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Numericals on Force and Laws of Motion for Class 9
Here you will find numericals on force and laws of motion for class 9.
Problem 1:
Find the force acting on a rocket and its acceleration if its velocity is 600 m s–1 at t = 60 s and 1000 ms–1 at t = 80s, presuming its mass remains unchanged during this time interval and its mass is 1.67 × 107 kg. The total weight of the rocket at t = 80 seconds is 1.6 × 107 kg. Calculate the upward force acting on the rocket at this time caused by the burning of fuel.
Solution:
(i) Calculate the acceleration (\(a\)) and force (\(F\)) at \(t=60\) seconds: \[ \begin{aligned} a &= \frac{v-u}{t} = \frac{1000 \mathrm{~m/s} – 600 \mathrm{~m/s}}{80 \mathrm{~s} – 60 \mathrm{~s}} = \frac{400 \mathrm{~m/s}}{20 \mathrm{~s}} = 20 \mathrm{~m/s^2} \\ F &= m \cdot a = 1.67 \times 10^7 \mathrm{~kg} \times 20 \mathrm{~m/s^2} = 3.34 \times 10^8 \mathrm{~Newton} \end{aligned} \] (ii) Calculate the force (\(F\)) at \(t=80\) seconds using the new total weight: \[ F = m \cdot a = 1.6 \times 10^7 \mathrm{~kg} \times 20 \mathrm{~m/s^2} = 3.2 \times 10^8 \mathrm{~Newton} = 320 \mathrm{~MN} \]Problem 2:
A body of 10 kg initially at rest is subjected to force of 20N. Calculate the kinetic energy acquired by the body at the end of 10 seconds.
Solution:
Given: \(m = 10 \mathrm{~kg}\), \(u = 0\), \(F = 20 \mathrm{~N}\), \(t = 10 \mathrm{~s}\) First, calculate the acceleration (\(a\)) using Newton’s second law: \[ F = m \cdot a \Rightarrow a = \frac{F}{m} = \frac{20 \mathrm{~N}}{10 \mathrm{~kg}} = 2 \mathrm{~m/s^2} \] Next, calculate the final velocity (\(v\)) using the equation of motion: \[ a = \frac{v – u}{t} \Rightarrow \frac{v – 0}{10 \mathrm{~s}} = 2 \mathrm{~m/s^2} \Rightarrow v = 20 \mathrm{~m/s} \] Finally, calculate the kinetic energy (\(KE\)) using the formula for kinetic energy: \[ KE = \frac{1}{2} m v^2 = \frac{1}{2} \cdot 10 \mathrm{~kg} \cdot (20 \mathrm{~m/s})^2 = 2000 \mathrm{~J} = 2 \mathrm{~kJ} \] So, the kinetic energy acquired by the body at the end of 10 seconds is \(2 \mathrm{~kJ}\).Problem 3:
A truck starts from the rest and rolls down a hill with constant acceleration. It travels 200 m in 20 s. Find its acceleration. Find the force acting on it, if its mass is 5 metric tonnes. (1 metric tonne = 1000 kg)
Solution:
Given: \(u = 0\), \(t = 20 \mathrm{~s}\), \(s = 20 \mathrm{~m}\), \(a = ?\), \(\mathrm{F} = ?\), \(m = 5000 \mathrm{~kg}\) Using the equation of motion: \[ s = ut + \frac{1}{2}at^2 \] Substituting the given values: \[ 20 \mathrm{~m} = 0 + \frac{1}{2} \cdot a \cdot (20 \mathrm{~s})^2 \] Solving for acceleration \(a\): \[ a = \frac{400}{400} = 1 \mathrm{~m/s^2} \] Now, using Newton’s second law (\(F = ma\)) to find the force (\(\mathrm{F}\)): \[ \mathrm{F} = m \cdot a = 5000 \mathrm{~kg} \cdot 1 \mathrm{~m/s^2} = 5000 \mathrm{~Newton} \] So, the acceleration (\(a\)) is \(1 \mathrm{~m/s^2}\) and the force (\(\mathrm{F}\)) is \(5000 \mathrm{~Newton}\).Problem 4:
A bullet of mass 10 g is horizontally fired with a velocity of 100 m s–1 from pistol of mass 1 kg. What is the recoil velocity of the pistol?
Solution:
Given: \(m = 10 \mathrm{~g} = \frac{10}{1000} \mathrm{~kg}\), \(v = 100 \mathrm{~m/s}\), \(\mathrm{M} = 1 \mathrm{~kg}\), \(\mathrm{V} = ?\)Using the formula for the velocity of the gun in the opposite direction to the bullet: \[ \mathrm{V} = -\frac{m v}{\mathrm{M}} = -\frac{\frac{10}{1000} \mathrm{~kg} \times 100 \mathrm{~m/s}}{1 \mathrm{~kg}} \] Solving for \(\mathrm{V}\): \[ \mathrm{V} = -1 \mathrm{~m/s} \] So, the velocity of the gun is \(-1 \mathrm{~m/s}\) in the opposite direction to the bullet being fired.
Force and Laws of Motion Class 9 Important Formula
1. \(1 \, \text{Newton} = 1 \, \text{kg} \, \text{m/s}^{-2}\); newton (\(\text{N}\)) is the SI unit of force.2. \(\vec{p} = \overrightarrow{m \mathbf{v}}\), where \(\vec{p}\) is momentum, \(m\) is mass, and \(\mathbf{v}\) is velocity.
3. \(\vec{F} = \frac{\overrightarrow{p_f} – \overrightarrow{p_i}}{t} = \frac{m \mathbf{v} – m \mathbf{u}}{t} = \frac{m(\mathbf{v} – \mathbf{u})}{t} = m \mathbf{a}\), where \(a\) = acceleration. \(\frac{v-u}{t} = a]\)
4. \(\overrightarrow{\mathbf{F}} = m \mathbf{a}\) Impulse \(= \text{force} \times \text{time} = \text{change in momentum}\)
5. \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\) : Law of conservation of momentum.
6. Recoil of gun: \(\mathbf{V} = \frac{-m \mathbf{v}}{\mathbf{M}}\), where \(m\) is the mass of the bullet, \(M\) is the mass of the gun, \(\mathbf{V}\) is the velocity of the gun, and \(\mathbf{v}\) is the velocity of the bullet.
7. \(\frac{\text{Rocket propulsion}}{\text{Velocity of rocket}} = V = \frac{-mV}{M}\), where \(m\) is the mass of burnt fuel, \(v\) is the velocity of burnt fuel, and \(M\) is the mass of the rocket at any instant.
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Practice More Numericals on Force and Laws of Motion
Numericals
Q1. A car of mass 1000 kg moving with a velocity of 36 km/h is brought to rest in 10 seconds. Calculate the force applied by the brakes and the distance it travels before stopping.
Q2. A truck of mass 5000 kg experiences a forward force of 8000 N and a frictional force of 2000 N. Calculate its acceleration.
Q3. An object of mass 2 kg is moving with a velocity of 3 m/s. How much force is required to bring it to rest in 2 seconds?
Q4. A bullet of mass 20 g is fired from a gun of mass 2 kg. If the bullet leaves the gun with a velocity of 400 m/s, find the recoil velocity of the gun.
Q5. A ball of mass 0.5 kg hits a wall with a velocity of 10 m/s and rebounds with the same speed. If the contact time is 0.04 s, calculate the force exerted by the wall on the ball.
Q6. A force of 15 N is applied to an object of mass 5 kg. Calculate the acceleration produced. If the same force is applied to a 10 kg mass, what will be the new acceleration?
Q7. A boy throws a ball of mass 500 g vertically upwards with a speed of 10 m/s. Calculate:
(i) the momentum of the ball, and
(ii) the change in momentum when it comes to rest at the highest point.
Q8. A scooter rider traveling at 54 km/h sees a child 30 meters ahead on the road. He applies brakes and decelerates uniformly at 4 m/s². Will the scooter stop before hitting the child? Support your answer with calculations.
Q9. A constant force acting on a body changes its velocity from 5 m/s to 15 m/s in 5 seconds. If the mass of the body is 4 kg, calculate the magnitude of the force applied.
Q10. A body of mass 10 kg is acted upon by two forces of 15 N and 25 N in opposite directions. Find the net force and acceleration produced in the body.
Answers
Q1.
Force = –1000 N (negative sign indicates deceleration)
Distance = 50 m
Q2.
Net Force = 8000 – 2000 = 6000 N
Acceleration = 6000 / 5000 = 1.2 m/s²
Q3.
Force = (0 – 3) × 2 / 2 = –3 N
Q4.
Recoil velocity of gun = –4 m/s
Q5.
Change in momentum = 0.5 × (10 + 10) = 10 kg·m/s
Force = 10 / 0.04 = 250 N
Q6.
For 5 kg: a = 15 / 5 = 3 m/s²
For 10 kg: a = 15 / 10 = 1.5 m/s²
Q7.
(i) Momentum = 0.5 × 10 = 5 kg·m/s
(ii) Change in momentum = 5 kg·m/s
Q8.
Stopping distance = (15)² / (2 × 4) = 28.125 m
Since 28.125 < 30 → Yes, the scooter stops in time.
Q9.
Acceleration = (15 – 5)/5 = 2 m/s²
Force = 4 × 2 = 8 N
Q10.
Net force = 25 – 15 = 10 N
Acceleration = 10 / 10 = 1 m/s²
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Also Check
Why are Physics Numericals Important for Class 9 Students?
- Application of Concepts: Physics numericals allow students to apply the theoretical concepts they learn in class to real-world situations. This practical application enhances their understanding of the subject.
- Problem-Solving Skills: Numerical problems require critical thinking and problem-solving skills. Students learn to analyze information, identify relevant formulas, and calculate solutions, skills that are valuable in many aspects of life.
- Enhanced Understanding: By solving numerical problems, students gain a deeper insight into the underlying principles of physics. It’s one thing to memorize formulas, but it’s another to comprehend how they are derived and applied.
- Preparation for Higher Classes: A strong foundation in physics at the Class 9 level prepares students for more advanced physics topics in higher classes. It paves the way for success in Class 10 board exams and beyond.
Tips for Solving Physics Numericals
- Understand the Concept: Before attempting numericals, ensure you have a clear understanding of the underlying physics concepts. Review your class notes and textbook.
- Identify Given Data: Carefully read the problem and identify the given data. This is crucial for selecting the appropriate formula.
- Choose the Right Formula: Select the formula that best fits the problem. Familiarize yourself with the relevant equations for each topic.
- Units and Conversions: Pay attention to units. Ensure all units are consistent throughout your calculations, and convert them if necessary.
- Organize Your Work: Show your work step by step. This not only helps you track your progress but also allows for partial credit if you make a mistake.
- Practice Regularly: Physics numericals improve with practice. The more problems you solve, the more confident and proficient you’ll become.
- Seek Help When Needed: If you’re stuck on a problem, don’t hesitate to seek help from your teacher, classmates, or online resources.
