AP 9th Class Physical Science Chapter 3 Solution – Laws of Motion

AP 9th Class Physical Science Chapter 3 Solution – Laws of Motion

Andhra Pradesh SCERT 9th Class Physical Science Chapter 3 Laws of Motion Solution for AP 9th Class Physics/Chemistry Exam. Lots of Students of Andhra Pradesh Board will search on internet for Andhra Pradesh Class 9 Physical Science Textbook Solution or Study Material for AP 9th exam. Here you search will end! Here in this page we have provided for all question answer for AP SCERT Chapter 3 Laws of Motion.

For any help regarding extra notes for AP 9th Class Physics/Chemistry Chapter 3 Laws of Motion you can comment us at comment box below.

AP 9th Class Physics Chapter 3 Laws of Motion Solution

1.) Explain the reasons for the following. (AS1)

a) When a carpet is beaten with a stick, dust comes out of it.

b) Luggage kept on the roof of a bus is tied with a rope.

c) A pace bowler in cricket runs in from a long distance before he bowls.

Ans: –a) An object always try stay in its rest or motion position because of inertia. In this case when we beat the carpet with the stick the carpet moves but the dust particle try to stay in position as a result it come out.

b) Luggage kept on the roof of a bus is tied with a rope as when the bus stops the inertia of luggage resist the force to change its motion of inertia , if we don’t tied with the rope the luggage moves forward.

c) A pace bowler in cricket runs in from a long distance before the bowls to get more inertia. When bowler came from long distance it get momentum with help of that the bowler bowl with more speed.

2.) Two objects have masses 8 kg and 25 kg. Which one has more inertia? Why? (AS1)

Ans:-As we all know that the moment of inertia is directly proportional to the masses of an object, so the object which is more masses comparing to other has more inertia and for that 25 kg object will have more inertia.

3.) Keep a small rectangular shaped piece of paper on the edge of a table and place anold five rupee coin on its surface vertically as shown in the figure below. Now givea quick push to the paper with your finger. How do you explain inertia with thisexperiment? (AS3)

Ans: –If we give a quick push to the paper with our finger the force of the paper will affect the coin to stay in its position. This will happened because of change in inertia of the coin as when the force applied in it its states of inertia changes to moment of inertia which force the coin to change its position but the coin will stay at its position.

4.) If a car is traveling westwards with a constant speed of 20 m/s, what is the resultantforce acting on it? (AS1, AS7)

Ans: –As we all know that when speed is constant the acceleration acting on an object is zero. So when the car travelling with a constant speed of 20m/s it doesn’t have any acceleration on it so the resultant force acting on it is zero as force = mass × acceleration.

5.) What is the momentum of a 6.0 kg bowling ball with a velocity of 2.2 m/s?(AS1)

Ans: – As we all know the momentum of any object is equal to the multiplication of its mass and velocity.

So the momentum of the ball is 6.0×2.2 = 13.2kg.m/s.

6.) Two people push a car for 3 s with a combined net force of 200 N. (AS1)

(a) Calculate the impulse provided to the car.

(b) If the car has a mass of 1200 kg, what will be its change in velocity?

Ans: –a) We know that impulse = force × time, so the impulse provided to the car is 3× 200 = 600N.s.

  1. b) As we know that impulse = change in momentum or m× (v-u), where v and u are final and initial velocity.

From question we get m= 1200kg and u= 0.So, 600= 1200× (v-0)

or, v= 600/1200 = 0.5 m/s.

7.) What force is required to produce an acceleration of 3 m/s2 in an object of mass 0.7 kg?

Ans: –We know that F= ma, where f, m, a are force, mass of the object and acceleration.

So F= 3 × 0.7 = 2.1kg.m/s^2.

8.) A force acts for 0.2 sec on an object having mass 1.4 kg initially at rest. The force stopsto act but the object moves through 4m in the next 2 sec. Find the magnitude of theforce? (AS1)

Ans: – Force = rate of change of momentum = mv/t.

Where m is mass and v is average velocity and t is time.

Here v=4/2=2m/s; m= 1.4kg; t = 0.2 s. Then f= 2×1.4/0.2=14N.

So magnitude of force is 14N.

9.) An object of mass 5 kg is moving with a velocity of 10 ms-1. A force is applied so that in20 s, it attains a velocity of 25 ms-1. What is the force applied on the object?

Ans: – From equation of motion we know that v= u + at, where v, uare final and initial velocity and a, t are acceleration and time.

So a= v – u/t = (25 – 10)/20 = 0.75 m/s^2.

So force applied on the object is (f=ma= 5×0.75= 3.75) 3.75N.

10) Find the acceleration of body of mass 2kg from the figures Shown.

Ans: – Let T as the tension in the string. Where m1=2g and m2=3g and a is acceleration.

So according to force diagram the for m2 the equation is,

m2g – T = m2a, or T= m2 (g – a).

For m1 the equation is,

T -m1g = m1a, or m2g – m2a – m1g = m1a

Or,a= (m2 – m1) g/ (m1 + m2)

Or, a = (3 -2) ×10/ (3 +2) =10/5= 2 (here g= 10)

So the acceleration of the body of mass 2kg is 2m/s^2.

11) Take some identical marbles. Make a path or a track keepingyour notebooks on either side so as to make a path in which marbles can move. Now use one marble to hit the other marbles. Take two, three marbles and make them to hit the other marbles. What can you explain from your observations? (AS5)

Ans: – At first when one marble hit the other both will move with same velocity. But in the second case when two hit their velocity Will increases so by this when three marble hit the other their momentum will increase.

12.) A man of mass 30 kg uses a rope to climb which bears only 450 N. What is the maximum acceleration with which he can climb safely? (AS1, AS7)

Ans: –We know that F= ma, where f, m, a are force,mass, Acceleration.

It is given that m=30kg, f= 450 N, so a = 450/30 = 15

The acceleration in which he can climb safely is 15m/s^2.

13.) An vehicle has a mass of 1500 kg. What must be the force between the vehicle and theroad if the vehicle is to be stopped with a negative acceleration of 1.7 ms-2?

Ans:- As F= ma .

Here given m=1500kg and a =-1.7 m/s^2.

So F = 1500 × (-1.7) =-2550N.

The force is 2550N and its direction is opposite to the vehicle.

14.) If a fly collides with the windshield of a fast-moving bus, (AS1, AS2)

(a) Is the impact force experienced, same for the fly and the bus? Why?

(b) Is the same acceleration experienced by the fly and the bus? Why?

Ans: –a) As the mass of the fly negligible as comparison to the bus so the fly will experienced more impact.

  1. b) No the acceleration would be different to both of them because of the wide difference of the mass.

15.) A truck is moving under a hopper with a constant speed of 20m/s. Sand falls on thetruck at a rate 20 kg/s. What is the force acting on the truck due to falling of sand?

Ans: – As we know that acceleration is change in velocity with respect to time.

Acceleration, a = 20/1= 20m/s^2.

From question it is given that mass is 20 kg, so by formula f= ma

Or, f = 20 × 20 = 400 N.

16.) Two rubber bands stretched to the standard length cause an object to accelerate at.2 m/s2. Suppose another object with twice the mass is pulled by four rubber bandsstretched to the standard length. What is the acceleration of the second object?

Ans: – We know that F=ma,

In first case F=F1, m=m1 and a= 2m/s^2.

So, F1 = 2m1.

In the second time four rubber bands stretched so F2=2F1 and m2=2m1.

So F2= m2a2. , or 2F1 = 2m1×a2, or a2 = 2F1/2m1 = 2×2m1/2m1=2ms^-2.

So the acceleration of the second object is 2m/s^2.

17.) Illustrate an example of each of the three laws of motion. (AS1)

Ans: –For the first law of motion is –in a running bus when it suddenly stop we came forward.

And for the second law –if we push a ball which is steady at its position in a floor we will see that it accelerates from rest.

For the third law of motion there are so much example which are flying of birds in the sky, movement of fishes in the water etc.

18.) Two ice skaters initially at rest, push of each other. If one skater whose mass is 60 kg has a velocity of 2 m/s. What is the velocity of other skater whose mass is 40 kg?

Ans: –The mass of skaters 1 are 60 kg and velocity is 2m/s.

And the mass of skaters 2 is 40kg.

As they push each other so their resultant momentum must be zero or m1v1 + m2v2 = 0.

60×2 + 40× v2 =0

Or, v2 = -120/40 = -3.

Negative sign show’s that velocity of the skaters 2 is opposite.

19.) A passenger in moving train tosses a coin which falls behind him. It means that themotion of the train is (AS7)

  1. a) Accelerated b) Uniform
  2. c) Retarded d) circular motion

Ans: – A passenger in moving train tosses a coin which falls behind him it shows that the train is accelerated.

20.) A horse continues to apply a force in order to move a cart with a constant speed. Explain. (AS1)

Ans: –The cart moves when the net momentum between these two are zero. As the horse applied force momentum create to moves the cart. But if the horse stop the cart will be stop.

21.) A force of 5N produces an acceleration of 8 ms-2 on a mass m1 and an accelerationof 24 ms–2 on a mass m2. What acceleration would the same force provide if both the masses are tied together? (AS1)

Ans: – We know that F=ma.

For acceleration 8 ms^-2 and F=5 m1 will be 5/8.

And for a=24ms^-2 the m2 is 5/24.

So the total mass is (5/8 +5/24) = (15 + 5)/24= 5/6.

So the new acceleration is a0,F=5/6×a0

Or, a0= 6ms^-2.

22.) A hammer of mass 400 g, moving at 30 m s-1, strikes a nail. The nail stops the hammerin a very short time of 0.01 s. What is the force of the nail on the hammer? (AS1)

Ans: – The hammer stops after 0.01 sec so final velocity is 0. Then the change in velocity is (30-0) 30m/s. And mass is 400gm or 0.4kg.

So acceleration will be (velocity/time or 30/0.01) =3000 ms^-2.

Hence the force of the nail on the hammer is (f=ma or f= 3000×0.4=1200) 1200N.

23.) System is shown in figure. (AS1)

Find the acceleration of the blocks and tension in the String. Take g=10m/s2.

Ans:-


24.) Three identical blocks, each of mass 10kg, are pulled as shown on the horizontal frictionless surface. If the tension (F) in the rope is 30N. What is the acceleration of each block? And what arethe tensions in the other ropes? (Neglect the masses of the ropes) (AS1)

Ans: – As there are three identical blocks so the total mass of the blocks are (10 +10 10)30kg.

The rope applied force on the blocks are 30N.

As we know F=ma, so a = F/m = 30/30 = 1 ms^-2.

Each block has 1 ms^-2 acceleration on them.

As the first rope pulls one block of mass 10 kg, 1ms^-2 acceleration so the tension will be 10×1=10 N.

And the second rope pulls two blocks of mass 10 kg each so the tension on it is (m=10 +10=20; a=1; 20×1=20) 20N.

25) A ball of mass ’m’ moves perpendicularly to a wall with a speed v, strikes it and rebounds with the same speed in the opposite direction. What is the direction and magnitude of the average force acting on the ball due to the wall? (AS7)

Ans: –Force = change in momentum/ time.

The ball exert force on the wall is f=mv/t (given mass is m and time is t and velocity is v).

So when the ball exert force on the wall the wall also exert reaction force according to Newton third laws of motion which will be in opposite direction is -mv/t. F(Ball to wall)=-F(wall to ball)

So the total force acting on the ball due to the wall is mv/t + mv/t 2mv/t.

26) Divya observed a horse pulling a cart. She thought that cart also pulls the horse with same force in opposite direction. As per third law of motion the cart should not move forward. But her observation of moving cart raised some questions in her mind. Can you guess what questions are raised in her mind? (AS2)

Ans: –The question which may raised in her mind be –

i) Why the cart is moving?

ii) Is there any effect of friction in that?

iii) Which force make the cart to moves? Etc.

27) How do you appreciate Galileo’s thought of “any moving body continues in the state only until some external force acts on it” which is a contradiction to the Aristatile’s belief of “any moving body naturally comes to rest”.(AS6)

Ans: – There are always some rumour about various things that happened in previous ancient times but the discoveries of different laws of science day by day makes some of them wrong by experiment.  Any moving body naturally comes to rest which was provided wrong by Galileo through experiment. And after coming out Newton laws of motion it also proven wrong.

Updated: November 20, 2021 — 1:05 pm

Leave a Reply

Your email address will not be published. Required fields are marked *