Physics : Kinematics : Book Back Important Questions, Answers, Solutions : Long Questions and Answer

__Physics : Kinematics__

__Long Answer Questions__

**1. Explain in detail the triangle law of addition.**

●
The two vectors are represented by two adjacent sides of a triangle taken in
the same order. Then the resultant is given by the third side of the triangle.

●
The head of the first vector is connected to the tail of second
vector .

●
Let θ is the angle between and .

●
R is the resultant vector connecting the tail of the first vector to
the head of the second vector .

●

**1)
Magnitude of resultant Vector :**

●
Consider the triangle ABN which is obtained by extending the side OA to ON.

●
ABN is a right angled triangle.

cos
θ = AN / B

∴ AN = B cos θ ……… (1)

cos
θ = BN / B

∴ AN = B cos θ ……… (2)

For
∆ OBN, OB^{2} = ON^{2} + BN^{2} ……… (3)

R^{2}
= (A+B cosθ)^{2} + (B sinθ)^{2}

R^{2}
= A^{2 }+ B^{2} cos^{2}θ + 2AB cos + B^{2}sin^{2}θ

R^{2}
= A^{2 }+ B^{2} (cos^{2}θ sin^{2}θ) + 2AB cosθ

R^{2}
= A^{2 }+ B^{2} + 2AB cosθ

R
= √[A^{2} + B^{2} + 2AB cosθ] ………… (4)

**2.
Direction of resultant Vectors :**

If θ is the angle between and , then

In
∆OBN

tan
α = BN / ON = BN / (OA + AN)

tan
α = Bsinθ / (A + B cosθ)

∴
α
= tan^{-1} (Bsinθ / [A + B cosθ])
……….(6)

**2. Discuss the properties of scalar and vector products.**

**Properties
of scalar Product:**

(i) The product quantity . is always a scalar. It is positive if the angle between the vectors is acute and negative if the angle between them is obtuse.

(ii)
The scalar product is commutative

. = **.**

(iii)
The vectors obey distributive law i,e,

(iv) The angle between the vectors

(v)
The scalar product of two vectors will be maximum when cosθ = 1 i.e. θ = 0°
When the vectors are parallel ( . )_{max} = AB

(vi)
The scalar product of two vectors will be minimum when cosθ = −1 i.e. θ = 180°
.

( . )_{mini} = −AB when the
vectors are anti parallel.

vii) If the vectors and are perpendicular to each other then their scalar product . = 0. Then the vectors and are said to be mutually orthogonal.

viii)
The scalar product of a vector with itself is termed as self dot product and is
given by

The
magnitude (or) norm of the vectors is **|****|** = A = √{.}

ix) In the case of a unit vector n^

x)
In the case of orthogonal unit vectors i^{^}, j^{^} and k^{^}

**Properties
of vector product:**

**3. Derive the kinematic equations of motion for constant acceleration.**

●
Consider an object moving in a straight line with uniform constant
acceleration.

●
Let u be the velocity of the object at time t = 0.

●
v be the velocity of the body at a time t.

**Velocity
- time relation :**

i)
The acceleration of the body at any instant is given by the first derivative of
velocity with respect to time.

a
= dv / dt (or) dv = a.dt ……. (1)

Integrating
on both sides

[v]_{u}^{v}
= a [t]_{0}^{t}

v
– u = at

**v = u + at** ……..(2)

**Displacement
- time relation :**

ii)
The velocity of the body is given by the first derivative of the displacement
with time.

v
= ds / dt

ds
= v.dt …………….(3)

since
v = u+at

ds
= (u+at) dt ……………. (4)

intergrating
on both sides

**s = ut + ½ at ^{2}** …………(5)

**Velocity
- displacement relation:**

iii)
The acceleration is given by the first derivative of velocity with respect to
time.

a
= dv / dt = dv/ds . ds/dt = v . dv/ds (ds / dt = v)

a
= ½ dv^{2}/ds

ds
= 1/2a d(v^{2}) ………..(6)

Integrating
on both sides

∴ s = 1/2a
(v^{2} – u^{2})

**v ^{2} = u^{2} + 2as** …………(7)

From
equation

at
= v - u …………….
(8)

Substitute
equation (8) in equation (5) we get,

s
= ut+ ½ (v-u) t

**s = [(u+v)t] / 2** …………(9)

Kinematic
equations,

v
= u+at

s
= ut+ ½ at^{2}

v^{2}
= u^{2 }+ 2as

s
= [(u+v)t ] / 2

**4. Derive the equations of motion for a particle (a) falling vertically (b) projected vertically**

**i)
A body falling vertically from a height h: **

●
Consider an object of mass in falling from a height h.

●
Let us choose the downward direction as positive y axis.

●
The object experience acceleration 'g' due to granty which is constant near the
surface of the earth. The acceleration** **

By
comparing the components we get, a_{x} =0, a_{z} =0, a_{y }=
g

a_{y}
= a = g

●
If the particle is thrown with initial velocity u downward which is in negative
y axis.

●
Then velocity and position of the particle at any time t is given by,

v
= u + gt …….......
(1)

y
= ut + ½ gt^{2 }.......... (2)

The
square of the speed of the particle is

v^{2
}= u^{2} + 2gy ......... (3)

Suppose
the particle starts from rest

Then
u = 0

v
= gt ..........
(4)

y
= ½ gt^{2 }.......... (5)

v^{2} = 2gy .......... (6)

**b)
A body thrown vertically upwards :**

●
An object of mass m thrown vertically up-wards with an initial velocity u.

●
The acceleration a = −g and g points towards the negative y axis.

●
The Kinematic equations for this motion are

v
= u − gt ……….
(7)

s
= ut – ½ gt^{2} ………. (8)

v^{2}
= u^{2 }-2gy

**5. Derive the equation of motion, range and maximum height reached by the particle thrown at an oblique angle θ with respect to the horizontal direction.**

● Consider an object thrown with initial velocity u at an angle θ with the horizontal.

Then

Where
u_{x} = u cos θ is the horizontal component.

u_{y}
= u sinθ is the vertical component of velocity.

After
the time t_{1} the velocity along horizontal motion

v_{x}
= u + a_{x }

t
= u_{x }= u cosθ

The
horizontal distance is

s_{x
}= u_{x}t + ½a_{x}t^{2}

Here
s_{x} = x, u_{x} = u cosθ, a_{x} = 0

x
= u cosθ.t

t
= x / ucosθ ………..(1)

For
vertical motion v_{y} = u_{y} + a_{y}t

Here
u_{y} = u sinθ, a_{y} = −g

v_{y}
= u sinθ – gt ………(2)

The
vertical distance travelled by the projectile is

s_{y}
= u_{y}t + ½ a_{y }t^{2}

Here
s_{y} = y, u_{y} = u sinθ, a_{x} = −g

than
y = u sinθ.t – ½ gt^{2} ………(3)

Substitute
the 't' value from the equation (1) in equation (3), we have

y
= u sinθ [ x / ucosθ ] – ½ g [ x^{2} / u^{2} cos^{2}θ
] ………(4)

The
path followed by the projectile is an inverted parabola.

**Maximum
(h _{max} )**

The
maximum vertical distance travelled by projectile during its journey is called
maximum height.

For
vertical part of motion,

v^{2}_{y}
= u^{2}_{y} + 2a_{y}S

Here,
u_{y} = u sinθ, a_{y }= −g, s = h_{max} ; v_{y}
= 0

O
= u^{2}sin^{2} θ − 2gh_{max}

**h _{max} = u^{2}sin^{2}θ
/ 2g**

…………(5)

**Horizontal
Range (R):**

The
maximum horizontal distance between the point of projection and the point on
the horizontal plane where the projectile hits the ground is called horizontal
range (R).

Range
R = Horizontal compoent of velocity × time of flight

R
= u cosθ × T_{f}

R
= ucosθ × [2u sinθ / g] = [ 2u^{2} sinθ cosθ ] / g

**R = [u ^{2} sin2θ] / g**

……………… (6)

The
maximum possible range is reached when sin2θ = l,

When 2θ = π /2 (or) θ = π /4

**R _{max} = u^{2} /
g**

…………..(7)

**6. Derive the expression for centripetal acceleration.**

●
The acceleration acting on an object towards the center of the circle in a
uniform circular motion is known as centripetal acceleration.

●
The centripetal acceleration is derived from a simple relationship between
position and velocity vector.

●
The directions of position and velocity vectors shift through the same angle θ
in small time ∆t as shown in the above figure

●
For uniform circular motion r = | | = | | and v = | | = | |

●
If the particle moves from position vector _{ }to_{ }

●
The displacement _{ }

●
Then change in velocity

then
∆r/r = - ∆v/v = *θ *…………. (1)

Here
negative sign implies that ∆v points radially inward, towards the centre of the
circle.

∆v
= −v (∆r / r)

a
= ∆v / ∆t = v/r (∆r / ∆t) = −v^{2} / r

For
uniform circular motion v = rω

Then
centripetal acceleration** a = −v ^{2}
/ r = - ω^{2}r**

** **

**7. Derive the expression for total acceleration in the non uniform circular motion.**

●
If the speed of the object in circular motion is not constant then it is called
non-uniform motion.

Example
: The bob attached to a string moves in vertical circle.

●
The speed of the bob is not the same at all time.

●
The speed is not same in circular motion, the particle will have both
centripetal and tangetial acceleration.

The
centripetal acceleration is v^{2}/r

The magnitude of resultant acceleration is

a_{R
}= √ [ a_{t}^{2 }+ (v^{2} / r)^{2 }]

The
resultant acceleration makes an angle θ with the radius vector.

The
angle is given by tan θ = a_{t} / (v^{2} / r)

Tags : Kinematics | Physics , 11th Physics : UNIT 2 : Kinematics

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