CHEAPTER
-2
VECTORS
AND SCALARS
VECTOR:
A vector is a quantity having both magnitude and
direction, such as displacement, velocity, force, and acceleration.
SCALAR:
A scalar is
a quantity having magnitude but no direction, e.g. mass, time, length,
temperature, and any real number.
UNIT
VECTOR:
A unit
vector is vector having unit magnitude. If A is a vector with magnitude A≠0, then A/A is a unit vector having the same direction as A.
Any vector A can be represented by a unit vector a in the direction of A
multiplied by the magnitude of A. In symbols, A=Aa.
NULL
VECTOR:
A vector
having initial and the terminal point co-incident is termed as a zero or null
vector. If the resultant is zero then it is called null or zero vector and it
is represented by o or simply 0.
Displacement:
A change of
position of a particle is called displacement.
o------------->p
o=origin or
initial point
p=terminal
point or terminus
Position
vector:
Any vector drawn from origin to a point is called
the position vector.
Reciprocal
vector:
Any vector
having its direction the same as that of a given vector A, but its magnitude is
the reciprocal of the magnitude of A_ is termed as the reciprocal vector of A
and written as 1/A.
The
dot or scalar product of two vectors A and B:
The dot or scalar product of two vectors A
and B is called scalar vector. The scalar product written as A.B
(spoken as “A dot B”) is defined as the product of the magnitude
of A and B and cosine of the angle θ between them.
In symbols A.B=ABcosϴ o<ϴ< π; since A and B are scalars and cosϴ is a pure number, scalar product of two vectors is a scalar and not a
vector. The scalar product of two vectors can be regarded as the product of one
vector and the component of other vector in the direction of the first.
The cross or Vector product of two
vectors A and B:
C
= A x B (Spoken as “A cross B”) the magnitude of A x B is
defined as the product of the magnitude of A and B and the sine of angle ϴ between them. In symbols, C=I A x B
I
= ABcosϴ 0=<ϴ=<1800
The
direction of the vector C = A x B is perpendicular to the plane
formed by A and B. Such that A, B and C for a
right-handed system.
C = A x
B =η^ABsinθ
Where η^is a unit vector including the
direction of A x B.
VECTOR PROBLEMS:
(1). Let two vectors be represented in terms
of their co-ordinates as
A=i^ Ax +j^ Ay + k^Az and B =
i^Bx +j^By
+ k^Bz show analytically that,
(a)
A.B= Ax
Bx+ Ay By+ Az Bz
(b)
A x B = І i^ j^ k^
І
І A x Ay A z І
І B x By B z І
(c)
A x B = - B x A
SOLUTION:
(a) Given
that,
A = i^Ax +j^ Ay
+ k^Az
and B = i^Bx
+j^By
+ k^Bz
A.B = (i^
Ax +j^
Ay + k^Az ).( i^Bx +j^By + k^Bz)
= Ax Bx(i^.i^)+
Ay By(j^.j^)+ Az Bz(k^.k^)+
Ax By(i^. j^) + Ax
Bz(i^. k^) +
Ay Bx(j^.i^)
+ Ay Bz(j^. k^)+ Az Bx(k^. i^) + Az By(k^. j^)
= Ax Bx+ Ay By+
Az Bz
(Proved)
(b) A x B = (i^ Ax +j^ Ay + k^Az ) x ( i^Bx +j^By + k^Bz)
= Ax Bx(i^
x i^)+ Ay By(j^ x j^)+ Az
Bz(k^ x k^)+ Ax By(i^ x j^) +
Ax Bz(i
^ x k ^)+Ay Bx(j^ x i^)
+ Ay Bz(j^ x k^)+ Az Bx(k^ x i^) + Az By(k^xj^)
= Ax By k^+ Ax Bz(-j^)+
Ay Bx(-k ^) +Ay Bz i^+
Az Bx j^+ Az By(-i ^)
= (Ay Bz
- Az By) i ^+( Az Bx -Ax Bz)
j^ +( Ax By - Ay Bx) k^
= (Ay Bz
- Az By) i ^- (Ax Bz- Az Bx) j^+( Ax By - Ay
Bx) k^
= І
i^ j^ k^ І
І A x Ay A z І
І B x By B z
І
(Proved)
(c) A
x B = - B x A :
Fig.(a)
Fig.(b)
AXB=C has magnitude ABSinѲ and direction such that
A, B and C form a right-handed system Fig. (a)
BXA=D has magnitude BASinѲ and direction such that B,A and D form a
right-handed system Fig.(b). Then D has some magnitude as C but is opposite in
direction i.e. C = -D or A x B
= -B x A. The commutative laws for cross product are not valid.
(2) Find the angle between A= 3i^+ 2j^ - 6k^and B=
4i^- 3j^ + k^.
SOLUTION: Given that,
A= 3i^+ 2j^ - 6k^and B= 4i^- 3j^ + k^.
We know that,
A.B =
ABcosϴ
0r, cosϴ =
Or, Cosϴ =
Or, Cosϴ =
Or, Cosϴ =0
Or, ϴ= 900
Answer: 900
(3) If A= 2i^+ j^ -3k^ and B= i^-2 j^ + k^ then
find a vector of magnitude 3 perpendicular both A and B.
SOLUTION: Given that,
A= 2i^+ j^ -3k^ and B= i^-2 j^ + k^
A x B is perpendicular to the plane of A and B ,
A x B = І i^ j^ k^ І
І 2 1
-3 І
І 1 2
1 І
=
(1-6)i^+ j^(2+3) + k^(-4-1)
= -5 i^-5 j^ -5
k^
= -5(i^+ j^ + k^)
Unit vector
parallel to A x B is,
A x B
= --------------------------
I A x B I
=
=
=
Another unit vector opposite to this direction
is,
=
A unit
vector perpendicular to both A and B is ±
So, a vector of magnitude 3 perpendicular to both
A and B is,
= ±3 x
=
Answer:
(4) Find a unit vector parallel to the
resultant of vectors of A= 2i^+4 j^ -5k^ and B=
i^+2
j^ +3 k^.
SOLUTION: Given that,
A= 2i^+4 j^
-5k^ and B= i^+2 j^ +3 k^.s
The resultant of vectors A and B
is,
A + B = ( 2i^+4 j^ -5k^) +( i^+2 j^
+3 k^)
=3 i^+6 j^ -2 k^
A unit vector parallel to the resultant is,
A x B
= ------------------
I A x
B I
=
=
=
=
Answer:
(5)
Determine the vector having initial point
P(x1, Y1,
Z1) and terminal point Q (x2, Y2,
Z2) and find its
magnitude.
SOLUTION:
Position
vector of p and Q with respect to the origin are:
R1= X1 i^+ Y1j^+ Z1k^
R2 = x2i^+ Y2j^+ Z2k^
From
vector addition we can write,
r1+ R = r2
Or, R = r2 – r1
= (x2i^+ Y2j^+ Z2k^)
– (X1 i^+ Y1j^+ Z1k^)
= (x2 – X1 )i^+ ( Y2 – Y1)j^+ (Z2 – Z1)k^
So, magnitude of R =I R I=
(6) Determine the vector having initial
point (2, -1, -3) and terminal point (3, 2, -4).
SOLUTION:
Let, r1 = 2i^- j^
-3 k^
r2
=3i^+2 j^ -4k^
From the vesctor addition we can write,
r1+ R
= r2
or, R = r2
– r1
= (2i^- j^ -3 k^) –( 3i^+2 j^ -4k^)
= i^+ 3j^ - k^
Answer:
i^+ 3j^ - k^
(7) Find the value of ‘a’ so that 2i^-3 j^ +5 k^ and
3 i^+a j^ -2 k^ .
SOLUTION:
Let, A=2i^- 3j^ + 5k^
B=3i^+ aj^
-2k^
Since, A and B are perpendicular so
we can write,
A.B =0
(2i^-3 j^ +5 k^).( 3 i^+a j^ -2 k^ )=0
Or, 6-3a-10=0
Or, a= - 4/3
Answer: - 4/3
(8) If A=
2i^+
j^ + k^ and B=
i^-2
j^ +2 k^ and C=3i^-4 j^ +2 k^ find
the projection of A+C in the direction of B .
SOLUTION: Given
that,
A= 2i^+ j^
+ k^
B= i^-2 j^
+2 k^
C=3i^-4 j^ +2 k^
The projection of A + C in the
direction of B is,
(A + C
)cosϴ=
=
=
Answer:
(9)Find
a unit vector in the direction of the resultant of vectors
A= 2i^-
j^+ k^ , B= i^+
j^ +2 k^, C=3 i^-2
j^ +4 k^ and also find A. (B x C).
SOLUTION:
Given that,
A= 2i^- j^+ k^
B= i^+ j^
+2 k^
C=3 i^-2 j^
+4 k^
The
resultant of vectors A, B and C is,
A+ B
+C = (2i^- j^+ k^ )+( i^+ j^ +2 k^)+(3 i^-2 j^
+4 k^)
=6i^-2 j^ +7 k^
A unit vector in the direction of the resultant
is,
A+ B +C
= ---------------
I A+ B +C I
=
=
s
A. (B x C) = (2i^- j^+ k^). І i^ j^ k^ І
І 1 1 2 І
І 3 -2 4 І
= (2i^- j^+ k^). (8i^+2 j^ -5 k^)
= 16 – 2 -5
= 9
Answer: 9
CHAPTER-3
COLLISION
Collision:
A collision is the impact of faces on two approaching bodies.
Conservation of momentum during collision:
Let us consider a collision between two particles of on masses m1
and m2.During the collision this particles exerted large force
on one another.
At any instant F1 is the force
exerted on particle 1 by particle 2 and F2 is the force
exerted on particle 2 by particle 1 also at any instant this forces are equal in
magnitude but oppositely directed so that
F2
(t) = - F1 (t).
The changing in momentum of particle 1 during collision is,
∆P1=∫tf
F1 dt
ti
=F1^∆t
Where F1^ is the average
value of force F during the time interval ∆t=tf
– ti. Similarly, the changing in momentum of particle 2 during collision is,
∆P2=∫tf
F2 dt
ti
=F2^∆t
Where F2^∆t is the average
value of force F during the time interval ∆t=tf
– ti. At any time,
F1 = -
F2
andF1^ = - F2^
And therefore, ∆P1=F1^∆t
= - F2^∆t
= - ∆P2
When two particles act as an isolated system the total amount of the
system is,
P
= P1+ P2
or, ∆P
= ∆P1+ ∆P2
or, ∆P
= - ∆P2 + ∆P2
=o.
Thus the total change in momentum of the system as a result of
collision is zero. Hence if there are no external forces the total momentum of
the system is not changed by the collision that means the momentum is
conversed.
Collision in one-dimension:
When kinetic energy is conserved the collision is said to be elastic
collision and when kinetic energy is not conserved, it is called inelastic collision.
Let us imagine two smooth non-rotating moving along the line joining
their centered, then colliding head-on and moving along the same direction or
straight line without rotation after collision. These bodies exert forces on
one each other during the collision.
(Fig: Two sphere before and after an elastic collision. The
velocity V1i - V2i
of m1 relative to m2 before collision is equal to the
velocity V2f - V1f
of m1 relative to m2 after collision)
Let masses of the spheres are m1
and m2, velocity being V1i and V2i
before collision and V1f and V2f after collision. We consider the
positive direction of the momentum and velocity to be along the positive
x-axis. Then from the conservation of momentum,
m1 V1i + m2V2i
= m1V1f + m2V2f
or, m1 (V1i
- V1f) = m2(V2f - V2i)……………………………………(1)
Because the collision is elastic, so the kinetic energy is
conserved
1/2
m1 V1i2 +1/2 m2V2i2
=1/2 m1 V1f2+1/2 m2V2f2
Or, m1
(V1i2 - V1f2) = m2(V2f2-
V2i2)…………………………………………(2)
So, if we know the masses and the initial velocities then we can
calculate the two final velocities V1f and V2f
from equation. If we assume V2f
≠ V2i and
V1i ≠ V1f and dividing (2) by (1) we get,
V1i + V1f = V2f +V2i
Or, V1i - V2i = V2f - V1f …………………………………………….. (3)
This tells us that in an elastic one-dimensional collision, the relative
velocity of approach before collision is equal to the relative velocity of
separation after collision. From equation (3),
V2f = V1f+ V1i - V2i
Putting this value in equation (1) we get,
m1
(V1i - V1f) = m2 (V1f+ V1i - 2V2i)
or, V1f (
m1 + m2 ) = ( m1 - m2)
V1i + 2mV2i
or, V1f = =(
)V1i +(
)V2i -------------------------------(4)
Similarly, from equation (3),
V1f
= V2f + V2i - V1i
Putting this value in equation (1) we get,
m1
(V1i –V2f -V2i +V1i) = m2 (V2f
-V2i)
Or, 2m1V1i
– m1V2f - m1V2i = m2V2f
- m2V2i
Or, V2f
(m1 + m2) =2 m1V1i + ( m2 -
m1) V2i
Or, V2f
=(
V1i
+ (
)V2i ----------------------- (5)
From equation (4) and (5) we can see several interesting properties:
Case -1: When the colliding
particles have the same masses i.e. m1 = m2 gives
V1f =
V2i and V2f = V1i
That is in a one –dimensional elastic collision of two particles of
equal masses, the particles simply exchange their velocities during collision.
Case -2: If one particle is stationary say m2 or initially
at rest, then V2i equals zero and we obtain V1f and V2f
from equations (4) and (5),
V1f = (
) V1i and V2f =
V1i
---------------- (6)
Again, from equation (6),
(i)If m1 = m2 gives V1f = 0 and
V2f = V1i
(ii) If m1 >> m2 , we obtain,
V1f =῀- V1i
and V2f =῀0
(iii) If m1 >> m2, we obtain,
V1f =῀ V1i
and V2f =῀ 2V1i ------------------------- (7)
Equation (7) means that the velocity of the massive incident particle
is virtually unchanged by the collision with the light stationary particle, but
the light particle rebounds with approximately twice the velocity of the
incident particle.
Impulse and momentum
Figure. Shows the magnitude of the force exerted on a body during
collision.
Let us consider two particles
collide each other. If we assume that the force has a constant direction, then
the plot of force Vs shows that the impulsive force F(t) might vary
with time during a collision starting at time ti and ending at time
tf .From Newton’s 2nd law motion we can write the change in
momentum dp =Fdt. The change in momentum of the body during a collision by
intergrading over the time of collision gives the impulse,
Pf – Pi =∆P =∫pf dp
= ∫pf Fdt
pi pi
So, the integral of the force over the time interval during which the
force acts is called the impulse of the force.
PROBLEMS:
(1)A
baseball weighing 0.35lb is struck by a bat while it is in horizontal flight
with a speed of 90ft/sec. After leaving the bat the ball travels with a speed
of 110ft/sec in a direction opposite to its original motion. Determine the
impulse of the collision.
SOLUTION:
We know that the
change in momentum of a particle acted on by an impulsive force is equal to the
impulse.
Hence, Impulse, j
= change in momentum = Pf – Pi
= m Vf
– m Vi
= m (Vf
- Vi)
= (
) (Vf
- Vi)
Assuming arbitrarily that the direction of Vi
is positive, the impulse is then,
J =
) (-110ft/sec -90ft/sec)
= - 2.2 lb.sec
The minus sign
shows that the direction of the impulse acting on the ball is opposite that of
the original velocity of the ball.
[If the bat and
ball were in contact for 0.0010sec the average force during this time would be
F = ∆p/∆t
= -2.2/0.0010
= -2200lb]
(2)(a)By what fraction is the kinetic energy of a neutron
(mass m1) decreased in head-on elastic collision with the atomic
nucleus (mass m2) initially at rest?
(b)Find the fractional decrease in the kinetic energy of a
neutron when it collides in this way with a lead nucleus, a carbon nucleus, and
a hydrogen nucleus. The ratio of nuclear mass to neutron mass (=m1/m2)
is 206 for lead, 12 for carbon, and 1 for hydrogen.
SOLUTION:
(a) The initial kinetic energy of the neutron ki is m1v1i2.Its
final kinetic energy kf is m1V1f2.The
fractional decrease in kinetic energy is,
=1-
But for such a collision,
V1f = (
) V1i
Or,
=
So that,
=1 – (
=
(b) For lead, m2 =206m1
= 0.02
=2%
For carbon, m2=12m1
= 0.28
=28%
For hydrogen, m2=m1
= 1
=100%
(3)A
cue strikes a billiard ball, exerting an average force of 50N over a time of
10millisecands. If the ball has mass 0.20kg, what speed does it have after
impact?
SOLUTION:
Here,s
F=50N
T=10milliseconds
=o.o1seconds
M=0.20kg
V=?
We know
that,
P = mv
Or, Ft = mv
Or, v =
=
= 2.5m/s
Answer:
2.5m/s
(4)
A 1.0kg ball drops vertically on to the floor with a speed of 25m/s. It
rebounds with an initial speed of 10m/s.
(a)What
impulse act on the ball during contact?
(b)
If the ball is in contact for 0.020sec,what is the average force exerted on the
floor?
SOLUTION:
Impulse, j =
change in momentum = Pf – Pi
= m Vf
– m Vi
= m (Vf
- Vi)
= 1(-25-10)
= - 35kg/s
Average force F =∆p/∆t
= -35/0.020
= -1700N
Answer: -1700N
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