# Class 9 Physics Chapter 3 Numerical Problems

Class 9 Physics Chapter 3 numerical problems are according to the new syllabus of the Punjab Board. Chapter 3 is related to Dynamics which is the branch of Physics that deals with the study of motion with discussing its effects.

## Class 9 Physics Chapter 3 Numerical Problems

The problems have been solved with easy-to-use methods using the following formulae:

F = ma

ma = T – mg

T = 2m1m2g / (m1 + m2)

a = (m1 – m2)g / (m1 + m2)

T = m1m2g / (m1 + m2)

a =  m1g / (m1 + m2)

F = (Pf – Pi) / t

Fµ R = µmg                          (where R = mg)

Fc = mv2/r

### Numerical Problems Chapter 3 – Dynamics

#### 3.1 A force of 20 N moves a body with an acceleration of 2 ms – 2. What is its mass?(10kg)

Solution:

Force = F = 20 N

Acceleration = a = 2 ms – 2

Mass = m = ?

F = ma

Or

m = F / a

m = 20 / 2

m = 10 kg

#### 3.2 The weight of a body is 147 N. What is its mass? (Take the value of g as 10 ms – 2 ) (14.7 kg)

Solution:

Weight = w = 147 N

Acceleration due to gravity = g = 10 ms – 2

Mass = m =?

w = mg

or

m = w / g

m = 147 / 10

m = 14. 7 kg

#### 3.3 How much force is needed to prevent a body of mass 10 kg from falling?(100 N)

Solution:

Mass = m = 50 kg

Acceleration = a = g = 10 ms – 2

Force = F =?

F = ma

F = 10 x 10

F = 100 N

#### 3.4 Find the acceleration produced by a force of 100 N in a mass of 50 kg. (2 ms – 2 )

Solution:

Force = F = 100 N

Mass = m = 50 kg

Acceleration = a =?

F = ma

Or

a = F / m

a = 100 / 50

a = 2 ms – 2

#### 3.5 A body has a weight of20 N. How much force is required to move it vertically upward with an acceleration of 2 ms – 2?(24 N)

Solution:

Weight = w = 20 N

Acceleration = a = 2 ms – 2

Vertically upward force (Tension) = T =?

Fnet  = T – w

Or                     ma = T – mg

Or                     ma + mg = T

Or                     T = m (a + g) ……………………(i)

Now,                 m = w / g

m = 20/10

m = 2 kg

Putting the value of m in Eq.(i), we get

T = 2(2 + 10)

T = 2(12)

T = 24 N

#### 3.6 Two masses 52 kg and 48 kg are attached to the ends of a string that passes over a frictionless pulley. Find the tension in the string and acceleration in the bodies when both the masses are moving vertically.(500 N, 0.4 ms – 2  )

Solution:

m1 = 52 kg       and       m2 = 48 kg

Tension = T = ?

Acceleration = a =?

T = 2m1m2g / (m1 + m2)

T = 2 x 52 x 48 x 10 / (52 + 48)

T = 49920 / 100

T = 499.20 ≈ 500 N

a = (m1 – m2)g / (m1 + m2)

a =   (52 – 48) x 10 / (52 + 48)

a =   40 / 100

a  = 0.4 ms – 2

#### 3.7 Two masses 26 kg and 24 kg are attached to the ends of a string which passes over a frictionless pulley. 26 kg is lying over a smooth horizontal table. 24 N mass is moving vertically downward. Find the tension in the string and the acceleration in the bodies.  (125 N, 4.8 ms – 2)

Solution: :

m1 = 24 kg       and       m2 = 26 kg

Tension = T =?

Acceleration = a =?

T = m1m2g / (m1 + m2)

T = 24 x 26 x 10 / 24 + 26

T = 6240 / 50

T = 124.8 ≈ 125 N

To find a:

a =  m1g / (m1 + m2)

a = 24 x 10 / 24 + 26

a = 240 / 50

a  = 4.8 ms – 2

#### 3.8 How much time is required to change 22 Ns momentum by a force of 20 N?(1.1s)

Solution:

Change in momentum = Pf – Pi = 22 Ns

Force = F = 20 N

Time = t = ?

F = (Pf – Pi) / t

t = (Pf – Pi) / N

t = 22 / 20

t =  = 1.1 S

#### 3.9 How much is the force of friction between a wooden block of mass 5 kg and the horizontal marble floor? The coefficient of friction between wood and marble is 0.6.(30 N)

Solution:

Mass = m = 5 kg

Coefficient of friction = µ = 0.6

Force of friction = FS =?

Fµ R                               (where R = mg)

Fµ mg

F= 0.6 x 5 x 10 = 30 N

#### 3.10 How much centripetal force is needed to make a body of mass 0.5 kg to move in a circle of radius 50 cm with a speed of 3 ms – 1?(9 N)

Solution:

Mass = m = 0.5 kg

Radius of the circle = r = 50 cm =  = 0.5 m

Speed = v = 3 ms – 1

Centripetal force = Fc = ?

Fc = mv2/r

Fc =0.5 x 32 / 0.5

Fc = 9N

Class 9 Physics Full Book Numerical Problems with Solution

#### By Ahsa.Pk

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