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Kinematics And Laws Of Motion

Solution for Class 12, Physics, Kinematics And Laws Of Motion

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A particle is moving with constant acceleration 'a'. Following graph shows v2 versus x (displacement) plot. The acceleration of the particle is ____ m/s². [NA, Aug. 31, 2021 (II) v²(m/s)² 20 80 60 40- 20 0 10 230

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Kinematics and Laws of Motion

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Q. W ball is bit by batsmon at on ange of

3 7^{\circ}

ot The man standing at

P

showed ruan et wat min-udbeity se that he cateres the bale befere it strik an groend stsume that height of man is

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Kinematics and Laws of Motion

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2. નીચેના પ્રશ્રોના બે કे ત્રણા વાક્યમાં જવાબ આ (1) નાણાકીય પત્રકો એટલે શું ?

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Kinematics and Laws of Motion

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A block of mass

50 kg

can slide on a rough horizontal surface, when coefficient of friction between the block and surface is 0.6. The least force of pull acting at an angle of

3 0^{\circ}

to the upward drawn vertical, which causes the block to just slide is

$29.43 N$
$219.6 N$
$21.96 N$
$294.2 N$

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= \frac{100 \times 1 0 ^{3} N}{3.14 \times ( 1 0 ^{- 2} m ) ^{2}} = 3.18 \times 1 0^{8} N m^{- 2}

The elongation, Example

9.2

A copper wire of length

2.2

m

and a steel wire of length

1.6 m

, both of diameter

3.0 mm

, are connected end to end. When stretched by a load, the net elongation is found to be

0.70 mm

. Obtain the load applied. nswer The copper and steel wires are unc tensile stress because they have the sa nsion (equal to the load

W

) and the same a cross-section A. From Eq. (9.7) we have str train

\times

Young's modulus. Therefore

W / A = Y_{c} \times (Δ L_{c} / L_{c}) = Y_{s} \times (Δ L_{s} / L_{s})

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Kinematics and Laws of Motion

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If on jupitor, unt of force is i tencmekar, unt of Acceleratin is 1 Mahi, \& unit of tim 1 I/vi. suen that

1 Ten = 10 N 1 Mahi = 5 m / s^{2} 1 Yeri = 0.1 Sec

mit of energy is 1 Gyyle then find of the value of 8 scugle?! \[ \begin{array}{l} E=F^{a} A^{b} T^{c} \\ \left(M L^{2} T^{-2}\right)=\left(M L T^{-2}\right)^{a}\left[L T^{-2}\right)^{b} T^{

Topic:

Kinematics and Laws of Motion

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7. किसी कमानीदार तुला का पैमाना 0 से

50 kg

तक के पाठ्यांक पढ़ता है। पैमाने की लम्बाई

20 cm

है। इस तुला से लटकाया कोई पिण्ड, दिं जब विस्थापित करके मुक्त किया जाता है,

0.6 s

के आवर्तकाल से दोलन करता हैं। पिंड का भार कितना है? 8. चित्र में दर्शाए अनुसार

3.0 kg

द्रव्यमान की ट्राली को

600 Nm^{- 1}

कमानी-स्थिरांक की दो कमानियों से जोड़ा गया है। यदि ट्राली को अपनी साम्यावस्था की स्थिति

5.0 cm

विस्थापित करके मुक्त छोड़ दिया जाता है, तो

(a)

होने वाले दोलनों का आवर्तकाल, तथा (b) ट्राली की अधिकतम चाल क्या होगी ? जितने समय में अवमंदन बलों के कारण ट्राली विराम अवस्था में आती है, उतने समय में कितनी ऊर्जा ऊष्मा के रूप में क्षय हो जाती है ? 5. सरल लोलक (Simple Pendulum)

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Kinematics and Laws of Motion

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A particle of mass

m

x

-axis as follows. It starts from rest at

t = 0

and comes to rest at

t = 1

x = 1

. No other information is available about its motion. intermediate times

(0 < t < 1)

α

denote the instantaneous acceleration of the particle, then

$α$ cannot remain positive for all t in the interval $0 \leq t \leq 1$
$∣ α ∣$ cannot exceed 2 at any point in its path
$∣ α ∣$ must be $\geq 4$ at some point or points in its path
$α$ must change sign during the motion, but no other change sign during the motion, but no $cts_{er}$ assertion can be made with the information given

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Kinematics and Laws of Motion

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A particle is projected from the ground with velocity

u

θ

with horizontal. The horizontal rane. maximum height and time of flight are R.

H

T

respectively. They are given by.

R = \frac{u ^{2} s i n 2 θ}{g}, H = \frac{u ^{2} s i n ^{2} θ}{2 g}

T = \frac{2 u s i n θ}{g} sin θ, 1

Now keeping u as fixed,

θ

3 0^{\circ}

6 0^{\circ}

\frac{1}{2}, \frac{1}{2}, \frac{3}{2}, \frac{1}{2}, \frac{3}{2}, \frac{1}{2}, \frac{1}{2}

$R$ will first increase then decrease, $H$ will increase and T will decrease
$R$ will first increase then decrease while $H$ and $T$ both will increase
$R$ will decrease while $H$ and $T$ will increase
$R$ will increase while $H$ and $T$ will increase

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Question 10

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Paragraph-2 shown in figure. A body is projected along the inclined plane with the speed

10 m / s

at an angle 53 with

x

-axis from origin

O [g = 10 m / s^{2}]

(39) Speed of body after

t = 2

$210 m / s$
$42 m / s$
$13 m / s$
$213 m / s$

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Kinematics and Laws of Motion

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Question 11

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Velocity of the body, when it crosses the

x

$6 \hat{i} - 8 \hat{j}$
$6 \hat{i} + 8 \hat{j}$
$- 6 \hat{i} - 8 \hat{j}$
$- 6 \hat{i} + 8 \hat{j}$

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Question 12

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Time after which body is moving perpendicular to initial direction is (approx)

$2.1 sec$
$1.8 sec$
3 sec
$3.6 sec$

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Kinematics and Laws of Motion

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Question 13

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(e5) Figure shows a uniform rod of length 1 and mass

M

which is pivoted at end

A

such that it can rotate in a vertical plane. The free end of the rod '

B

' is initially vertically above the pivot and then released. As the rod rotates about

A

B

of the rod will hit the ground with a linear speed:

$2 g ℓ$
$5 g ℓ$
$3 g ℓ$
$\frac{2 g}{ℓ}$

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Kinematics and Laws of Motion

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Question 14

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(d) The angular momentum of the particle increases Rotational Kinctic Energy Figure shows a uniform rod of length

I

M

which is pivoted at end

A

such that it can rotate in a vertical plane. The free end of the rod '

B

' is initially vertically above the pivot and then released. As the rod rotates about

A

\frac{6 g}{L} sin θ

\frac{6 g}{L} cos \frac{θ}{2}

B

of the rod will hit the ground with a linear speed: (a)

2 g ℓ

5 g ℓ

\frac{m v ^{2}}{2 g}

3 g ℓ

\frac{2 g}{ℓ}

49. A particle of mass

m

46. A uniform rod of mass

M

L

lies radially on uniform speed

v = 32

a disc rotating with angular speed

w

in a horizontal the line

y = x + 4

. W plane about its axis. The rod does not slin se the

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Kinematics and Laws of Motion

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Question 15

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(56.) The correct acceleration-time graph of the particle is shown as

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Kinematics and Laws of Motion

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Question 16

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Paragraph-4 Two inclined planes

OA

OB

having inclinations

3 0^{\circ}

6 0^{\circ}

with horizontal respectively, interest each other at ' 0 '. A particle is projected from point

P

with a velocity

v = 103 m / s

along a direction perpendicular to plane

OA

. If the particle strikes plane

OB

perpendicularly at

Q

g = 10 m / I^{2}

) 45. Velocity with which particle strikes plane

OB

$103 m / s$
$53 m$ 's
$10 m / s$
None of these

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Kinematics and Laws of Motion

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Question 17

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Time it takes to travel from point

P

Q

25
Is
$23$
None of these

Topic:

Kinematics and Laws of Motion

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Question 18

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Three Marks Question 19. Derive the equation of motion given below (a)

v = u + a t

s = u t + \frac{1}{2} a t^{2}

Where symbols have their usual meanings 20. Derive the relation by calculus method (a)

v^{2} = u^{2} + 2 a s

s = u t + \frac{1}{2} a t^{2}

21. Derive the equation of motion by graphical method. (a)

v^{2} = u^{2} + 2 a s

5 = u t + \frac{1}{2} a t^{2}

v = u + a t

Where symbols have their usual meanings 22. (a) Plot

yy □ t, v □ t

a □ t

graph for a body falling freely under gravity. (b) Write the expression for the relative velocity of two bodies when their moving in sa direction and moving in opposite direction. (c) Give one example where a particle having zero speed but non zero acceleration at t instant. 23. The displacement (in meter) of a particle moving along

X

-axis is given by

x = 18 t + 5 t^{2}

. Calculate: (a) The instantaneous velocity at

t = 2 s

, (b) Average velocity between

t = 2 s

t = 3 s

, (c) Instantaneous acceleration. 24. What type of motion does the following velocity-time curve represent? Convert the velocity and time graph into

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Kinematics and Laws of Motion

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Question 19

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Time after which particle velocity is perpendicular to its initial velocity in projectile motion ground to ground

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Kinematics and Laws of Motion

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Question 20

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Pate 1. Write equation of trajectory, w.r I) stone, Raleen (aladin). Princess At what time at usme leud at will say stone is it wasprojected 1i1. At what time Aladin will soy stone was direct ly onerhis head 1v. P. T. JabAladin Mo stone Apre Sarpar dikh raha hai Tab kya Tasmine

Al

stone Ussimament Be. Aladinka sarpar dikhega.

sin 2 3^{\circ} = \frac{5}{13}

Topic:

Kinematics and Laws of Motion

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