RECTILINEAR MOTION MECHANICS Mechanics is the branch of physics which deals with the cause and effects of motion a particle, rigidtwo objects andnamely deformable bodies etc. Mechanics is of classified under streams Kinematics and Dynamics.
“E pur, si muove.”
Mechanics
(Translated: “Even so, it does move.”) When Galileo was declared a heretic by the Catholic Church and chose to recant his beliefs in order
The word kinematics means ‘science of motion’.branch of the mechanics which deals with
to spare his life, he is known to have said: “I must altogether that the Sun is the center
study of motion without going into the cause of motion, i.e.
of
force, torque etc.
abandon the false opinion the
world
and
immovable, and that the Earth is not the center of
1.
the world, and moves, and that I must not hold, defend, or teach... the said doctrine.” But afterwards, as he rose from his kneeling position, his body weary from seventy years of age, legend has it that he murmured, in Italian the above statement.
Dynamics( or Kinetics)
Kinematics
2.
It is branch of mechanics which is concerned about the causes (i.e. force , torque) that cause motion of bodies.
MOTION AND REST Motion is a combined property of the object and the observer. There is no meaning of rest or motion without the observer. Nothing is in absolute rest or in absolute motion. An object is said to be in motion with respect to a observer, if its position changes with respect to that observer. It may happen by both ways either observer moves or object moves.
RECTILINEAR MOTION Rectilinear motion is motion, along a straight line or in one dimension. It deals with the kinematics of a particle in one dimension.
2.1
Position
The position of a particle refers to its location in the space at a certain moment of time. It is concerned with the question − “ where is the particle at a particular moment of time ? ”
2.2
Displacement
The change in the position of a moving object is known as displacement . It is the vector joining the initial position of the particle to its final position during an interval of time.
2.3
Distance
The length of the actual path travelled by a particle during a given time interval is called as distance. The distance travelled is a scalar quantity which is quite different from displacement. In general, the distance travelled between two points may not be equal to the magnitude of the displacement between the same points.
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Example 1.
Ram takes path 1 (straight line) to go from P to Q and Shyam takes path 2 (semicircle). 2
(a) Find the distance travelled by Ram and Shyam? (b) Find the displacement of Ram and Shyam? Sol.
(a)
P
Distance travelled by Ram = 100 m
1 100 m
Q
Distance travelled by Shyam = π(50 m) = 50π m (b)
Displacement of Ram = 100 m Displacement of Shyam = 100 m
2.4
Average Velocity (in an interval) :
The average velocity of a moving particle over acertain time interval is defined as the displacement divided by the lapsed time. Average Velocity = displaceme nt time int erval for straight line motion, along x−axis, we have ∆x
vav = v =
=
∆t
=
xf − xi tf − ti
The average velocity is a vector in the direction of displacement. For motion in a straight line, directional aspect of a vector can be taken care of by +ve and -ve sign of the quantity.
2.5
Average Speed
(in an interval)
Average speed is defined as the total path length travelled divided by the total time interval during which the motion has taken place. It helps in describing the motion along the actual path. Average Speed =
distance travelled time interval
NOTE : (a) Average speed is always positive in contrast to average velocity which being a vector, can be positive or negative. (b)
If the motion of a particle is along a straight line and in same direction then, average velocity = average speed.
(c)
Average speed is, in general, greater than the magnitude of average velocity. The dimension of velocity and speed is [LT-1] and their SI unit is meters per second (m/s)
Example 2.
Sol.
(a)
In the example 1, if Ram takes 4 seconds and Shyam takes 5 seconds to go from P to Q, find (a) Average speed of Ram and Shyam? (b) Average velocity of Ram and Shyam? Average speed of Ram =
100
m/s = 25 m/s
4
50π Average speed of Shyam = (b)
Average velocity of Ram =
100
Average velocity of Shyam = Example 3. Find Sol.
m/s = 10 π m/s
5
m/s = 25 m/s
4
100 5
m/s = 20 m/s
A particle travels half of total distance with speed v 1 and next half with speed v2 along a straight line. out the average speed of the particle? Let total distance travelled by the particle be 2s. s
Time taken to travel first half =
v1 s
Time taken to travel next half =
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v
2
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Average speed =
Q.1
Total distance covered Total time taken
=
2s s v1
+
s v2
2v1v 2
= v +v 1 2
3 1 of total distance with speed v 1 and next with v2. Find the average speed of the 4 4
A particle covers particle?
Ans. Q.2.
4v 1v 2 v 1 + 3v 2 A car is moving with speed 60 Km/h and a bird is moving with speed 90 km/h along the same direction as shown in figure. Find the distance travelled by the bird till the time car reaches the tree?
Ans.
240 m
360 m
2.6
Instantaneous Velocity (at an instant) : The velocity at a particular instant of time is known as instantaneous velocity. The term “velocity” usually means instantaneous velocity.
dx ⎛ ∆x ⎞ ⎟ = ⎝ ∆t ⎠ dt
⎜ Vinst. = ∆lim t →0
In other words, the instantaneous velocity at a given moment (say , t) is the limiting value of the average velocity as we let ∆t approach zero. The limit as ∆t → 0 is written in calculus notation as dx/dt and is called the derivative of x with respect to t.
2.7
Average acceleration (in an interval): The average acceleration for a finite time interval is defined as : Average acceleration = change in velocity
time int erval
Average acceleration is a vector quantity whose direction is same as that of the change in velocity. ∆v r
r
a av =
vf − vi r
=
r
∆t ∆t Since for a straight line motion the velocities are along a line, therefore
aav =
∆v
vf − vi
= tf − ti ∆t (where one has to substitute vf and vi with proper signs in one dimensional motion)
2.8
Instantaneous Acceleration (at an instant):
The instantaneous acceleration of a particle is its acceleration at a particular instant of time. It is defined as the derivative (rate of change) of velocity with respect to time. We usually mean instantaneous acceleration when we say “ acceleration”. For straight motion we define instantaneous acceleration as : a=
dv ⎛⎜ ∆v ⎞⎟ = ∆lim t →0 ⎝ ∆t ⎠ dt r
dv
⎛ ∆v ⎞ r
⎜ ⎟ and in general a = = lim ∆t →0 ⎝ ∆t ⎠ dt r
The dimension of acceleration is [LT-2] and its SI unit is m/s2.
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Example 4.
Sol.
Position of a particle as a function of time is given as x = 5t 2 + 4t + 3. Find the velocity and acceleration of the particle at t = 2 s?
V elocity;
v=
At t = 2 s
dx = 10t + 4 dt ⇒
v=10(2)+4
Acceleration;
a=
d2 x dt 2
v = 24 m/s
= 10
Acceleration is constant, so at t = 2 s a = z10 m/s2
Ans.
The position of a particle moving on X-axis is given by x = At3 + Bt 2 + Ct + D. The numerical values of A, B, C, D are 1, 4, -2 and 5 respectively and SI units are used. Find (a) the dimensions of A, B, C and D, (b) the velocity of the particle at t = 4 s, (c) the acceleration of the particle at t = 4s, (d) the average velocity during the interval t =0 to t = 4s, (e) the average acceleration during the interval t = 0 to t = 4 s. [(a) [A] = [LT -3], [B] = [LT-2], [C] = [LT-1] and [D] = [L] ; (b) 78 m/s ; (c) 32 m/s 2 ; (d) 30 m/s ; (e) 20 m/s 2]
3.
MO TION WITH UNIF ORM VEL OCIT Y
Q.3
Consider a particle moving along x −axis with uniform velocity u starting from the point x = x i at t = 0. Equations of x, v, a are : x (t) = x i + ut ; v (t) = u ; a (t) = 0 • • •
x− t graph is a straight line of slope u through xi. as velocity is constant, v − t graph is a horizontal line. a−t graph coincides with time axis because a = 0 at all time instants. x xi
slo pe =
u
u is negative t
O
v u
v
positive velocity
t
O
4.
t
O u
negative velocity
UNIF ORML Y AC CE LE RATED MO TION If a particle is accelerated with constant acceleration in an interval of time, then the motion is termed as uniformly accelerated motion in that interval of time. For uniformly accelerated motion along a straight line (x−axis) during a time interval of t seconds, the following important results can be used. (a) (b)
v=u+at s = ut + 1/2 at 2 s = vt − 1/2 at2 xf = x i + ut + 1/2 at 2 (c) v2 = u2 + 2as (d) s = 1/2 (u + v) t (e) sn = u + a/2 (2n − 1) u = initial velocity (at the beginning of interval) a = acceleration v = final velocity (at the end of interval) s = displacement (xf − xi) xf = final coordinate (position) xi = initial coordinate (position) sn = displacement during the n th sec
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Example 5. A particle moving rectilinearly with constant acceleration is having initial velocity of 10 m/s. After some time, its velocity becomes 30 m/s. Find out velocity of the particle at the mid point of its path? Sol. Let the total distance be 2x. ∴ distance upto midpoint = x Let the velocity at the mid point be v and acceleration be a. From equations of motion v2 = 10+2 2ax ____ (1) 302 = v+2 2ax ____ (2) (2) - (1) gives v2 - 302 = 102 - v2
v2 = 500
⇒
⇒
v = 10 5 m/s
Example 6. A police inspector in a jeep is chasing a pickpocket an a straight road. The jeep is going at its maximum speed v (assumed uniform). The pickpocket rides on the motorcycle of a waiting friend when the jeep is at a distance d away, and the motorcycle starts with a constant accelerationa.Showthatthepick Sol.
pocketwillbecaughtif
v
≥
2ad .
Suppose the pickpocket is caught at a time t after motorcycle starts. The distance travelled by the motorcycle during this interval is
s=
1 2 at 2
____ (1)
During this interval the jeep travels a distance
s + d = vt By (1) and (2),
1 2 at + d = vt 2
or,
t=
____ (2)
v ± v 2 − 2ad a
The pickpocket will be caught if t is real and positive.
v 2 ≥ 2ad or,
This will be possible if
v≥
2ad
Q.4
A car deccelerates from a speed of 20 m/s to rest in a distance of 100 m. What was its acceleration, assumed constant? Ans. [- 2 m/s 2]
Q. 5
A 150 m long train accelerates uniformly from rest. If the front of the train passes a railway worker 50 m away from the station at a speed of 25 m/s, what will be the speed of the back part of the train as it passes the worker? Ans. [50 m/s]
5.
GRAPHS IN UNIF ORMLY ACCE LER ATED MO TION (a ≠ 0) ACC ELE RA •
x is a quadratic polynomial in terms of t. Hence x − t graph is a parabola. x
x a<0
xi
xi
a>0 t
0
•
t
0
x-t graph v is a linear polynomial in terms of t. Hence v −t graph is a straight line of slope a. v e op sl
=
v
a
u
slo pe =
a
u a is positive 0
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t
a is negative 0
t
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•
v-t graph a−t graph is a horizontal line because a is constant. a a
a
positive acceleration
0 a
t
0
negative acceleration
a-t graph
6.
REACTION TIME When a situation demands our immediate action. It takes some time before we really respond. Reaction time is the time a person takes to observe, think and act.
7.
DIRECTIONS OF VECTORS IN STRAIGHT LINE MOTION In straight line motion, all the vectors (position, displacement, velocity & acceleration) will have only one component (along the line of motion) and there will be only two possible directions for each vector. •
For example, if a particle is moving in a horizontal line (x −axis), the two directions are right and left. Any vector directed towards right can be represented by a positive number and towards left can be represented by a negative number. + line of motion
•
+
For vertical or inclined motion, upward direction can be taken +ve and downward as−ve
-
•
n o ti o m f o e n li
+
-
e lin
n io ot m f o
For objects moving vertically near the surface of the earth, the only force acting on the particle is its weight gravitational pull2 of thesign, earth. Hence the acceleration this type ofare motion will always (mg) be a i.e. = −gthei.e. a = − 9.8 m/s (−ve because force andfor acceleration directed downwards, If we select upward direction as positive).
NOTE : (a) If acceleration is in same direction as velocity, then speed of the particle increases. (b) If acceleration is in opposite direction to the velocity then speed decreases i.e. the particle slows down. This situation is known as retardation. Example 7.
Sol.
(a)
Mr. Sharma brake his car with constant acceleration from a velocity of 25 m/s to 15 m/s over a distance of 200m. (a) How much time elapses during this interval? (b) What is the acceleration? (c) If he has to continue braking with the same constant acceler ation, how much longer would it take for him to stop and how much additional distance would he cover? We select positive direction for our coordinate system to be the direction of the velocity and choose the srcin so that x i = 0 when the braking begins. Then the initial velocity is ux = +25 m/s at t = 0, and the final velocity and position are vx = +15 m/s and x = 200 m at time t. Since the acceleration is constant, the average velocity in the interval can be found from the average of the initial and final velocities. ∴
vav, x =
1 2
(ux + vx) =
1
2
(15 + 25) = 20 m/s.
The average velocity can also be expressed as v av, x =
∆x ∆t
. With ∆x = 200 m and ∆t = t − 0,
we can solve for t: t=
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Δx 200 v av,x = 20 = 10 s.
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(b)
We can now find the acceleration using v
x
= ux + axt
v x − ux 15 − 25 ax = = = − 1 m/s2. t 10 The acceleration is negative, which means that the positive velocity is becoming smaller as brakes are applied (as expected). (c)
Now with known acceleration, we can find the total time for the car to go from velocity u x = 25 m/s to vx = 0. Solving for t, we find t=
v x − ux 0 − 25 = = 25 s. ax −1
The total distance covered is x = xi + uxt + 1 axt2 = 0 + (25)(25) + 1 (−1)(25)2 = 625 − 312.5 = 312.5 m.
2
2
Additional distance covered = 312.5 – 200 = 112.5 m. Example 8. Sol.
A particle is dropped from a tower. It is found that it travels 45 m in the la st second of its journey. Find out the height of the tower?
Let the total time of journey be n seconds.
sn
Using;
=
a ( 2n − 1) 2
u+
45 = 0 +
10 ( 2 n − 1) 2
n = 5 sec
Height of tower; h=
1 2 gt 2
=
1 2 × 10 × 5 = 125 m 2
Example 9.
A particle is dropped from height 100 m and another particle is projected vertically up with velocity 50 m/s from the ground along the same line. Find out the position where two particle will meet?
Sol.
Let the upward direction as positive. Let the particles meet at a distance y from the ground. For particle A, y0 = + 100 m
y=100m A
u=0 m/s
u = 0 m/s y=0m
a = − 10 m/s2 y = 100 + 0(t) − = 100 - 5t 2 For particle B, y0 = 0 m
1 × 10 × t2 2
[y = y0 + ut +
u=50 m/s
B
1 2 at ] 2
---- (1) ⇒
u = + 50 m/s
⇒
a = − 10 m/s2
y = 50(t) − 1 × 10 × t2 2 = 50t − 5t2 ---- (2) According to the problem; 50t − 5t2 = 100 − 5t2 t = 2 sec Putting t = 2 sec in eqn. (1), y = 100 − 20 = 80 m Hence, the particles will meet at a height 80 m above the ground.
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Q. 6
A particle is thrown vertically with velocity 20 m/s . Find (a) the distance travelled by the particle in first 3 seconds, (b) displacement of the particle in 3 seconds. Ans. [25m, 15m]
Q.7
A stone is dropped from a balloon going up with a uniform velocity of5 m/s. If the balloon was50 m high when the stone was dropped, find its height when the stone hits the ground. Take g = 10 m/s 2. Ans. [68.5 m] NOTE:- As the particle is detached from the balloon it is having the same velocity as that of balloon, but its acceleration is only due to gravity and is equal to g.
8.
GRAPHIC AL INTERPRET ATION OF SOME QU ANTITIES 8.1
Average Velocity
If a particle passes a point P (xi) at time t = ti and reaches Q (xf) at a later time instant t = tf , its average velocity in the interval PQ is V av =
∆x ∆t
=
x f − xi t f − ti
x xf
This expression suggests that the average velocity is equal to the slope of the line (chord) joining the points corresponding to P and Q on the x −t graph.
Q
xi
P
O
8.2
ti
t
tf
Instantaneous Velocity
Consider the motion of the particle between the two points P and Q on the x −t graph shown. As the point Q is brought closer and closer to the point P, the time interval between PQ ( ∆t, ∆t ′, ∆t′′,......) get progressively smaller. The average velocity for each time interval is the slope of the appropriate dotted line (PQ, PQ ′, PQ′′......). As the point Q approaches P, the time interval approaches zero, but at the same time the slope of the dotted line approaches that of the tangent to the curve at the point P. As ∆t → 0, Vav (=∆x/∆t) → Vinst.
Geometrically, as ∆t → 0, chord PQ → tangent at P.
Hence the instantaneous velocity at P is the slope of the tangent at P in the x − t graph. When the slope of the x − t graph is positive, v is positive (as at the point A in figure). At C, v is negative because the tangent has negative slope. The instantaneous velocity at point B (turning point) is zero as the slope is zero.
8.3
Instantaneous Acceleration
The derivative of velocity with respect to time is the slope v
of the tangent in velocity time (v −t) graph. a=0 a>0
a<0
a=0 0
t a=0
Displanncement from v - t graph graph 8.4 Displacement = ∆x = area under v−t graph. Since a negative velocity causes a negative displacement, areas below the time axis are taken negative. Insimilar way, can see that∆ v = a ∆t leads to the conclusion that area under a − t graph gives the change in velocity ∆v during that interval. RESONANCE
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Example 10.
Describe the motion shown by the following velocity-time graphs.
(a)
Sol.
(b)
(a)
During interval AB: velocity is +ve so the particle is moving in +ve direction, but it is slowing down as acceleration (slope of v-t curve) is negative. During interval BC: particle remains at rest as velocity is zero. Acceleration is also zero. During interval CD: velocity is -ve so the particle is moving in -ve direction and is speeding up as acceleration is also negative.
(b)
During interval AB: particle is moving in +veindire with constant velocity and is zero. During interval BC: particle is moving +vection direction as velocity is +ve, butacceleration it slows down until it comes to rest as acceleration is negative. During interval CD: velocity is -ve so the particle is moving in -ve direction and is speeding up as acceleration is also negative.
Important Points to Remember •
For uniformly accelerated motion (a ≠ 0), x −t graph is a parabola (opening upwards if a > 0 and opening downwards if a < 0). The slope of tangent at any point of the parabola gives the velocity at that instant.
•
For uniformly accelerated motion (a≠ 0), v−t graph is a straight line whose slope gives the acceleration of the particle.
•
In general, the slope of tangent in x−t graph is velocity and the slope of tangent in −vt graph is the acceleration.
•
The area under a−t graph gives the change in velocity.
•
The area between the v−t graph gives the distance travelled by the particle, if we take all areas as positive.
•
Area under v−t graph gives displacement, if areas below the t −axis are taken negative.
Example 11. For a particle moving along x-axis, velocity-time graph is as shown in figure. Find the distance travelled and displacement of the particle? Sol.
Distance travelled = Area under v-t graph (taking all areas as +ve.) ∴ Distance travelled = Area of trapezium + Area of triangle =
1 2
(2 + 6 )× 8
+
1 2
× 4×5
=32+10 =42m Displacement = Area under v-t graph (taking areas below time axis as -ive.) ∴ Displacement = Area of trapezium − Area of triangle =
1 2
(2 + 6 )× 8
−
1 2
× 4×5
= 32 −1 0 =22m Hence, distance travelled = 42 m and displacement = 22 m. Q.8
For a particle moving along x-axis, following graphs are given. Find the distance travelled by the particle in 10 s in each case?
Ans. [(a) 100m; (b) 50m]
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9.
Interpretation of some more Graphs 9.1
Position vs Time graph 9.1.1 Zero Velocity As position of particle is fix at all the time, so the body is at rest.
dx
Slope;
dt
= tanθ = tan 0º = 0
Velocity of particle is zero 9.1.2
Uniform Velocity
Here tanθ is constant
∴ ∴
dx dt
tan θ =
dx dt
is constant.
velocity of particle is constant.
9.1.3 Non uniform uniform velocity velocity (increasing with time) In this case; As time is increasing, θ is also increasing. ∴
dx dt
= tanθ is also increasing
Hence, velocity of particle is increasing.
9.1.4 Non uniform velocity (decreasing with time) In this case; As time increases, θ decreases. ∴
dx
= tanθ also decreases.
dt Hence, velocity of particle is decreasing.
9.2
Velocity vs time graph 9.2.1 Zero acceleration Velocity is constant. tanθ = 0 ∴
dv dt
=0
Hence, acceleration is zero. 9.2.2
Uniform acceleration
tanθ is constant. ∴
dv dt
= constant
Hence, it shows constant acceleration. 9.2.3
Uniform retardation
∴
Since θ > 90º tanθ is constant and negative.
∴
dv dt
= negative constant
Hence, it shows constant retardation.
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9.3
Acceleration vs time graph 9.3.1
Constant acceleration
tanθ = 0
da
∴
dt
=0
Hence, acceleration is constant. 9.3.2
Uniformly increasing acceleration
is constant. 0º < θ < 90º ⇒ tanθ > 0 θ
da
∴
dt
= tanθ = constant > 0
Hence, Uniformly acceleration is uniformlyacceleration increasing with time. 9.3.3 decreasing ∴
∴
Since θ > 90º tanθ is constant and negative.
da dt
= negative constant
Hence, acceleration is uniformly decreasing with time Example 12.
The displacement vs time graph of a particle moving along a straight line is shown in the figure. Draw velocity vs time and acceleration vs time graph.
Sol:
x = 4t2
⇒
v=
dx = 8t dt
Hence, velocity-time graph is a straight line having slope i.e. tanθ = 8.
a=
dv =8 dt
Hence, acceleration is constant throughout and is equal to 8.
Problem 10. The displacement vs time graph of a particle moving along a straight line is shown in the figure. Draw velocity vs time and acceleration vs time graph.
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Page # 11
11.
MO TION WITH NON-UNIF ORM ACCELERA TION (use of def init e int egrals) tf ∆x
∫ v( t )dt
=
(displacement in time interval t = t i to tf)
ti
The expression on the right hand side is called the definite integral of v(t) between t = ti and t = tf. . Similarly change in velocity tf
∆v
∫ a(t )dt
= vf − vi =
ti
Table 2 : Some quantities defined as derivatives and integrals.
dx v = slope of x −t graph
v(t) = dt
dv
a (t) =
dp
F (t) =
∆x
∫ dx
=
a = slope of v−t graphs
dt
F = slope of p−t graph (p = linear momentum)
dt
tf
∫ v ( t ) dt
=
∆x = area under v−t graph
ti
∆v
tf
∫
=
∫ a ( t ) dt
dv =
∆v = area under a −t graph
ti
∆p
∫
=
tf
∫ F ( t ) dt
dp =
∆p = area under F−t graph
ti
W=
∫ dW =
xf
∫ F (x) dx
W = area under F−x graph
xi
12.
SO LVING PR OBLEMS WHICH INV OLVES NONUNIF NONUNIF ORM ACCELERA TION 12.1
Acceleration depending on velocity v or time t
dv . If a is in terms of t, dt
By definition of acceleration, we have a = v
terms
ofv,
∫ v0
x
∫ dx
using
t
v0
0
t
∫ dt . On integrating, we get a relation between v and t, and then 0
t
=
x
∫ v ( t ) dt , x and t can also be related. 0
0
12.2
dv = a (v )
v
∫ dv = ∫ a ( t ) dt . If a is in
Acceleration depending on velocity v or position x a=
dv
⇒
dt
a=
dv dx
⇒
dx dt
a=
dx dv dt dx
⇒
a =v
dv dx
This is another important expression for acceleration. x
v
If a is in terms of x,
∫ v dv
v0
RESONANCE
=
∫ a ( x ) dx
.
x0
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Page # 12
v
v dv
x
∫ a(v ) = ∫ dx
If a is in terms of v,
v0
x0
x
On integrating, we get a relation between x and v. Using
∫ x0
t
dx =
v (x)
∫ dt , we can relate x and t. 0
An object starts from rest at t = 0 and accelerates at a rate given by a = 6t. What is (a) its velocity and (b) its displacement at any time t? As acceleration is given as a function of time,
Example 13.
Sol.
v(t)
t
dv =
∴
∫
a(t)dt
∫
v(t 0 )
t0
Here t0 = 0 and v(t0) = 0 t
∴
v(t) =
∫ 6tdt 0
So,
⎛ t2 ⎞ t ⎟⎟ ⎝ 2 ⎠0
=6(
⎛ t3 ⎞ t ⎟⎟ ⎝ 3 ⎠0
= 3⎜⎜
= 6⎜⎜
t2 2
- 0)
=3t
2
2
v(t)=3t t
As
∆x =
∫ v(t)dt t0 t
∴
∫
2
∆x = 3t dt
= 3⎜⎜
0
⎛ t3 ⎝3
⎞
− 0⎟⎟
⎠
= t3
Hence, velocity v(t) = 3t2 and displacement ∆x = t 3 2
For a particle moving along x-axis, acceleration is given as a = 2v . If the speed of the particle is v at 0 x = 0, find speed as a function of x.
Q.9.
Ans. [ v = v0 e Q . 10.
2x
]
For a particle moving along x-axis, velocity is given as a function of time as v = 2t 2 + sint. At t = 0, particle is at srcin. Find the position as a function of time? Ans. [ x =
RESONANCE
2 3
t 3 − cos(t) + 1 ]
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Page # 13
SUMMARY Rectilinear motion is motion, along a straight line or in one dimension.
Rectilinear Motion Motion:
Displacement : The vector joining the initial position
of the particle to its final position during an interval of time. Distance : The length of the actual path travelled by
Maxima and and Minima Minima
Conditions for maxima are:−
dy
<0 dx 2 Conditions for minima are:−
dx
a particle during a given time interval
dy
xf − xi displaceme nt Average Velocity = = tf − ti time int erval
dx
distance travelled Average Speed = time interval Instantaneous Velocity : Vinst. =
d2 y dx 2
dx ⎛ ∆x ⎞ ⎟ = ⎝ ∆t ⎠ dt
=
∫ v( t )dt
ti
lim ⎜
∆t → 0
>0
tf
tf
∆v
= vf − vi =
∫ a(t )dt ti
v f − vi change in velocity = = time int erval t f − ti Acceleration
a=
= 0(b)
Motion with Non-Uniform Acceleration ∆x
Average Acceleration
Instantaneous
d2 y
= 0 (b)
dv dt
Solving Problems which Involves Nonuniform Acceleration
:
= lim
∆t →0
v
⎛ ∆v ⎞ ⎜ ⎟ ⎝ ∆t ⎠
If a is in terms of t,
v
(a) v = u + at (b) s = ut + 1/2 at 2 s = vt − 1/2 at2 xf = x i + ut + 1/2 at2 (c) v2 = u2 + 2as (d) s = 1/2 (u + v) t (e) sn = u + a/2 (2n − 1)
If a is in terms of v,
Important Points to Remember
If a is in terms of v,
•
For uniformly accelerated motion (a≠ 0), x−t graph is a parabola (opening upwards if a > 0 and opening downwards if a < 0). The slope oftangent at any point of the parabola gives the velocity at that instant.
•
For uniformly accelerated motion (a≠ 0), v−t graph is a straight line whose slope gives the acceleration of the particle.
•
In general, the slope of tangent in x−t graph is velocity and the slope of tangent in v − t graph is the acceleration.
•
The area under a−t graph gives the change in velocity.
•
The area between the v−t graph gives the distance travelled by the particle, if we take all areas as positive.
•
Area under v−t graph gives displacement, if areas below the t−axis are taken negative.
t
∫ v0
If a is in terms of x,
∫ a ( t ) dt
=
v0
Equations of Motion
RESONANCE
∫ dv
0
dv = a (v )
v
∫ v dv
t
∫ dt 0
x
∫ a (x ) dx
=
x0
v0 v
v dv
x
∫ a(v ) = ∫ dx
v0
x0
TM
Page # 14
.
E xer ci se 1 OBJECTIVE PROBLEMS 1.
(RECTILINEAR MOTIN)
BA T CH - CC
A motor car is going due north at a speed of 50 km/h. It makes a 90º left turn without changing the speed. The change in the velocity of the car is about (A)50km/htowardswest
(B)50
(C) 50 2 k m/h towards north-west
2 km/h towards south-west
(D)zero
2.
A particle has a velocity u towards east at t = 0. Its acceleration is towards west and is constant, Let x A and xB be the magnitude of displacements in the first 10 seconds and the next 10 seconds. (A) xA < xB (B) xA = xB (C) xA > xB (D) the information is insufficient to decide the relation of x with x .
3.
A ball takes t seconds to fall from a height h 1 and 2t seconds to fall from a height h 2. Then h1/h2 is (A) 2 (B) 4 (C) 0.5 (D) 0.25
4.
A body starts from rest and is uniformly acclerated for 30 s. The distance travelled in the first 10 s is x 1, next 10 s is x2 and the last 10 s is x 3. Then x1 : x2 : x3 is the same as (A) 421 : (B) 5 2 1: (C)5 3 1: (D)9 3 1:
5.
A stone is released from an elevator going up with an acceleration a. The acceleration of the stone after the release is (A)a upward (B) (g-a) upward (C) (g-a) downward (D) g downward
6.
A person standing near the edge of the top of a building throws two balls A and B. The ball A is thrown vertically downward and the ball B is thrown vertically upward with the same speed. The ball A hits the ground with a speed υ and the ball B hits the ground with a speed υ . We have A B (A) υ A > υ B (B) υ A < υ B (C) υ A = υ B (D) the relation between A and B depends on height of the building above the ground.
7.
Figure shows position-time graph of two cars A and B. (A) Car A is faster than car B. (B) Car B is faster than car A. (C) Both cars are moving with same velocity. (D) Both cars have positive acceleration.
A
8.
9.
B 5 x
0
v
v
(B)
D E
F
C t
v
C)
(D)
S
S
S
S
The displacement time graphs of two particles A and B are straight lines making angles of respectively 300 and 600 vA with the time axis. If the velocity of A is v A and that of B is v B, then the value of is vB (A) 1/2
(B) 1/
(C) 3
3
(D) 1/3
The initial velocity of a particle is u (at t=0) and the acceleration f is given by (f = at). Which of the following relations is valid? (A) v = u + at
12 .
A
A particle starts from rest and moves along a straight line with constant acceleration. The variation of velocity v with displacement S is :
(A)
11.
x(m)
The displacement–time graph of a moving particle is shown below. The instantaneous velocity of the particle is negative at the point : (A) C (B) D (C) E (D) F
v
10.
B
2
(B) v = u +
at 2 2
(C)v=u+at
(D)v=u
A stone is dropped into a well in which the level of water is h below the top of the well. If v is velocity of sound, the time T after which the splash is heard is given by (A)T=2h/v
RESONANCE
(B)
T
=
2h g
+
h v
(C) T
=
2h g
+
h 2v
(D) T
=
h 2g
+
2h v
TM
Page # 15
13 .
A body is released from the top of a tower of height h metre. It takes T seconds to reach the ground. Where is the ball at the time T/2 seconds ? (A) at h/4 metre from the ground (B) at h/2 metre from the ground (C) at 3h/4 metre from the ground (D) depend upon the mass of the ball
14 .
A stone is thrown vertically upward with an initial velocity 3u. The height of the tower is: (A)
15 .
3u2 g
(B)
velocity u from the top of a tower, reaches the ground with a
4u2 g
(C)
A particle start s from rest with uniform acceleration a. Its body in the last two seconds is : (A)
2v(n - 1) n
(B)
v(n - 1) n
(C)
6u2 g
(D)
9u2 g
velocity after n seconds is v. The displacement of the
v(n + 1) n
(D)
2v(2n + 1) n
16.
Consider the motionisofzero the tip of the minute hand of (B) a clock. In onecovered hour is zero (A) the displacement the distance (C) the average speed is zero (D) the average velocity is zero
17.
A particle moves along the X-axis as x = u(t – 2) + a(t – 2)2 (A) the initial velocity of the particle is u (B) the acceleration of the particle is a (C) the acceleration of the particle is 2a (D) at t =2s particle is at the origin.
18.
Mark the correct statements for a particle going on a straight line: (A) If the velocity and acceleration have opposite sign, the object is slowing down. (B) If the position and velocity have opposite sign, the particle is moving towards the srcin. (C) If the velocity is zero at an instant, the acceleration should also be zero at that instant. (D) If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.
19.
The velocity of a particle is zero at t = 0 (A) The acceleration at t = 0 must be zero (B) The acceleration at t = 0 may be zero. (C) If the acceleration is zero from t = 0 to t = 10 s, the speed is also zero in this interval. (D) If the speed is zero from t = 0 to t = 10 s the acceleration is also in the interval.
20.
Mark the correct statements: (A) The magnitude of the velocity of a particle is equal to its speed. (B) The magnitude of average velocity in an interval is equal to its average speed in that interval. (C) It is possible to have a situation in which the speed of a particle is always zero but the average speed is not zero (D) It is possible to have a situation in which the speed of the particle is never zero but the average speed in an interval is zero.
21.
v(m/s)
The velocity-time plot for a particle moving on a straight line is shown in fig. (A) The particle has constant acceleration (B) The particle has never turned around. (C) The particle has zero displacement (D) The average speed in the interval 0 to 10s is the same as the average speed in the interval 10s to 20s.
10
t(s)
0 10
-10
20 30
-20
SUBJECTIVE PROBLEMS 1.
A particle covers each 1/3 of the total distance with speed v 1, v 2 and v3 respectively. Find the average speed of the particle ?
2.
The position of a particle moving on x-axis is given by x = 4t3 + 3t2 + 6t + 4. Find (a) The velocity and acceleration of particle at t = 5 s. (b) The average velocity and average acceleration during the interval t = 0 to t = 5 s, x = 4t 3 + 3t2 + 6t + 4 (a) The velocity and acceleration of particle at t = 5 s. (b) The average velocity and average acceleration during the interval t = 0 to t = 5 s.
3.
A train starts from rest and moves a constant acceleration of 2.0 m/s for half abyminute. The(b) brakes are then applied and the train comes to rest with in one minute. Find (a) the total distance moved the train, the maximum speed attained by the train and (c) the position(s) of the train at half the maximum speed.
4.
A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/s and one second later the speed becomes 150 m/s. Find (a) the acceleration and (b) the distance travelled during the (t+1) th second. v(m/s)
5.
For a particle moving along x-axis, following graphs are given. Find the distance travelled by the particle in 10 s in each case.
2
x(m)
10
10
2 10
(A)
RESONANCE
t(s)
0
10
t(s)
(B)
TM
Page # 16
6.
For a particle moving along x-axis, velocity-time graph is as shown in figure. Find the distance travelled and displacement of the particle? Also find the average velocity of the particle?
7.
The acceleration of a cart started at t = 0, varies with time as shown in figure . Find the distance travelled in 30 seconds and draw the positiontime graph.
8.
For a particle moving rectilinearly, acceleration as a function of speed is given as a = 8v2. Find the speed as a function of x if the particle is having a speed of v 0 at x = 0?
9.
Under what conditions does the magnitude of the average velocity equal to the average speed.
10.
Can an object have increasing speed but its acceleration decreases? If yes, give an example; if not, explain why?
11.
Figure shows four paths along which obejcts move from a starting point to a final point (particle is moving along the same straight line), all in the same time. The paths pass over a grid of equally spaced straight lines. Rank the paths according to (a) the average velocity of the objects and (b) the average speed of the objects, greatest first.
12 .
A man walking with a speed ' v ' constant in magnitude and direction passes under a lantern hanging at a height H above the ground. Find the velocity with which the edge of the shadow of the man's head moves over the ground, if his height is ' h '.
13.
An elevator is descending with uniform acceleration. To measure the acceleration , a person in the elevator drops a coin at the moment the elavator starts. The coin is 6 ft above the floor of the elevator at time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator. [Take g = 32 ft/s2]
14 .
When a model rocket is launched, the propellant burns for a few seconds, accelerating the rocket upward. After burnout, the rocket moves upward for a while and then begins to fall. A parachute opens shortly after the rocket starts down. The parachute slows the rocket to keep it from breaking when it lands. The figure here shows velocity data from the flight of the model rocket. Use the data to answer the following. (a) How fast was the rocket climbing when the engine stopped? (b) For how many seconds did the engine burn? (c) When did the rocket reach its highest point? W hat was its velocity then? (d) When did the parachute open up ? How fast was the rocket falling then? (e) How long did the rocket fall before the parachute opened? (f) When was the rocket's acceleration greatest? (g) When was the acceleration constant? What was its value then (to the nearest interger)?
1.
A particle of mass 10 F(x) = (i ) (ii)
2.
−
−2
kg is moving along the positive x
− axis
under the influence of a force
K where K = 10 2 N m 2. At time t = 0 it is at x = 1.0 m and its velocity is v = 0. Find 2x 2 its velocity when it reaches x = 0.50 m the time at which it reaches x = 0.25 m. [ JEE '98, 8 ] −
In 1.0 sec. a particle goes from point A to point B moving in a semicircle of radius 1.0 m. The magnitude of average velocity is: [JEE '99, 2] (A)3.14m/sec (B)2.0m/sec (C)1.0m/sec (D)zero
RESONANCE
TM
Page # 17
A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height d/2. Neglecting subsequent motion and air resistance, its velocity v varies with the height h above the ground as [ JEE '2000, 3 ]
3.
(A)
(B)
(C)
(D)
A block is moving down a smooth inclined plane starting from rest at time t = 0. Let S n be the distance travelled by
4.
the block in the interval t = n – 1 to t = n. The ratio
(A)
2n − 1 2n
(B)
Sn is Sn+1
2n − 1 2n + 1
(C)
[JEE Scr. 2004, 3]
2n + 1 2n − 1
(D)
5.
A particle is initially at rest, It is subjected to a linear acceleration a , as shown in the figure. The maximum speed attained by the particle is [JEE Scr. 2004; 3] (A)605m/s (B)110m/s (C)55m/s (D)550m/s
6.
The velocity displacement graph of a particle moving along a straight line is shown. The most suitable acceleration-displacement graph will be [JEE Scr. 2005; 3]
(A)
(B)
(C)
EXERCISE # 1 OBJECTIVE PROBLEMS 1. B 2. D 5. D 6. C B 9. 10. D 13 . C 14. B 17. C,D 18. A,B,D
3. D 7. C 11. B 15. A 19. B,C,D
4. C 8. C 12. B 16. A,D 20. A 21.A,D
2. 3. 4. 6.
7.
v 1v 2
(D)
When the distance travelled is equal to the displacement of the particle, i.e. particle moves along the straight line in the same direction without reversing the direction of motion. 10 . Yes. Speed of object will increase if acceleration is in the direction of velocity, yet its magnitude may decrease.
SUBJECTIVE PROBLEMS
1.
2n 2n − 1
a=
3v 1v 2 v 3 v 2 v 3 + v 1v 3
dv = slope of the curve. dt
acceleration (slope of the curve) is decreasing with time yet the speed is increasing.
+
(a) vt = 5 s = 336 units, at = 5 s = 126 units ; (b) = 121 units, = 66 units (a) 2.7 km , (b) 60 m/s, (c) 225 m and 2.25 km (a) 50 m/s2; (b) 125 m 5. (A) 0; (B) 60m distance travelled = 10 m; displacement = 6 m; average velocity = 1.2 m/s 1000 ft. ,
11. (a)
r
<
v1
>
=
r
<
v2
>
=
r
<
v3
>
=
r
<
v4
>
(b) v4, avg > v1, avg = v2, avg > v3, avg 12 .
H ⎜⎜ ⎟⎟ v ⎝ H−h
13. 20 ft/s 2
1 4 . ((d) a ) 10.8 190 ft/s (c) 8 s, s0 ft/s s, 90(b) ft/s 2 s (e) 2.8 (f) greatest acceleration happens 2 s after launch (g) constant acceleration between 2 and 10.8 s, – 32 ft/s2.
parabolic curve (i j oyh; oØ )
straight line (l j y j s [ kk)
EXERCISE # 2 paraboliccurve(
i joyh; oØ)
1.
( i ) V = − 1 ˆi m/s
2.
B
r
8.
v=v
0
e8x
r
9. < v > =
∴
displacmen t dis tan ce travelled = time time
RESONANCE
3. A
4. B
(ii) t =
π +
3
5. C
3 4
6. B
TM
Page # 18
RELA TIVE MO TION RELA TIVE MO TION Motion is a combined property of the object under study and the observer. Motion is always relative, there is no such term like absolute motion or absolute rest. Motion is always defined with respect to an observer or reference frame. Reference frame :
relative concept.
Reference frame is an axis system from which motion is observed. A clock is attached to measure time. Reference frame can be stationary or moving. There are two types of reference frame:
•
(i) Inertial reference frame : A frame of reference in which Newton’s first law is valid is called as inertial reference frame.
•
Impossibility is a
If my Theory of
relativity is proven successful, Germany
(ii) Non-inertial reference frame : A frame of reference in which Newton’s first law is not valid is called as non-inertial reference frame.
will claim me as a German and France will declare that I am a citizen of the world. Should my theory prove untrue, France will say that I am a German and
Note : Earth is by definition a non-inertial reference frame because of its centripetal acceleration towards sun. But, for small practical applications earth is assumed stationary hence, it behaves as an inertial reference frame.
RELA TIVE VEL OCIT Y Definition : Relative velocity of a particle (object) A with respect to B is defined as the velocity with which A appears to move is B if considered to be at rest. In other words, it is the velocity with which Aappears to move as seen by the B considering itself to be at rest. Relative motion along straight line -
Germany will declare that I am a Jew.
-Einstein
vA =
dx A , dt
vB =
dx B dt
xBA = xB – xA vBA =
dx B dx A – dt dt
vBA = vB – vA
⇒
vAA = vA – vA = 0 (velocity of A with respect to A)
Note : velocity of an object w.r.t. itself is always zero. TM
RESONANCE
Page # 19
Ex.1
An object A is moving with 5 m/s and B is moving with 20 m/s in the same direction. (Positive x-axis) (i) Find velocity of B with respect to A. vB = 20 ˆi m/s
Sol.
⇒
vA = 5 ˆi m/s
⇒
vB – vA = 15 ˆi m/s
(ii) Find velocity of A with respect to B vB = 20 ˆi m/s, vA = 15 ˆi m/s
Sol. Not:e
Ex.2
v
BA
⇒
vAB = vA – vB = – 15 ˆi m/s
= – vAB
Two objectsA and B are moving towards each other with velocities 10 m/s and 12 m/s respectively a shown.
(i) Find out velocity of A with respect to B. vAB = vA – vB = (10) – (–12)
Sol.
= 22 m/s towards right.
(ii) Find out velocity of B with respect to A vBA = vB – vA = (–12) – (10)
= –22 m/s towards left.
Velocity of Approach It is the rate at which a separation between two moving particles decreases. If separation decreases velocity of approach is positive, Velocity of approach = 22 m/s Velocity of approach = 15 m/s If separation increases, velocity of approach is negative. It is mainly called velocity of separation.
Velocity of separation It is the rate with which separation between two moving object increases. Velocity of separation = 2 m/s Velocity of separation = 15 m/s
Illustration :
Two balls A and B are moving in the same direction with equal velocities, find out their relative velocity.
r
Velocity of A with respect to B (v AB ) = 0
TM
RESONANCE
Page # 20
Illustration :
A and B are thrown vertically upward with velocity, 5 m/s and 10 m/s respectively (g = 10 m/s2. Find separation between them after one second SA = ut –
Sol.
1 2 gt 2
= 5t –
= 5 × 1 – 5 × 12 SB = ut –
= 5 –
1 2 gt. 2
510 –=
1 × 10 × t2 2 = 0 1 × 10 × 1 2 2
=10×1– 5=
∴ SB – SA = separation = 5m. Alter : r
By relative
aBA
r
r
= aB – a A
= (–10) – (–10) = 0 r
r
r
Also v BA = v B – v A= 10–5 r
=5m/s r
∴ sBA ( in1sec)
= v BA×
t
1 × 5 =
m 5 =
∴ Distance between A and B after 1 sec = 5 m. Illustration :
A ball is thrown downwards with a speed of 20 m/s from top of a building 150 m high and simultaneously another ball is thrown vertically upwards with a speed of 30 m/s from the2 foot of the building. Find the time when both the balls will meet. (g = 10 m/s )
Sol.
(I)
S1 = 20 t + 5 t2 + S2 = 30 t – 5 t 2 ____________________ 150 = 50 t
⇒
t = 3 s.
(II)
Relative acceleration of both is zero since both have acceleration in downward direction r
a AB
r
r
= a A – aB =g–g=0
r
v BA
= 30 – (–20) = 50
sBA = vBA × t t=
sBA 150 = =3 s v BA 50
TM
RESONANCE
Page # 21
Ex.5
Two cars C1 and C2 moving in the same direction on a straight road with velocities 12 m/s and 10 m/s respectively. When the separation between the two was 200 m C2 started accelerating to avoid collision. What is the minimum acceleration of car C 2 so that they don’t collide.
Sol.
By relative r
r
r
a C1C 2 = a C1 – aC2 r
r
v
=0–a
=(–a)
r
v v = C1 – C 2 =12–10 So by relativity we want the car to stop. C1C 2
= 2 m/s.
∴ v2 – u2 = 2as. ⇒ 0 – 22 = – 2 × a × 200 = 0.1 m/s2
⇒ a=
1 m/s 2 100
= 1 cm/s2.
∴ Minimum acceleration needed by car C2 = 1 cm/s2
RELA TIVE MO TION IN LIFT Illustration :
A lift is moving up with acceleration a. A person inside the lift throws the ball upwards with a velocity u relative to hand. (a) What is the time of flight of the ball? (b) What is the maximum height reached by the ball in the lift? So l .
(a)
r
r
r
aBL = aB – aL 1 r a t2 2 BL
r
r
∴
T =
2u ( g + a)
(b)
v2 – u=2
s = ut +
=(g+a)
as 2
downwards
0 = uT –
u – 0
2
1 (g + a)T2 2
= –2(g + a) H
2
H=
u 2(g + a )
Projectile motion in a lift moving with acceleration a upwards (1) (2)
Initial velocity = u Velocity at maximum height = u cos θ
(3)
T=
(4)
Maximum height (H) =
(5)
Range=
2u sin θ g+a
u2 sin2 θ 2( g + a )
u 2 sin 2θ g+a
TM
RESONANCE
Page # 22
RELA TIVE MO TION IN T W O DIMENSION DIMENSION r
rA = position of A with respect to O r
rB = position of B with respect to O
r
rAB = position of A with respect to B. r
r
r
rAB = rA − rB r
∴
r
r
d ( rAB ) d ( rA ) d ( rB ) = – . dt dt dt
⇒
v AB = v A − v B
d ( vr AB ) d ( vr A ) d ( vr B ) = – dt dt dt
⇒
a AB = a A − a B
r
r
r
r
r
r
Note : These formulae are not applicable for light. Illustration :
Object A and B has velocities 10 m/s. A is moving along East while B is moving towards North from the same r
point as shown. Find velocity of A relative to B ( v AB )
r
r
r
v AB = v A – v B
Sol.
r
∴ v AB = 10√2
Note :
r
r
v A − vB =
v 2A + v B2 − 2v A v B cos θ
Illustration :
Two particles A and B are projected in air. A is thrown horizontally, B is thrown vertically up. What is the separation between them after 1 sec. r
Sol.
∴ ∴
r
r
a AB
= a A – aB = 0
vr AB
=
sAB
10 2 + 10 2
= vABt
= 10 2 (10 =
2 )t
=10
2 m
Consider the situation, shown in figure
TM
RESONANCE
Page # 23
Ex.23
(i)
Find out velocity of B with respect to A r
r
r
= 20 ˆj – 20 ˆi
v BA = v B − v A
(ii)
Find out velocity of A with respect to B r
r
r
= 20 ˆi – 20 ˆj
v AB = v A − v B
Ex.24
Sol.
(1)
Find out motion of tree, bird and old man as seen by boy.
(2)
Find out motion of tree, bird, boy as seen by old man
(3)
Find out motion of tree, boy and old man as seen by bird.
(1)
With respect to boy :
(2)
(3)
vtree
= 16 m/s (←)
vbird
= 12 m/s (↑)
vold man
= 18 m/s (←)
With respect to old man : vBoy
= 18 m/s (→)
vTree
= 2 m/s ( →)
vBird
= 18 m/s (→) and 12 m/s (↑)
With respect to Bird : vTree
= 12 m/s (↓)
and
16m/s( ←)
vold man
= 18 m/s (←)
and
12m/s( ↓).
vBoy
= 12 m/s (↓).
MO TION OF A TRAIN MO VING ON EQU AT OR : If a train is moving at equator on the earth’s surface with a velocity vTE relative to earth’s surface and a point on the surface of earth with velocity v E relative to its centre, then r
r
r
v TE = v T − v E
So, if the train moves from west to east (the direction of motion of earth on its axis)
or
r
r
r
v T = v TE + v E
and if the train moves from east to west (i.e. opposite to the motion of earth)
TM
RESONANCE
Page # 24
vT = vTE + vE
vT = vTE – vE
r traing : r Relative Motion on a moving If a boy is running with speed v BT on a train moving with velocity v T relative to ground, the speed of the boy r
relative to ground v B will be given by: r
r
r
v BT = v B – v T r
or
r
r
v B = v BT + v T
so, if the boy is running in the direction of train vB = u + v and if the boy is running on the train in a direction opposite to the motion of train vB = u – v
RELA TIVE MO TION IN RIVER FL O W : r
If a man can swim relative to water with velocity v mR and water is following relative to groun d with velocity r
r
v R , velocity of man relative to ground v m will be given by : r
r
mR
r
or
r
= v
v
– v m
r
R
r
v m = v mR + v R
So, if the swimming is in the direction of flow of water, vm = vmR + vR
and if the swimming is opposite to the flow of water, vm = vmR – vR
Illustration :
A swimmer capable of swimming with velocity v relative to water jumps in a flowingiver r having velocity u. The man swims a distance d down stream and returns back to the original position. Find out the time taken in complete motion. Sol.
t = tdown + tup =
d 2dv d + = 2 v −u v +u v − u2
Ans.
TM
RESONANCE
Page # 25
CROSSING RIVER A boat or man in a river always moves in the direction of resultant velocity of velocity of boat (or man) and velocity of river flow.
1.
Shortest Time : r
The person swims perpendicular to the river flow crossing a river : consider a river having flow velocity v R and swimmer jump into the river from a point A, from one bank of the river, in a direction perpendicular to the direction of river current. Due to the flow velocity of river the swimmer is drifted along the river by a distance r
BC and the net velocity of the swimmer will be v m along the d irection AC. If we find the components of velocity of swimmer along and perpendicular to the flow these are. y B
Velocity along the river, v x = vR. Velocity perpendicular to the river, vf = vmR
drift = x
C
<
The net speed is given by vm =
v 2mR + v R2
d
vr vmR
vm θ
v mR at an angle of tan θ = (down stream with the direction of flow). vR
x
A
d d Velocity of vy is used only in crossing the river, time taken to cross the river is t = v = v . y mR Velocity vx is only used to drift the motion of the swimmer in the river, drift is along the river flow, x = (v(t) ) x
or
v=x
R
d v mR
Illustration :
A river 400 m wide is flowing at a rate of 2.0 m/s.A boat is sailing at a velocity of 10 m/s with respect to the water, in a direction perpendicular to the river. (a)
Find the time taken by the boat to reach the opposite bank.
(b)
How far from the point directly opposite to the starting point does the boat reach the opposite bank.
(c)
In what direction does the boat actually move.
(a)
time taken to cross the river
Sol.
d 400 m t= v = 10 m / s y
=
(b)
drift (x) = (v x)(t) = (2m/s) (40s) = 80 m
(c)
Actual direction of boat,
40 s
Ans. Ans.
⎛ 10 ⎞
θ = tan–1 ⎜ 2 ⎟ = tan–1 5, (downstream) with the river flow. ⎝ ⎠ TM
RESONANCE
Page # 26
2.
SHORTEST PA TH : When the person crosses the river perpendicularly (along the shortest path). It should swim up stream r
making an angle θ with AB such that the resultant velocity v m , of man must be perpendicular to the flow of river along AB. If we find the components of velocity of swimmer along and perpendicular to the flow, these are, velocity along the river, vx = 0
and
2 v mR − v R2
velocity perpendicular to river v y = 2
2
v mR − v R at an angle of 90º with the river direction. The net speed is given by vm =
d velocity vy is used only to cross the river, therefore time to cross the river, t = v = y and
d 2 v mR − v R2
velocity v x is zero, therefore, in this case the drift (x) should be zero . x=0 or
vx = vR – vmR sin θ = 0
or
vR = vmR sin θ
or
θ = sin–1 ⎜⎜
⎛ vR ⎞ ⎟⎟ ⎝ v mR ⎠
Hence, to cross the river perpendicular (along the shortest path) the man should swim at an angle of
π 2
⎛ vR ⎞ ⎟⎟ upstream from the direction of river flow. ⎝ v mR ⎠
+ sin −1 ⎜⎜
further, since
sin θ < 1,
Swimmer can cross the river perpendicularly only when v mR > vR ie Practically it is not possible to reach at B if the river velocity (vR) is too high. Illustration :
A man can swim at the rate of 5 km/h in still water. A river 1 km wide flows at the rate of 3 km/h. The man wishes to swim across the river directly opposite to the starting point.
Sol.
(a)
Along what direction must the man swim?
(b)
What should be his resultant velocity?
(c) How much time the would take to cross? The velocity of man with respect to river v mR= 5 km/hr, this isgreater than the river flow velocity, therefore, he can cross the river directly (along the shortest path). The anlge of swim must be
θ=
π 2
⎛ vr + sin–1 ⎜⎜ v ⎝ mR
⎞ ⎛ v ⎟⎟ = 90º + sin-1 ⎜⎜ r ⎠ ⎝ v mR
⎛3⎞ = 90º + sin–1 ⎜ ⎟ = 90º + 37 º ⎝5⎠
⎞ ⎟⎟ ⎠
= 127º, with the river flow (upstream)
Ans.
TM
RESONANCE
Page # 27
(b)
Resultant velocity will be v
m
=
2 v mR − v R2
=
52 − 32
= 4 km/hr along the direction perpendicular to the river flow. (c)
time taken to cross the d t=
2 v mR − v R2
1 km 1 = 4 km / hr = h = 15 min 4
Ex.
The velocity of about in still water is 5 km/h it crosses 1 km wide river in 15 minutes along the shortest possible path. Determine the velocity of water in the river in km/h
Ans.
3km/h
Illustration :
A man wishes to cross a river flowing with velocity u jumps at an anlge θ with the river flow. Find out the net velocity of the man with respect to ground if he can swim with speed v. Also find How far from the point directly opposite to the starting point does the boat reach the opposite bank. in what direction does the boat actually move. If the width of the river is d. Sol.
Velocity of man = vM = tan φ =
u2 + v 2 + 2v u cos θ
Ans.
v sin θ u + v cos θ
(v sinθ) t = d
Ans.
⇒ t=
x = (u + v cos θ)t
d v sin θ
= (u+vcos
θ)
d v sin θ
Ans.
Illustration :
Sol.
A boat moves relative to water with a velocity which is n times less than the river flow velocity. At what anlge to the stream direction must the boat move to minimize drifting? In this problem, one thing should be carefully meted that the velocity of boat is less than the river flow velocity. In such a case, boat cannot reach the point directly opposite to its starting point. i.e. drift can never be zero. Thus, to minimize the drift, boat starts at an angle θ from the normal direction up stream as shown. Now, again if we find the components of velocity of boat along and perpendicular to the flow, these are, velocity along the river, v x = u – v sin θ. and velocity perpendicular to the river, v y = v cos θ. d time taken to cross the river is t =
= vy
In this time, drift x = (v x)t = (u – v sin θ) or
x=
d
.
v cos θ
d v cos θ
ud sec θ – d tan θ v
The drift x is minimum, when
dx = 0, dθ
TM
RESONANCE
Page # 28
or
⎛ ud ⎞ ⎜ ⎟ (sec θ . tan θ) – d sec 2 θ = 0 ⎝ v ⎠
or
u sin θ = 1 v
or
sin θ =
v 1 = u n
(as v =
u ) n
⎛v⎞ so, for minimum drift, the boat must move at an angle θ = sin–1 ⎜ ⎟ from normal direction or ⎝u⎠
π
an angle
2
⎛v⎞ + sin–1 ⎜ ⎟ from stream direction. ⎝u⎠
RAIN PROBLEMS r
r
If rain is falling vertically with a velocity v R and on observer is moving horizontally with velocity v m , the velocity of rain relative to observer will be : r
r
r
v Rm = v R – v m
or
vRm =
2 v R2 + v m
⎛ vm ⎞ and direction θ = tan–1 ⎜⎜ v ⎟⎟ with the vertical as shown in figure. ⎝ R⎠
Illustration :
Rain is falling vertically and a man is moving with velocity 6 m/s. Find the angle with which umbrella should be hold by man to avoid getting wet.
Sol.
r
ˆ v rain = –10 j
r ˆ v man = 6 i
r Velocity of rain with respect to man = v r m = –10 ˆj – 6 ˆi
tan θ =
6 10
⎛3⎞
θ = tan–1 ⎜ 5 ⎟ ⎝ ⎠
Where θ is angle with vertical TM
RESONANCE
Page # 29
Illustration :
A man moving with 5m/s observes rain falling vertically at the rate of 10 m/s. Find the speed and direction of the rain with respect to ground.
vRM = 10 m/s, v M = 5 m/s r
r
r
r
r
r
v RM = v Ru – v M
⇒
v Ru = v RM – v M
⇒
vR = 5 5
r
tan θ =
1 1 , θ = tan–1 . 2 2
Illustration :
A man standing, observes rain falling with velocity of 20 m/s at an angle of 30º with the vertical. (1) Find out velocity of man so that rain appears to fall vertically. (2) Find out velocity of man so that rain again appears to fall at 30º with the vertical. r
Sol.
(1)
v m = –v ˆi (let) r
v R = –10 ˆi – 10√3 ˆj r
v RM = –(10 – v) ˆi – 10√3 ˆj
⇒
– (10 – v) = 0
(for vertical fall, horizontal component must be zero)
or
v=10m/s
Ans.
(2)
r v R = –10 ˆi – 10√3 ˆj r v m = –vx ˆi r ˆ ˆ v RM = – (10 – v x) i – 10√3 j
Angle with the vertical = 30º 10 − v x
⇒
tan 30º =
⇒
vx = 20 m/s
− 10 3
TM
RESONANCE
Page # 30
WIND AIRPLANE This is very similar to boat river flow problems the only difference is that boat is replaced by also plane and river is replaced by wind. Thus, velocity of aeroplane with respect to wind r
r
r
v aw = v a − v w r
r
r
v a = v aw + v w
or
r
where,
v a = absolute velocity of aeroplane
and,
v w = velocity of wind.
r
Illustration :
An aeroplane flies along a straight path A to B and returns back again. The distance between A and B is l and the aeroplane maintains theconstant speed v. There is a steady wind with a speed u at an angleθ with line AB. Determine the expression for the total time of the trip.
Sol.
A to B : Velocity of plane along AB = v cos α – ucos θ , and for no-drift from line AB : v sinα = usinθ
⇒
time taken fromAto B :
sinα = t
AB
u sin θ v l
=
v cos α − u cos θ
B to A : velocity of plane along BA = vcosα + u cosθ and for no drift from line AB : vsinα = usinθ
⇒
sin α =
u sin θ v
time taken from B to A : t BA =
l
v cos α + u cos θ
total time taken = tAB + t BA =
=
l
v cos α + u cos θ
+
l
v cos α + u cos θ
2v l 1 −
2vl cos α 2
2
2
2
=
v2
2
v cos α + u cos θ Ex.
u2 sin 2 θ 2
.
v −u
Find the time an aeroplane having velocity v, take to fly around a square with side a and the wind blowing at a velocity u, in the two cases, (a) if the direction of wind is along one side of the square, (b) If the direction of wind is along one of the diagonals of the square
Ans.
(a)
2a 2
v −u
2
⎛ v + v 2 − u2 ⎞ ⎜ ⎟ ⎝ ⎠
⎛ 2v 2 − u 2 ⎞ ⎜ ⎟ (b) 2 2 a ⎜ . v 2 − u2 ⎟ ⎝
⎠
TM
RESONANCE
Page # 31
Condition to collide or to reach at the same point When the relative velocity of one particle w.r.t. to other particle is directed towards each other then they will collide. (If there is a zero relative acceleration). Illustration :
Sol.
There are particles A, B and C are situated at the vertices of an equilateral triangle ABC of side a at t = 0. Each of the particles moves with constant speed v. A always has its velocity along AB, B along BC and C along CA. At what time will the particle meet each other? The motion of the particles is roughly sketched in figure. By symmetry they will meet at the centroid O of the triangle. At any instant the particles will from an equilateral triangle ABC with the same
Centroid O. All the particles will meet at the centre. Concentrate on the motion of any one particle, say B. At any instant its velocity makes angle 30º with BO. The component of this velocity along BO is v cos 30º. This component is the rate of decrease of the distance BO. Initially. BO =
a a/2 = = displacement of each particle. cos 30º 3
Therefore, the time taken for BO to become zero 2d 2d d/ 3 = = . 3v × 3 3v v cos 30º Alternative : Velocity of B is v along BC. The velocity of C is along CA. Its component along BC is v cos 60º = v/2. Thus, the separation BC decreases at the rate of approach velocity. =
∴ v 3v = 2 2 Since, the rate of approach is constant, the time taken in reducing the separation BC from a to zero is
∴ approach velocity = v +
a 2a = 3v 3v 2 Six particles situated at the corners of a regular hexagon of side a move at a constant speed v. Each particle maintains a direction towards the particle at the next corner. Calculate the time the particles will take to meet each other. 2 a/v. ‘A’ moves with constant velocity u along then ‘x’ axis. B always has velocity towards A. After how much time will B meet A if B moves with constant speed v. What distance will be travelled by A and B. t=
Q.
Ans. Q.
Ans.
distance travelled by A =
distance travelled by B =
v 2l 2
v − u2
,
uvl v 2 − u2
TM
RESONANCE
Page # 32
Illustration :
Two cars A and B are moving west to east and south to north respectively along crossroads. A moves with a speed of 72 kmh–1 and is 500 m away from point of intersection of cross roads and B moves with a speed of 54 kmh–1 and is 400 m away from point of intersection of cross roads. Find the shortest distance between them ? Sol. Method – I (Using the concept of relative velocity)
In this method we watch the velocity of A w.r.t. B. To do this we plot the resultant velocity Vr. Since the accelerations of both the bodies is zero, so the relative acceleration between them is also zero. Hence the relative velocity will remain constant. So the path of A with respect to B will be straight line and along the direction of relative velocity of A with respect to B. The shortest distance between A & B is when A is at point F (i.e. when we drop a perpendicular from B on the line of motion of A with respect to B). From figure VB 15 3 = = ........................(i) VA 20 4 This θ is the angle made by the resultant velocity vector with the x-axis. Also we know that from figure tanθ =
x = 3 ..............................(ii) 500 4 From equation (i) & (ii) we get x = 375 m EB = OB – OE = 400 – 375 = 25 m But the shortest distance is BF. OE =
∴
From magnified figure we see that BF = EB cosθ = 25 ×
∴
4 5
BF = 20 m Method II (Using the concept of maxima – minima)
A & B be are the initial positions and A’,B’ be the final positions after time t. B is moving with a speed of 15 m/sec so it will travel a distance of BB’ = 15t during time t. A is moving with a speed of 20 m/sec so it will travel a distance of AA’ = 20t during time t. So OA’ =500 – 20 t OB’ = 400 – 15 t ∴ A’B’2 = OA ’2 + OB’2 = (500 – 20t)2 + (400 – 15t) 2 ..................(i) For A’B’ to be minimum A’B’2 should also be minimum TM
RESONANCE
Page # 33
∴
d(A'B'2 ) d( 400 − 15t )2 + (500 − 20 t )2 = =0 dt dt = 2(400 – 15t) (–15) + 2(500 – 20t) (–20) = 0 = – 1200 + 45t = 2000 – 80 t 125 t = 3200
∴
t=
∴
128 128 s. Hence A and B will be closest after s. 5 5
Now
d 2 A ' B'
comes out to be positive hence it is a minima. dt 2 On substituting the value of t in equation (i) we get
∴
⎛
A’B’2 = ⎜ 400 − 15 ×
⎝
128 ⎞
2
128 ⎞ ⎛ ⎟ + ⎜ 500 − 20. ⎟ 5 ⎠ 5 ⎠ ⎝
2
= 16 2 + ( −12)2 = 20 m ∴
Minimum distance A’B’ = 20 m.
Method III (Using the concept of relative velocity of approach) After time t let us plot the components of velocity of A & B in the direction along AB. When the distance between the two is minimum, the relative velocity of approach is zero. ∴ VA cosαf + VB sinαf = 0 (where αf is the angle made by the line A’B’ with the x-axis) 20 cosαf = – 15 sinαf 20 4 = − 15 3 Here do not confuse this angle with the angle θ in method (I) because that θ is the angle made by the resultant with x-axis. Here αf is the angle made with x-axis when velocity of approach in zero, 400 − 15t 4 ∴ 500 − 20 t = − 3 128 ∴ t= So, OB’ = 16 m and OA’ = – 12m 5 tanαf = −
∴
A’B’ = 16 2 + ( −12) 2 = 20 m Qus.
Two ships are 10 km apart on a line running south to north. The one farther north is steaming west at 20 km h –1. The other is steaming north at 20 km h–1. What is their distance of closest approach ? How long do they take to reach it ?
Ans.
5 2 km/h ; 1/4 h = 15 min consider the situation shown in figure for the two particle A and B.
Qus.
(1) (2)
Ans.
(1) The particles will not collide
Will the two particle will collide Find out shortest distance between two particles
(2) 4 5 m. Note : Muzzle Velocity is th e velocity of bul let with resp ect to the gun i.e. it is Rel ative Velocity.
TM
RESONANCE
Page # 34
E xer ci se 1 OBJECTIVE PROBLEMS A stone is thrown upwards with a velocity 50 ms–1. Another stone is simultaneously thrown downwards from 1. the same location with a velocity 50 ms –1. When the first stone is at the highest point, the relative velocity of the second stone w.r.t. the first stone is : –1 (A)Zero (B)50ms (C) 100 ms–1 (D) 150 ms–1 2.
A thief is running away on a straight road in a jeep moving with a speed of 9 m s –1. A police man chases him on a motor cycle moving at a speed of 10 m s –1. If the instantaneous separation of the jeep from the motorcycle is 100m, how long will it take for the police man to catch the thief? (A) 1s (B) 19s (C) 90s (D) 100s
3.
Two cars are moving in the same direction with a speed of 30 km h–1. They are separated from each other by 5 km. Third car moving in the opposite direction mee ts the two cars after an interval of 4 minutes. What is the speed of the third car? (A) 30 km h–1 (B) 35 km h–1 (C) 40 km h–1 (D) 45 km h–1
4.
Shown in the figure are the displacement time graph for two children going home from the school. Which of the following statements about their relative motion is true? Their relative velocity : (A) first increases and then decreases (B) first decreases and then increases (C) is zero (D) is non zero constant.
5.
A person standing on the escalator takes time t1 to reach the top of a tower when the escalator is moving. He takes time t2 to reach the top of the tower when the escalator is standing. How long will he take if he walks up a moving escalator? (A) t2 – t1 (B) t1 + t2 (C) t1 t2 /(t1 – t2) (D) t 1 t2/(t1 + t2)
6.
Shown in the figure are the velocity time graphs of the two particles P 1 and P2. Which of the following statements about their relative motion is true? Their relative velocity : (A) is zero (B) is non-zero but constant (C) continuously decreases (D*) continuously increases
7.
Two particles P1 and P2 are moving with velocities v 1 and v2 respectively. Which of the statement about their relative velocity vr 12 is true? (A) vr 12 > (v1 + v2) v (B) r 12 cannot be greater than v 1 – v2 (C) vr 12 cannot be greater than v 1 + v2 (D) vr 12 < (v1 + v2)
8.
Two identical trains take 3 sec to pass one another when goi ng in the opposite direction but only 2.5 sec if the speed of one is in creased by 50 %. The time on e would take to pass the other whe n going in the same direction at their srcinal speed is : (A)10sec (B)12sec (C)15sec (D)18sec
9.
Two billiard balls are rolling on a flat table. One has velocity compon ents vx = 1m/s, vy = 3 m/s and the other has components vx = 2m/s and vy = 2 m/s. If both the balls start moving from the same point, the angle between their path is (A) 60° (B) 45° (C) 22.5° (D*) 15°
10.
A battalion of soldiers is ordered to swim across a river 500 ft wide . At what minimum rate should they swim perpendicular to river flow in order to avoid being washed away by the waterfall 300 ft downstream. The speed of current being 3 m.p.h. : (A) 6m.p.h. (B*) 5m.p.h. (C) 4m.p.h. (D) 2m.p.h.
11.
A boat, which has a speed of 5 km/hr in still water, crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of the river water in km/hr is 1 (A)
RESONANCE
(B*) 3
TM
4 (C)
(D)
41
Page # 35
12.
A bucket is placed in the open where the rain is falling vertically. If a wind begins to blow at double the velocity of the rain, how will be rate of filling of the bucket change? (A)Remainunchanged (B)Doubled (C)Halved (D)Becomefourtimes
13.
A car with a vertical wind shield moves along in a rain storm at the speed of 40 km/hr. The rain drops fall vertically with a terminal speed of 20 m/s. The angle with the vertical at which the rain drop strike the wind shield is (A) tan–1( 5/9) (B)tan–1(9/5) (C)tan –1( 3/2) (D)tan–1(3)
SUBJECTIVE PROBLEMS 14.
Men are running along a road at 15 km/h behind one another at equal intervals of 20 m. Cyclist are riding in the same direction at 25 km/h at equal intervals of 30 m. At what speed an observer travel along the road in opposite direction so that whenever he meets a runner he also meets a cyclist? s
15.
Two perpendicular rail tracks have two trains A & B respectively. Train A moves north with a speed of 54 km h–1 and train B moves west with a speed of 72 km h –1 . Assume that both trains starts from same point. Calculate the (a) rate of separation of the two trains (b) relative velocity of ground with respect to B (c) relative velocity of A with respect to B.
16.
A man is swimming in a lake in a direction of 30° East of North with a speed of 5 km/hr and a cyclist is going on a road along the lake shore towards East at a speed of 10 km/hr. In what direction and with what speed would the man appear to swim to the cyclist.
17.
A motor boat has 2 throttle position on its engine. The high speed position propels the boat at 10 km hr −1 in still water and the low position gives half the higher speed. The boat travels from its dock downstream on a river with the throttle at low position and returns to its dock with throttle at high position. The return trip took 15 % longer time than it did for the downstream trip. Find the vel ocity of the water current in the river.
18.
(Ι) A man can swim with a speed of 4 km h –1 in still water. How long does he take to cross a river 1 km wide if the river flows steadily at 3 km h –1 and he makes his strokes normal to the river current ? How far down the river does he go when he reaches the other bank ? ΙΙ) Find ((a) If he keeps himself always at anthe angle of (b) 120° withpoint the river flowopposite while swimming. the time he takes to cross river. AtCwhat on the bank will he arrive ?
19.
A river is flowing from west to east at a speed of 5 m/min. A man on the south bank of the river, capable of swimming at 10 m/min in still water, wants to swim across the river in shortest distance. In what direction should he swim ?
20.
An airplane is flying with velocity 50 2 km/hour in north-east direction. Wind is blowing at 25 km/hr from north to south. What is the resultant displacement of airplane in 2 hours ?
21.
When a train has a speed of 10 m s –1 eastward, raindrops that are falling vertically with respect to the earth make traces that are inclined 30º to the vertical on the windows of the train. (a) What is the horizontal component of a drop's velocity with respect to the earth ? With respect to the train ? (b) What is the velocity of the raindrop with respect to the earth ? With respect to the train ?
22.
To a man walking at 7 km/h due west, the wind appears to blow from the north-west, but when he walks at 3 km/h due west, the wind appears to blow from the north. What is the actual direction of the wind and what is its velocity ?
23.
When a motorist is driving with velocity 6 ˆi + 8 ˆj , the wind appears to come from the direction
ˆi . When
he doubles his velocity the wind appears to come from the direction ˆi + ˆj . Then the true velocity of the wind expressed in the form of a ˆi + b ˆj is ______. 24.
' n ' numbers of particles are located at the vertices of a regular polygon of ' length ' a '. They all start moving simultaneously with equal constant speed ' other all the time. How long will the particles take to collide?
25.
Two ships are 10 km apart on a line running south to north. The one further north is streaming west at 40 km/hr. The other is streaming north at 4 0 km/hr. What is their distance of closest appr oach and how long do they take to reach it?
RESONANCE
TM
n ' sides having the edge v ' heading towards each
Page # 36
26.
A ship is sailing towards north at a speed of 2 m/s. The current is taking i t towards East at the rate of 1 m/s and a sailor is climbing a vertical pole on the ship at the rate of 1 m/s. Find the velocity of the sailor in space.
27.
A motorboat is observed to travel 10 km hr –1 relative to the earth in the direction 37º north of east. If the velocity of the boat due to the wind only is 2 km hr–1 westward and that due to the cu rrent only is 4 km hr–1 southward, what is the magnitude and direction of the velocity of the boat due to its own power ?
28.
A person P sitting on a wooden block (which does not move relative to water) in a flowing river sees two swimmers A and B. A and B both have constant speed v m relative to water. P observes that A starts from one point of the river bank and appears to move perpendicular to the river flow. P also observes that B starts from some point on the other bank at the same time and moves downstream. The width of the river is ' d ' and it flows with velocity v r . If A and B both reach a point at the same time, than find the initial separation b etween A and B.
29.
A motorboat going down stream at afrom pointtheM. 60 minutes it turned and after some time passed the float at a overcome distance ofa float 6 km point M. Findlater the velocity ofback the stream assuming a constant velocity for the motorboat in still water.
30.
2 swimmers start from point A on one bank of a river to reach point B on the other bank, lying directly opposite to point A. One of them crosses the river along the straight line AB, whil e the other swims at right angles to the stream and then walks the distance which he has been carried away by the stream to get to point B. What was the veloci ty (assumed uniform) of his walking if both the swimmers reached point B simultaneously . Velocity of each swi mmer in still water is 2.5 km hr −1 and the stream velocity is 2 km hr −1.
31.
An airplane pilot sets a compass course due west and maintains an air speed of 240 km. hr–1. After flying for 1 hr, he finds himself over a town that is 150 km west and 40 km south of his starting point. 2 (a) Find the wind velocity, in magnitude and direction. (b) If the wind velocity were 120 km. hr –1 due south, in what direction should the pilot set his course in order to travel due west ? Take the same air speed of 240 km. hr –1.
32.
Two straight AOB and COD meet each other right angles. A person walking at a speed of 5 km/hr along AOB is at the crossing O at noon. Another person walking at the same speed along COD reaches the crossing O at 1:30 PM. Find at what time the distance between them is least and what is its value?
1.
An aeroplane is flying vertically upwards with a uniform speed of 500 m/s. When it is at a height of 1000 m above the ground a shot is fired at it with a speed of 700 m/s from a point directly below it. What should be the acceleration of the aeroplane so that i t may escape from being hit? [REE ’94, 6 ]
2.
The width of a river is 25 m and in it water is flowing with a velocity of 4 m /min. A boatman is standing on the bank of the river. He wants to sail the boat to a point at the other bank which is directly opposite to him. In what time will he cross the river, if he can sail the boat at 8 m/min, relative to the water. [ REE '95, 6 ]
3.
On a frictionless horizontal surface, assumed to be
the x − y plane a small trolley A is moving along a
straight line parallel to the y − axis as shown in the figure with a constant velocity of ( 3 – 1) m/s. At a particular instan t, when the line OA makes an angle of 45º with the x − axis, a ball is thrown along the φ surface from the srcin O. Its velocity makes an angle with the x − axis when it hits the trolley . (a) The motion of the ball is observed from the frame of the trolley. Calculate the angle θ made by the velocity of the ball wi th the x − axis in this frame. (b) Find the speed of the ball with respect to the surface, if φ = 4 θ /3. [ JEE 2002, 2 + 3 ]
RESONANCE
TM
Page # 37
EXERCISE # 1
22. Coming from 5 km/hr, 53°N of E
OBJECTIVE PROBLEMS
1. C
2. D
3. D
4. D
5.
D
6. D
7. C
8. C
9.
D
10. B
11. B
12. A
13. A
23.
25.
(4 ˆi + 8 ˆj) 10 2
,
1 8
24.
a 2π v ⎛⎜1−cos n ⎞⎟ ⎝ ⎠
hr
SUBJECTIVE PROBLEMS
14. 5 km/h 15. (a) 25 m/s or 90 k m/hr
26. 2m/s in a direction making an angle of 60º with E, 45º with N and 60º with the vertical (b) 20 m/s or 72 km/hr due
east (c) 25 m/s or 90 km/hr at 37°N of E 16. 30º N of W at 5 3 km/hr.
17. 3 km/hr.
1 2 3
28.
2d
29. 3 km/hr
30. 3 km/hr towards B
18. (Ι) 0.75 km (ΙΙ) (a)
27. 10 2 km/hr, 45º N of E
h
(b)
31. (a) 1 00 km/ hr, 37° W of S (b) 3 0° N of W
1 2 3
km.
32. 12 : 45 PM
19. At an angle 30° west of north.
EXERCISE # 2
20. 50 5 km 21. (a) 0, 10 m/s West
RESONANCE
(b) 10
TM
3 m/s, 20 m/s
1.
a > 10 m/s 2
3.
(a) θ =45º ;
2. 3.6 minute (b) 2m/s
Page # 38
PROJECTILE MO TION
1.
BASIC CONCEPT : 1.1
I mentally conceive of some movable projected on a horizontal plane all impediments being put aside. Now it is evident ... that the equable motion on this plane would be perpetual if the plane were of infinite extent; but if we assume it to be ended, and [situated] on high, the movable ... , driven to the end of the plane and going on further, adds on to its previous equable and indelible motion that downward tendency which it has from its own heaviness. Thus there emerges a certain motion, compounded from equable horizontal and from naturally accelerated downward [motion], which I call projection. - Galileo (Two New Sciences 1638)
PROJECTILE
Any object that is given an initial velocity obliquely and that subsequently follows a path determined by the gravitational force acting on it, is called a Projectile. A projectile may be a football, a cricket ball, or any other object. 1.2
TRAJECTORY
The path followed by a particle (here projectile) during its motion is called its Trajectory. NOTE : 1. We shall consider only trajectories that are of sufficiently short range so that the gravitational force can be considered constant in both magnitude and direction. 2. All effects of air resistance will be ignored; thus our results are precise only for motion in a vacuum on flat, non rotating Earth. 1.3
PROJECTILE MOTION
(i) The motion of projectile is known as projectile motion. (ii) It is an example of two dimensional motion with constant acceleration. (iii) Projectile motion is considered as combination of two simultaneous motions in mutually perpendicular directions which are completely independent from each other i.e.horizontal motion and vertical motion.
=
+
Parabolic path = vertical motion + horizontal motion.
TM
RESONANCE
Page # 39
2.
PR OJECTILE THR O WN AT AN ANGLE WITH HORIZONT
AL
(i)
Consider a projectile thrown with a velocity u making an angle θ with the horizontal.
(ii)
Initial velocity u is resolved in components in a coordinate system in which horizontal direction is taken as x-axis, vertical direction as y-axis and point of projection as srcin. ux = u cos θ uy = u sin θ
(iii)
Again this projectile motion can be considered as the combination of horizontal and vertical motion. Therefore, Horizontaldirection
2.1
Verticaldirection
(a)
Initial velocity u x = u cos θ
Initial velocity uy = u sin θ
(b)
Acceleration a=x
Acceleration a
(c)
Velocity after time t, v
0 x
= u cos θ
y
=g
Velocity after time t,v y = u sin θ – gt
TRAJECTORY EQUATION : If we consider the horizontal direction, x = ux.t x = u cos tθ.
...(1)
For vertical direction : y = uy . t – 1/2 gt 2 = u sin θ . t – 1/2 gt2
...(2)
Substituting the value x equation (1)
y = u sin θ .
x ⎛ x ⎞ 1 ⎟ – g ⎜ u cos θ 2 ⎝ u cos θ ⎠
2
This is an equation of parabola called as trajectory equation of projectile motion. 2.2
TIME OF FLIGHT : Since the displacement along vertical direction does not occur. So, Net displacement = 0 (u sin θ) T –
T=
1 gT2 = 0 2
2u sin θ g
TM
RESONANCE
Page # 40
2.3
HORIZONTAL RANGE : R = ux .T 2u sin θ g
R = u cos θ.
u2 sin 2θ g
R=
2.4
MAXIMUM HEIGHT : Using 3rd equation of motion i.e. 2
2
v = u + 2as we have for vertical direction 0 = u2 sin2 θ – 2gH
H=
2.5
u 2 sin2 θ 2g
RESULTANT V ELOCITY :
v = v x ˆi + v y ˆj r
= u cos θ ˆi + (u sin θ – gt) ˆj where
NOTE: •
| v| = v
u 2 cos 2 θ + (u sin θ − gt )2
and tan θ = vy / vx.
Results of article 2.2,2.4 and 2.4 are valid only for complete flight, that is when proejctile lands at same horizontal level from which it has been projected.
•
Vertical component of velocity is zero when particle moves horizontally, i.e., at the highest point of trajectory.
•
Vertical component of velocity is positive when particle is moving up and vertical component of velocity is negative when particle is coming down if vertical upwards direction is taken as positive. Any direction upward or downward can be taken as positive and if downward direction is taken as positive then vertical component of velocity coming down is positive.
2.6
GENERALRESULT : (i)
For maximum range
θ = 45º
u2 R
max
= g
In this situation Hmax = (ii)
R max 2
We get the same range for two angle of projections α and (90 – α) but in both cases, maximum heights attained by the particles are different.
TM
RESONANCE
Page # 41
Ιf
(iii)
H = R
i.e.
u2 sin 2θ g
⇒
tan θ = 4
u2 sin2 θ 2g
=
Range can also be expressed as
(iv)
R=
u 2 sin 2θ g
=
2u sin θ.u cos θ g
2u x u y = (v)
g
CHANGE IN MOMENTUM : (a)
Initial velocity ui = u cos θ ˆi + u sin θ ˆj
(b)
Final velocity u f = u cos θ ˆi – u sin θ ˆj
r
r
Change in velocity for complete motion
∆ u = u f − u i = – 2u sin θ ˆj r
(c)
r
r
Change in momentum for complete motion
∆ P = Pf − Pi = m ( u f − u i ) = m(–2u sin θ) ˆj r
r
r
r
r
= – 2mu sin θ ˆj Velocity at the highest point = u f = u cos θ ˆi r
(d)
Change in momentum at highest point m ( u f − u i ) = m [u cos θ ˆi – (u cos θ ˆi + u sin θ ˆj ) = – mu sin θ ˆj r
r
Ex.1
A body is projected with a speed of 30 ms –1 at an angle of 30º with the vertical. Find the maximum height, time of flight and the horizontal range of the motion. [ Take g = 10 m/s 2 ]
Sol.
Here
u = 30 ms
–1
, Angle of projection, θ = 90 – 30 = 60º
Maximum height, H =
Time of flight,
T=
900 3 u2 sin2 θ 30 2 sin2 60º 135 × = = = m 20 4 4 10 2g
2u sin θ 2 × 30 × sin 60 0 = = 3 3 sec. g 10
Ex.2
A stone is thrown with a velocity v at angle θ with horizontal. Find its speed when it makes an angle β with the horizontal.
Sol.
Since horizontal component of velocity remains constant. Therefore, v cosθ = u cos β
u=
v cos θ cos β
TM
RESONANCE
Page # 42
Ex.3
A projectile is thrown in the upward direction making an angle of 60 0 with the horizontal with a speed of 147 m/s. Find the time after which its inclination with the horizontal is 450 ?
Sol.
ux = 147 × cos600 =
uy = 147 × sin600 =
vy = uy + ay t =
v x = ux =
147 2
147 3 2
147 3 – gt 2
147 2
When angle is 450 ,
tan45
0
=
vy vx
⇒
vy = vx
⇒
147 3 – gt 2
⇒
147 ( 3 − 1) = gt 2
⇒
t=
=
147 2
147 ( 3 − 1) s 2 × 10
Ex.4
A large number of bullets are fired in all directions with the same speed v. What is the maximum area on the ground on which these bullets will spread ?
Sol.
Maximum distance upto which a bullet can be fired is its maximum range, therefore 2
Rmax = v g
Maximum area = π(Rmax)2 = Ex.5
g2
.
The velocity of projection of a projectile is given by : u = 5ˆi + 10 ˆj . Find r
(a) Time of flight, Sol.
πv 4
(b) Maximum height,
(c) Range
We have ux = 5 uy = 10 (a)
Time of flight =
2u y 2 × 10 2u sin θ = = =2s g 10 g
u 2 sin2 θ (b)
Maximum height =
(c)
Range=
2g
uy
2
10 × 10
= 2g = 2 × 10 = 5 m
2u sin θ.u cos θ 2 × 10 × 5 = = 10 m 10 g
HEIGHT AND RANGE : Ex.6
A batter hits a baseball so that it leaves the bat with an initial speed v0 = 37.0 m/s at an initial angle α0 = 53º, at a location where g = 10.0 m/s2
TM
RESONANCE
Page # 43
(a)
Find the position of the ball, and the magnitude and direction of its velocity, when t = 2.0 s.
(b)
Find the time when the ball, reaches the highest point of flight and the find its height h at this point
(c)
Find the horizontal range R - that is, the horizontal distance from the starting point to the point at which the ball hits the ground. For each part, treat the baseball as a projectile
Sol.
y
The initial velocity of the ball has components t = 2.00 s
x= 46 m
v0x = v0 cos α0 = (37.0 m/s) cos 53º = 22.3 m/s
v t1 = 3 s v1
v0y = v0 sin α0 = (37.0 m/s) sin 53º = 29.5 m/s (a)
x=v
0x
y = 40 m
t = (22.3 m/s) (2.00 s) = 44.6 m
h = 45 m
v 0 = 37.0 m/s
y = v0y t –
α0
1 2 gt 2
t=6s R = 138 m
= 53º
x v2
1 (10 m/s2) (2 s)2 2
= (29.5 m/s) (2 s) –
= 59.0 – 20 = 39.0 m vx = v0x = 22.3 m/s vy = v0y – gt = 29.5 m/s – (10 m/s 2) (2.00 s) = 9.5 m/s The y-component of velocity is positive, which means that the ball is still moving upward at this time (Figure). The magnitude and direction of the velocity are v=
v 2x + v 2y =
( 22.3 m / s) 2 + (9.5 m / s) 2
= 24.2 m/s
⎛ 10.0 m / s ⎞ ⎟⎟ = tan–1 0.4 α = tan–1 ⎜⎜ ⎝ 22.3 m / s ⎠ (b)
At the highest point, the vertical velocity v
y
is zero at time t1 ; then
vy = 0 = v 0y – gt1 t1 =
v 0y g
29.5 m/s
=
10 m/s 2
= 3.0 s
The height h at this time is the value of y when t = t1 = 3 s; h = v0y t1 –
1 gt 2 2 1
= (29.5 m/s) (3.0 s) –
1 (10.0 m/s2) (3.0 s)2 2
= 88.5 – 45 = 43.5 m
(c)
v = 0y
t =
0y 2
1 1 2 ⎛ ⎞ gt = t2 ⎜ v 0 y − gt 2 ⎟ 2 ⎝ ⎠ 2 2
This is a quadratic equation for t2. It has two roots, t2 = 0
and t
2
=
2 v 0y g
2( 29.5 m / s) =
10 m / s 2
= 5.9 s
There are two times at which y = 0; t2 = 0 is the time the ball leaves the ground, and t2 = 6 s is the time of its return. This is exactly twice the time to reach the highest point, so the time of descent equals the time of TM
RESONANCE
Page # 44
ascent. (This is always true if the starting and end points are at the same elevation and air resistance can be neglected). The horizontal range R is the value of x when the ball returns to the ground, that is, at t = 6.64 s; R = v0xt2 = (22.3 m/s) (5.9 s) = 131.6 m The vertical component of velocity when the ball hits the ground is vy = v0y – gt2 = 29.5 m/s – (10 m/s2)(5.9s)
=–29.5m
That is, vy has the same magnitude as the initial vertical velocity v0y but the opposite direction (down). Since vx is constant, the angle α = – 53º )(below the horizontal) at the point is the negative of the initial angle α0 = 53º. Ex.7
A clever monkey escapes from the zoo. The zoo keeper finds him in a tree. After failing to entice the monkey and shoots. Theand clever monkey letsthat go at the same instant dart leaves regardless the gun barrel, intending to land on the ground escape. Show the dart always hitsthe the monkey, of the dart’s muzzle velocity (provided that it gets to the monkey before he hits the ground).
Sol.
The monkey drops straight down, so x monkey = d at all times. For the dart, x dart = (v0 cos a0)t. When these xcoordinates are equal, d = (v cos α )t, or 0
0
d t= v 0 cos α 0 To have the dart hit the monkey, it must be true that ymonkey = ydart at this same time. The monkey is in oney monkey = d tan α0 –
dimensional free fall
Forthedart
y
dart
= (v0 sin α0) t –
1 2 gt . 2 1 2 gt 2
So if d tan α0 = (v0 sin α0) t at the time when the two x-coordinates are equal, then ymonkey = ydart and we have a hit. Ex.8
A ball is thrwon from ground level so as to just clear a wall 4 m high at a distanc e of 4 m and falls at a distance of 14 m from the wall. Find the magnitude and direction of the ball.
Sol.
The ball passes through the point P(4, 4). So its range = 4 + 14 = 18 m. The trajectory of the the ball is, y = x tan θ ( 1 –
x ) R
Now
x=4m,y=4mandR=18m
∴
4⎤ ⎡ 7 4 = 4 tan θ ⎢1 − ⎥ = 4 tan θ . 9 ⎣ 18 ⎦ TM
RESONANCE
Page # 45
9 7
or
tanθ =
And
= R
2u2 sin θ cos θ g
or
18 =
2 × u2 × 9 .8
or
u2 =
or
=u
θ = tan–1
⇒
9 130
×
9 7
7 130
18 × 9.8 × 130 = 182 2× 9× 7 and θ = tan–1
182
9 . 7
Qus.
Two projectiles are thrown with different speeds and at different angles so as the cover the same maximum height. Find out the sum of the times taken by each to the reach to highest point, if time of flight is T.
Ans.
Total time taken by either of the projectile.
Qus.
A particle is projected with speed 10 m/s at an angle 600 with horizontal. Find : (a)timeofflight (b)range (c) maximum height (d) velocity of particle after one second. (e) velocity when height of the particle is 1 m
Ans.
(a)
3s
ec.
(b)5 3 m
(c)
15 m 4
(d)5.17m/s
(e) v = 5ˆi ± 55 ˆj r
3.
PR OJECTILE THR O WN PARALLEL T O THE HORIZONT AL : Consider a projectile thrown from point O at some height h from the ground with a velocity u. Now we shall deal the characteristics of projectile motion with the help of horizontal and vertical direction motions.
Horizontaldirection
Verticaldirection
(i)
Initial velocity u=x
u
Initial velocity u
(ii)
Acceleration a=x
3.1
TRAJECTORY EQUATION : The path traced by pr ojectile is called the trajectory.
0
Acceleration a
y y
=0
= g (downward)
After time t, = xut y= From equation(1)
....(1)
−1 2 gt 2
....(2)
t = x/u
Put value of t in equation (2)
TM
RESONANCE
Page # 46
x2 −1 g. 2 2 u
y=
This is trajectory equation of the projectile. 3.2
Velocity at a general point P(x, y) v=
u2x + u2y
Here horizontal velocity of the projectile after time t vx = u velocity of projectile in vertical direction after time t vy = 0 + (–g)t = –gt = gt (downward)
∴ 3.3
v=
u2 + g2 t 2
and
tan θ = vy/vx
DISPLACEMENT : The displacement of the particle is expressed by S = x ˆi + y ˆj 1 = (ut) ˆi + ( gt2) ˆj 2 where
3 .4
|S|=
x2 + y2
TIME OF FLIGHT : This is equal to the time taken by the projectile to return to ground. From equation of motion 1 2 at 2
S = ut +
Thereforeforvertical direction
At highest point vy = 0
t=± 3.5
2h g
–h=v
⇒
⇒
h=
t=
t+
y
1 (–g)t2 2
1 2 gt 2 2h g
HORIZONTAL RANGE : Distance covered by the projectile along the horizontal direction between the point of projection to the point on the ground. R = ux . t
2h g
R =u
TM
RESONANCE
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3.6
VELOCITY AT VERTICAL DEPTH h : v y2 = 02 + 2 . (–h) (–g)
Along vertical direction
vy =
2gh
EXAMPLES BASED ON PROJECT ILE PROJECTED HORIZONTALLY Ex.9
A projectile is fired horizontally with a speed of 98 ms –1 from the top of a hill 490 m high. Find (i) the time taken to reach the ground (ii) the distance of the target from the hill and (iii) the velocity with which the projectile hits the ground. (take g = 9.8 m/s2)
Sol.
(i) The projectile is fired from the top O of a hill with speed u = 98 ms–1 along the horizontal as shown as OX. It reaches the target P at vertical depth OA, in the coordinate system as shown, OA = y = 490 m
As,
= y
∴
1 2 gt 2 1 × 9.8 t2 2
490 =
or
=
t
100 = 10 s.
(ii)
Distance of the target from the hill is given by,
(iii)
AP = x = Horizontal velocity × time = 98 × 10 = 980 m. The horizontal and vertical components of velocity v of the projectile at point P are vx = u = 98 ms –1 vy = uy + gt = 0 + 9.8 × 10 = 98 ms –1 V=
v 2x + v 2y =
98 2 + 98 2 = 98 2 = 139 ms–1
Now if the resultant velocity v makes air angle β with the horizontal, then
tan β = Ex.10
vy vu
=
98 =1 98
∴
β = 45º
A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with magnitude 9.0 m/s. Find the motorcycle’s position, distance from the edge of the cliff and velocity after 0.5 s.
Sol.
TM
RESONANCE
Page # 48
At t = 0.50 s, the x and y-coordinates are x = v0xt = (9.0 m/s) (0.50 s) = 4.5 m y=–
1 2 1 gt = – (10 m/s2) (0.50 s)2 = –1.2 m 2 2
The negative value of y shows that a this time the motorcycle is below its starting point. The motorcycle’s distance from the srcin at this time 2
r=
x2 + y2 =
⎛ 45 ⎞ ⎛ 12 ⎞ m⎟ + ⎜ − m⎟ ⎜ ⎝ 10 ⎠ ⎝ 10 ⎠
( 4.5 m)2 + ( −1.2m)2 =
2
=
3 (15 )2 + ( 4)2 ~ – 5 sec. 10
The components of velocity at this time are vx = v0x = 9.0 m/s vy = – gt = (–10 m/s2) (0.50 s) = –5 m/s. The speed (magnitude of the velocity) at this time is
v 2x + v 2y
v=
=
(9.0 m / s )2 + ( −5 m / s)2 = 10.2 m/s
The angle α of the velocity vector is
= tan–1
⎛ − 5 m/s ⎞ = tan–1 ⎜⎜ 9.0 m / s ⎟⎟ ⎝ ⎠
vy vx
Qus.
Two tall buildings face each other and are at a distance of 180 m from each other. With what velocity must a ball be thrown horizontally from a window 55 m above the ground in one building, so that it enters a window 10.9 m above the ground in the second building.
Ans.
60 m/s.
Qus.
Two paper screens A and B are separated by a distance of 100 m. A bullet pierces A and then B. The hole in B is 10 cm below the hole in A. If the bullet is travelling horizontally at the time of hitting the screen A, calculate the velocity of the bullet when it hits the screen A. Neglect the resistance of paper and air.
Ans.
700 m/s
4.
PROJECTILE FROM A TOWER
Case (i) :
Horizontal projection ux;u=
Case (ii) :
u
a
y
=
−g
Projection at an angle θ above horizontal ux = ucosθ ;
Case (iii) :
=y ; 0
uy = usinθ;
ay =
−g
Projection at an angle θ below horizontal ux = ucosθ;
uy =
− usinθ;
ay =
−g
In all the above three cases, we can calculate the velocity of projectile at the instant of striking the ground by using v =
v 2x + v 2y and tanφ =
vy vx
, where
φ is the angle at which the p rojectile strikes the ground.
TM
RESONANCE
Page # 49
Ex.11
From the top of a 50m high tower a stone is projected with speed 10 m/s, at an angle of 37º as shown in figure. Find out (a) velocity after 3s (b)time of flight. (c) horizontal range. (d) themaximum height attained by the particle.
Sol.
(a)
Initial velocity in horizontal direction = 10 cos 37 = 8 m/s Initial velocity in vertical direction = 10 sin 37º = 6 m/s Velocity after 3 seconds v = vx ˆi + vy ˆj = 8 ˆi + (uy + ay t) ˆj = 8 ˆi – 24 ˆj
(b)
Sy = uy t +
1 a t2 2 y
5t2 – 6t – 50 = 0
(c)
Range=8×(
⇒
– 50 = 6 × t +
⇒
t=
1 × (–10) t2 2
6 ± 1036 10
6 ± 1036 ) 10
(d)
vy = uy + ay t 0 = 6 – 10t t = 0.6 or 0= 6–2× 10×h h = 1.8 maximum height = 50 + 1.8 = 51.8 m. Qus.
Two stones A and B are projected simultaneously from the top of a 100 m high tower. Stone B is projected horizontally with speed 10 m/s, and stone A is dropped from the tower. Find out the following (a) time of flight of the two stone. (b) distance between two stones after 3 sec. (c) angle of strike with ground. (d) horizontal range of particle B.
Ans.
(a) 2 5 s ec.
5.
PR OJECTION FR OM A MO VING BOD Y
(b) xB = 30 m, y B= 45
(c) tan−1 2 5
(d) 20 5 m
Consider a boy standing on a trolley who throws a ball with speed u at an angle θ with the horizontal. Trolley moves horizontally with constant speed v . Case (i) : When ball is projected in the direction of motion of the trolley, horizontal component of ball’s velocity = u cos θ + v Initial vertical component of ball’s velocity = u sin θ Case (ii) :
The ball is projected opposite to the direction of motion of the trolley Horizontal component of ball’s velocity = u cosθ
− v
Initial vertical component of ball’s velocity = u sinθ TM
RESONANCE
Page # 50
Case (iii) :
The ball projected upwards from a platform moving with speed v upwards. Horizontal component of ball’s velocity = u cos θ Initial vertical component of ball’s velocity = u sin θ + v
Case (iv) :
The ball projected upwards from a platform moving with speed v downwards. Horizontal component of ball’s velocity = u cosθ Initial vertical component of ball’s velocity = u sinθ − v
Ex.12
Sol.
A particle is projected at an angle of 30 0 with speed 20 m/s : (i)
Find out position vector of the particle after 1s
(ii) (i)
Find out angle betwen velocity vector and position vector at t = 1s sx = u cos θ t
3 × 2
= 20 ×
sy = u sin θt +
t
= 10
1 × 10 × t 2 2
1 × 1 – 5 (1) 2 2
= 20 ×
3m
= 5m
Position vector = 10 3 ˆi + 5 ˆj (ii)
vx = 10 3 ˆi vy = uy + ayt = 10 – 10 = 0 v = 10
r
r
v . s = |v| |s| cos θ
3 ˆi
r
10 3 × 10 3 10 3 cos θ = 10 3 × 325 = 325
3 = 2 13
⎛ 3 ⎞⎟ −1 θ = cos ⎜⎜ 2 ⎟ 13 ⎝ ⎠ Ex.13
A boy standing on a long railroad car h t rows a ball straight upwards. The car is moving on the horizontal road with an acceleration of 1 m/s2 and the projection speed in the vertical direction is 9.8 m/s. How far behind the boy will the ball fall on the car ?
Sol.
Let the initial velocity of car be ‘u’. where
t=
2u ⊥ =2 g
u ⊥ = component of velocity in vertical direction xc = u × 2 +
1 × 1 × 22 2
= 2u + 2 where xc = distance travelled by car xb = distance travelled by ball xb = 2u xc – x b = 2u + 2 – 2u = 2m
Ans.
TM
RESONANCE
Page # 51
Qus.
A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8m. Find the speed and the angle of projection (a) as seen from the truck , (b) as seen from the road. [Ans : (a) 19.6 m/s upward (b) 24 .5 m /s a t 53 º wi th h orizontal]
6.
PROJECTION ON AN INCLINED PLANE To solve the problem of projectile motion on an incline plane we can adopt two types of axis system as shown in the figures
Case (i) : Up the incline
axisystem 1 Here αi sangleofprojection withthehorizontal. In thiscase: ax = 0 ay =
−g
axisystem 2 Here α is angle of projection withtheinclinedplane In this case:
ux = ucosα
ax = - gsinβ
uy = usinα
ay = - gcos β uy = usinα Time of flight (T) :
ux = ucosα
when the particle strikes the inclined plane y coordinate becomes zero 1 2 at 2 y 1 ⇒ 0 = usinαT – gcosβT2 2 2u sin α 2u ⊥ ⇒ T= = g cos β g⊥ Maximum height (H) : when half of the time is elasped y coordinate is equal to maximum height of the projectile y = uyt +
u sin α ⎛ ⎞ ⎟⎟ – 1 g sin β⎜ u sin α ⎟ ⎜ g cos β ⎟ 2 ⎝ g cos β ⎝ ⎠
2
H = u sinα ⎜⎜
⇒
H=
u2 ⊥ u 2 sin 2 α = 2g ⊥ 2g cos β
Range along the inclined plane (R): When the particle strikes the inclined plane x coordinate is equal to range of the particle 1 x = uxt + ax t2 2
⎛ 2u sin α ⎞ ⎛ 2u sin α 1 ⎟⎟ – g cos β ⎜⎜ ⎟⎟ 2 ⎝ g cos β ⎝ g cos β
⇒
R = ucosθ ⎜⎜
⇒
R=
2
2u 2 sin α cos(α + β) g cos 2 β
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RESONANCE
Page # 52
Case (ii) : Down the incline
axisystem 1 In this case: ax = 0 ay =
−g
axisystem 2 In this case :
ux = ucosα
ax = gsinβ
uy = usinα
ay =
ux = ucos α
−gcosβ
uy = usinα
Time of flight (T) : when the particle strikes the inclined plane y coordinate becomes zero y = uyt +
1 2 at 2 y 1 gcosβT2 2
⇒
0 = usinαT –
⇒
2u 2u sin α T = g cos β = g ⊥ ⊥
Maximum height (H) : when half of the time is elasped y coordinate is equal to maximum height of the projectile
⎛ u sin α ⎞ ⎛ u sin α ⎞ 1 ⎟⎟ – g sin β⎜⎜ ⎟⎟ 2 ⎝ g cos β ⎠ ⎝ g cos β ⎠
2
H = u sinα ⎜⎜
⇒
H=
u2 sin 2 α u2 ⊥ = 2g cos β 2g⊥
Range along the inclined plane (R): When the particle strikes the inclined plane x coordinate is equal to range of the particle x = uxt +
1 a t2 2 x
⎛ 2u sin α ⎞ ⎛ 2u sin α ⎞ 1 ⎟⎟ + g cos β ⎜⎜ ⎟⎟ 2 ⎝ g cos β ⎠ ⎝ g cos β ⎠
⇒
R = ucosθ ⎜⎜
⇒
R=
2
2u2 sin α cos( α − β) g cos 2 β
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RESONANCE
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Table 1 : Standard results for projectile motion on an inclined plane
Range
Up the Incline
Down the Incline
2u 2 sin α cos( α + β )
2u 2 sin α cos( α − β )
g cos 2 β
g cos 2 β
Time of flight
2u sin α g cos β
2u sin α g cos β
Angle of projection for maximum range
π β − 4 2
π β + 4 2
Maximum Range
u2 g(1 + sin β )
u2 g(1 − sin β )
Here α is the angle of projection with th e incline and
β is the angle of incline.
NOTE : For a given speed, the direction - which gives the maximum range of the projectile on an incline, bisects the angle between the incline and the vertical, for upward or downward projection. Ex.14 Sol.
A particle is projected horizontally with a speed u from the top of a plane inclined at an angle θ with the horizontal. How far from the point of projection will the particle strike the plane? Take X,Y-axes as shown in figure. Suppose that the particle strikes the plane at a point P with coordinates (x,y) . Consider the motion between A and P.
Motion in x direction : Initial velocity = u Acceleration = 0 x = ut......................(i) Motion in y direction : Initial velocity = u Acceleration = g
y
=
1 gt 2.......................(ii) 2
Eliminating t from (i) and (ii)
Also
y
=
y
=xtan
1 x2 g 2 u2
θ
2
2 Thus , gx 2 = xtan θ giving x = 0 or, 2u tan θ gu g
Clearly the point P corresponds to x =
then y = xtan θ =
2u 2 tan 2 θ g
The distance AP =
x2 + y2
2u 2 tan θ g
TM
RESONANCE
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= Ex.15
Sol.
2u2 2 tan θ 1 + tan θ g
=
2u2 tan θ sec θ g
A projectile is thrown at an angle θ with an inclined plane of inclination β as shown in figure. Find the relation between β and θ if : (a) projectile strikes the inclined plane perpendicularly, (b) projectile strikes the inclined plane horizontal. (a) If projectile strikes perpendicularly. vx = 0 when projectile strikes vx = ux + axt 0 = u cos θ – g sin β T u cos θ T = g sin β 2u sin θ we also know that T = g cos β
⇒ (b)
u cos θ 2u sin θ = g sin β g cos β
⇒
2 tan θ = cot β
If projectile strikes horizontally, then at the time of striking the projectile will be at the maximum height from the ground. Therefore : 2u sin θ t0P = g cos β t0P =
2u sin(θ + β) 2× g
⇒
2u sin θ 2u sin(θ + β) = g cos β 2g
⇒
2 sin θ = sin (θ + β) cos β .
Elastic collision of a projectile with a wall : Suppose a projectile is projected with speed u at an angle θ from point O on the ground. Range of the projectile is R. If a wall is present in the path of the projectile at a distance x from the point O. The collision with the wall is elastic, path of the projectile changes after the collision as described below. R 2 Direction of x component of velocity is reversed but its magnitude remains the same and y component of velocity remains unchanged, therefore the remaining distance (R – x) is covered in the backward direction and projectile falls a distance (R – 2x) ahead of the point O as shown in figure. Case I : If x ≥
R 2 Direction of x component of velocity is reversed but its magnitude remains the same and y component of velocity remains unchanged, therefore the remaining distance (R – x) is covered in the backward direction Case II : If x <
and projectile falls a distance (R – 2x) behind the the point O as shown in figure.
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RESONANCE
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E xer ci se 1 OBJECTIVE PROBLEMS 1.
Two stones are projected from the same point with same speed making angles 45º + θ and 45º– θ with the horizontal respectively. If θ < 45º, then the horizontal ranges of the two stones are in the ratio of (A) 1:1 (B) 2:1 (C) 3:1 (D) 4:1
2.
A hunter takes an aim at a monkey sitting an a tree and fires a bullet. Just when the bullet leaves barrel of the gun, it so happens that the monkey begins to fall freely. The bullet will (A)goabovethemonkey (B)gobelowthemonkey (C) hit the monkey (D) may or may not hit the monkey. It will depend upon the velocity of the bullet.
3.
5π It was calculated that a shell when fired form a gun with a certain velocity and at an angle of elevation rad 36 should strike a given target. In actual practice, it was found that a hill just prevented the trajectory. At what angle of elevation should the gun be fired to hit the target. (A)
5π rad 36
(B)
11π rad 36
(C)
7π rad 36
(D)
13π rad. 36
4.
A projectile is thrown with a speed v at an angle θ with the vertical. Its average velocity between the instants it crosses half the maximum height is (A) v sin θ, horizontal and in the plane of projection (B) v cos θ, horizontal and in the plane of projection (C) 2v sin θ, horizontal and perpendicular to the plane of projection (D) 2v cos θ, vertical and in the plane of projection.
5.
Two bullets are fired horizontally, simultaneously and with different velocities from the same place. Which bullet will hit the ground earlier? (A) It would depend upon the weights of the bullets (B) The slower one.
6.
7.
(C)Thefasterone (D)Bothwillreachsimultaneously. A stone is thrown upwards. It returns to ground describing a parabolic path. Which of the following remains constant? (A)speedoftheball (B)kineticenergyoftheball (C) vertical component of velocity (D) horizontal component of velocity. A body is thrown horizontally with a velocity
2gh from the top of a tower of height h. It strikes the level
ground through the foot of the tower at a distance x from the tower. The value of x is (A) h
(B)
h 2
(C) 2h
(D)
2h 3
8.
A particle, with an initial velocity v0 in a plane, is subjected to a constant acceleration in the same plane. Then, in general, the path of the particle could be (A)acircle (B)astraightline (C)aparabola (D)ahyperbola.
9.
A ball is projected from a certain point on the surface of a planet at a certain angle with the ho rizontal surface. The horizontal and vertical displacement x and y vary with time t in second as: x = 10
2
3t and y = 10t – t The maximum height attained by the ball is (A) 100 m (B) 75 m (C) 50 m 10.
(D) 25 m.
A bag is dropped from an aeroplane flying horizontally at a constant speed. If air resistance is ignored, where will the aeroplane be when the bag hits the ground? (A)aheadofthebag (B)directlyabovethebag (C)farbehindthebag (D)dataisnotsufficient.
RESONANCE
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Page # 56
11.
The path of one projectile in motion as seen from another moving projectile is (A)astraightline (B)acircle (C)anellipse (D)aparabola.
12.
A plane surface is inclined making an angle θ with the horizontal. From the botto m of this inclined plane, a bullet is fired with velocity v. The maximum possible range of the bullet on the inclined plane is (A)
13.
v2 g
v2 g(1 + sin θ)
(B)
(C)
v2 g(1 − sin θ)
(D)
v2 + g(1 cos θ)
A ball is projected horizontal with a speed v from the top of a plane inclined at an angle 45º with the horizontal. How far from the point of projection with the ball strike the plane? (A)
v2 g
v2 2 g
(B)
(C)
2v 2 g
(D)
⎡ 2v 2 ⎤ 2 ⎢ g ⎥ ⎣⎢ ⎦⎥
14.
The time of flight of a projectile on an upward inclined plane depends upon (A) angle of inclination of the plane (B) angle of projection (C) the value of acceleration due to gravity (D) all of these.
15.
A ball rolls of the top of a stairway horizontally with a velocity of 4.5 m s–1. Each step is 0.2 m high and 0.3 m wide. If g is 10 ms –1, then the ball will strike the nth step where n is equal to (A) 9 (B) 10 (C) 11 (D) 12
16.
The velocity of projection of a projectile is (6 ˆi + 8 ˆj ) ms–1.. The horizontal range of the projectile is (A)4.9m (B)9.6m (C) 19.6 m (D)14 m.
17.
If R and h represent the horizontal range and maximum height respectively of an obliquie projectile, then R2 + 2h represents 8h (A)maximum horizontal range (C)timeofflight
18.
A particle move along the parabolic path x = y vector remain 5m/s during the motion. The (A) 50 m/s 2 (B) 100 m/s 2
19.
(B)maximum verticalrange (D)velocityofprojectileathighestpoint 2
+ 2y + 2 in such a way that the y-component of velocity
magnitude of the accleration of the particle is (C) 10 2 m/s 2 (D) 0.1 m/s 2
A ball is projected from point A with a velocity 10 m/s perpendicular to figure. Range of the ball on the inclined plane is : 40 (A) m 3
20 (B) m 13
13 (C) m 20
13 (D) m 40
:
the inclined plane as shown in
SUBJECTIVE PROBLEMS 20.
In order to project a body for maximum range, what is the condition ?
21.
What is the angle between the directions of velocity & acceleraiton at the highest point of a projectile path ?
22.
At what point of the projectile path the speed is minimum ?
23.
Two bodies are projected at anglesθ and (90 – θ) to the horizontal with the same speed. Find the ratio of their times of flight?
24.
In above question find the ratio of the maximum vertical heights ?
25.
What should be the angles of projection to obtain maximum height and maximum time of flight ?
RESONANCE
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Page # 57
26.
A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 ms –1. What is the trajectory of the bob ; if the string is cut (a) when the bob is at one of its extreme position (b) at its mean position
27.
A projectile can have the same range R for two angles of projections. If t1 & t2 be the times of flight in two cases, then find out relation between t1t2 and R ?
28.
The height y and the distance x along the horizontal plane of a projectile on a certain planet (without no surrounding atmosphere) are given by y = (8t – 5t 2) metere & x = 6t m. Then what will be the velocity of projection ?
29.
A glass marble projected horizontal from the top of a table falls at a distance x from the edge of the table. If h is the height of the table, find the velocity of projection ?
30.
A projectile is fired horizontally with a velocity of 98 m/s from the top of a hill 490 m high. Find : (i) the time taken to reach the ground (ii) the distance of the target from the hill (iii) the velocity with which the particle hits the ground
31.
The equation of a projectile is y =
32.
Consider a boy on a trolley who throws a ball with speed 20 m/s at an angle 370 w.r.t. trolley which moves horizontally with speed 10 m/s.
3x–
gx 2 , find the angle of projection ? 2
(a) Find horizontal and vertical components of initial velocity of ball when ball is projected in direction of motion of trolley. (b) Find horizontal and vertical components of initial velocity of ball when ball is projected opposite to direction of motion of trolley 0
33.
Consider a boy on a 10 platform who throws a ball withand speed 20 m/s at an angle 37 w.r.t. platform which moves upwards with speed m/s. (a) Find the horizontal vertical component of balls velocity. (b) Find horizontal and vertical components of balls velocity when ball is projected downwards from the platform.
34.
A bomb is dropped from an aeroplane moving horizontally at a certain height from the ground. Does the time taken by the bomb to reach the ground depend on the velocity of the aeroplane ?
35.
A particle is projected at an angle θ with an inclined plane making an angle β with the horizontal as shown in figure, speed of the particle is u, after time t find : (a) x component of acceleration ? (b) y component of acceleration ? (c) x component of velocity ? (d) y component of velocity ? (e) x component of displacement ? (f) y component of displacement ? (g) y component of velocity when particle is at maximum distance from the incline plane ?
36.
On an inclined plane of inclination 30º, a ball is thrown at an angle of 60º with the horizontal from the foot of the incline with a velocity of 10 3 ms–1. If g = 10 ms–2, then find the time in which ball will hit the inclined plane?
37.
The direction of motion of a projectile at a certain instant is inclined at an angle α to the hor izon. After t seconds it is inclined an angle β . Find the horizontal component of velocity of projection in terms of g, t , α and β.
RESONANCE
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Page # 58
r
varies with time t as r = at ˆi − bt 2 ˆj , where a and b ˆ ˆ are positive constants and i and j are the unit vectors of the x and y axes. Fi nd: (i) the equation of the point's trajectory y (x) ; plot this function (ii) the time dependence of the velocity v and acceleration a vectors as well as of the moduli of these quantities .
38.
A radius vector of a point A relative to the srcin
39.
Two particles are projected simultaneously with the same speed V in the same vertical plane with angles of elevation θ and 2 θ, where θ < 45º . A t what time will their velocities be parallel.
40.
If 4 seconds be the time in which a projectile reaches a point P of its path and 5 seconds the time from P till it reaches the horizontal plane through the point of projection . Find the height of P above the horizontal plane. [ g = 9.8 m/sec 2 ]
41.
A man is travelling on a flat car which is moving up a plane inclined at cos θ = 4/5 to the horizontal with a speed 5 m/s. He throws a ball towards a stationary hoop located perpendicular to the incline in such a way that the ball moves parallel to the slope of the incline while going through the centre of the hoop. The centre of the hoop is 4 m high from the man's hand cal culat e the time taken by the ball to rea ch the hoop.
4m
/s 5m •
θ
•
1.
A ship is approaching a cliff of height 105 m above sea level. A gun fitted on the ship can fire shots with a speed of 110 ms −1. Find the maximum distance from the fo ot of the cliff from where the gun can hit an object on the top of the cliff. [ g = 10 m/s 2 ] [REE '94, 6]
2.
Two towers A B and CD are situated a distance ' d ' apart as shown in the fig. AB is 20 m high and CD is 30 m high from the ground. An object of mass ' m ' is thrown from the top of AB horizontall y with a velocity of 10 m/s towards CD. Simultaneously another object of mass 2 m is thrown from the top of CD at an angle of 60º to the horizontal towards AB with the same magnitude of initial velocity as that of the first object. The two objects move in the same vertical plane , collide in mid air and stick to each other. (i) calculate the distance ' d ' between the towers. (ii) find the position where the objects hit the ground.
3.
[JEE ' 94, 6]
A building 4.8 m high 2b meters wide has a flat roof. A ball is projected from a point on the horizontal ground 14.4 m away from the building along its width. If projected with velocity 16 m/s at an angle of 45º with the ground, the ball hits the roof in the middle, find the width 2b. Also find the angle of projection so that the ball just crosses the
roof if projected with velocity 10
3 m/s.
[REE '95, 6]
4.
3 m/s Two guns situated on the top of a hill of height 10 m, each fired shots with the same speed 5 at some interval of time. One gun fires horizontally and other fires upwards at an angle of 60º with the horizontal. The shot collide in air at a point P. Find: [JEE '96, 5] (i) the time interval between the firings and (ii) the coordinates of the point P. Take srcin of the coordinate system at the foot of the hill right below the muzzle and trajectories in xy plane.
5.
A vertical pole has a red mar k at some height. A stone is projected from a fixed point on the ground. When projected at an angle of 45º it hits the pole orthogonally 1 m above the mark. When projected with a different velocity at an angle of tan −1(3/4), it hits the pole orthogonally 1.5 m below the mark. Find the velocity and angle of projection so that it hits the mark orthogonally to the pole. [ g = 10 m/sec 2 ] [REE '96, 6 ]
RESONANCE
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Page # 59
6.
The trajectory of a projectile in a vertical plane is y = ax − bx2, where a, b are constants and x and y are respectively the horizontal and vertical distances of the projectilefrom the point of projection. The maximum height attained is _______ and the angle of projection from the horizontal is _______. [JEE '97, 2]
7.
The coordinates of a particle moving in a plane are given by x (t) = a cos (p t) and y (t) = b sin (p t), where a, b (< a) and p are positive constants of appropriate dimensi ons. (A) the path of the particle is an ellipse (B) the velocity and acceleration of the particle are norm al to each other at t = π/2p (C) the ac celeration of t he p article is a lways di rected t owards a f ocus (D) the distance travelled by the particle in time interval t = 0 to = π/2p is a. [JEE '99, 2]
8.
An object A is kept fixed at the point x = 3 m and y = 1.25 m on a plank P raised above the ground. At time t = 0 the plank starts moving along the + x direction with an acceleration 1.5 m/s 2. At the same instant a stone is projected from the srcin with a velocity u as shown. A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of 45º to the horizontal. All the motions are in x-y plane. Find ‘u’ and the time after which the stone hits the object. [ Take g = 10 m/s 2 ] [JEE20005 ,] Shots fired simultaneous ly from the top and foot of a vertical cliff at elevations of 30 0 and 60 0 respectively, strike an object simultaneously which is at a height of 100 meters from the ground and at a horizontal
9.
distance of 200 3 meters from the cliff. Find the height of the cliff, the velocities of projection of the shots and the time taken by the shots to hit the object. [ REE ’2000, 5 ]
EXERCISE # 1
36. 2 s
37.
OBJECTIVE PROBLEMS
1.
A
2. C
3. D
4. A
5.
D
6. D
7. C
8. B,C
9.
D
10. B
11. A
12. B
13. D
14. D
15. A
16. B
17. A
18. A
19. A
38. (i) y = −
21.
r
900 .
0
0
25. (a) θ = 90 , (b) θ = 90 . 26. (a) vertically downwards (b) parabolic path
30. (i) 10 sec.
28. 10 m/s
(ii) 980 m
29. x
ω = 2b
v ⎛θ ⎛ 3θ cos ⎜ ⎟ cos ec ⎜ ⎟ 40. 98 meters g 2 2
41. 1 s
1.
1100 m
2.
(i) 17.32 m (ii) combined mass strik es at 11.55 m from B and 5.77 m from D
3. 4.
width of the roof is 9.6 m; (i) 1 s (ii) ( 5 3 ,5 )
5.
9 3620 ⎟ m/s , tan −1 ⎜ ⎝ 10 3
g 2h
(iii) 138.59 m/s
31. tan θ =
−1
32. (a) 26, 12 (b) –6, 12
3 33. (a) 16, 22 (b) 16, 2
34. No
35. (a) – g sinβ, (b) –g cos β, (c) u cos θ – g sinβ × t, (d)
u sinθ – gcosβ × t, (e) u cos θ × t – u sinθ × t –
r
2 2 2 a +4b t ,
EXERCISE # 2
24. sin2 θ : cos2 θ
27. t1t2 = 2R/g
39.
23. sin θ : cos θ
22. At the highest point.
r
(ii) v = ai$ − 2b tj $ , ω = − 2 b $j ,
v =
Angle of projection = 450 .
a
2
2
r
SUBJECTIVE PROBLEMS
20.
bx
gt tanα − tanβ
1 g sinβ × t2, (f) 2
θ = tan
6.
⎛3 ⎜ ⎟ ⎝2
a2 , tan −1 a 4b
7.
A, B
9.
400 m, VT = 40 3 m/s, VF = 40 m/s, T = 10 s.
8. u = 7.29 m/s, t = 1 s.
1 g cosβ × t2, (g) zero. 2
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