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 = *2*m _{1}*m

_{2}g / (m

_{1}+ m

_{2})

a = *(m _{1} – m_{2})*g / (m

_{1}+ m

_{2})

*T = m _{1}*m

_{2}g / (m

_{1}+ m

_{2})

a = *m _{1}*g / (m

_{1}+ m

_{2})

F = (P_{f} – P_{i}) / t

F_{S }= *µ* R = *µ*mg (where R = mg)

F_{c} = mv^{2}/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)**

^{ – 2}. What is its mass?

**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)**

^{ – 2 })

**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 })

^{– 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)**

^{– 2}?

**Solution: **

Weight = w = 20 N

Acceleration = a = 2 ms ^{– 2}

Vertically upward force (Tension) = T =?

F_{net } = 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 })

^{– 2 })

**Solution: **

m_{1} = 52 kg and m_{2} = 48 kg

Tension = T = ?

Acceleration = a =?

*T = *2*m _{1}*m

_{2}g / (m

_{1}+ m

_{2})

*T = *2 x 52 x 48 x 10 / (52 + 48)

T = 49920 / 100

T = 499.20 ≈ 500 N

a = *(m _{1} – m_{2})*g / (m

_{1}+ m

_{2})

** **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})

^{– 2})

**Solution: : **

m_{1} = 24 kg and m_{2} = 26 kg

Tension = T =?

Acceleration = a =?

*T = m _{1}*m

_{2}g / (m

_{1}+ m

_{2})

*T = *24 x 26 x 10 / 24 + 26

T = 6240 / 50

T = 124.8 ≈ 125 N

To find a:

a = *m _{1}*g / (m

_{1}+ m

_{2})

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 = P_{f} – P_{i} = 22 Ns

Force = F = 20 N

Time = t = **?**

F = (P_{f} – P_{i}) / t

t = (P_{f} – P_{i}) / 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 = F_{S} =?

F_{S }= *µ* R (where R = mg)

F_{S }= *µ* mg

F_{S }= 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)**

^{– 1}?

**Solution: **

Mass = m = 0.5 kg

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

Speed = v = 3 ms ^{– 1}

Centripetal force = F_{c} = **?**

F_{c} = mv^{2}/r

F_{c} =0.5 x 3^{2} / 0.5

F_{c} = 9N

**Class 9 Physics Full Book Numerical Problems with Solution**