Physics questions and answers for class 9 cbse icse

Last updated on July 13th, 2023 at 10:15 am

Physics questions and answers for class 9 for CBSE and ICSE boards – selected Numericals and fundamental questions are being published under this super set.

Assignments

1.) A ball is thrown vertically upwards. It goes to a height of 20 m and then returns to the ground. Taking acceleration due to gravity g to be 10 m S-2, find :

(a) the initial velocity of the ball

(b) the final velocity of the ball on reaching the ground and

Solution:
The maximum height reached 20 m
h = 20 m g = 10 m/s^2

h = u^2/(2g)
i.e., the initial velocity with which it was thrown upwards = u = √(2gh)
So here u = √(2 * 10 * 20) = 20 meter/sec

the final velocity of the ball on reaching ground v
v = √(2gh= √(2 * 10 * 20) = 20 meter/sec

Total time of journey= time for upward journey + time for downward journey = 2 u/g,
Here u is the initial velocity with which the ball was thrown upwards.
T = 2 u/g = 2 *20/10 = 4 secs

2.) A body is dropped from the top of a tower. It acquires a velocity 20 m S-l on reaching the ground. Calculate the height of the tower.

(Take g = 10 m S-2)

solution:
final velocity during the downward movement =V= 20 m/s

Here height h = (V^2)/(2g) = (20*20)/(2*10) m= 20 m

3.) A ball is thrown vertically upwards. It returns 6 s later. Calculate (i) the greatest height reached by the ball, and (ii) the initial velocity of the ball. (Take g = 10 m s-2)

Solution:
(i) Max or greatest height = H = (u ^2) /(2g)……….(a)

We have to find out the value of the initial velocity of upward movement first.
The time to reach the max height and then come back is given as t=6 sec.
So time to reach max height = 6/2 = 3 secs

From the equation of upward movement, v = u – gt, where v =0 we get u = gt.
So the value of initial velocity u = gt = 10.3 m= 30 m/s

from equation (a) now we get the max height = H = (30*30)/(2*10) m= 45 m

(ii) Initial velocity we have already found out as 30 m/s

4.) A pebble is thrown vertically upwards with a speed of 20 m/s. How high will it be after 2 s? (Take g = 10 m S-2)

5.) (a) How long will a stone ground from the top of a building 80 m high and (b) what will be the velocity of the stone on reaching the ground? (Take g = 10 m S-2)

6.) A body falls from the top of a building and reaches the ground 2·5 s later. How high is the building? (Take g = 9·8 m S-2)

7.) A ball is thrown vertically upwards with an initial velocity of 49 m/s. Calculate :

(i) the maximum height attained, (ii) the time taken by it before it reaches the ground again.(Take g = 9·8 m S-2).

8.) A stone is dropped freely from the top of a tower and it reaches the ground in 4 s. Taking g = 10 m S-2, calculate the height of the tower.

9) What is the relationship between displacement and time in uniform motion?

10) What do we get from the slope of displacement – time graph?

11) What do we get from the slope of the velocity-time graph?

12) Velocity-time graphs of 2 cars A and B are analyzed. Car A has a steeper slope than that of car B. What does this mean?

13) How can we get displacement from the velocity-time graph?

14) What is linear motion?

15) When is the distance traveled equal to the magnitude of displacement?

16) give one example where displacement is zero while distance traveled is non-zero positive.

17) Can we get the first law of motion from the second law of motion? Briefly show how.

18) What is the relation between the rate of change of momentum of a system and the force applied to that system?

19) Show the forces acting on a simple pendulum bob suspending vertically with a string.

20) Mass is the measure of inertia. Briefly explain.

21) What does G stand for? What is the unit, dimension, and value of G? [hint: chapter Gravitation]

22) A 10 kg block is resting on a plain surface. What is the thrust applied on the surface?

23) If the surface area is doubled, what can we do to retain the same pressure on the surface?

24) Can you derive the equation of force from Newton’s 2nd law of motion? If yes, pls derive it.

25) What is the difference between Kgf and Kg?

26) What is the relation between Kgf and Newton? What is the relation between Newton and Dyne?

27) Define the least count? What is the least count of the ruler you generally use?

28) Can you differentiate between g and G ? Which one will change if we move to the moon?

29) Can you draw a velocity-time graph for non-uniform acceleration?

30) If we throw a ball vertically upwards with a speed of 90 m/s then what will be its speed just before touching the ground? What is the total time taken for the to and fro flight? g =10 m/s^2.

31) Can we get the value of acceleration from the velocity-time graph? If yes, how?

Answer hint: yes, from the slope of the graph

32) Can we get the value of distance travelled from the velocity-time graph? If yes, how?

Answer hint: yes, from the area covered by the graph

33) A train travels at a speed of 60 km/hr for 0·52 hr, at 30 km/h for the next 0·24 hr, and then at 70 km/h for the next 0·71h. What is the average speed of the train?

Solution:

(i) In the first case, the train travels at a speed of
60 km/h for a time of 0·52 h.
Now, Speed =distance/ time
Distance = speed x time =60 × 0·52 = 31·2 km …(1)

(ii) In the second case, the train travels at a speed
of 30 km/h for a time of 0·24 hr.
Distance = = speed x time =30 × 0·24 = 7·2 km …(2)

(iii) In the third case, the train travels at a speed of
70 km/h for a time of 0·71 hr.
Distance = = speed x time = 70 × 0·71 = 49·7 km …(3)

From equations 1, 2 & 3 we get,
Total distance travelled = (31·2 + 7·2 + 49·7) km = 88·1 km.
Total time taken = (0·52 + 0·24 + 0·71) hr =1·47 hr.

Average speed =Total distance travelled/Total time taken= (88·1/1.47) km/hr
So, Average speed = 59·9 km/hr

34) Which law of motion can best describe the following :
(a) Force applied while pulling a lawn mower.
(b) Shoulder bone fracture from the recoil of a gun.
(c) Coin remains on the table when the table cloth is suddenly removed.

Answer:
(a) According to Newton’s II law of motion
(b) According to Newton’s III law of motion
(c) According to Newton’s I law of motion.

35) State three characteristics of action-reaction
forces.

Ans. (i) Equal in magnitude.
(ii) Opposite in direction.
(iii) Acts simultaneously on two different bodies.

36)A scooter is moving with a velocity of 25 m/s and it
takes 5 s to stop after the brakes are applied. If the
mass of the scooter along with the rider is 180 kg,
find the change in momentum in this case.

Solution:
u = 25 m/s, v = 0, m = 180 kg
t = 5 s (this time data is not required here)
Initial momentum = p1 = mu = 180 × 25
= 4500 kg-m/s 1
Final momentum = p2 = mv = 180 × 0 = 0 kgm/s 1
So change in momentum= p2– p1 = 0 – 4500
= – 4500 kg-m/s

37) Look at the diagram above and answer the following questions :

(a)
(i) When a force is applied through the free end of the spring balance A the reading on the spring balance A is 20 gwt. What will be the reading shown by the spring balance B?
(ii) Write reasons for your answer.

(b)
If a balloon is filled with air and its mouth untied, the air is released from its mouth in the downward direction. Write the other observations made by you. Justify your
answer.

(a)
(i) Spring balance B will also show 20 gwt.

(ii) It is because of Newton’s Third law of motion. When spring balance A exerts a pulling force on balance B then balance B also pulls balance A with an equal force of 20 gwt, but in the opposite direction.
Balance A exerts a force of action on balance B then balance B exerts an equal and opposite force of reaction on balance A.

(b) It is observed that the balloon moves in an upward direction as air comes out from the balloon. As the air comes out in the downward direction, it performs an action in the downward direction. As a result, there will be a reaction (equal in magnitude but opposite in direction) in the form of the upward motion of the balloon.

many more questions are coming …watch this place

for physics problems with ready solutions-class 9 (link below)
selected set for class 9 -Motion