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Physics problems for class 9 – numericals with solutions


Physics problems for class 9 (CBSE)–numericals set1

This set of Physics problems for class 9 students of CBSE/ICSE/HS boards cover Newton’s Laws of motion, Force, linear motion equations and Gravitation. Hope the students will like this set and try to solve the numerical problems themselves. I am providing the solutions as well, but you should try first yourself. the difficulty level is medium. But this set of Physics problems for class 9 will help to test your basics.

Physics BLOG for K12, High School-discussions and problem solving

physics problems and questions - solution

physics problems and questions – solution

 

numerical problems in Physics for class 9


Physics problems for class 9- set 1 Q

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

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

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

Physics problems .. continued

 

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 second. What was the force applied on the ball when it hits the wall?

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

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

Physics problems.. continued

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

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

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

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

 

Solutions- Physics problems for class 9
(CBSE/ICSE/state/etc) set 1

physics problems for class 9 solutions

Physics problems for class 9 – solutions

 

1) 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

Physics problems for class 9 – solutions: continued

 

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

Solution:

first 3 secs (Force is present)  –> next 3 secs (F is not there, so uniform velocity)

In the next 3 seconds the mass travels with uniform velocity (as there is no force à 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 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

Physics problems for class 9 – solutions: continued

 

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

Solution:

Mass = 50 kg
u = 400 m/s
v= 50 m/s
F = 40000 N
t = unknown
Distance travelled 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

Physics problems for class 9 – solutions: continued

 

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 second. What was the force applied on the ball when it hits the wall?

Solution:

physics problems numericals solutions                                                                                           Physics problems for class 9 with solutions

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.

Physics problems for class 9 – solutions: continued

 

5) Momentum of an object changes from 100 kgm/s to 200 kgm/s in 2 seconds. What is the force applied on 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

Physics problems for class 9 with solutions

6) A force of 200 N is applied on 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/acc.
here acc =a = (10-5)/1 = 5 m/s2
so, mass  = 200/5 kg = 40 kg

Physics problems for class 9 – solutions: continued

 

7) A mass of 1 kg is moving from east to west with a velocity 10 m/s. A force is applied on 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 acc = 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 direction of force will be west to east.

Physics problems for class 9 – solutions: continued

 

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 = ¼so the final gravitational force will be ¼ th 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 = 4/1so the final gravitational force will be 4 times of 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 of the initial gravitational force

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