Class 9 - Unit # 3 : Dynamics - Solved Numericals

 

Physics Numerical Sheet

Unit: Dynamics (Class 9 / Matriculation System)

Momentum

1. b) Find the momentum of a body of mass $6\text{ kg}$ moving with a velocity of $25\text{ ms}^{-1}$.
c) What will be its velocity if the momentum becomes $200\text{ Ns}$?

Data:

Part (b):

• Mass of the body ($m$) = $6\text{ kg}$
• Velocity ($v$) = $25\text{ ms}^{-1}$
• Momentum ($P$) = ?

Part (c):

• New Momentum ($P'$) = $200\text{ Ns}$
• New Velocity ($v'$) = ?

Solution:

b) According to the definition of momentum:

$$P = m \times v$$ $$P = 6\text{ kg} \times 25\text{ ms}^{-1}$$ $$P = 150\text{ Ns}$$

c) Now momentum becomes $200\text{ Ns}$, so:

$$P' = m \times v' \implies v' = \frac{P'}{m}$$ $$v' = \frac{200\text{ Ns}}{6\text{ kg}}$$ $$v' = 33.33\text{ ms}^{-1}$$

RESULT: b) The momentum of the body is $150\text{ Ns}$. c) The velocity of the body will be $33.33\text{ ms}^{-1}$.

2. A body of mass $10\text{ kg}$ is moving with a velocity of $10\text{ ms}^{-1}$. A force acts for $5\text{ seconds}$ to reduce its velocity to $2\text{ ms}^{-1}$. Find the momentum of the body before and after the application of the force on it.

Data:

• Mass of the body ($m$) = $10\text{ kg}$
• Initial Velocity ($v_i$) = $10\text{ ms}^{-1}$
• Final Velocity ($v_f$) = $2\text{ ms}^{-1}$
• Time interval ($t$) = $5\text{ s}$
• Initial Momentum ($P_i$) = ?
• Final Momentum ($P_f$) = ?

Solution:

According to the definition of momentum ($P = mv$):

Initial Momentum ($P_i$) is calculated as:

$$P_i = m \times v_i$$ $$P_i = 10\text{ kg} \times 10\text{ ms}^{-1}$$ $$P_i = 100\text{ Ns}$$

Final Momentum ($P_f$) is calculated as:

$$P_f = m \times v_f$$ $$P_f = 10\text{ kg} \times 2\text{ ms}^{-1}$$ $$P_f = 20\text{ Ns}$$

RESULT: The initial momentum of the body is $100\text{ Ns}$ and the final momentum is $20\text{ Ns}$.

Laws of Motion

5. b) A force of $3400\text{ N}$ is applied on a body of mass $850\text{ kg}$. Find the acceleration produced by the force.
c) How much force should be applied on a body of mass $425\text{ kg}$ to produce the same acceleration calculated in part (b)?

Data:

Part (b):

• Mass ($m_b$) = $850\text{ kg}$
• Force ($F_b$) = $3400\text{ N}$
• Acceleration ($a$) = ?

Part (c):

• Mass ($m_c$) = $425\text{ kg}$
• Required Acceleration ($a$) = same as part b
• Force ($F_c$) = ?

Solution:

b) According to Newton's Second Law of Motion ($F = ma$):

$$a = \frac{F_b}{m_b}$$ $$a = \frac{3400\text{ N}}{850\text{ kg}}$$ $$a = 4\text{ ms}^{-2}$$

c) Now mass of the body becomes $425\text{ kg}$ keeping acceleration constant:

$$F_c = m_c \times a$$ $$F_c = 425\text{ kg} \times 4\text{ ms}^{-2}$$ $$F_c = 1700\text{ N}$$

RESULT: b) The acceleration produced is $4\text{ ms}^{-2}$. c) The force that should be applied is $1700\text{ N}$.

6. b) Find the mass of a body which is accelerated by applying a force of $200\text{ N}$, that speeds it up at $36\text{ ms}^{-2}$.
c) What should be the acceleration of the same body if the applied force changes to $280\text{ N}$?

Data:

Part (b):

• Force ($F_1$) = $200\text{ N}$
• Acceleration ($a_1$) = $36\text{ ms}^{-2}$
• Mass ($m$) = ?

Part (c):

• New Force ($F_2$) = $280\text{ N}$
• New Acceleration ($a_2$) = ?

Solution:

b) According to Newton's Second Law ($F = ma$):

$$m = \frac{F_1}{a_1}$$ $$m = \frac{200\text{ N}}{36\text{ ms}^{-2}} = \frac{50}{9} \approx 5.56\text{ kg}$$

c) Now force changes to $280\text{ N}$ for the same body ($m = \frac{50}{9}\text{ kg}$):

$$F_2 = m \times a_2 \implies a_2 = \frac{F_2}{m}$$ $$a_2 = \frac{280}{\left(\frac{50}{9}\right)} = \frac{280 \times 9}{50} = \frac{252}{5} = 50.4\text{ ms}^{-2}$$

RESULT: b) The mass of the body is $5.56\text{ kg}$. c) The new acceleration becomes $50.4\text{ ms}^{-2}$.

7. An empty car has a mass of $1200\text{ kg}$. Its engine can produce an acceleration of $4\text{ ms}^{-2}$. If a $300\text{ kg}$ load is added by passengers and luggage, what acceleration will the same engine produce?

Data:

• Mass of empty car ($m_1$) = $1200\text{ kg}$
• Initial acceleration ($a_1$) = $4\text{ ms}^{-2}$
• Total mass with load ($m_2$) = $1200\text{ kg} + 300\text{ kg} = 1500\text{ kg}$
• New acceleration ($a_2$) = ?

Solution:

First, calculate the constant accelerating force ($F$) exerted by the engine:

$$F = m_1 \times a_1$$ $$F = 1200\text{ kg} \times 4\text{ ms}^{-2} = 4800\text{ N}$$

Since the force from the engine remains constant, the new acceleration ($a_2$) for the loaded mass is:

$$a_2 = \frac{F}{m_2}$$ $$a_2 = \frac{4800\text{ N}}{1500\text{ kg}} = 3.2\text{ ms}^{-2}$$

RESULT: The acceleration produced by the engine with the added load is $3.2\text{ ms}^{-2}$.

8. The mass of an object is $60\text{ kg}$. Find its weight on (i) Earth, (ii) Moon, and (iii) Mars.
Assume the acceleration due to gravity is: $g_{\text{earth}} = 9.8\text{ ms}^{-2}$, $g_{\text{moon}} = 1.6\text{ ms}^{-2}$, and $g_{\text{mars}} = 3.7\text{ ms}^{-2}$.

Data:

• Mass of the object ($m$) = $60\text{ kg}$
• Gravity on Earth ($g_e$) = $9.8\text{ ms}^{-2}$
• Gravity on Moon ($g_m$) = $1.6\text{ ms}^{-2}$
• Gravity on Mars ($g_a$) = $3.7\text{ ms}^{-2}$

Solution:

According to the definition of Weight ($w = mg$):

(i) On the surface of Earth:

$$w_e = m \times g_e$$ $$w_e = 60\text{ kg} \times 9.8\text{ ms}^{-2} = 588\text{ N}$$

(ii) On the surface of Moon:

$$w_m = m \times g_m$$ $$w_m = 60\text{ kg} \times 1.6\text{ ms}^{-2} = 96\text{ N}$$

(iii) On the surface of Mars:

$$w_a = m \times g_a$$ $$w_a = 60\text{ kg} \times 3.7\text{ ms}^{-2} = 222\text{ N}$$

RESULT: The weight of the object is $588\text{ N}$ on Earth, $96\text{ N}$ on the Moon, and $222\text{ N}$ on Mars.

Circular Motion

9. A car is running on a circular part of a highway having a $1000\text{ m}$ radius. The mass of the car is $600\text{ kg}$ and its velocity is $72\text{ km/h}$. Find (i) Centripetal force exerted on the car, and (ii) Centripetal acceleration of the car.

Data:

• Radius of highway ($r$) = $1000\text{ m}$
• Mass of car ($m$) = $600\text{ kg}$
• Velocity ($v$) = $72\text{ km/h} = \frac{72 \times 1000\text{ m}}{3600\text{ s}} = 20\text{ ms}^{-1}$
• Centripetal Force ($F_c$) = ?
• Centripetal Acceleration ($a_c$) = ?

Solution:

(i) According to the definition of Centripetal Force:

$$F_c = \frac{m v^2}{r}$$ $$F_c = \frac{600\text{ kg} \times (20\text{ ms}^{-1})^2}{1000\text{ m}}$$ $$F_c = \frac{600 \times 400}{1000} = 240\text{ N}$$

(ii) According to the definition of Centripetal Acceleration:

$$a_c = \frac{v^2}{r}$$ $$a_c = \frac{(20\text{ ms}^{-1})^2}{1000\text{ m}} = \frac{400}{1000} = 0.4\text{ ms}^{-2}$$

RESULT: The centripetal force is $240\text{ N}$ and the centripetal acceleration is $0.4\text{ ms}^{-2}$.

Friction

10. A block is placed on a wet slippery floor. The mass of the block is $15\text{ kg}$. When it is pulled through a string using a spring balance, it shows a limiting friction force equal to $3\text{ N}$. Find the coefficient of friction.

Data:

• Mass of block ($m$) = $15\text{ kg}$
• Limiting Friction Force ($F_s$) = $3\text{ N}$
• Acceleration due to gravity ($g$) = $9.8\text{ ms}^{-2}$ (standard baseline)
• Coefficient of friction ($\mu$) = ?

Solution:

The equation for static friction is given by:

$$F_s = \mu R = \mu mg$$

Isolating the coefficient of friction ($\mu$):

$$\mu = \frac{F_s}{m \times g}$$ $$\mu = \frac{3\text{ N}}{15\text{ kg} \times 9.8\text{ ms}^{-2}}$$ $$\mu = \frac{3}{147} \approx 0.0204$$

Note: If the textbook curriculum problem rounds gravity $g \approx 10\text{ ms}^{-2}$ for simplicity, the calculation yields $\mu = \frac{3}{15 \times 10} = 0.02$. Both answers are fundamentally matching.

RESULT: The coefficient of friction between the block and the slippery floor is $0.02$.