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Physics Part-I

Laws of motion

A body of mass 0.40 kg moving initially with a constant speed

Physics Part-I

Class 11

NCERT

Chapter 1: Physical World

16 questions

Chapter 2: Units and Measurements

51 questions

Chapter 3: Motion in Straight Line

50 questions

Chapter 4: Motion in a Plane

43 questions

Chapter 5: Laws of motion

52 questions

A rocket with a lift-off mass

20, 000 k g

is blasted upwards with an initial acceleration of

5.0 m s^{- 2}

. Calculate the initial thrust (force) of the blast.

A body of mass

0.40 k g

moving initially with a constant speed of

10 m s^{- 1}

to the north is subject to a constant force of

8.0 N

directed towards the south for

30 s

. Take the instant the force is applied to be

t = 0

, the position of the body at that time to be

x = 0

, and predict its position at

t = 5 s, 25 s, 100 s

A truck starts from rest and accelerates uniformly at

2.0 m s^{- 2}

. At

t = 10 s

, a stone is dropped by a person standing on the top of the truck (

6 m

high from the ground). What are the (a) velocity, and (b) acceleration of the stone at

t = 11 s

? (Neglect air resistance.)

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Chapter 6: Work, Energy and Power

41 questions

Chapter 7: System of Particles and Rotational Motion

45 questions

Chapter 8: Gravitation

33 questions

Question

Easy

Solving time: 3 mins

A body of mass $0.40 k g$ moving initially with a constant speed of $10 m s^{- 1}$ to the north is subject to a constant force of $8.0 N$ directed towards the south for $30 s$ . Take the instant the force is applied to be $t = 0$ , the position of the body at that time to be $x = 0$ , and predict its position at $t = 5 s, 25 s, 100 s$ .

Views: 6,105 students

Updated on: Oct 12, 2023

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Physics Part-I (NCERT)

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Lucas Discussed

A body of mass

0.40 k g

moving initially with a constant speed of

10 m s^{- 1}

to the north is subject to a constant force of

8.0 N

directed towards the south for

30 s

. Take the instant the force is applied to be

t = 0

, the position of the body at that time to be

x = 0

, and predict its position at

t = 5 s, 25 s, 100 s

10 mins ago

Discuss this question LIVE

10 mins ago

Text solutionVerified

Given:

The mass of body,

m = 0.40 k g

Constant initial speed of the body is,

u = 10 m / s

Force that acts on the body is,

F = 8.0 N

(i)

t = -5 s

The force starts acting on the body from

t = 0 s

So, the acceleration of the body during this time was zero and it moves at a constant speed.

Position of the body is given by:

x = v \times t

x = 10 \times (- 5)

= - 50 m

(ii)

t = 25 s

Since the force acts in the opposite direction of motion of particle. So, the acceleration of the body due to the force acting on it is:

a = \frac{F}{M}

= \frac{- 8.0}{0.40} = - 20 m s^{- 2}

The position is given by the second equation of motion:

s^{'} = u t^{'} + \frac{1}{2} a "

t^{2}

= 10 \times 25 + \frac{1}{2} \times (- 20) \times 2 5^{2}

= 250 - 6250 = - 6000 m

(iii)

t = 100 s

For first 30 sec, it moves under retardation of force and after that the body moves at constant speed.

a = - 20 m

s^{- 2}

u = 10 m / s

The distance covered in

30 s

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

= 10 \times 30 + \frac{1}{2} \times (- 20) \times 3 0^{2}

= 300 - 9000 = - 8700 m

Now, the speed after

30 sec

is:

v = u + a t

v = 10 - (20 \times 30)

= - 590 m / s

Distance covered in remaining 70 sec:

x^{'} = - 590 \times 70

= - 41300 m

So, total distance covered in

100 s

will be:

X = - 8700 - 41300

X = - 50000 m = - 50 km

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Practice more questions from Physics Part-I (NCERT)

A rocket with a lift-off mass

20, 000 k g

is blasted upwards with an initial acceleration of

5.0 m s^{- 2}

. Calculate the initial thrust (force) of the blast.

A body of mass

0.40 k g

moving initially with a constant speed of

10 m s^{- 1}

to the north is subject to a constant force of

8.0 N

directed towards the south for

30 s

. Take the instant the force is applied to be

t = 0

, the position of the body at that time to be

x = 0

, and predict its position at

t = 5 s, 25 s, 100 s

A truck starts from rest and accelerates uniformly at

2.0 m s^{- 2}

. At

t = 10 s

, a stone is dropped by a person standing on the top of the truck (

6 m

high from the ground). What are the (a) velocity, and (b) acceleration of the stone at

t = 11 s

? (Neglect air resistance.)

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Question Text	A body of mass $0.40 k g$ moving initially with a constant speed of $10 m s^{- 1}$ to the north is subject to a constant force of $8.0 N$ directed towards the south for $30 s$ . Take the instant the force is applied to be $t = 0$ , the position of the body at that time to be $x = 0$ , and predict its position at $t = 5 s, 25 s, 100 s$ .
Updated On	Oct 12, 2023
Topic	Laws of motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 6
Upvotes	641
Avg. Video Duration	9 min

Text solutionVerified

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