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
FS = µ 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 of 20 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 =?
FS = µ R (where R = mg)
FS = µ mg
FS = 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
