicular to the plane OABC, then find its displacement.
2.4 Distance
The distance covered by the particle increases with time whereas the magnitude of the displacement (the shortest distance between the initial and final position) increase, decrease or may be zero (e.g. when it passes through its initial position). Therefore the distance travelled is never smaller than the magnitude of the displacement.
d S.
Illustration 3: A car moves from A to B on a straight road and returns from B to the mid-point of AB. AB = 10 m. Find the displacement and distance covered.
Solution: Let A be the origin and C be the mid-point of A B. The displacement ) 0 r ( r r r r s C A C A iˆ ) BC AB ( iˆ ) AC ( iˆ ) r ( s C s(105)iˆ 5iˆm The distance covered = d= AB + BC = 10+5 = 15 m. Y A rA C P B X
Exercise 3 : Referring to previous illustration, what are the magnitudes of the displacement and distance covered when the car returns to A? 2.5 Speed
(a) Average Speed
A particle covers a distance by s = 10 m during a time interval t = 5 seconds. Therefore s/t is the ratio of distance covered s and the time interval t. This ratio is termed as time rate of change of distance. It is common known as Speed of the particle over that particular time interval t. That is why this speed is known as “Average Speed” uav denoted mathematically as av u = 2m/sec sec 5 m 10 t s (b) Instantaneous Speed
If the particle is not moving with uniform speed, the average speed completed in the above section depends on the time interval t and the instant when it was computed. It becomes necessary to define a quantity, called instantaneous speed which does not depend on the time interval t. We take the time interval t as it becomes very small (infinitesimal) and the corresponding distance travelled s (which is also
s 0 (infinitesimal distance) As t 0, s 0 t=(2+0.1)sec. t 0.1 sec. t = (2 + 0.01)sec t 0.01sec t=0 t=2 sec s = 2m s = .25m t = (2 + 0.001)sec t 0.001sec s = .015m t = (2 + 0.0001)sec t 0.0001sec s = .001m t = (2 +t 0)sec t 0 s 0 t time at 0 t dt v ds t s Lim
(or instantaneous velocity)
v v dt ds insant .
Here v at the time t = 2 sec is represented by
2 t dt ds .
Instantaneous speed is equal to the magnitude of instantaneous velocity.
The ratio of the infinitesimal distance covered and the corresponding infinitesimal time yields a finite value. This finite value is known as the instantaneous speed of the particle.
t s Lim u 0 t (c) Uniform Speed
If the particle covers equal distances in equal intervals of time, it is said to be moving with uniform speed.
Student Notes
The slope of s-t graph at any time t gives the speed at that time t.
The slope of x-t graph at any time t gives the instantaneous velocity (velocity at that time)
For positive slope v is positive, for negative slope v is negative.
The area of u-t graph during time t (= 2t0, say) gives the distance traversed during
the time 2t0.
The area of v-t graph during time t (= 2t0, say) gives the displacement of the
displacement) = (s+ - s-) We see that snet = vt0 + (-vt0) = 0 where as d = 2ut0
during time t = 2t0. 2.6 Velocity
(a) Average Velocity
As the displacement of a particle changes with time, the ratio of change s of displacement vector s in a time interval t to that interval of time is known as time rate of change of displacement (vector), this is known as the Velocity of the particle averaged over the time interval t or the average velocity, vav
av v = t s (b) Instantaneous Velocity
Following the previous argument of instantaneous speed, instantaneous velocity t s Lim v 0 t dt s d v = kˆ dt dz jˆ dt dy iˆ dt dx
* Instantaneous speed is the magnitude of instantaneous velocity v. Since a=
dt v d ,
The slope of the v-t graph gives the acceleration. If the slope is positive the acceleration is positive; if the slope is negative then the acceleration is negative (retardation or deceleration).
Since v =
0ta dt(discuss later),the area of a-t graph gives the change in velocity. If the area remains (above x- axes) positive, the change in velocity is positive and vice versa.
(c) Uniform Velocity
A particle is said to be moving with uniform velocity if its velocity is independent of time. Illustration 4 : A particle moves in a straight line obeying the relation x = t (t-1) where
x = displacement in m and t = time in sec. Find the velocity of the particle when its displacement is zero.
Solution: x = t (t-1).
If x = 0, t(t-1) = 0 t = 0 and t = 1 The instantaneous velocity
1 t 2 dt dx v
Putting t = 0 and t = 1 we obtain sec / m 1 v & sec / m 1 v t 1 0 t .
Exercise 4 : Referring to the previous illustration,
(a) What is the significance of positive and negative values of instantaneous velocities?
(b) What are the distance covered and displacement during t = 2 sec. ? 2.7 Acceleration
(a) Average Acceleration
If the velocity (magnitude, direction or both) of a particle changes with time, its motion is said to be non-uniform.
Suppose that the velocity of particle changes by v over a time interval t, the time rate of change of velocity is given by v/t, this is known as the average acceleration of the particle over the time interval t.
t v aav . (b) Instantaneous Acceleration
For an instant (infinitely small or infinitesimal time interval), the change in velocity of the particle is infinitely small, but the ratio of infinitesimal change in velocity and the infinitesimal time is finite. This finite ratio is known as instantaneous acceleration:
dt v d t v Lim a 0 t (c) Uniform Acceleration
When a particle undergoes constant acceleration (vector) for some time interval we say that it is moving with constant or uniform acceleration over that time interval.
2.8 Average speed and Velocity
Average Speed
When a particle moves with different uniform speeds u1, u2, u3 etc. in different (finite) time
intervals t1, t2, t3 etc. respectively, its average speed over the total time of journey is
given as elapsed time total ered cov ce tan dis total uav = ... t t t ... s s s 3 2 1 3 2 1 where s1 = u1 t1, s2 = u2t2 etc. uav = ... t t .... t u t u 2 1 2 2 1 1 uav =
i n 1 i i n i 1 i i i t t u Average VelocityWhen a particle moves with different velocities v1,v2, v3
etc. in different time intervals t1, t2, t3 etc. respectively, its average velocity over the total time of motion can be
given as t s time Total vector nt displaceme Net vav
= ... t t t ... s s s 3 2 1 3 2 1
Where the displacement of the particle during time interval t1, t2 etc. are given by
s1v1t1, s2 v2t2 etc. v v tt vt t... ... 2 1 2 2 1 1 av
i n 1 i i n i 1 i i i av v t t v Illustration 5: A particle moves with a velocity v(t) = (1/2)kt2 along a straight line. Find
the average speed of the particle in a time t.
Solution : uav =
t 0 dt ) t ( v T 1 =
t 0 2 kt 2 1 T 1 dt = kT2 6 1 .Exercise 5 : Find the average speed of a particle whose velocity is given by v = v0 sin t. Time Average
T 0 T 0 av v dt T 1 T dt v ) t ( v dt a T 1 ) t ( a T 0 av
Here, v and a are the functions of t.
Illustration 6: A passenger is standing ‘d’ m away from a bus. The bus begins to move with constant acceleration a. To catch the bus, the passenger runs at a constant speed v towards the bus. What must be the minimum speed of the passenger so that he may catch the bus ?
Solution : Let the passenger catch the bus after time t, the distance travelled by the bus
s1 = 0 +
2 1
at2 . . . . (1)
and the distance travelled by the passenger s2 = ut . . . . (2)
Now passenger will catch the bus if d + s1 = s2 . . .. (3)
we get d + 2 1 at2 = ut i.e. 2 1 at2 – ut + d = 0 or t = a ad 2 u u 2
So the passenger will catch the bus if t is real, i.e. u2 2ad or u 2ad
So the minimum speed of passenger for catching the bus is 2ad.
Exercise 6 : A man standing 9m behind a bus starts running uniformly to catch the bus. At the same instant the bus starts from rest with an acceleration of 2m/s2. Find the speed of the man if he is just able to get in.
Illustration 7: A float is overcome by a motor boat going downstream at a point A. T hours later it turns back and after some time it passes the float at a distance d km from point A. What is the velocity of stream if the speed of motor remains constant w.r.t. water ?
Solution : Let u = speed of stream, v = speed of motor boat ( v > u) Let t1 = time of motion of motor boat
during up stream.
The motion of float and motor boat is shown in the following velocity time graph. (V+u) u – (V– u) T + t1 t V
. As float has no speed of its own hence it will move for the complete time (T + t1) hrs. with the flow speed.
For motor boat, velocity down stream = (v + u) and velocity up stream = – v – u)
Negative sign represents the opposite direction.
As area of velocity time graph gives the displacement and from the question,
displacement of float = d km displacement of boat = d km
Hence, we get the following equations.
For float, u (T + t1) = d …(i)
For motor boat (v + u) (T) + { – (v – u) (t1)} = d …(ii)
Solving (i) & (ii), we get u = km/hr
T 2
d
Exercise 7 : A car accelerates from rest with = 2 m/sec2 in a straight track, then it
distance of 100 m. Using a graphical method, find the maximum speed of the car.
3.
Accelerated Motion
3.1 Uniformly accelerated motion
Suppose that a particle passes the origin of the coordinate system with a velocity v0 at
the time t = 0. Suppose that its displacement is s and its velocity v after time t.
Then, applying the definitions of average acceleration for constant value, average value, we get the following expressions.
v = v0 + at s = t 2 v v0 = v0t+ 2 t a 2 1 v2 = v2 2a.s 0
t 1
a v vt1 0 , s(tth sec) = v0 + (2t – 1) 2 aIllustration 8 : A particle moves in a straight line with constant acceleration. If it covers 10 m in first second and extra 10 m in next second, find its initial velocity.
Solution: Let the initial speed be V0 and the acceleration be a 2 0 at 2 1 t v s
Putting t = 1 sec. and s = 10 m, we obtain
10 = v0 + 5 a …(1)
Again in next second, that means t = (1+1) sec [t is the time measured from the initial position (when t = 0, s = 0)] the displacement of the particles = 10 + 10 = 20 m.
Putting t = 2 sec and s = 20 m, we obtain 10 = 2 v0 + 2 a
v + a = 5 …(2)
Solving (1) and (2), we obtain 10 = v0 + 5 (5-v0) 10 = 3.75m/sec 4 15 v 25 v 4 0 0
Exercise 8 : Referring to the previous illustration, (a) What is the acceleration of the particle?
(b) What is the velocity of the particle at the end of 3rd second? (c) What is the displacement of the particle in 3 seconds? (d) What is the displacement of the particle in 3rd second?
3.2 Non-uniformly accelerated motion in a straight line
If the acceleration a is a function of time (t) or function of distance (s) the equation of instantaneous velocity and acceleration can be written as
a = f(t)
t 0 vdt s dt ds v and
t 0 0 v dt a v dt dv a 2 2 dt s d dt ds dt d a a = f(s) Again dt ds . ds dv dt dva , considering the magnitudes we obtain a scalar equation. ds dv v a v dt ds . Therefore,
S 0 v v ads vdv 0
S 0 2 0 2 ads 2 v v
s 0 2 0 2 v ads v .Illustration 9: The acceleration of a particle moving rectilinearly varies with the magnitude of its velocity as a = -v. Find its initial speed, if it stops after t0 = 1 sec from starting.
Solution: a = - v v dt dv dt v dv
v v0 dt v dv 2( v v0) t t 2
v0 v
At t = t0, it stops; Putting v = 0, we obtain t0 2 v0
; 2 t v 2 0
0 putting t0= 1, we obtain v0 = 0.5 m/sec.
Exercise 9 : Referring to the previous illustration, (a) Find the velocity-time equation. (b) Find the displacement-time equation.
Consider a body P projected vertically upwards from the surface of the earth with an initial velocity, u (see figure). The body rises to a maximum height at B and then returns (motion BC).
Even through we have shown the motion of the body as ABC, the actual path is always along the same line AB. The picture has been slightly modified for clarity.
B P g A C u +ve
Let the displacement of the body (at P) at time t, measured from its initial position A, be denoted by h. We can now apply the equations before. The acceleration, a= -g (note the positive direction in the figure, any vector in the opposite direction is negative);
x = h and the initial velocity is u. After time t, v = u - gt . . . . (1) h = ut + 2 1 (-g)t2 = ut -2 1 gt2 . . . . (2) v2 = u2 + 2 (-g)h = u2 - 2gh . . . . (3)
Other interesting questions that may be posed are: (a) What's the maximum height to which it rises? (b) What’s the time of flight?
Let us note that at the point of maximum elevation, B, vB = 0 (it's got to be zero, if it were
not the body would have risen further).
vB = u - gtAB = 0 ; tAB represents the time taken for motion AB.
or, tAB = u/g . . . . (4)
If the maximum elevation is H (at the pt. B of course)
2 B v = 0 = u2 - 2gH or, H = g 2 u2 . . . . (5)
When the body reaches the ground again (at the pt C), we can write, h = 0 = utAC -
2 1
g 2 AC
t ; Where tAC represents the time taken for motion AC
= (u - 2 1 gtAC) tAC u - 2 1 gtAC = 0 or tAC = g u 2 = 2tAB . . . . (6)
The velocity at the point C, VC = u - gtAC = u - g g
u 2
= -u, which is equal in magnitude to u (the velocity of projection) but opposite in direction. In both the examples considered above, the acceleration is constant (or uniform). This may not always be true.
For the motion of a particle projected vertically up or down from some height (h) we have assign the directions of displacement vector, velocity vector and acceleration vector with reference to a system of coordinates.
Let us consider the following cases :
(i) When the particle is projected with an initial velocity v0 in the upward direction and its
vertical distance from the point of projection O at a time t be h, then h = v0t - gt2
2 1
(if it is above the point O) -h = v0t - gt2
2 1
(if it is below the point O)
We obtain the value of time by putting the given values of h and v0.
If the origin is shifted to the ground level, which is at a depth H below the point of projection, then h and –h mentioned above are replaced by (H + h) and (H – h) respectively.
Similar procedure is applicable for the following equations v = u gt
v2 = u2 2gh
Now using the following equations h t v gt 2 1 0 2 t = 2 g h 8 g v 4 g v 2 2 2 0 0 = g h 2 g v g v 2 2 0 0
If the particle is released from rest then v0 = 0.
Illustration 10: A stone is dropped from a balloon ascending with v0 = 2 m/sec, from a
height h = 4.8 m. Find the time of flight of the stone (g = 10 m/sec).
Solution: h = gt v t 2 1 0 2 0 g h g g v 2 t2 0 1 v gh 2 1 g v t 0 2
Putting the values of v0 and h etc.
we obtain, 1 1.2sec. 2 x 2 ) 8 . 4 ( ) 10 ( 2 1 10 2 t h S t = 0 v0 v0 t = t
Exercise 10 : Referring to the previous illustration, if the stone reaches the ground after t = 2 second from same initial height of release, find the (a) speed of the balloon at the time of releasing the stone, (b) total distance covered by the stone till it reaches the ground level, (c) the average speed and (d) average velocity of the stone for the total time of its flight, one from the bottom and the other from the top.
(i) Particle from the Top Projected Down and Particle at the Bottom Projected Up
Refer Figure . We see that s1 is downward s1 = 1 gt2 2 1 t v …(1) s2 = 2 gt2 2 1 t v …(2) (1) + (2) gives s1 + s2 = (v1 + v2)t t = 2 1 v v h …(3) v s1 s2
v1 + v2 = Relative velocity and h = initial relative distance of separation. Now putting t=h/
(v2 – v1) in (a) or in (b) find the position of meeting in terms of distance s1 from the top or
from the bottom (ground) in terms of distance s2.
Illustration 11 : A stone is released form the top of a cliff. Another particle is simultaneously projected with v = 10 m/sec. from the bottom of the cliff. If they meet after 2 seconds, find the height of the cliff.
Solution: Let after a time t, they meet si is downward 2 I gt 2 1 s …(1) Since, 2 2 2 gt2 2 1 t v s , upward is s Putting 2 2 2 gt 2 1 vt s obtain we , v v …(2) (1) + (2) s1+s2 = v t h = v t.
Putting v = 20 m/sec and t = 2 sec. We obtain h = (20) (2) = 40 m. Exercise 11 : A body is released from a height and falls freely towards the earth. Exactly 1 sec. later another body is released. What is the distance between the bodies 2 sec. after the released of the second body if g = 9.8 m/s2 ?
(ii) Particle from the Top Projected Up and Particle at the Bottom Projected Up
Refer Figure.
Case 1: Let the particles meet above the top of the cliff. Now s1&s2 are upwards. Therefore, 2 1 1 gt 2 1 t v s …(1) and s2 = 2 gt2 2 1 t v …(2) s2 - s1 = (v2 - v1)t h = (v2 - v1)t t = 1 2 v v h .
Now put the value of t in any one equation (a) or (b) to find the position of meeting.
h S22 s1 v1 t =t t = 0 v2 h Ss22 v1 t = 0 v2 t = 0 s1 t = t
Case II: Let the particles meet below the point of projection as shown in Figure Hence
the s1 is downwards and s2 is upwards. s1 = gt v t 2 1 1 2 …(1) and s2 = 2 gt2 2 1 t v …(2) (a) + (b): s1 + s2 = (v2 - v1)t, Setting s1 + s2 = h.
We obtain t = h/(v2 – v1) The expression for time remains equal for any point of
meeting.
From the previous analysis we know that t = h/(v2 – v1) for any point of meeting. Therefore
we should not bother about the position of point of meeting at the beginning of analysis of a problem. We may take it above or below the top of the cliff. Then find t and substitute in the equation s2 = v2t –
2 1
gt2 . If it is greater than h, meeting point is above the top of the
cliff and vice versa.
Illustration 12: Two particles are simultaneously projected upwards from the top and bottom of a cliff of height h = 20 m. If the speed of one is double that of the other and they meet after a time t = 2 second, find their speed of projection.
Solution: From the obtained general formula, 1 2 v v h t Putting v2 = 2v1, we obtain 1 1 1 v h v v 2 h t 10m/sec. 2 20 t h v1 Therefore, v1 = 2 v1 = 20 m/sec.
Exercise 12: Referring to the previous illustration, what is the position of meeting of the particles?
Non-Simultaneous Projection of Particles
Take any case, as shown given in the Figure. Let a particle be projected up from the top of the cliff and another particle be projected up from the bottom of the cliff t0 second later. Let they meet after a
time t from the instant of projection of the first particle. Therefore times of flights of the particles are t and (t-t0) respectively.
s1 = gt v t 2 1 1 2 …(a) and s2 = 2 0 g(t t0)2 2 1 ) t t ( v …(b) h Ss22 v1 t = t0 v2 t = 0 s1 t = t t =(t - t0) (a) + (b) s1 + s2 = h h =
1 0 20 2 0 2 2gt v t 1 t gt ) v v ( t = 0 1 2 2 0 2 gt ) v v ( ) gt ½ t v h ( …(3)Now obtain the position of meeting point by substituting t in (a) or (b). Remember that, since equation (a) is simpler, substitute t in equation (a) for quicker result. Similarly you can use this concept for other cases as discussed earlier.
Illustration 13: A body is dropped from a height h = 50 m. Another body is projected with vertically up a speed V = 10 m/sec after a time t0 = 2 sec from the
instant of release of the first body. Find the time of their meeting.
Solution: Let the time of fall of the first body be t, before meeting at time = t. The displacement of the first body
2 ` 1 gt 2 1 s …(a)
The 2nd body is projected from the ground
after t0 = 2 sec from the instant of
projection of the first body. Therefore the time of its motion before its meeting =(t - t0)
= (t-2) sec. v1 = 0 s1 s2 t =2 h v2 = v t = (t-t0) t = t
2 2 g(t 2) 2 1 ) 2 t ( v s …(b) (a) + (b)
2 2
2 l g t (t 2) 2 1 ) 2 t ( v s s h = (2)(2t 2) 2 g ) 2 t ( v Putting h = 50 m, g = 10 m/sec2 and v = 10 m/sec., we obtain
50 = (t 1) 2 20 ) 2 t ( 10 5 = t2 + 2t2 = 3t4 3t = 9 t = 3 sec.
Exercise 13 : Referring to the previous illustration, (a) Where do the particles meet?
(b) What are the speeds of the particle sat the time of meeting?
5.
Relative Velocity
Suppose that two particles A & B are at the two points as shown in the figure. The position vectors of the particles with respect to an inertial reference frame be
B A & r
r
respectively.
The relative separation between the particles is given by rBA rB - rA
Differentiating both sides w.r.t time, we obtain dt r d -dt r d dt r dBA B A vBA vB vA where rBA
is the position of B with respect to A & vBA is the velocity of B relative to A. Z A B Y Z 0 A r A r B r
Physical Significance of Relative Velocity
Two persons A & B are in the two vehicles moving in the same direction as shown in the figure.
Assume, vA = 10m/sec & vB = 4m/sec
The person A notices the person B moving towards him with a speed of (10 - 4) m/s = 6 m/sec. This is the velocity of B with respect to ( or relative to) A.
BA
v is directed from B to A.
VA VB
Similarly A seems to move towards B with a speed of 6 m/sec. Therefore the velocity of A relative to B
v
AB
has a magnitude of 6 m/sec & is directed from A to B as shown in the figure.
Therefore vAB = -vBA . In general, vBA = vB - vA vBA vAB BA v AB v
)
cos
v
v
(
)
sin
v
(
tan
&
cos
v
v
2
v
v
v
B A B 1 B A 2 B 2 A AB A v A B v v B v A v Illustration 14 : A train moves due east with a velocity v1 = 20m/sec. and a car moves
due north with a velocity v2 = 15 m/sec. Find the velocity of the car as
observed by a passenger sitting in the train.
Solution : The passenger observes the velocity of the car w.r.t. himself. That is, the velocity of the car relative to the train is
t c ct v - v v v21 v2 -v1 Magnitude of v21 v21 =
v
2
-
v
1
N E S W .vc v2 .vt v1 v v v2 2 2 1 21 (20)2(15)2 v21 = 25 m/sec Direction of v21 : 3 4 tan 15 20 tan1 -1 west of north N E S W .v21 .v2 . v1 Relative Motion between Rain and Man We know that
vr vrg = velocity of rain w.r.t. ground, vm vmg velocity
of man w.r.t. ground m r rm v v v vr vrm vm
That means the vector addition of the velocity of rain with respect to man (vrm
) and the velocity of man (vehicle)
m
v yields the actual velocity of rain vr
. The magnitude and direction of vr can be given as Vm Direction of rain fall as seen by the moving man rm
v
Vm
m rm rm 1 m rm 2 m 2 rm rv
cos
v
sin
v
tan
cos
v
v
2
v
v
v
with horizontal (vm ) VmDirection of rain fall as seen by the moving man (Actual direction of rainfall) r v Vm= 0 r v rm v m v r m rm v v v
Illustration 15: A man holds his umbrella vertically up while walking due west with a constant velocity of magnitude vm = 1.5 m/sec in rain. To protect himself
from rain, he has to rotate his umbrella through an angle = 30° when he stops walking. Find the velocity of the rain.
Solution : vr vrm vm .Since vrm vm , we obtain, r m v v sin vr = vm cosec
vr = (1.5 m/sec) cosec 30° = 3m/sec .
Exercise 14 : A boy is running on a horizontal road with a velocity of 5 m/s. At what angle should he hold his umbrella in order to protect himself from the rain, if it is raining with a velocity of 10 m/s vertically downward ?
Relative Motion of a Swimmer in Flowing Water m
v can be found by the velocity addition of w mw&v v . Since vmw vm vw . vm vmw vw mw v m v w mw v v w v
Illustration 16: A man swims at an angle = 120° to the direction of water flow with a speed vmw = 5km/hr relative to water. If the speed of water vw = 3km/hr,
find the speed of the man.
Solution : vmw vm vw w mw m v v v mw v vm vw vm vmw vw v2mw v2w 2vmw .vwcos
v 5232 2(5)(3)cos120
m
vm 25915 19m/sec.
Exercise 15 : To a man walking at 7 km/ hr due west the wind appears to blow from north west, but when he walks at 3 km/hr due west the wind appears to blow from the north. What is the actual direction of the wind & what is its velocity ?
Crossing of the river with Minimum drift CASE -1 : vmw vw
A man intends to reach the opposite bank at the point directly opposite to the stationary point. He has to swim at angle with a given speed vmw w.r.t.
water, such that his actual velocity vm
will direct along AB, that is perpendicular to the bank (or velocity of water vw
).
For minimum drift, vm vw
) v v ( mw w mw v A d B w v
You can realise the situation by a simple example. If you want to reach the directly opposite point or cross the river perpendicularly, a man, that is to say, Hari, must report you that, you are moving perpendicular to the shore. What does this report signify? Since Hari observes your actual velocity (vm
) to be perpendicular to the bank vm
is perpendicular to vw
.
Observing the vector-triangle vw = vmw sin & vm = vmw cos
mw w 1 v v sin 2 w 2 mw m v v v & The time of crossing = t =
m v d 2 w 2 mw v v d t
Illustration 17: The speed of a man in a pond is double that of the water in a river. The man starts swimming from a point P on the bank. What is the angle at which the man should swim so as to get directly to the opposite bank.
Solution : Let the speed of the man w.r.t. water be vmw
When the man swims in a river (moving water), the velocity of the man is directly across the river, i.e.
vm vw
mw w v v sin mw w 1 v v sin …(1)
Since the speed of water is half of that of the man relative to water (speed of the man in still water) mw v m v w v River vmw = 2vw 2 1 v v mw w …(2)
Using (a) & (b) we find = sin-1 30
2 1
The angle of swimming = = 90 + = 90° + 30° = 120° to the direction of flow of water.
CASE 2 : vmw < vw
Let the man head at an angle with normal to the bank for minimum drift. Suppose the drift is equal to zero. For zero drift, the velocity of the man along the bank is zero. vm = vw – vmw sin = 0 This gives, mw w mw w ,since v v v v sin
sin > 1 which is impossible. Therefore the drift can not be zero.
mw v A d B w v t = 0 m v C t = t x y
Now let the man head at an angle with normal to the bank to experience minimum drift. Suppose that the drifting of the man during time t when he reaches the opposite bank is BC = x.
x = (vm)x (t) …(1)
Where t = (v )ABcos v cosd m y
m
…(2)
& (vm)x = component of velocity of man along the water flow
(vm)x = vw – vmw sin …(3)
Using (1), (2) & (3) we obtain
d tan sec v v cos v d ) sin v v ( x mw w mw mw w
Exercise 16: Referring to the previous illustration, find the velocity of the swimming man, if speed of water is 3km/hr.
d tan -sec v v x mw w …(4) For x to be minimum, 0 d sec tan sec v v d dx 2 mw w w mw mw w v v sin ) (sec tan v v w mw 1 v v
sin . Substituting the value of
in (4) we obtain, d v v v x mw 2 mw 2 w mw v A d B w v t = 0 m v C t = t x y
Illustration 18: The speed of water is double that of (swimmer) relative to water. What should be the direction of the swimmer so as to experience minimum displacement assuming vm = 5m/sec ?
Solution: Putting the value of
2 1 v v w mw
in the derived expression
w mw 1 v v sin we obtain 30 2 1 sin 1
Crossing of the River in the Minimum Possible Time
Case 1: To reach the opposite bank for a given vmw
Let the man swim at an angle with AB. We know that the component of the velocity of man along shore is not responsible for its crossing the river. Only the component of velocity of man (vm) along AB is responsible for its crossing along AB.
The time of crossing = cos v AB t mw
Time is minimum when cos is maximum The maximum value of cos is 1 for = 0.
That means the man should swim perpendicular to the shore vmw vw Then mw 0 mw min v d cos v d t
Exercise 17 : Referring to previous illustration, What is the time of crossing of the man assuming width of the river is 500 m?
mw min v d t
Illustration 19: A man crosses the river in shortest time at an angle = 60° to the direction of flow of water. If the speed of water is vw = 4km/hr, find the
speed of the man.
Solution : Referring to the theory, we know that for minimum time of crossing the man should head perpendicular to the shore
vmw vw Therefore, m w v v cos m 0 v 4 60 cos vm = 8km/hr mw v w v m v
CASE 2: To reach directly opposite point on the other bank for a given vmw &
velocity v of walking along the shore.
To attain the direct opposite point B in the minimum time. Let the man swim at an angle with the direction AB. The total time of journey t = the time taken from A to C+ the time taken from C to B t = tAC + tCB Where tAC = v cos AB mw & v BC
tCB where v = walking speed of the man from C to B.
BCv cos v AB t mw Again BC = (vm)xt cos v AB ) sin v (v BC mw mw w
Using (a) & (b) we obtain,
) cos v(v AB ) sin v (v cos v AB t mw mw w mw mw v A d B w v m v C Exercise 18 : Referring to previous illustration,
(a) If the width of the river is ½ km, find the displacement of the man in crossing the river.
v
tan
v
sec
v
v
1
AB
t
mw w
v
tan
v
sec
v
v
1
v
d
t
mw w mw Putting 0 d dt for minimum t we obtain
v
tan
v
sec
v
v
1
v
d
d
d
d
dt
mw w mw0
v
sec
v
v
1
v
tan
sec
2 w mw
v sec v v 1 v tan w mw w mw v v v sin w mw 1 -v v v sin .This expression is obviously true when vmw < v + vw.
Illustration 20: A man can walk on the shore at a speed v1 = 6km/hr & swim in still water
is a speed v2 = 5km/hr. If the speed of water is v3 = 4 km/hr, at what angle
should he head in the river in order to reach the right opposite shore in shortest time including his swimming & walking?
Solution : Directly using the previous result we obtain the angle of swimming
w mw 1 -v v v sin Putting v = v1 = 6km/hr, vmw = v2 = 5km/hr vw = v3 = 4km/hr we obtain = sin-1 ½ = 30°
The man should head to an angle = 90 + = 120° with the direction of flow of water.
6.
Projectile Motion
6.1 Oblique projection on a horizontal surface
Let a particle (body) be projected with certain velocity v0
at an angle 0 to the horizontal.
The horizontal component of its velocity v0
= (vo)x = v0 cos0 and the vertical component
of v0
= (v0)y = v0sin0. The particle moves simultaneously in both horizontal and vertical
directions under earths gravitational field (no other external forces like wind drag are small and therefore their effect neglected.
Exercise 19 : Referring to previous illustration, What is the time elapsed for swimming, if the width of the river is ½ km?
Change in Position Vector (Displacement)
Let the particle acquire a position P having the coordinates (x, y) just after time t from the instant of
projection. The
corresponding position vector of the particle at time t is
r
as shown in the Figure y x P(x,y) v r 0 v0 t = 0 O r x iˆ y jˆ …(1)Since no external force acts upon the particle horizontally, its horizontal acceleration is zero, that means, the particle moves horizontally with constant velocity of magnitude (v0)x
= v0cos0.
vx = v0cos0. …(2)
The horizontal distance covered during time t is given as x = (v0)xt x = (v0cos0)t …(3)
If we consider the vertical motion of the particle, the external force acting on the particle is gravitational force mg. Consequently the particle accelerates downwards (towards the centre of earth) with an acceleration of magnitude g = 9.8 m/sec2. In the other words we
can say that the particle decelerates upward with g = 9.8 m/sec2. Consequently the
vertical velocity of the particle at time t is given as vY = (v0)y – gt, putting (v0)y = v0 sin 0 we obtain
vy =v0 sin0-gt …(4)
The vertical velocity of the particle at any height y is given as vy2 = (v02) – 2gy, putting (v0)y = v0 sin we obtain
gy 2 sin v v 2 2 0 0 y …(5)
Now the vertical displacement y is given as y = (v0)y t –1/ 2 gt2 y = 0 0 gt2 2 1 t ) sin v ( …(6)
Putting the values of x and y from equation (3) and equation (6) in equation (1) we obtain the position vector at any time t as
jˆ gt 2 1 t ) sin v ( iˆ t ) cos v ( r 2 0 o 0 0 r = 2 2 0 0 2 0 0 gt ) 2 1 ) sin t v ( ) cos t v ( r = 3 0 0 4 2 0 2 2 2 0 0 2 2 2 0 g t v gt sin 4 1 sin t v cos t v
0 0 2 0 0 v sin gt v 2 gt 1 t v r
and = tan-1(y/x)
= ) cos t v ( ) gt 2 / 1 sin t v ( tan 0 0 2 0 0 1 0 0 0 0 1 cos v 2 gt sin v 2 tan .
Illustration 21: A boy throws a stone with an speed V0 = 10 m/sec at an angle 0 = 30o to
the horizontal. Find the position of the stone w.r.t. the point of projection just after a time t = 1/2 sec.
Solution: The position of the stone is given by r xiyj where x=(v0 cos 0)t = 2 1 ) 30 cos 10 ( = 4.33 m. and y = 2 0 0 gt 2 1 t ) sin v ( x y O y u0 t = t P (x, y) 0 r x t = 0 = 2 2 1 10 2 1 2 1 ) 30 sin 10 ( = 1.25 m r (4.33i 1.25j)m.
Exercise 20 : Referring to the previous illustration find the position of the particle as t = v0/g.
Average Velocity
Therefore the average velocity of the particle during time t can be found as vav
=
r
/t. We have assumed the point of projection as the origin of the coordinate system. That means, the initial position vector of the particle has a magnitude equal to zero
r
=
r
. Putting t = t we obtain, t r vav . Substituting the obtained value of |
r
| we obtain.And 0 0 0 0 1 0 2 0 0 av cos v 2 gt sin v 2 tan v sin gt v 2 gt 1 v v . Instantaneous Velocity
The velocity vof the particle at time t is equal to the vector sum of the velocity components along x and y axis-+
v vx ivy j …(7)
Putting the values of vx and vy using equations (2) and (3) in Equation (7), we obtain
j ) gt sin v ( i cos v v 0 0 0 0 v = 2 0 0 2 0 0 cos ) (v sin gt) v (
v = v cos v sin g2 t2 2v0 gtsin 0 2 2 0 0 2 2 0 v = 2 2 2 0 0 0 g t 2v gtsin v 0 0 2 0 0 sin v gt 2 v gt 1 v v and = cos v gt sin v tan v v tan 0 0 0 1 x y 1 = tan-1 {tan 0 – (gt/v0)sec0}
Substituting the values of vx and vy by equations (2) and (5) in equation (7) we obtain
j ) gy 2 sin v ( i ) cos v ( v 2 0 2 0 0 0 v = (v cos ) v2sin2 0 2gy 0 2 0 0 v v2 2gy 0 . and 0 0 0 2 2 0 1 x y 1 cos v gy 2 sin v tan v v tan . Equation of trajectory
Substituting ‘t’ by equation (3) in equation (6), we obtain Y = 2 0 0 0 0 0 0 cos v x g 2 1 cos v x ) sin v ( 2 2 0 0 2 0 sec v 2 gx x ) (tan y
This is the equation of a parabola. Therefore the path of the particle is parabolic when the particle passes the level of projection. The horizontal distance covered is known as its range R and the time of motion is known as time of flight T. For this range R = Vxt,
t = T; X = R, y = 0 (as shown in Figure). R O y v0 0 t = 0 y H t = T/2 (R/2, H) V0cos (R, 0) 0 v t = T
Illustration 22: A ball is projected so that it follows a curve y = ax – bx2. Find the initial
Solution: Comparing the given equation y = ax – bx2 with the general trajectory equation y = (tan 0) x – 2 0 2 v gx 2 1 (1 + tan2 0), we obtain, tan 0 = a and 2 0 v 2 g (1 + tan2 0) = b 0 = tan-1 a and 2 0 v 2 g (1 + a2) = b (since tan 0 = a) v0 = b 2 ) a 1 ( g 2
and 0 = tan-1 (a).
Time of Flight
Substituting these values in equation (6), we obtain.
0 = 2 0 0 gT 2 1 T ) Sin V ( T 2 V0gSin0 . Maximum Height
When the particle is at the highest position, Vy = 0 and y = H, putting Vy = 0 and y = H in
equation (5), we obtain 0 = V Sin 0 2gH 2 2 0 H = g 2 Sin V2 2 0 0 H is maximum when 0 = 90o Hmax = V02/2g Range
When x = R, y =0, putting this value in locus equation or equation of trajectory we obtain, 0 = (tan0) R - 0 2 2 0 2
Cos
V
2
R
g
tan 0 = 0 0 2 0 Sin Cos V 2 gR R = g Cos Sin V 2 0 0 2 0 g 2 Sin V R 0 2 0 R is maximum when Sin2o is maximum 0 = 45o.
Illustration 23: A body is projected up such that its position vector varies with time as r
= 6t
iˆ + (8t-5t2) jˆ . Find the (a) initial velocity (b) time of flight (c)
range of the body.
Solution: (a) The position of the body at any time t is given as r
= 6t iˆ + (8t-5t2) ˆj . When t = 0, r = 0. That means the body is projected from the origin of the coordinate system. Differentiating both sides w.r.t. time ‘t’, we obtain
dt r d
= 6 iˆ + (8-10t) j
v = 6 iˆ + (8-10t) j. Putting t = 0, we obtain the initial velocity
(velocity of projection) given as v|t=0 = v
v0 = 10 m/sec;
(b) The time of flight T = g Sin v 2 0 0 T = g ) v ( 2 y 0 where (vy)0 = 8 T = 10 8 2 = 1.6 sec.
Exercise 21: Referring to the previous illustration,
(a) Find the maximum height attained by the body. (b) Find the equation of trajectory of the body.
Angle of Projection for Given Ratio of Range and Maximum Height
When the range R is times greater than the maximum height H. Then R = H g 2 2 sin v g cos sin v 2 2 0 0 0 2 0 cot0 = 4 tan0 4/ 0 = tan-1 76o R H 4 (when H = R or = 1). Therefore, the angle of projection will be 76° for H = R.
Illustration 24: A cricketer of height 2.5 m throws a ball at an angle 30° with the horizontal such that it is received by another cricketer of same height standing at a distance of 50 m from the first one. Find the maximum height attained by the ball.
Solution: Let the ball rise to an additional height H. The distance between the cricketer is equal to R = 50 m. The angle of projection, 0 = 30°, we know that u0 h 2.5 m H 0 tan 0 = R H 4 tan 30° = 50 H 4 H = 3 4 50 m
7.2 m.Therefore the maximum height attained by the ball =H =2.5m + 7.2m. H´ = 9.7 m.
Exercise 22: Referring to the previous illustration, find the time of flight of the ball.
If the particle passes two points situated at equal height y at t = t1
and t = t2, then referring to the
Figure we obtain y = (v0 sin0) t1 2 1 gt12 = (v0sin0) t2 - 2 1 gt22 …(1) Y X 0 O V t1 t2 y y h H v0 sin0 (t2 – t1) = 2 1 g (t22-t12) v0sin0 = T g sin v 2 Since ; 2 ) t t ( g 1 2 0 0 t1+ t2 = T Substituting v0sin0 = 2 ) t t ( g 1 2 in y = (v0 sin)
2
gt
t
2 1 1
we obtain gt1t2 2 1 ySpeed and Angle of Projection so that projectile Passes through Two Given Points
Referring the Figure, we see that the particle passes through two points P and Q having coordinates (x1, y1) and (x2, y2)
respectively. Setting these values of the coordinates in the equation of locus of the particle (trajectory equation) we obtain, y1 = (tan0) x1 - 0 2 2 0 2 1
cos
v
gx
2
1
Y X O v0 0 = ? P(x1, y1) Q(x2, y2) …(1) and y2 = (tan0 ) x2 - 0 2 2 0 2 2cos
v
2
gx
…(2) y1 x22 - y2 x12 = tan0 (x1 x22 – x2 x12) 2 1 2 2 2 1 2 1 2 2 2 1 1 0 x x x x x y x y tan .The speed of projection
Similarly y1x2 – y2x1 =
22 1 21 2
0 2 2 0 x x x x cos v 2 g v02 = ) y x y x ( 2 ) tan 1 ( ) x x ( x x g 2 1 1 2 0 2 1 2 2 1 Putting the value of tan0 =
2 1 2 2 2 1 2 1 2 2 2 1x
x
x
x
x
y
x
y
, we can obtain V0.Minimum Velocity of Projection required to Pass through a given point
When the particle is projected with velocity v at an angle 0 so as to pass
through a point P(x,y) as shown in figure, the trajectory equation is written as y = (tan0) x - 2 2 v 2 gx (1+tan2 0) gx2 (1+tan2 0) – 2v2x tan0 + 2v2y = 0 x P(x,y) x y tan2 0 - 0 gx ) gx y v 2 ( tan gx v 2 2 2 2 0 2
For real value of 0, the discriminant of this quadratic equation is greater than or equal
to zero. = 0 gx ) gx y v 2 ( 4 gx v 2 2 2 2 2 2 0 gx ) y v 2 gx ( x g v 2 2 2 2 2 4 v4 = g2 x2 – 2v2y g 0 Putting v2 = K we obtain k2 –(2gy) k - g2x2 0 k 2 x g 4 y g 4 ) gy 2 ( 2 2 2 2 k g (y + x2 y2) v2 g (y + x2 y2) v g(y x2 y2) v g(y x2 y2) min .
To find the corresponding angle of projection, known as critical angle of projection (0).
We have to differentiate the trajectory equation w.r.t. to obtain.
dy /d = tan0 (x ) d d v 2 ) tan 1 ( ) tan 1 ( d d v 2 gx ) (tan d d x d dx 2 2 2 0 2 2 2 0
Setting dy/d = 0 for constant y and 0for x d dx Constant we obtain 0 = x sec2 0 - (2tan sec ) v 2 gx 0 2 0 2 2 tan = gx v2
Putting 0 = (0) critical for v = vmin = g(y x2y2)
we obtain, tan(0)c = gx ) y x y ( g 2 2 x y x y tan ) ( 2 2 1 c 0
Position, Time and Speed at Any Angular Elevation
Let a particle be projected with v0 at an
angle 0. After a time t it moves with V at
an angle with horizontal as shown in the Figure. Since the horizontal components of velocity of a projectile remains constant (vx)0 = V0 cos0 = (Vx)
= Vcos cos cos v v 0 0 Y X O v0 0 = ? P(x, y) Q(x2, y2) vy vx v As we know that v2 = v
02 – 2gy, substituting the obtained value of V in this equation we
obtain y = 2g cos cos v v 2 0 0 2 0 02 22 0 cos cos 1 g 2 V y We know that Vy = (Vy)0 - gt
V sin = V0sin0 – gt, substituting V =
cos cos V0 0 we obtain, gt sin v sin cos cos V 0 0 0 0 t = cos sin cos sin g v 0 0 0 cos g ) sin( v t 0 0
The horizontal distance x covered during the time t is given as x = (V0 cos0) t x = cos g ) sin( v ) cos v ( 0 0 0 0 cos g cos ) sin( v x 20 0 0
Referring to the adjoining figure
When the velocity vector V becomes perpendicular to the initial velocity vector V0
0 V .
V 0 where V0 = V0 cos0 iˆ +V0 sin0 j
ˆ
and V = (V0cos0) iˆ + (gt – V0 sin0) (-ˆj )
{(V0cos0) iˆ + (V0sin0-gt) ˆj }.
{(V0cos0 iˆ + V0sin0 ˆj } = 0
(V0cos0)2 + (V0 sin0 – gt) sin0 = 0
Y X O v0 0 = ? P 900 v vx vy = /2+0
0 0 gSin V t .
Illustration 25 : A particle is projected at an angle 0 = 60° with horizontal. If after 3
seconds the angle between its velocity vector is reduced to half of its initial value, find the speed of projection (g = 10 m/sec2).
Solution: Directly from the derived expression, t =
cos g ) ( sin v0 0 ; since is reduced to half of 0, putting = 0/2, we obtain t =
) 2 / ( cos g ) 2 / ( sin v 0 0 0 v0 = gt tan (0/2). Putting 0 = 60° and t = 3,
we obtain v0 = 10 m/sec.
6.2 Horizontal Projection from a given height
Referring to the Figure, let a particle be projected with a horizontal velocity V0, which remains
constant along horizontal line due to the absence of any horizontal force. Due to earth’s gravitation the particle acquires vertical velocity Vy at any time t and at any position P(x,y).
Let the position vector of this point be
r
. Vy = (Vy)0 + gt.Since there is no vertical component of V0 initially, (Vy)0 = 0.
Vy = gt
and the vertical displacement is y = (Vy)0t + gt2 2 1 = gt2 2 1 Again, Vy2 = (Vy)02 + 2gy Putting (Vy)0 = 0 Vy = 2gy . Displacement
Now the horizontal displacement x = V0t and the vertical displacement y = 1/2 gt2. Since
the position vector r xiˆyjˆ, putting the values of x and y, we obtain,
jˆ ) gt 2 1 ( iˆ t V r 2 0 .
Therefore, 0 1 0 2 1 2 0 o 2 2 2 0 2V gt tan ) t V ( ) 2 / gt ( tan & V 2 gt 1 t V 2 gt ) t V ( r Again g y 2 t putting jˆ y iˆ t V r 0 we obtain, iˆ yjˆ g y 2 V r 0 2 2 0 y g y 2 V r 0 1 0 1 2 0 V 2 gy tan g / y 2 V y tan & 1 gy V 2 y r Velocity V = Vxiˆ - Vyˆj V = V0 iˆ – Vyˆj V V0iˆ gt jˆ ) gt V ( y . Therefore 0 1 2 0 0 2 2 0 V gt tan & V gt 1 V ) gt ( V V Again V V0iˆ 2gy jˆ (Vy 2gy). Therefore 0 1 2 0 V gy 2 tan & gy 2 V V Range
If y = H (height of the cliff or height of fall of the projectile), the corresponding horizontal distance (Range R) can be found by putting the values of time of fall t =
g H 2 in the equation x = V0t. x R V0 2gH Putting x = R and t = g H 2 etc.
We can find the average velocity and displacement of the particle during the motion when projected horizontally from the top of a cliff of height H with a speed V0.
Equation of Trajectory
The locus of the path of the particle is given as y = 0 2 V x t where gt 2 1 2 0 2 V 2 x g y It is a parabola.
Illustration 26: A ball is thrown from ground level so as to just clear a wall 4 metres high at a distance of 4 metres and falls at a distance of 14 metres from the wall. Find the magnitude and direction of the velocity of the ball.
Solution : Referring to figure, let P be a point on the trajectory whose coordinates are (4, 4). As the ball strikes the ground at a distance 14 metre from the wall, the range is 4 + 14 = 18 metre.
The equation of trajectory is u u sin v sin v cos h u cos X y = x tan - 2 1 g 2 2 2 cos u x or y = x tan tan . cos u 2 gx 1 2 2 or y = x tan g / cos sin u 2 x 1 2 = x tan 1 Rx . . . (1) Here x = 4, y = 4 and R = 18. 4 = 4 tan [1 – 4/18] = 4 tan (7/9) or tan = 9/7, sin = 130 9 and cos = 130 7 Again R = (2u2 sin cos /g)
= 130 7 130 9 u 8 . 9 2 2 u2 = 182. u = (182) m/s.
Exercise 23 : A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If and be the base angle and the angle of projection, prove that tan = tan + tan .
6.3 Calculation of Radius of curvature at any point on the path of a projectile Let at any time t, the velocity vector V be
inlined at an angle with horizontal at a point, say, P as shown in Figure . Since gravitational acceleration g is always acting vertically downwards, the components of g perpendicular to the velocity vector V can be treated as a radial acceleration towards, the centre of a circular path of radius (let ) as shown in Figure; is known as radius of curvature of the parabola at P. Y X O v0 0 P
y=x tan 0- gx2 (1+tan20)/2v02
vx v ar = g cos Effective circle r v a : a v 2 r r , when ar = g cos
is given as r 2 a V where ar = 2 V = r 2 a V where ar = g cos cos g V2
where v and can be found in terms of t as
v = 2 2 0 0 0 gt 2V gt sin v and = 0 0 0 1 cos v gt sin v tan
or v and can be found in terms of y as v = V2 2gy
0
and = tan
v 2sin 2gy v cos
0 0
2 0
1 .
Illustration 27: A boy is standing inside a train moving with a constant velocity of magnitude 10 m/sec. He throws a ball vertically up with a speed 5 m/sec relative to the train. Find the radius of curvature of the path of the ball just at the time of projection.
Solution: The ball continues to move horizontally with (vx)0 = 10 m/sec. It begins
to move up with (vy)0 = 5 m/sec. Therefore 0 is given as, 0 = tan-1
) v ( ) v ( 0 x 0 y 0 = tan-1 (5/10) = tan-1 (1/2).
Now the required radius of curvature is given as = cos g v2 putting v = v0 = y 02 2 0 x) (v ) v (
= 10252 = 5 5 m/sec, g = 10 m/sec2 and = tan-1 (1/2),
we obtain
14 m.Exercise 24: Referring to the previous illustration, if the boy releases the ball from rest, what will be the radius of curvature of the path at the instant of its release?
6.4 Projectile on an inclined plane
0 y v 0 x g 0 g cos 0
