1. Laws of Motion

9th class science chapter 1 laws of motion exercise answers

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Question And Answer :

 

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Content :

#  Q.1. Fill in the blank and rewrite the completed statements.                                 

# Q. 2 Find the odd one out 

  #Q.3. Find co-related terms                 

#Q.4. Match the pair :                             

#Q.5.True or false

#Q.6. Name the  following                     

#Q.7.mcq             

 #Q.8. solve numerical problem 

#Q.9. write shrot notes.   

#Q.10. attempt the following     

#Q.11. dinstinguish between 

#Q.12.give scientific reason     

   #Q.13. Give example 

#Q.14. Give explanation using the given statement 

#Q.15. attempt the following 

#Q.16. solve the numerical problem  

 

 

Q.1. Fill in the blank and rewrite the completed statements: 

1 .Is the displacement that occurs in unit time.

Ans Velocity is the displacement that occurs in unit time.

2. The distance travelled by an object in a particular direction in unit time is called …………. .

Ans The distance travelled by an object in a particular direction in unit time is called velocity.

3. If an object covers unequal distance in equal time intervals, it is said to be moving with speed.

Ans If an object covers unequal distance in equal time intervals, it is said to be moving with non-uniform speed.

4. The product of mass and velocity is called …………. .

Ans The product of mass and velocity is called momentum.

5. When the velocity of an object increase, the acceleration is …………. .

Ans When the velocity of an object increase, the acceleration is Positive.

6. When a bullet is fired from the gun, the gun moves in backward direction. This motion is called as …………. .

Ans When a bullet is fired from the gun, the gun moves in backward direction. This motion is called as recoil.

7. When the velocity of an object decrease, the acceleration is …………. .

Ans When the velocity of an object decrease, the acceleration is negative.

8. The force causes the change in the state of an object at rest or in uniform motion.

Ans The unbalanced force causes the change in the state of an object at rest or in uniform motion.

9. The third equation of motion gives the relation between displacement and …………. .

Ans The third equation of motion gives the relation between displacement and velocity.

10. A body is said to be in if it changes its position with respect to its surrounding.

Ans A body is said to be in motion if it changes its position with respect to its surrounding.

11. The first equation of motion gives the relation        between………… and time.

Ans The first equation of motion gives the relation between Velocity and time.

12. …………… is the length of the actual path covered by an object in motion while going from one point to another.

Ans Distance is the length of the actual path covered by an object in motion while going from one point to another.

13. When an object moves with constant speed along a circular path, the motion is called motion.

Ans When an object moves with constant speed along a circular path, the motion is called uniform circular motion.

14. is a relative concept.

Ans Motion is a relative concept.

15. Newton’s first law of motion explains the phenomenon of …………. .

Ans Newton’s first law of motion explains the phenomenon of Inertia.

15. The distance covered by an object in unit time is called ……………. .

Ans The distance covered by an object in unit time is called speed.

16. In CGS system, the unit of force is …………. .

Ans In CGS system, the unit of force is dyne.

17. The second equation of motion gives the relation between and time.

Ans The second equation of motion gives the relation between Displacement and time.

18. The S.I. unit of speed is ………….

Ans The S.I. unit of speed is m/s.

19. A body is said to be at if it does not change its position with respect to its surrounding.

Ans A body is said to be at rest if it does not change its position with respect to its surrounding.

20. Motion of an object was studied by …………. .

Ans Motion of an object was studied by Sir Isaac Newton.

21. The rate of change of velocity is called …………………….

Ans The rate of change of velocity is called acceleration.

22. If an object covers equal distance in equal time intervals, it is said to be moving with …………………….speed.

Ans If an object covers equal distance in equal time intervals, it is said to be moving with uniform speed.

Q. 2 Find the odd one out 

1. Force, Momentum, Acceleration, Mass.

Ans Mass

It is scalar quantity whereas remaining are vector quantities.

2. Motion of vehicles on a crowded street, a man going for a stroll on a beach, soldiers marching, motion of fishes in water.

Ans Soldiers marching

This is an example of uniform motion whereas the rest are examples of non-uniform motion.

3. newton, joule, kg m/s2, dyne

Ans joule

4. It is a unit of energy whereas remaining are units of force.

Q.,3. Find co-related terms 

1. 1 m/s2 : 102 cm/s2 :: 1 N : ………….

Ans 1 m/s2 : 102 cm/s2 :: 1 N : 105 dynes.

                   1 N = 1 kg × 1 m/s2 = 103 g × 102 cm/s2

                                                   = 105 g cm/s2

                                                   = 105 dynes.

2. Velocity : m/s :: m/s2

Ans Velocity : m/s :: Acceleration : m/s2

3. The tendency of a body to resist change in a state of rest or state of motion : Inertia :: Product of mass and velocity of an object :

Ans Tendency of a body to resist change in a state of rest or state of motion : Inertia :: Product of mass and velocity of an object : Momentum.                                                                Tendency of a body to resist change in a state of rest or state of motion is termed as inertia, similarly product of mass and velocity of an object is termed as momentum.

4. S.I. unit of momentum: Kg.m/s2:: C.G.S unit of momentum: ……………

Ans S.I. unit of momentum: Kg.m/s2:: C.G.S unit of momentum: g.cm/s2

5. Speed : Distance :: Velocity: ……………

Ans Speed : Distance :: Velocity: Displacement

6. Speed of light : 3 x 108m/s :: Speed of sound : ……………

Ans Speed of light : 3 x 108m/s :: Speed of sound : 343.2m/s

    7.   1kg × 1 m/s2 : 1 N :: 1 g × 1 cm/s2 :                                     Ans 1 kg × 1 m/s2 : 1 N :: 1 g × 1 cm/s2 : 1 dyne.                      Product of 1 kg and 1 m/s2 is equivalent to 1 N, similarly product of 1 g and 1 cm/s2 is equivalent to 1 dyne.

8. S.I. unit of force: : : C.G.S unit of force: dyne

Ans S.I. unit of force: newton : : C.G.S unit of force: dyne

Q.4. Match the pair :

1

Column “A”

Column “B”
 

a. Describes the relationship between the forces on two interacting object

i. Newton’s first law of motion.
 

ii. Newton’s second law of motion

  

b. Also called the law of inertia
 

iii. Newton’s third law of motion

c. Gives the idea of effects of force

Ans

i. Newton’s first law of motion

Also called the law of inertia

ii. Newton’s second law of motion

  

Gives the idea of effects of force
 

iii. Newton’s third law of motion

Describes the relationship between the forces on two interacting object

2

Column “A”

Column “B”

i. Newton’s first law of motion

a. Motion of rocket

ii. Newton’s second law of motion

b. Only the carom coin at the bottom of a pile moves on been hit by the striker

iii. Newton’s third law of motion

c. In a high jump athletic event, the athletes are made to fall on a sand bed

Ans

i. Newton’s first law of motion

Only the carom coin at the bottom of a pile moves on been hit by the striker

ii. Newton’s second law of motion

In a high jump athletic event, the athletes are made to fall on a sand bed

iii. Newton’s third law of motion

  

Motion of rocket
 

3

Column “A”

Column “B”

i. Inertia

a. Body at rest

ii. Rate of change of momentum

b. Newton’s first law
 

c. Newton’s second law
 

d. Mass x Acceleration

Ans

i. Inertia

Newton’s first law

ii. Rate of change of momentum

Newton’s second law

Q.5. State True or False :

 1.Action and reaction forces act simultaneously.

Ans Action and reaction forces act simultaneously. – True

2. In S.I. system, the unit of force is dyne.

Ans False – In S.I. system, the unit of force is newton.

3. During the collision the total momentum remains constant.

Ans During the collision the total momentum remains constant. – True

4. The distance and displacement are equal only if motion is along a straight line.

Ans The distance and displacement are equal only if motion is along a straight line. – True

5. The balanced force causes a change in the state of rest of an object.

Ans False – The unbalanced force causes a change in the state of rest of an object.

6. A body moving with a uniform velocity is said to have uniform acceleration.

Ans False – In uniform acceleration, velocity changes by equal amounts in equal time interval.

7. Distance is vector quantity.

Ans False – Displacement is a vector quantity as it has both magnitude and direction.

8. The inertia of the body is measured in terms of mass.

Ans The inertia of the body is measured in terms of mass. – True

9. Positive acceleration is called deceleration.

Ans False – negative acceleration is called deceleration.

10. In uniform circular motion, the direction of motion is tangential.

Ans In uniform circular motion, the direction of motion is tangential. – True

11. An object covers unequal distance in equal time intervals then it is moving with non-uniform speed.

Ans An object covers unequal distance in equal time intervals then it is moving with non-uniform speed. – True

12. The velocity of a body is given by the distance covered by it in unit time in a given direction.

Ans The velocity of a body is given by the distance covered by it in unit time in a given direction. – True

Q.6. Name the following : 

1.The scientist who summarized motion in a set of equations of motion.

Ans Sir Isaac Newton

2. When an object comes to rest at the end of its motion, what is its final velocity?

Ans Final velocity (v)=0

3. Negative acceleration is also called.

Ans Deceleration

4. The motion that covers unequal distance in equal interval of time.

Ans Non-uniform motion

5. S.I. unit of velocity.

Ans m/s

6. S.I. unit of momentum

Ans kg m/s

7. The motion that covers equal distance in equal interval of time.

Ans Uniform motion

8. S.I. unit of acceleration.

Ans m/s2

9. What is the backward motion of the gun called?

Ans recoil

10. Motion of an object along a circular path with uniform speed.

Ans Uniform circular motion

11. When an object is at rest in the beginning of its motion, what is its initial velocity?

Ans Initial velocity (u)= 0

12. What are the two types of acceleration for an object in motion?

Ans Uniform acceleration and Non-uniform acceleration.

Q.7 . Multiple Choice Questions : 

 

1. A car is moving with a velocity of 50km/hr for 5 hours, is an example of …………….

         a. zero acceleration                         b. positive acceleration

          c. negative acceleration                 d. retardation

Ans Option a.

2. v 2 = u2 + 2as is the relation between …………….

        a. speed and velocity               b. displacement and velocity

        c. displacement and time         d. velocity and time

Ans Option b.

3. The force necessary to cause an acceleration of 1m/s2 in an object of mass 1kg is called …………… .

    a.  dyne            b. 1 newton          c. acceleration        d. inertia

Ans Option b.

4. An object continues to remain at rest or in a state of uniform motion along a straight line unless an ……………

acts on it

         a. external balanced force        b. external unbalanced force

        c. internal balanced force           d. internal unbalanced force

Ans Option b.

5. is an example of negative acceleration.

        a.  coin thrown vertically upwards                                                    b. A coin falls freely towards the earth                                            c. A train starting from the station                                                    d. An object is speeding up.

Ans Option a.

6. If the velocity changes by equal amounts in equal time intervals, the object is said to have …………… .                                    a. uniform velocity             b. non-uniform velocity

   c. non-uniform motion       d. uniform acceleration.

Ans Option d.

7.  The displacement that occurs in unit time is called …………… .

a. direction b. velocity c. peed d. acceleration

Ans Option b.

12. Even if the displacement of an object is zero, the actual       distance traveled by the object …………… .

     a. will be constant                  b. will be zero

     c. may not be zero                 d. will be negative

Ans Option c.

8. The velocity of a body changes when …………… .

a. Speed changes but direction is constant

b.direction changes but speed is constant

c. Speed and direction changes

d. All of the three

Ans Option d.

9. v = u + at is the relation between …………….

a.speed and velocity b. displacement and velocity

c. displacement and time d. velocity and time

Ans Option d.

Q.8. Solve Numerical problems 

1.  A farmer moves along the boundary of a square field of side 10m in 40s.

i. What will be the magnitude of displacement and distance of           the farmer at the end?

ii. What will be the magnitude of displacement and distance if he covers from point A to C?                                                                Ans:  i.  Given : side of the square field = 10m                                                                 So, the perimeter = 10x 4 = 40m                     ii. Hence, the distance covered by the farmer is 40m while the displacement is zero as he comes back to his original position.

Distance covered from point A to C is 10 +10 = 20m Displacement       =  shortest path                                                            AB2 + BC2   =  AC2                                                                             102 + 102  =   AC2                                                                              √100 +   √100  =   AC2                                                                                 10√2m =  AC

2. Balls have masses of 50gm and 100gm and they are moving along the same line in the same direction with velocities of 3m/s respectively. They collide with each other and after collision, the first ball moves with a velocity of 2.5m/s. Calculate the velocity of the other ball after collision.

Ans Given: The mass of first ball (m1) = 50g   = 0.05 kg

The mass of the second ball (m2) = 100g = 0.1 kg                       Initial velocity of the first ball (u1) = 3m/s                                 Initial velocity of the second ball (u2) = 1.5m/s Final velocity of                                the second ball (v1) = 2.5m/s

To find: Final velocity of the second ball (v2) = ?

Formula: m1u1 + m2u2 = m1v1 + m2v2

Solution: According to the law of conservation of momentum, total initial momentum = total final momentum

                m1u1 + m2u2 = m1v1 + m2v2

(0.05 x 3) + ( 0.1 X 1.5) = (0.05 x 2.5) + (0.1 x v2)

                 (0.15 + 0.15) = (0.125 + 0.1v2)

                                 0.3 = 0.125 + 0.1v2

                                   v2 = 0.175/0.1 = 1.75m/s

The final velocity of the second ball is 1.75m/s.

3. A body of mass 5kg moves with an acceleration of 2m / s2. Find the change in momentum of the body in 2 seconds.

Ans Given: a = 2m/s2 u = 0

                    m = 5kg t = 2s

To find: change in momentum = ?

Formula: a = v-u/t Momentum = mxv

Solution: Momentum = m x v (v= at +u)

                                     = 5 x 4 (v=2×2 + 0 =4m/s)    = 20 kg m/s

Ans: The change in momentum is 20 kg m/s

4. An object of mass 16 kg. is moving with an acceleration of 3m/s2. Calculate the applied force. If the same force is applied on an object of mass 24 kg. how much will be the acceleration?

Ans Given: Mass of 1st body (m1) = 16 kg

          Acceleration of 1st body (a1) = 3m/s2                                                                   Mass of 2nd body (m1) = 24 kg

To find: Force on 1st body (F1) =?                                                     Acceleration of 2nd body (a2) = ?

Formula: F = m.a

Solution: F1 = m.a

                      = 16 × 3   F1 = 48N

                  a2 = F2/m2

                       = 48/24        ;  a2 = 2m/s2

Ans: The force acting on the first body is 48N and the acceleration of the second body is 2m/s2.

5. A car starting from rest acquires a speed of 20m/s in covering a distance of 100m. Calculate the acceleration of the car.

Ans Given: Initial velocity (u) = 0

Final velocity (v) = 20m/s distance covered (s) = 100m

To find: a = ?

Formula: v2 = u2 + 2as

Solution: v2 = u2 + 2as a = v2 – u2/2s

= (20)2-(0)2/2 x 100

= 400/200 = 2m/s2

Ans: The acceleration of the car is 2m//s2.

6. A bullet has a muzzle velocity of 300m/s. the gun of mass 3kg has a recoil velocity of ‘v’. Calculate ‘v’ if the mass of the bullet is 30gm.

Ans Given: The mass of gun (m2) = 300m/s

The mass of the bullet (m1) = 30g = 0.03 kg Initial velocity of the gun (u2) = 0m/s

Initial velocity of the bullet (u1) = 0m/s Final velocity of the bullet (v1) = 300 m/s

To find: Recoil velocity of the gun (v2) = ?

Formula: m1v1 + m2v2 v2

v2 = 0

    = -m1/m2 x v1

    = -0.03/3 x 300

     = 3m/s

Ans: The recoil velocity of the gun is 3m/s

7. A person travels a distance of 72 km in 4 hours. Calculate the average speed.

Ans Given: Total distance = 72km

= 72 x 1000m = 72000m

Total time taken = 6 hours

= 4 x 3600 (1 hour = 3600 sec)

= 14400 sec

To find: Average speed = ?

Formula: Average speed = Total distance covered/Total time taken

Solution: Average speed = Total distance covered/Total time taken

= 72000/14400

= 10/2 = 5m/s

Ans: The person travels with average speed of 5m/s

8. A force of 10 N acts on a body of mass 2 kg for 3s, initially at rest. Calculate: 1. Velocity acquired by the body

2. Change in momentum of the body.

Ans Given: F =10N, m =2 kg t =3s

u =0

To find: v =? momentum =?

Formula: a =v-u/t momentum =mx v

Solution: F =mx a

a F/m = 10/2 =5m/s2 a v-u/t

v at+u

=5×3 + 015m/s v =15m/s

Momentum =m x v

=2 x 15 = 30 kg m/s

Ans: The velocity of the body is 15m/s and the momentum is 30 kg m/s.

9. If the momentum of a body of mass 10kg is 20kg m/s. find its velocity.

Ans Given: mass of a body = 10kg, momentum = 20 kg m/s

To find: velocity = ? Formula: momentum = m × v Solution: Momentum = m × v

V = momentum/m

= 20/10 = 2m/s

Ans: The velocity of a body is 2m/s.

10. The mass of cannon is 500kg and it recoils with a speed of 0.25m/s. What is the momentum of the cannon?

Ans Given: Mass of the cannon = 500 kg

recoil speed = 0.25 m/s

To find: Momentum = ?

Formula: momentum = m x v

Solution: Momentum = m x v

= 500 x 0.25

= 125 kg m/s

Ans: The momentum of the cannon is 125 kg m/s.

12. A car acquires a velocity of 72km/hr in 10seconds starting from rest. Find the acceleration and distance travelled in this time.

Ans Given: Initial velocity (u) = 0

Final velocity (v) = 72km/hr = 72×5/18 = 20m/s time (t) = 10s

To find: acceleration (a) = ?

Distance (s) = ?

Formula: 1. a = v-u/t

Formula: 2. S = ut +1/2at2

Solution: 1.

Solution: 2.

a = v-u/t

= 20-0/10

a = 2m/s2

S = ut +1/2at2

= 0 x 10 + ½ x2 x 100

= 0 + 100

= 100m

Ans: The acceleration is 2m/s2 and the distance travelled is 100m

13. A motorboat starts from rest and moves with a uniform acceleration. If it attains the velocity of 15m/s in 5s. Calculate the acceleration and the distance travelled in that time.

Ans Given: Initial velocity (u) =0,

Final velocity (v) =15m/s time (t) =5s

To find: acceleration (a) =?

Distance (s) =?

Formula: 1. a=v-u/t

Formula: 2. S=ut +1/2at2

Solution: 1. =

a=v-u/t

15-0/5

a 3m/s2

Solution: 2.

S=ut +1/2at2

=0 x 5 + ½ x3 x 25

=0 + 75/2

=37.5m

Ans: The acceleration is 3m/s2 and the distance travelled is 37.5m.

14. A bullet having a mass of 10g and moving with a speed of 1.5m/s, penetrates a thick wooden plank of mass 90g. The plank was initially at rest. The bullet gets embedded in the plank and both move together. Determine their velocity.

Ans Given: Mass of bullet (m1) = 10g = 0/1000kg = 0.01kg Mass of the plank (m2) = 90g = 90/1000kg=0.09kg Initial velocity of the bullet (u1) = 1.5m/s

To Find: Common velocity = ?

Formula: m1u1 + m2u2 = m1v1 + m2v2

Solution: let v1 and v2 be the common velocities of the bullet and the plank

m1u1 + m2u2 = m1v1 + m2v2

(0.01 x 1.5) + (0.09 x 0) = (0.01 x v) + ( 0.09 x v)

0.015 + 0 = v (0.01 + 0.09)

0.015 + 0 = 0.1 v

v = 0.015/0.1

v = 0.15m/s Ans: The plank moves with a velocity of 0.15m/s.

15. A car starts from rest and moves with a uniform acceleration of 2m/s2. How much time will it take to cover the distance of 49m?

Ans Given: a = 2m/s2 s = 49m u = 0

To find: t = ? Formula: s = ut + ½ at2 Solution: s = ut + ½ at2

= 0 x t + ½ x 2 x t2

49 = 0 + t2

7 sec = t

Ans: The car will take 7 seconds to cover the distance of 49m.

16. A person runs 100m in the first 50s, 80m in the next 30s and 45m in the last 35s. What is the average speed?

Ans Given: Total distance = 100 + 80 + 45 = 225m Total time taken = 50 + 30 + 35 =115 sec

To Find: Average speed = ?

Formula: Average speed = Total distance covered / Total time taken

Solution: Average speed = Total distance covered / Total time taken

= 225/ 105

= 1.95 m/s

Ans: The person runs with an average speed of 1.95m/s.

17. A Kangaroo can jump 2.5m vertically. What must be the initial velocity of the kangaroo?

Ans Given: a = 9.8m/s2 , s= 2.5m, v = om/s,

To find: u = ? Formula: v2 = u2 + 2as Solution: v2 = u2 + 2as

u2 = v2 – 2as

02 – 2 x (- 9.8) x (Negative acceleration shows that the acceleration is in the direction opposite

= 2.5

u2 = 49m/s u = 7m/s

to that of velocity)

Ans: The initial velocity of the kangaroo must be 7m/s.

18. An aero plane taxies on the runway for 30 sec with an acceleration of 3.2m/s2 before taking off. How much distance would it have covered on the runway?

Ans Given: a = 3.2m/s2 , t = 30 s, u = 0

To find: s = ?

Formula: s = ut + ½ at2

s = ut + ½ at2

= 0 x 30 + ½ x 3.2 x 302

= 1440m

Ans: The distance covered by the aero plane on the runway is 1440m.

19. A person swims 100m in the first 40s, 80m in the next 40s and 45 m in the last 20s. What is the average speed?

Ans Given: Total distance = 100 + 80 + 45 = 225m Total time taken = 40 + 40 + 20 = 100sec

To Find: Average speed = ?

Formula: Average speed = Total distance covered / Total time taken

Solution: Average speed = Total distance covered / Total time taken

= 225/100

= 2.25 m/s

Ans: The person swims with an average speed of 2.25m/s.

20. An athlete is running on a circular track. He runs a distance of 400m in 25 s before returning to his original position. What is his average speed and velocity?

Ans Given: Total distance travelled = 400m

Total displacement = 0, as the athlete returns to his original position. Total time taken = 25 seconds

To find: Average speed = ?

Average velocity = ?

Formula: Average speed = Total distance covered/ total time taken.

Average velocity = Total displacement/ total time taken.

Solution: Average speed = Total distance covered/ time

= 400/25 = 16m/s

Average velocity = Total displacement/ total time taken

= 0/25 = 0m/s

Ans: The average speed is 16m/s and the average velocity is 0m/s.

22. An object moves 18m in the first 3s, 22m in the next 3s and 14m in the last 3s. What is the average speed?

Ans Given: Total distance = 18 + 22 + 14 = 54m Total time taken = 3 + 3 + 3 = 9 sec.

To find: Average speed=? = ?

Formula: Average speed = Total distance covered / Total time taken

Solution: Average speed = Total distance covered / Total time taken

= 54/9

= 6m/s

Ans: The object moves with an average speed of 6m/s.

Q.9. Write Short Notes :

1. Newton’s third law of motion (Action and Reaction forces)

Ans i. When the air rushes out of a balloon, the opposite reaction is that the balloon flies up.

i.When you walk, you push the street i.e. you apply a force on the street and the reaction force moves you forward.

ii. When someone fires the gun, the reaction force pushes the gun backwards.

iii. While rowing a boat, the boatman pushes the water backwards with the oars. The water exerts equal and opposite push on the boat which makes the boat move forward.              

 2. Newton’s first law of motion (Law of Inertia)

Ans i. If a car is moving forward, it will continue to move forward unless friction or the brakes are applied.

If a person jumps from a moving bus, his body is still moving in the direction of the bus. This is because

    1. when the person hits the ground, his feet has come to rest but the upper part of his body is still in the state of motion.
    2. When you stir tea or coffee, the swirling motion still continues due to inertia.
    3. When pulling a Band-Aid off, it is better to pull it fast because the skin will remain at rest due to inertia and the force pulls the Band-Aid off.3. Newton’s second law of motion (momentum)

Ans i. A fielder pulls his hands backwards, while catching a cricket ball coming with a great speed, to reduce the momentum of the ball with a delay.

    1. For athletes of long and high jump, sand bed or cushioned bed is provided to allow a delayed change of momentum.
    2. It is easier to push an empty shopping cart than a full one because the full shopping cart has more mass.
    3. The batsman requires more force to hit a ball to change its direction.

Q.10. Attempt the following. 

1. You are travelling in a bus. Is the person sitting next to you in motion? What do you take into consideration to decide if an object is moving or not?

Ans The person sitting in the bus is at rest if I am the observer while the same person is motion for an observer outside the bus. The motion of the object with respect to the observer will decide if the object is in motion or at rest.

2. Explain positive and negative acceleration.

Ans i. Positive acceleration: When the velocity of a body increase, the acceleration is in direction of velocity and hence acceleration is said to be positive.

ii.Negative acceleration: When the velocity of a body decrease, the acceleration is in opposite direction of velocity and hence acceleration is said to be negative.

3. In which of the following examples can you sense motion? How will you explain the presence and absence of motion?

i..The flight of a bird

ii. A stationary train

iii. Leaves flying through air

iv. A stone lying on the hill

Ans In example I and III we can see motion. This is because, if the body is moving in a certain interval of time then we can say that it is in motion.

4. Define word equation and give example.

Ans The simple way of representating a chemical reaction in word is called word equation. Zinc dust + aqueous solution of copper sulphate → aqueous solution of zinc sulphate + copper.

  1. .                                               5. Name and explain the device shown in the picture and state its principle.

Ans i. The device that converts electrical energy into mechanical energy is called an electric motor.

ii. It works on the principle that a current carrying conductor placed in a magnetic field experiences a force.

6. Give the formula for each.

   

Speed = distance/time

i.

Average speed
   

ii.

Average velocity

……………
 

Relation between initial, final velocity acceleration and displacement in a uniform accelerated straight line motion

  

iii.

……………
   
 

Relation between initial, final velocity acceleration and time in a uniform accelerated straight line motion

  

iv.

……………
   
 

Relation between initial velocity acceleration, displacement and time in a uniform accelerated straight line motion

  

v.

……………
   

Ans

i.

Average speed

Speed = distance/time

ii.

Average velocity

Velocity=displacement/time
 

Relation between initial, final velocity acceleration and displacement in a uniform accelerated straight line motion

  

iii.

V2= u2 + 2as
   
 

Relation between initial, final velocity acceleration and time in a uniform accelerated straight line motion

  

iv.

V= u +at
   
 

Relation between initial velocity acceleration, displacement and time in a uniform accelerated straight line motion

  

v.

s= ut+ 1/2at2
   

7. Observe the velocity of the ball as it rolls down along the channel. Is its velocity the same at all points?

Ans No, the velocity will not be the same at all the points. As the ball moves down, the velocity will go on increasing.

8. Observe the following figures. If you increase the number of sides of the polygon and make it infinite, how many times will you have to change the direction? What will be the shape of the path?             

     

Ans i. If we go on increasing the number of sides of the polygon and make it infinite, then we will have to change the direction an infinite number of times.

ii. The shape of the path thus obtained will be a circle.

9. Give the mathematical expression used to determine the velocity of an object moving with uniform circular motion.

Ans Velocity = Circumference / Time                                                             ∴ V = 2π / t

This is the expression used to determine the velocity of a body moving with uniform circular motion.

10. What do you infer if ?

i.

Distance-time graph is a straight line

……………

ii.

Velocity –time graph is curved

……………

iii.

Displacement-time graph is

……………

iv.

Displacement-time graph is zig-zag

……………

Ans

i.

Distance-time graph is a straight line

Speed is constant

ii.

Velocity –time graph is curved

Acceleration is not uniform

iii.

Displacement-time graph is straight

Uniform velocity

iv.

Displacement-time graph is zig-zag

Non-uniform velocity

Q.11. Distinguish between :

1.Balanced force and Unbalanced force

Ans:

 

Balanced force

Unbalanced force
 

Balanced forces are the two equal forces

Unbalanced forces

are

the

two

unequal

forces

i.

applied on a body in the opposite direction.

applied on the body

          
 

ii.

Balanced force does not change the state of rest or the state of uniform motion.

Unbalanced force can change the state of rest or the state of uniform motion of a body in a straight line.
 

2. Uniform and Non-uniform motion

Ans

 

Uniform motion

Non-uniform motion
 

If an object covers equal distances in equal intervals of time it is said to be in uniform motion.

If an object covers unequal distances in equal intervals of time it is said to be in non-uniform motion

i.
 

ii.

In uniform motion, acceleration is zero

In non-uniform motion, acceleration is not zero.
 

Distance-time graph for uniform motion is a straight line.

Distance-time graph for non-uniform motion is not a straight line.

iii.
 

3. Speed and Velocity

Ans

 

Speed

Velocity
 

Speed is the distance covered by the body per unit time

Velocity is the displacement that occurs in unit time.

i.
 

ii.

It is a scalar quantity.

It is a vector quantity.

iii.

Speed is always positive.

Velocity can be positive or negative.

iv.

Speed= Distance/time

Velocity= Displacement/time

4. Distance and Displacement.

Ans

 

Distance

Displacement
   

Displacement is the minimum distance between the starting and finishing points.

Distance is the length of the actual path travelled by an object in motion while going from one point to another.

i.
 
 

ii.

It is a scalar quantity

It is a vector quantity
   

Displacement may be positive, negative or zero

iii.

Distance travelled is always positive
   
   

It is either equal to or less than distance.

iv.

It is either equal to or greater than displacement
   

Q.12. Give scientific reasons :

1. When an object falls freely to the ground, its acceleration is uniform.

Ans i. If the velocity changes by equal amounts in equal time interval s, the object is said to be in uniform acceleration.

i] When the body falls freely on the ground, there are equal changes in velocity of the body in equal intervals of time.

ii] Thus the acceleration of the body is constant.

iii] Hence, it possesses uniform acceleration.

iv] It is easier to stop a tennis ball as compared to a cricket ball,

 

2. when both are travelling with the same velocity.

Ans i. The above example is based on Newton’s second law of motion.

i.Momentum of an object depends on its mass as well as its velocity.

ii. Cricket ball is heavier than a tennis ball. Although they are thrown with the same velocity, cricket ball has more momentum and inertia due to its greater mass than a tennis ball.

iii. Hence, force required to stop cricket ball is more than a tennis ball.

iv. Therefore, it is easier to stop a tennis ball than a cricket ball moving with same velocity.

 

3. The velocity of an object at rest is considered to be uniform.

Ans i. The velocity of an object is said to be in uniform motion when its speed is constant all the time.

ii. For an object at rest, its speed is zero all the time.

iii. Hence, an object in a state of rest is an example of uniform motion.                                                                                   Motion is relative.

i.A body is said to be in motion if it changes its position with respect to its surrounding.

ii. Motion of an object depends on the observer, hence a body may appear to be moving for one person and at the same time at rest for another.

iii. Hence, motion is relative.

Heavier objects offer more inertia.

i. According to Newton’s first law of motion, ‘An object continues to remain at rest or in a state of uniform motion along a straight line unless an external unbalanced forces acts on it’

ii. As mass is the quantity of matter in a body, we need to exert more force to push a heavier body.

iii.Inertia is related to the mass of the object.

iv. Hence, heavier objects offer more inertia.       

The launching of a rocket is based on Newton’s third law of motion.

i.Newton’s third law of motion states that,’ Every action force has an equal and opposite reaction force which acts simultaneously ’.

ii. When the fuel in a rocket is ignited, it burns as a result of chemical reaction.

iii. The exhaust gases escape with a great force through a small opening at the tail end of the rocket.

iv. It exerts an equal and opposite reaction force on the rocket, due to which the rocket moves in the forward direction. Thus, the principle of launching of rocket is based on Newton’s third law of motion.

4. Even though the magnitudes of action force and reaction force are equal and their directions are opposite, their effects do not get cancelled.

Ans i. This is based on Newton’s third law of motion.

  1. Forces are always applied in pairs.
  2. When one object applies a force on another object, the latter object also simultaneously applies a force on the former object.
  3. Action and reaction forces act on different bodies.
  4. They don’t act on the same body, hence they don’t cancel each other’s effect.

Therefore, even though the magnitude of action force and reaction force are equal, they do not cancel each other.

Q.13. Give examples 2

1 Give examples of circular motion. (any 4)

Ans i. Movement of artificial satellite around the earth.

  1. Motion of moon around the sun.
  2. Motion of earth around the sun.
  3. Motion of seconds hands of a watch.
  4. Athlete moving on a circular path.

Q.14. Give explanation using the given statement: 15

  1. Complete the sentences and explain them

The minimum distance between the start and finish points of the motion of an object is called the of

the object.

Ans The minimum distance between the start and finish points of the motion of an object is called the

displacement of the object.

Displacement is the minimum distance between the start and the finish points of the motion of the object whereas distance is the actual path travelled by the object. Even if the displacement of an object is zero, the actual distance traversed by it may not be zero.

Complete the sentences and explain them

Deceleration is acceleration.

Ans Deceleration is negative acceleration.

In deceleration, the velocity of the body goes on decreasing. Hence it is called negative acceleration or deceleration.

Complete the sentences and explain them

When an object is in uniform circular motion, its changes at every point.

Ans When an object is in uniform circular motion, its direction changes at every point.

In uniform circular motion, the speed is constant along the circumference but its direction at every point is tangential.

Complete the sentences and explain them.

During collision remains constant.

Ans During collision total momentum remains constant.

It is the law of conservation of momentum which states that the magnitude of the total final momentum is equal to the magnitude of the total final momentum if no external forces are acting on two objects.

Complete the sentences and explain them

The working of a rocket depends on Newton’s law of motion.

Ans The working of a rocket depends on Newton’s third law of motion.

According to Newton’s third law, “every action force has an equal and opposite reaction force which acts simultaneously”.

In this case, the escaping gases exert an equal and opposite reaction on the rocket so that it gets propelled in the forward direction.

Q.15 Attempt the following. 3

1 A train is moving with a uniform velocity of 60km/hr for 5hours. The velocity-time graph for this uniform motion is shown in figure.

  1. With the help of the graph, how will you determine the distance covered by the train between 2 and 4 hours?
  2. Is there a relation between the distance covered by the train between 2 and 4 hours and the area of a particular quadrangle in the graph? What is the acceleration of the train?

Ans i. The distance covered by the train between 2 and 4 hours is given by the equation Distance covered = Velocity x time

= 60 x 2

= 120 km

ii. Yes, there is a relation between the distance covered by the train between 2 and 4 hours and the area of the quadrangle in the graph.

Distance covered = Area of the quadrangle (ABDC)

= l x b

= 60 x 2 = 120km

Acceleration = ( final velocity – initial velocity)/time|

= (60 -60)/2

= 0/2

= 0m/s2

Since the acceleration is zero, the train is moving with uniform velocity. This is an example of zero acceleration.

Q.16. Explain with the help of examples 3

1 Explain Uniform Circular Motion with example.

Ans i. When an object is moving with a constant speed along a circular path, the change in velocity is only due to the change in direction. Hence, it is accelerated motion.

  1. When an object moves with constant speed along a circular path, the motion is called uniform circular motion.
  2. If an object, moving along a circular path of radius ‘r’, takes time ‘t’ to come back to its starting position, its speed can be determined using the formula given below:

Speed = Circumference/time V= 2 ( r= radius of the circle)

  1. In uniform circular motion, the speed is constant along the circumference but its direction at every point is tangential.
  2. For example: when a coin is placed on a circular disc and the disc is rotated at a high speed, the coin will be thrown off in the direction of the tangent which is perpendicular to the radius of the disc. Thus, as the

coin moves along a circular path the direction of its motion is changing at every point.

Q.17. Solve Numerical problems 6

1

u(m/s)

a(m/s2)

t(sec)

v=u + at(m/s)

2

4

3

……………

……………

5

2

20
 

u(m/s)

a(m/s2)

t(sec)

s=ut+1/2at2(m)

5

12

3

……………

7

……………

4

92
 

u(m/s)

a(m/s2)

t(sec)

V2=u2 +2as(m/s)2

4

3

……………

8

……………

5

8.4

10

Ans

u(m/s)

a(m/s2)

t(sec)

v=u + at(m/s)

2

4

3

14

10

5

2

20
 

u(m/s)

a(m/s2)

t(sec)

s=ut+1/2at2(m)

5

12

3

69

7

8

4

92
 

u(m/s)

a(m/s2)

t(sec)

V2=u2 +2as(m/s)2

4

3

8

8

4

5

8.4

10

2

Every morning, Swaralee walks around the edge of a circular field having a radius of 100m. As shown in

  1. figure (a), if she starts from the point A and takes one round, how much distance has she walked and what is her displacement?
  2. If a car, starting from point P, goes to point Q (see figure b) and then returns to point P, how much distance has it travelled and what is its displacement?

Ans 1.(a)Radius of the circular field = 100m

(b)Distance covered = circumference of the circle

(c) = 2π r

=2 x 3.14 x 100 = 628m

(d)Displacement will be zero ( Displacement is the shortest distance between initial and final position) 2.(a)Distance covered = PQ + QP

= 360 + 360 = 720m

(b)Displacement = 0m ( shortest distance between initial and final position is zero).

Q.18. Complete the sentences in paragraph 3

  1. Complete the paragraph:

(moving, increases, decreases, tangential, accelerated, direction, magnitude, opposite.)

The velocity of object changes with time. This change in velocity can be in terms of

or of velocity or both.

When the velocity of a body with time, its acceleration is positive. Here acceleration is in the direction of velocity.

When the velocity of a body with time, its acceleration is negative. Here acceleration is to direction of velocity.

Ans The velocity of accelerated object changes with time. This change in velocity can be in terms of direction

or magnitude of velocity or both.

When the velocity of a body increases with time, its acceleration is positive. Here acceleration is in the direction of velocity.

When the velocity of a body decreases with time, its acceleration is negative. Here acceleration is opposite

to direction of velocity.

Q. 19. Write answers based on given diagram 6

1

  1. Measure the distance between points A and B in different ways as shown in the figure(a)
  2. Now measure the distance along the dotted line. Which distance is correct according to you and why?
  3. Sheetal first went to her friend Sangeeta’s house on her way to school. Prashant went straight from house to school. Both are walking with the same speed. Who will take less time to reach the school and why?

Ans i. The distances measured will be on different lengths depending upon the path taken.

  1. Dotted line shows the shortest way of reaching from A to B
  2. Prashant will take less time as the path followed by him is the shortest. While Sheetal took the longest route to reach school.
  3. Amar, Akbar and Anthony are travelling in different cars with different velocities. The distances covered by them during different time intervals are given in the following table.
 

Distance covered by

Time in the clock

Amar in km

Akbar in km

Anthony in km
 

5.00

0

0

0

5.30

20

18

14

6.00

40

36

28

6.30

60

42

42

7.00

80

70

56

7.30

100

95

70

8.00

120

120

84
    1. What is the time interval between the noting of the distances made by Amar, Akbar and Anthony?
    2. Who has covered equal distances in equal time intervals?
    3. Considering the distances covered by Amar, Akbar and Anthony in fixed time intervals, what can you say about their speeds?

Ans i. The time interval between the noting of the distances made by Amar, Akbar and Anthony is of 30 minutes.

  1. Amar and Anthony have covered equal distances in equal intervals of time.
  2. Amar and Anthony are travelling with uniform speed while Akbar is travelling with non-uniform speed.

Answer the following 24

  1. State the three equations of motion and give the relationship explained by them.

Ans i.

ii.

v = u + at

s = ut + 1 at2

2

This is the relation between velocity and time.

This is the relation between displacement and time.

iii. v2 = u2 + 2as This is the relation between displacement and velocity.

  1. A body goes around the sun with constant speed in a circular orbit. Is the motion uniform or accelerated?

Ans i. When a body moves along a circular path, then its direction of motion keeps on changing continuously

  1. Since the velocity changes (due to change in direction) therefore the motion along a circular path is said to be accelerated.
  2. The above example is an accelerated one because the motion in a circle with constant speed is an example of accelerated motion.
  3. Explain recoil and recoil velocity and derive its expression.

Ans i. Let us consider the example of a bullet fired from a gun. When a bullet of mass m1 is fired from a gun of mass m2, its velocity becomes v1 and its momentum becomes m1v1. Before firing the bullet, both the gun and the bullet are at rest. Hence the total momentum is zero.

  1. According to the above law, the total final momentum has to be zero. Thus, the forward moving bullet causes the gun to move backward after firing.
  2. This backward motion of the gun is called its recoil.
  3. The velocity of recoil, v2 is v. m1v1 + m2v2 = 0

vi. v2 = -m1/m2 × v1

  1. What is velocity? State its units and types.

Ans i. The velocity of a body is the distance travelled by a body in a particular direction in unit time. Thus, rate of change of displacement is called velocity.

  1. Velocity = Displacement/time

V= s/t

  1. The units of velocity: M.K.S.(S.I.) unit = m/s and C.G.S. unit = cm/s
  2. There are two types of velocities.
    1. Uniform Velocity: If there is equal displacement taking place in equal interval of time, it is uniform velocity.
    2. Non-uniform Velocity: If there is unequal displacement taking place in equal interval of time, it is non- uniform velocity.
  3. What happens to speed, velocity and acceleration when an object moves in a circle with uniform speed?

Ans i. Speed remains constant

  1. With velocity, magnitude remains constant but the direction changes continuously.
  2. With acceleration, the magnitude remains constant but the direction changes continuously.
  3. What is speed? State its units and types.

Ans i. The speed of a body is the distance travelled in unit time.

  1. The unit of speed : M.K.S. (S.I.) unit = m/s and C.G.S. unit = cm/s
  2. Speed = Distance/ time
    1. Uniform speed: When a body moves equal distance in equal interval of time throughout its motion, it is said to have uniform speed.
    2. Non-uniform speed or variable speed: A body is said to move with variable speed if it covers unequal distance in equal intervals of time.
  3. Explain the three different ways to change the velocity of an object.

Ans As velocity is related to speed and direction, it can be changed by :

    1. Changing the speed while keeping the direction constant.
    2. Changing the direction while keeping the speed constant.
    3. Changing both the speed as well as the direction of motion.
  1. What is acceleration? State its unit and types.

Ans i. Acceleration is the rate of change of velocity. It is a vector quantity.

  1. ( a = v-u/t) where v= final velocity, u= initial velocity, t= time, a= acceleration.
  2. Units of acceleration : M.K.S.(S.I.) unit= m/s2 and C.G.S. unit = cm/s2
  3. There are two types of acceleration:
    1. Uniform acceleration: If the change in the velocity is equal in equal interval of time, the acceleration is uniform acceleration.
    2. Non-uniform acceleration: If the change in the velocity is unequal in equal interval of time, the acceleration is uniform acceleration.

Q.21. Extra data (Not to be Use) 5

1 Why is there a thick bed of sand for a high jumper to fall on after the jump?

Ans i. When a sand bed is used for high jump, the rate of change of momentum is reduced. Thereby reducing the impact.

ii. Thus the force experienced by the high jumper on landing is less.

Q.22. Answer the following in detail 50

  1. Obtain the equations of motion by graphical method: – Newton’s Third law of Motion

i.

ii.

iii.

The velocity-time graph shows the change in velocity with time of a uniformly accelerated object. The object starts from the point D in the graph with velocity v. Its velocity keeps increasing and after time t, it reaches the point B on the graph.

The initial velocity of the object = u = OD The final velocity of the object = v = OC Time t = OE

Acceleration (a) = Change in Velocity

Time

(Final velocity− Initial velocity) Time

=

  1. ∴

=

= at … (i) (OC – OD = CD)

  1. From the graph … BE =
  2. ∴
  3. ∴
  4. ∴

v = … ( AB = CD and AE = OD)

v = at + u … ( from i)

v = u + at … (i)

  1. We can determine the distance covered by the object in time t from the area of the quadrangle DOEB. DOEB is a trapezium. So we can use the formula for its area.
  2. ∴
  3. ∴
  4. ∴
  5. ∴

s = area of trapezium DOEB

s = ½ x X distance between the parallel sides.

s = ½ x (OD + BE) x But, OD = u, BE = v, OE = t s = ½ x (u + v) x t … (ii)

  1. But, a =
  2. ∴
  3. ∴

t = … (iii)

s = ½ x ( u + v ) x ( v – u)/a

  1. ∴ s =
  2. ∴

= ( u + v) ( v – u ) = v2 – u2`

  1. ∴

v2 =

Ans

This is Newton’s third law of motion.

The velocity-time graph shows the change in velocity with time of a uniformly accelerated object. The

  1. object starts from the point D in the graph with velocity v. Its velocity keeps increasing and after time t, it reaches the point B on the graph.

The initial velocity of the object = u = OD

  1. The final velocity of the object = v = OC Time t = OE

Acceleration (a) = Change in velocity/time

iii. = (Final velocity- Initial velocity)/time

= (OC-OD)/t

  1. ∴ CD = at … (i) (OC – OD = CD)
  2. From the graph … BE = AB + AE
  3. ∴
  4. ∴
  5. ∴

v = CD + OD … ( AB = CD and AE = OD)

v = at + u … ( from i)

v = u + at … (i)

  1. We can determine the distance covered by the object in time t from the area of the quadrangle DOEB. DOEB is a trapezium. So we can use the formula for its area.
  2. ∴
  3. ∴
  4. ∴
  5. ∴

s = area of trapezium DOEB

s = ½ x sum of the lengths of parallel sides X distance between the parallel sides. s = ½ x (OD + BE) x OE But , OD = u, BE = v, OE = t

s = ½ x (u + v) x t … (ii)

  1. But, a = (v – u)/t
  2. ∴
  3. ∴
  4. ∴

t = (v-u)/a … (iii)

s = ½ x ( u + v ) x ( v-u)/a s = ( u + v) (u – v)/2a

  1. ∴

2as = ( u + v) ( v – u ) = v2 –u2`

  1. ∴

v2 = u2 + 2as

This is Newton’s third law of motion.

  1. Explain Newton’s second law of motion and derive the formula.
    1. Newton’s second law explains about change in momentum. It states that ‘The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the of the force.’

ii Suppose an object of mass ‘m’ has an initial velocity ‘u’. When a force ‘F’ is applied in the direction of its velocity for time‘t’, its velocity becomes ’v’. then, the total initial momentum of the body = Its final momentum after time t =

  1. So the rate of change of momentum = Change in momentum/ time

=

=

= ma … (a = )

  1. Hence by Newton’s second law of motion, the rate of change of momentum is proportional to the applied force.
  2. ∴
  3. ∴
  4. ∴

ma F

F α

F= k ma (k = constant of proportionality and its value is )

  1. ∴ F= ma

Ans i. Newton’s second law explains about change in momentum. It states that ‘The rate of change of momentum is proportional to the applied force and the change of momentum occurs in the direction of the force.’

ii Suppose an object of mass ‘m’ has an initial velocity ‘u’. When a force ‘F’ is applied in the direction of its velocity for time‘t’, its velocity becomes ’v’. then, the total initial momentum of the body = ‘mu’. Its final momentum after time t = ‘mv’.

  1. So the rate of change of momentum = Change in momentum/ time

mv − mu t

=

m(v − u) t

=

= ma … (a = v − u )

t

  1. Hence by Newton’s second law of motion, the rate of change of momentum is proportional to the applied force.
  2. ∴
  3. ∴

ma α F F α ma

  1. ∴ F= k ma (k = constant of proportionality and its value is 1)
  2. ∴ F= ma
  3. State the law of conservation of momentum and derive the formula.
    1. Suppose an object A has mass m1 and its initial velocity is u1. An object B has mass m2 and initial velocity u2.
    2. According to the formula for momentum, the initial momentum of A is and that of B is
    3. Suppose these two objects collide.
    4. Then, let the force on A due to B be F1. This force will cause acceleration in A and its velocity will become v1.
    5. ∴ Momentum of A after collision =

According to Newton’s third law of motion, A also exerts an equal force on B but in the opposite direction.

    1. This will cause a change in the momentum of B and its velocity after collision is v2. Thus, the momentum of B after collision =
    2. If F2 is the force that acts on object B,
    3. F2 =
    4. ∴
    5. ∴
    6. ∴

m2a2 = … F=ma)

m2 x (v2-u2)/t = … (a= (v-u)/t) m2 (v2-u2) =

    1. ∴

m2v2 – m2u2 =

(m1v1 + m2v2 ) =

    1. The magnitude of total final momentum = The magnitude of total initial momentum. Thus, if no external force is acting on two objects, then their total initial momentum is equal to their total final momentum.
    2. When no external force acts on two interacting objects, their total momentum remains constant. It does not change.

Ans i. Suppose an object A has mass m1 and its initial velocity is u1. An object B has mass m2 and initial velocity u2.

  1. According to the formula for momentum, the initial momentum of A is m1u1 and that of B is m2u2.
  2. Suppose these two objects collide.
  3. Then, let the force on A due to B be F1. This force will cause acceleration in A and its velocity will become v1.
  4. ∴

Momentum of A after collision = m1v1

According to Newton’s third law of motion, A also exerts an equal force on B but in the opposite direction.

  1. This will cause a change in the momentum of B and its velocity after collision is v2. Thus, the momentum of B after collision= m2v2.
  2. If F2 is the force that acts on object B,
  3. F2= -F1
  4. ∴

m2a2 = -m1a1

… F=ma)

  1. ∴
  2. ∴

m2 x (v2-u2)/t = – m1 x (v1-u1)/t … (a= (v-u)/t) m2 (v2-u2) = – m1 (v1-u1)

  1. ∴

m2v2 – m2u2 = – m1v1+ m1u1

(m1v1 + m2v2 )= (m1u1 + m2u2)

The magnitude of total final momentum = The magnitude of total initial momentum. Thus, if no

  1. external force is acting on two objects, then their total initial momentum is equal to their total final momentum.

When no external force acts on two interacting objects, their total momentum remains constant. It does not change.

  1. Obtain the equations of motion by graphical method. Equation for velocity –time relation.

    1. The velocity-time graph shows the change in velocity with time of a uniformly accelerated object. The object starts from the point D in the graph with velocity v. Its velocity keeps increasing and after time t, it reaches the point on the graph.
    2. The initial velocity of the object = u = The final velocity of the object = v = Time t =
    3. Acceleration (a) =

(Final velocity− Initial velocity) Time

=

iv ∴

=

CD = … (i) (OC-OD = CD)

  1. From the graph … BE =
  2. ∴
  3. ∴

v = … ( AB = CD and AE = OD)

v = … (from i)

  1. ∴ v = u + at
  2. This is the first equation of motion.

Ans i. The velocity-time graph shows the change in velocity with time of a uniformly accelerated object. The object starts from the point D in the graph with velocity v. Its velocity keeps increasing and after time t, it reaches the point B on the graph.

  1. The initial velocity of the object =u = OD The final velocity of the object = v = OC Time t = OE
  2. Acceleration (a) = Change in Velocity

TIme

(Final velocity− Initial velocity) Time

=

=

iv ∴

(OC−OD)

t

CD= at … (i) (OC-OD = CD)

  1. From the graph … BE= AB+ AE
  2. ∴
  3. ∴

v= CD + OD … ( AB= CD and AE = OD)

v= at + u … (from i)

  1. ∴ v= u + at
  2. This is the first equation of motion.
  3. In the velocity time graph for uniform acceleration. The changes in the velocity of a car in specific time are given in the following table.

    1. The above graph represents which type of motion?
    2. How much the velocity change in every 5 minutes?
    3. For all uniformly accelerated motions, what is the shape for velocity-time graph?
    4. For all non-uniformly accelerated motions, what is the shape for velocity-time graph?
    5. Calculate the distance covered by the car between the 10th and the 20th seconds.

Ans i. In the above graph, the velocity changes by equal amounts in equal time intervals. Thus this is an example of uniformly accelerated motion.

  1. The velocity changes by 8m/s in every 5 minutes.
  2. For all uniformly accelerated motions, the velocity-time graph is a straight line.
  3. For all non-uniformly accelerated motions, the velocity-time graph may have any shape depending on how the acceleration changes with time.
  4. Distance s = Average Velocity x Time

= (16+ 32 ) m/s x (20s -10s)

2

= 24m/s x 10 s

= 240 m

Ans. Distance covered by the car = 240m

  1. A train is moving with a uniform velocity of 60km/hr for 5hours. The velocity-time graph for this uniform motion is shown in figure.
    1. With the help of the graph, how will you determine the distance covered by the train between 2 and 4 hours?
    2. Is there a relation between the distance covered by the train between 2 and 4 hours and the area of a particular quadrangle in the graph? What is the acceleration of the train?

Ans i. (a) The distance covered by the train between 2 and 4 hours is given by the equation

(b) Distance covered = Velocity x time

= 60 x 2 = 120 km

ii. (a) Yes, there is a relation between the distance covered by the train between 2 and 4 hours and the area of the quadrangle in the graph.

(b) Distance covered = Area of the quadrangle (ABDC)

(c) = l x b

(d) = 60 x 2 = 120km

(e) Acceleration = ( final velocity – initial velocity)/time (f) = (60 -60)/2

(g) = 0/2 = 0m/s2

(h) Since the acceleration is zero, the train is moving with uniform velocity. This is an example of zero acceleration.

  1. Take 5 examples from your surroundings and give explanation based on Newtons laws of motion.

Ans Example 1:

When we are travelling by a bus, we experience backward jerk as the bus starts moving from rest. This is in accordance with Newton’s first law of motion.

Explanation:

    1. A body resists change in its state of rest or of motion due to inertia.
    2. When the bus is at rest, inside the bus we are also at rest.
    3. As the starts moving, the portion of the our body which is in contact with the bus acquires velocity, but the upper part of the body tries to remain at rest.
    4. As a result, we exert an inertia of rest and get a backward jerk, when the bus moves in the forward direction.

Example 2:

A carpet when lifted up and given jerks, dust falls out of it. This is in accordance with Newton’s first law of motion.

Explanation:

  1. When we dust a carpet, by lifting it, the carpet is set into motion.
  2. While the dust remains in its state of rest due to inertia of rest.

Example 3:

While catching a ball, cricketer moves his hands backwards. This is in accordance with Newton’s second law of motion.

Explanation:

  1. In the act of catching the ball, by drawing hands backward, cricketer allows longer time for his hands to stop the ball.
  2. By Newton’ s second law of motion, force applied depends on the rate of change of momentum. iii.Taking longer time to stop the ball ensures smaller rate of change of momentum.

iv.Due to this, the cricketer can stop the ball by applying smaller amount of force and thereby not hurting his hands.

Example 4:

A book kept on a table remains stationary. This is in accordance with Newton’s third law of motion.

Explanation:

  1. A book kept on a table has some weight. This weight is the force acting on the table.
  2. By Newton’s third law of motion, every action has an equal and opposite reaction. iii.Thus, the table also exerts an upward force on and balances weight of the book..

iv.Thus, both the forces are balanced and there is no displacement. Hence, a book kept on a table remains stationary.

Example 5:

An air filled ballon held in hand when released, moves forward. This is in accordance with Newton’s third law of motion.

Explanation:

i. As air is released out in downward direction, it applies equal and opposite force on balloon pushing it forward.

  1. Explain how can you calculate the distance covered by the body in a given time from its velocity – time graph?

Ans

  1. If we want to calculate the distance the body has covered between say time t1 perpendiculars from points corresponding to the time t1 and t2 on the above graph.

and t2. We draw

The two perpendiculars enclose a rectangle ABCD between the graph and the time axis. The sides AC

  1. and BD are equal to 60km/h while AB and CD is 2hours. We know that the distance s covered by a body is given by the area under the rectangle.
  2. s= l Xb = v x t
  3. s =60 x 2 = 120km
  4. Hence the distance covered by the body between 2 to 4hours id 120km.

  1. If we want to calculate the distance the body has covered between say time t1 and t2. We draw perpendiculars from points corresponding to the time t1 and t2 on the above graph.

The two perpendiculars enclose a quadrangle ABCD between the graph and the time axis. From the

graph, the sides AE and DC are equal to 10seconds while AD and CE is 16m/s. Also form the triangle, AEB, sideAE is 10s and Side BE is 16m/s. We know that the distance s covered by a body is given by the area under the quadrangle.

  1. s= Area of AEDC + Area of AEB

ix. = (AD x DC) + (1/2 x AE x BE) x. =(16 x 10) + (1/2 x 10 x 16)

xi. = 160 + 80

S= 240m

  1. Hence the distance covered is 240m.
  2. In the velocity time graph for uniform acceleration. The changes in the velocity of a car in specific time are given in the following table.

    1. In the above graph represents which type of motion?
    2. How much the velocity change in every 5 minutes?
    3. For all uniformly accelerated motions, what is the shape for velocity-time graph?
    4. For all non-uniformly accelerated motions, what is the shape for velocity-time graph?
    5. Calculate the distance covered by the car between the 10th and the 20th seconds.

Ans i. In the above graph, the velocity changes by equal amounts in equal time intervals. Thus this is an example of uniformly accelerated motion.

  1. The velocity changes by 8m/s in every 5 minutes.
  2. For all uniformly accelerated motions, the velocity-time graph is a straight line.

vi. For all non-uniformly accelerated motions, the velocity-time graph may have any shape depending on how the acceleration changes with time.

v. Distance s = average velocity × time

= (16+32/2)m/sx(20s -10s) 24m/s × 10s = 240 m

Ans. Distance covered by the car = 240m

  1. Law of conservation of momentum

Ans i. Suppose an object A has mass m1 and its initial velocity is u1. An object B has mass m2 and initial velocity u2.

  1. According to the formula for momentum, the initial momentum of A is m1u1 and that of B is m2u2.
  2. Suppose these two objects collide.
  3. Then, let the force on A due to B be F1. This force will cause acceleration in A and its velocity will become v1.
  4. Momentum of A after collision = m1v1

According to Newton’s third law of motion, A also exerts an equal force on B but in the opposite direction.

  1. This will cause a change in the momentum of B and its velocity after collision is v2. Thus, the momentum of B after collision= m2v2.
  2. If F2 is the force that acts on object B,
  3. F2= -F1
  4. m2a2 = -m1a1 (F=ma)

x. m2 x (v2-u2)/t = – m1 x (v1-u1)/t (a= (v-u)/t)

xi. m2 (v2-u2) = – m1 (v1-u1)

  1. m2v2 – m2u2 = – m1v1+ m1u1
  2. (m1v1 + m2v2 )= (m1u1 + m2u2)

The magnitude of total final momentum = The magnitude of total initial momentum. Thus, if no

  1. external force is acting on two objects, then their total initial momentum is equal to their total final momentum.

When no external force acts on two interacting objects, their total momentum remains constant. It does not change.

P H Dhakne
http://phdsciencegyan.com

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