 # Physics Numericals Class 9 Motion & other Chapters solved

Last updated on July 21st, 2023 at 11:01 am

Solved Numerical Problems in Physics Class 9 Motion & other chapters – This page is based on Physics Numericals Class 9. Find 101+ solved numerical problems, Questions & answers from class 9 motion & other chapters, for CBSE, ICSE, & UPSC.

This page contains solved numerical problems from Motion Physics (kinematics), and other topics and chapters as per CBSE & ICSE syllabus for class 9 physics. This superset of physics questions includes 3 worksheets to solve [links given at the end of this post] as well.

## Solved Numerical Problems in Physics class 9 Motion | Motion Numericals class 9

The next set of numerical questions (set 1) covers questions from the Motion chapter (kinematics) of class 9 physics and these can be easily solved using motion equations (Suvat equations).

1) A speedboat has an acceleration of 2 m/s2. What would the final velocity of the speedboat be after 5 seconds if the initial velocity of the speedboat is 4 m/s?

Solution:

a = 2 m/s2
u = 4 m/s

t = 5 s

final velocity = v

v = u + at = 4 + 2×5 = 14 m/s

2) A vehicle has an acceleration of 5 m/s2. What would the final velocity of the vehicle be after 10 seconds if the initial velocity of the vehicle is 20 m/s? What is the displacement of the vehicle during the 10 s time interval?

Solution:

a = 5 m/s2
u = 20 m/s

t = 10 s

final velocity = v

v = u + at = 20 + 5×10 = 70 m/s
final velocity = 70 m/s

Displacement during the 10 s time interval = s

s = ut + (1/2)at2

=> s = 20×10 + (½)x5x102

=> s= 200 + 250 = 450 m

Displacement during the 10 s time interval = 450 m

3) A child on a toboggan starts from rest and accelerates down a snow-covered hill at 0.8 m/s2. How long does it take the child to reach the bottom of the hill if it is 25.0 m away?

Solution: initial velocity u = 0
acceleration a = 0.8 m/s2
distance = s = 25 m
time taken t = ?

s= ut + (1/2)at2

As here, u = 0, so we can write:

s= (1/2)at2

=> t2 =(2 s /a ) = (2×25)/0.8 = 62.5
time t = 7.9 sec

4. A car accelerates uniformly from a velocity of 21.8 m/s to a velocity of 27.6 m/s. The car travels 36.5 m during this acceleration.

a) What was the acceleration of the car?

b) Determine the time interval over which this acceleration occurred.

Solution:

a) final velocity v = 27.6 m/s
initial velocity u = 21.8 m/s
distance traveled s=36.5 m
We will use this equation: v2 = u2 + 2as
a = ( v2 – u2) /(2s) = (27.62 – 21.82)/(2×36.5) = 3.92 m/s2
Acceleration = 3.92 m/s2

b) time interval over which this acceleration occurred = t
v = u + at
=> t = (v-u)/a = (27.6-21.8) / 3.92 = 1.48 seconds.

5. A motorcycle starts from rest and accelerates at +2.50 m/s2 for a distance of 150.0 m. It then slows down with an acceleration of –1.50 m/s2 until the velocity is +10.0 m/s. Determine the total displacement of the motorcycle.

Solution:

First part:
displacement = s = 150 m
We need to find out the final velocity(v) of this accelerating part.
v2 = u2 + 2as
=> v2 = 0 + 2×2.5×150 = 750
=> v = 27.4 m/s
This velocity becomes the initial velocity of the second part of the journey.

Second part:

u = 27.4 m/s
a = -1.5 m/s2
v= 10 m/s
v2 = u2 + 2as
102 = 27.42 + 2.(-1.5).s
=> 100 = 750 – 3s
s= 650/3 = 216.7 m
Second part displacement = 216.7 m

Considering a straight-line journey without a change of direction,
total displacement = (150 + 216.7) m = 366.7 m

## Physics Numericals class 9

So let’s start with sets of Physics Numericals Class 9 which contains Numerical problems & questions from various chapters of Class 9 physics with solutions & answers.

### Motion Numerical problems – on force, momentum & acceleration

1) A force of 1000 Newton is applied on a 25 kg mass for 5 seconds. What would be its velocity?

Solution:
F =1000 N
m=25kg
t= 5 sec
u=0
v=?

Acceleration = a = F/m = 1000/25 = 40 m/s

v = u + at =0 + 40X5 = 200 m/s

2) A force is applied to a mass of 16 kg for 3 seconds. As the force is removed the mass moves 81 meters in 3 seconds. What was the value of the force applied?

Solution:

First 3 secs: Force is present, and in the next 3 secs Force is not there, so uniform velocity for the last 3 seconds is mentioned.

In the last 3 seconds, the mass travels with uniform velocity (as there is no force that means no acceleration).

So the velocity in the next 3 secs:
V = 81/3 = 27 m/s

This velocity was attained in the first 3 seconds when the force was present.

So in the first 3 seconds, velocity changes from 0 to 27m/s.

Therefore the acceleration in the first 3 seconds caused by the force present = (27-0)/3 = 9 m/s².

So the force applied on the mass for first 3 secs= mass x acc = 16 x 9 = 144 N

3) A mass of 50 kg was moving with a velocity of 400 m/s. A force of 40000 N is applied to the mass and its velocity is reduced to 50 m/s after some time. What is the distance traveled by the mass during this period?

Solution:

Mass = 50 kg
u = 400 m/s
v= 50 m/s
F = 40000 N
t = unknown
Distance traveled in time t =?

Acceleration = a = F/m = 40000/50 = 800 m/s2

Again, acceleration a = change of velocity /t
or, t = change of velocity/a = 350/800 sec

Distance travelled: s = ut – (1/2)a t2

= 400 * (350/800) – (1/2) 800. (350/800)2
= 175 – 76.5 = 98.5 m

4) A ball of 150 g mass moves with 12 m/s and bounces back after hitting a wall with 20 m/s after a small duration of 0.01 seconds. What was the force applied to the ball when it hits the wall?

Solution:

Physics problems for class 9 with solutions – physics problems solved

Mass = 150 gram = 150/1000 kg = 0.15 kg

Change of velocity = 20 – (-12) = 32 m/s

T = 0.01 sec

Acceleration = change in velocity / time = 32/0.01 = 3200 m/s2

Force = mass X acc = 0.15 X 3200 N= 480 N.

5) Momentum of an object changes from 100 kg m/s to 200 kg m/s in 2 seconds. What is the force applied to it?

Solution:

Note:Force = ma = m(v-u)/t = (mv – mu)/t = change of momentum /time = rate of change of momentum

Force= change of momentum /time = (200-100)/2 = 100/2 = 50 N

6) A force of 200 N is applied to a body and its velocity changes from 5m/s to 10m/s in a second. What is the mass of the body?

Solution:

mass = force/acceleration
here acceleration =a = (10-5)/1 = 5 m/s2
so, mass  = 200/5 kg = 40 kg

7) A mass of 1 kg is moving from east to west with a velocity of 10 m/s. A force is applied to it for 2 seconds and its velocity becomes 5 m/s. What is the value and direction of the force?

Solution:

Mass = 1kg
velocity = 10 m/s (east to west)
Because of the force, the velocity reduces to 5 m/s in 2 secs.
acceleration = change of vel/time=5/2 = 2.5 m/s2

Force = mass X acceleration= 1 X 2.5 = 2.5 N.
As the velocity reduces because of force, (negative acc i.e. retardation) hence the direction of the force would be opposite to the initial direction. So the direction of the force will be west to east.

### gravitation – solved questions

8) If the distance between 2 objects is doubled then how will the gravitational force between them change?

Solution:

F1 = GMm/r2
F2 = GMm/(2r)2
= GMm/4r2
so F2/F1 = ¼ ,
so the final gravitational force will be ¼th of the initial gravitational force

9) If the distance between 2 objects is halved then how will the gravitational force between them change?

Solution:

F1 = GMm/r2
F2 = GMm/(r/2)2
=4 GMm/r2

so F2/F1 = 4/1
so the final gravitational force will be 4 times the initial gravitational force

10) If the masses of the 2 objects are doubled then how will the gravitational force between them change?

Solution:

F1 = GMm/r2
F2 = (G2M2m)/(r)2
=4 GMm/r2
so F2/F1 = 4/1
so the final gravitational force will be 4 times the initial gravitational force

### force, & acceleration questions – solved

[solutions: for Q11 to Q71 – solution links are just beside the question above. Click that link to reach the page with a solution.] This set is again based on force and acceleration.

11) A force acts for 10 s on a stationary body of mass 100 kg after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate (i) the velocity acquired by the body, (ii) the acceleration produced by the force, and (iii) the magnitude of the force.
[ Solution: click this link for solution Q11 ]

12) A force acts for 0.1 sec on a body of mass 2 kg initially at rest. The force is then withdrawn and the body moves with a velocity of 2 m/s. Find the magnitude of the force.
Solution: click this link for solution Q12

13) A cricket ball of mass 100 g moving at a speed of 30 m/s is brought to rest by a player in 0.03 s. find the average force applied by the player.
Solution: click this link for solution Q13

14) A body of mass 500 g, initially at rest, is acted upon by a force that causes it to move a distance of 4 m in 2 sec. Calculate the force applied.

Solution: click this link for solution Q14

15) A force causes an acceleration of 10 m/s^2 in a body of mass 500 g. What acceleration would be caused by the same force in a body of mass 5 kg.

Solution: click this link for solution Q15

16) A bullet of mass 50 g moving with an initial velocity of 100 m/s, strikes a wooden block and comes to rest after penetrating a distance 2 cm in it.
Calculate:
(i) Initial momentum of the bullet (ii)final momentum of the bullet (iii) retardation caused by the wooden block, and (iv) resistive force exerted by the wooden block

Solution: click this link for solution Q16

### Motion Numerical in physics class 9

Find another set of Physics Numerical for Class 9 Motion.

17) From the velocity-time graph, find out the distance traveled in each section A, B, C and D. Also find the total distance traveled.

Solution: click this link for solution Q17

18) A car starts from rest and accelerates uniformly over a time of 7 seconds for a distance of 98 m. Determine the acceleration of the car.

Solution: click this link for solution Q18

19) A car traveling at 20 m/s skids to a stop in 3 s. Determine the skidding distance of the car (assume uniform acceleration).

Solution: click this link for solution Q19

20) An airplane accelerates down a runway at 4 m/s^2 for 30 s until it finally lifts off the ground. Determine the distance traveled before takeoff.

Solution: click this link for solution Q20

21) A feather is dropped on the moon from a height of 10 meters. The acceleration of gravity on the moon is one-sixth of that on the earth. Determine the time for the feather to fall to the surface of the moon.

Solution: click this link for solution Q 21

22) If a rocket-powered sled is accelerated to a speed of 400 m/s in 2 seconds, then what is the acceleration, and what is the distance that the sled travels?

Solution: click this link for solution Q22

23) An engineer is designing the runway for an airport. Of the planes that will use the airport, the lowest acceleration rate is likely to be 3 m/s2. The takeoff speed for this plane will be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the runway?

Solution: click this link for solution Q 23

24) Upton Chuck is riding the Giant Drop at Great America. If Upton free falls for 2.60 seconds, what will be his final velocity and how far will he fall?

Solution: click this link for solution Q 24

25) A bike accelerates uniformly from rest to a speed of 10 m/s over a distance of 30 m. Determine the acceleration of the bike.

Solution: click this link for solution Q25

26) A race car accelerates uniformly from 25 m/s to 50 m/s in 5 seconds. Determine the acceleration of the car and the distance traveled.

Solution: click this link for solution Q 26

27) A bullet leaves a rifle with a muzzle velocity of 600 m/s. While accelerating through the barrel of the rifle, the bullet moves a distance of 0.8 m. Determine the acceleration of the bullet (assume uniform acceleration).

Solution: click this link for solution Q 27

28) A bullet is moving at a speed of 350 m/s when it embeds into a lump of moist clay. The bullet penetrates for a distance of 0.05 m. Determine the acceleration of the bullet while moving into the clay. (Assume a uniform acceleration.)

Solution: click this link for solution Q 28

29) A plane has a takeoff speed of 100 m/s and requires 1500 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.

Solution: click this link for solution Q29

30)A baseball is popped straight up into the air and has a hang-time of 6 s. Determine the height to which the ball rises before it reaches its peak. (Hint: the time to rise to the max height is one-half the total hang time.)

Solution: click this link for solution Q30

31) If a box is pushed horizontally on the floor with a force of 10 N and it moves 5 meters along the line of action of the force, then what is the work done by the Gravity or earth’s gravitational pull on that box?

Solution: click this link for solution Q31

1) physics Questions & Answers for class 9 CBSE, ICSE, state

2) Class 9 Numericals set 2 on Laws of Motion and Force – hand-picked problems,

32) A train is moving at a velocity of 25 ms-1. It is brought to rest by applying the brakes which produce uniform retardation of 0.5 ms-2.
Calculate
(i) the velocity of the train after 10 s
(ii) If the mass of the train is 20000 kg then calculate the force required to stop the train

Solution: click this link for solution Q32

33) A spring balance is used to find the weight of a body X on the surface of the moon. The mass of the body X is 2 kg and its weight is recorded as 3.4 N. The weight of another body Y recorded by the same balance is found to be 7.65 N. Calculate the mass of the body Y.

Solution: click this link for solution Q33

### gravitation numerical (set 2)

34) Calculate the force of gravitation between two objects of masses 50 kg and 120 kg respectively kept at a distance of 10 m from one another. (Gravitational constant, G = 6.7 × 10 -11 Nm^2/kg^2 )

Solution: click this link for solution Q34

35) What is the force of gravity on a body of mass 150 kg lying on the surface of the earth?
(Mass of earth = 6 × 10^24 kg; Radius of earth = 6.4 × 10^6 m;
G = 6.7 × 10 -11 Nm^2/kg^2)

Solution: click this link for solution Q35

36) The mass of sun is 2 × 10^30 kg and the mass of earth is 6 × 10^24 kg. If the average distance between the sun and the earth is 1.5 × 10^8 km, calculate the force of gravitation between them.

Solution: click this link for solution Q36

### Vertical motion numerical

The following set on physics numerical for class motion specifically covers vertical motion questions.

37) A piece of stone is thrown vertically upwards. It reaches the maximum height in 3 seconds. If the acceleration of the stone is 9.8 m/s^2 directed towards the ground, calculate the initial velocity of the stone with which it is thrown upwards.

Solution: click this link for solution Q37

38) A stone falls from a building and reaches the ground 2.5 seconds later. How high is the building? (g = 9.8 m/s^2)

Solution: click this link for solution Q38

39) A stone is dropped from a height of 20 m.(i) How long will it take to reach the ground? (ii) What will be its speed when it hits the ground? (g = 10 m/s^2)

Solution: click this link for solution Q39

40) A stone is thrown vertically upwards with a speed of 20 m/s. How high will it go before it begins to fall? (g = 9.8 m/s^2)

Solution: click this link for solution Q40

41) When a cricket ball is thrown vertically upwards, it reaches a maximum height of 5 meters. (a) What was the initial speed of the ball? (b) How much time is taken by the ball to reach the highest point? (g =10 m/ s^2)

Solution: click this link for solution Q41

42) A force of 20 N acts upon a body whose weight is 9.8 N. What is the mass of the body and how much is its acceleration? (g = 9.8 m/ s^2).

Solution: click this link for solution Q42

43) A stone resting on the ground has a gravitational force of 20 N acting on it. What is the weight of the stone? What is its mass? ( g = 10 m/s^2).

Solution: click this link for solution Q43

### Physics sum for class 9

44)
(a) An object has a mass of 20 kg on Earth.
What will be its (i) mass, and (ii) weight, on the moon ?(g on moon = 1.6 m/s^2).
(b) Calculate the acceleration due to gravity on the surface of a satellite having a mass of 7.4 × 10^22 kg and a radius of 1.74 × 10^6 m (G = 6.7 × 10 -11 Nm^2/kg^2).

Solution: click this link for solution Q44

45) The mass of a planet is 6 × 10^24 kg and its diameter is 12.8 × 10^3 km. If the value of the gravitational constant is 6.7 × 10 -11 Nm2/kg2, calculate the value of acceleration due to gravity on the surface of the planet.

Solution: click this link for solution Q45

46) The values of g at six distances A, B, C, D, E, and F from the surface of the earth are found to be 3.08 m/s^2, 9.23 m/s^2, 0.57 m/s^2, 7.34 m/s^2, 0.30 m/s^2 and 1.49 m/s^2, respectively.

(a) Arrange these values of g according to the increasing distances from the surface of the earth (keeping the value of g nearest to the surface of the earth first)

(b) If the value of distance F be 10000 km from the surface of the earth(hint: check if possible or not), state whether this distance is deep inside the earth or high up in the sky.

Solution: click this link for solution Q46

47) A pebble is dropped freely in a well from its top. It takes 20 s for the pebble to reach the water surface in the well. Taking g = 10 m/s^2 and speed of sound = 330 m/s, find (i) the depth of the water surface, and (ii) the time when the echo is heard after the pebble is dropped.
**[answer Hint: 2000 m ; 26.06 secs ]

Solution: click this link for solution Q47

48) A ball is thrown vertically upwards from the top of a tower with an initial velocity of 19·6 m/s. The ball reaches the ground after 5 s. Calculate: (i) the height of the tower, (ii) the velocity of ball on reaching the ground. Take g = 9·8 m/s^2
**[answer Hint: 24.5 m ; 29.4 m/s ]

Solution: click this link for solution Q48

### Class 9th – Numericals from Pressure chapter – CBSE, ICSE

49) A hammer exerts a force of 1·5 N on each of the two nails A and B. The area of the cross-section of the tip of nail A is 2 mm^2 while that of nail B is 6 mm^2. Calculate the pressure on each nail in Pascal.

Solution: click this link for solution Q49

50) A block of iron of mass 7·5 kg and of dimensions 12 cm x 8 cm x 10 cm is kept on a tabletop on its base of side 12 cm x 8 cm. Calculate (a) thrust and (b) pressure exerted on the tabletop. Take 1 kgf=10 N.

Solution: click this link for solution Q50

51) A vessel contains water up to a height of 1·5 m. Taking the density of water 1()^3 kg /m^3, acceleration due to gravity 9·8 m/ S^2, and area of the base of vessel 100 cm^2, calculate: (a) the pressure and (b) the thrust, at the base of the vessel.

Solution: click this link for solution Q51

52) The area of the base of a cylindrical vessel is 300 cm^2. Water (density = 1000 kg /m^3) is poured into it up to a depth of 6 cm. Calculate (a) the pressure and (b) the thrust of water on the base. (g = 10 m/ s^2).

Solution: click this link for solution Q52

53)Calculate the height of a water column that will exert on its base the same pressure as the 70 cm column of mercury. The density of mercury is 13.6 g/cm^3.

Solution: click this link for solution Q53

### Numerical problems in physics class 9 – on Conservation of momentum

54) A boy of mass 50 kg running at 5 m/s jumps onto a 20 kg trolley traveling in the same direction at 1.5 m/s. What is their common velocity?

Solution: click this link for solution Q54

55) A girl of mass 50 kg jumps out of a rowing boat of mass 300 kg onto the bank, with a horizontal velocity of 3 m/s. With what velocity does the boat begin to move backward?

Solution: click this link for solution Q55

56)A truck of mass 500 kg moving at 4 m/s collides with another truck of mass 1500 kg moving in the same direction at 2 m/s. What is their common velocity just after the collision if they move off together?

Solution: click this link for solution Q56

57) A ball X of mass 1 kg traveling at 2 m/s has a head-on collision with an identical ball Y at rest. X stops and Y moves off. Calculate the velocity of Y after the collision.

Solution: click this link for solution Q57

58) A heavy car A of mass 2000 kg traveling at 10 m/s has a head-on collision with a sports car B of mass 500 kg. If both cars stop dead on colliding, what was the velocity of car B?

Solution: click this link for solution Q58

59) A man wearing a bulletproof vest stands still on roller skates. The total mass is 80 kg. A bullet of mass 20 grams is fired at 400 m/s. It is stopped by the vest and falls to the ground. What is then the velocity of the man?

Solution: click this link for solution Q59

60) A metallic sphere of mass 2 kg and volume 2.5 x 10-4 m3 is completely immersed in water. Find the buoyant force exerted by water on the sphere. Density of water = 1000 kg/m3

Solution: click this link for solution Q60

### pendulum problems

61) Two simple pendulums A and B have equal lengths, but their bobs weigh 50 gf and l00 gf respectively. What would be the ratio of their time periods? Give a reason for your answer.

Solution: click this link for solution Q61

62) Two simple pendulums A and B have lengths of 1·0m and 4·0 m respectively at a certain place. Which pendulum will make more oscillations in 1 minute? Explain your answer.

Solution: click this link for solution Q62

63)A simple pendulum completes 40 oscillations in one minute. Find its (a) frequency, (b) time period.

Solution: click this link for solution Q63

64) The time period of a simple pendulum is 2 s. What is its frequency? What name is given to such a pendulum?

Solution: click this link for solution Q64

65) A seconds’ pendulum is taken to a place where the acceleration due to gravity falls to one-fourth. How is the time period of the pendulum affected, if at all? Give reason. What will be its new time period?

Solution: click this link for solution Q65

66) Find the length of a seconds’ pendulum at a place where g = 10 m/s2 (Take  ∏= 3·14).

Solution: click this link for solution Q66

67) A pendulum completes 2 oscillations in 5 s. (a) What is its time period? (b) If g = 9·8 m /S^2, find its length.

Solution: click this link for solution Q67

68) Compare the time periods of two pendulums of length 1 m and 9 m

Solution: click this link for solution Q68

69) The time periods of two simple pendulums at a place are in a ratio of 2:1. What will be the ratio of their length?

Solution: click this link for solution Q69

70) It takes 0·2 s for a pendulum bob to move from the mean position to one end. What is the time period of the pendulum?

Solution: click this link for solution Q70

71) How much time does the bob of a seconds’ pendulum take to move from one extreme of its oscillation to the other extreme?

Solution: click this link for solution Q71

Related: worksheets

Motion worksheet (worksheet 2)

Mirrors & images numerical (worksheet 3)

We have 30 more numerical questions of this series for class 9 physics – these are listed as worksheets in the next 3 pages [links are given below]