Today’s post is going to cover the * Force and Laws of Motion Class 9 Numericals*.

Newton’s Laws of Motion and Force are tightly coupled.

The first law gives us the concept of force and its effect on inertia.

The second law gives us more concrete measurements of the Force and provides us the equation of force. Not only that, we also get the idea of momentum.

The third law introduces the reaction force. So solving numerical on force and laws of motion is very important to understand the underlying theory and its real-life application.

With this introduction, we will start off listing selected

*students.*

**numerical problem**s from the chapter**Force and laws of motion for class 9**These will not only help class 9 students to prepare their school exams but also help them to build a foundation for tougher exams in the future like IIT and other JEEs.

## Force and Laws of Motion Class 9 Numericals

### numerical problems in physics from force and laws of motion chapter – part 1

**1> A force of 10 N is applied for 0.1 second on a body of mass 1 kg initially at rest. The force then ceases to act. **

**What would be the velocity of the body just after this?**

After 2 more seconds with what velocity the body will move?

After 2 more seconds with what velocity the body will move?

**2> A force acts for 2 seconds on a body of mass 2 kg initially at rest. Just after the force stops to act the body moves 10 m in the next 2 seconds. Find the magnitude of the force.**

**3> Calculate the magnitude of force which when applied on a body of mass 0.5 kg produces an acceleration of 5 m/s**^{2}** **

**4> A ball of mass 0.01 kg is moving with a velocity of 50 m/s. On applying a constant force for 2 seconds on the ball, it acquires a velocity of 70 m/s.**

Calculate the initial and final momentum of the ball. **Also, calculate the rate of change of momentum and the acceleration of the ball. ****What would be the magnitude of the force applied?**

**5> An object of mass 1 kg is moving with a speed of 50 m/s. The object is brought to rest in 0.05 seconds by an external system. Find the change in momentum of the object and the average force applied by the external system.**

### numerical problems in physics from force and laws of motion chapter –part 2

**6> A car of mass 500 kg moving at a speed of 36 Km/hr is stopped by applying brakes in 10 s. Calculate the force applied by the brakes.**

**7> A bullet of mass 50 g moving with an initial velocity of 100 m/s, strikes a wooden block and comes to rest after penetrating a distance 2 cm in it. Calculate ****– initial momentum of the bullet -final momentum of the bullet -retardation caused by the wooden block and -resistive force applied by the wooden block**

**8> A force when applied on a block of 10 kg mass produces an acceleration of 5 m/s^2. When the same force is applied on a different block initially at rest then the block gains a velocity of 2 m/s in 1 second. **

**What is the magnitude of the force in Newton and what is the mass of the second block?**

**9> Velocity of an object changes from 2 m/s to 10 m/s in 4 seconds.**** If the mass of the object is 10 kg then how much force will be required to do this?**

**10> A car is moving with a uniform velocity of 30 m/s. it is stopped in 2 seconds by applying a force of 1500 N through its brakes. ****Calculate (a) the change in momentum of the car (b) the retardation produced in the car and (c) the mass of the car.**

### Force and laws of motion numerical problems –part 3

**11> A boy pushes a wall with a force of 20 N towards North. What force is exerted by the wall on the boy?**

**12> A block of mass 1.5 kg is hanging from a rigid support by a string. ****What is the force exerted by the (i) block on the string (ii) string on the block (g = 10 m/s^2)**

**13> A stone of size 1 kg is thrown with a velocity of 20 m/s across the frozen surface of a lake and it comes to rest after traveling a distance of 50 m. ****What is the force of friction between the stone and the ice?**

**14> A body of mass 0.5 kg is resting on a frictionless surface. ****When a force of 2000 dyne acts on it for 10 s, then calculate the distance traveled by it in 10 seconds.**

**15> A force of 10 N produces an acceleration of 5 m/s^ in mass m1 and 20 m/s^2 in mass m2.**** What will be the acceleration by this force, if both masses m1 and m2 are tied together? **

## Force and Laws of Motion Class 9 Numericals Solved – Solutions

**Solution of problem 1**

**A force will cause an acceleration a.**

a = force/mass = 10/1 m/s^2 = 10 m/s^2

The force acts for 0.1 seconds. That means the body will remain under accelerated motion for 0.1 sec.

The velocity acquired post this interval of 0.1 s is V

And, V =u + at = 0 + 10. 0.1 = 1 m/s

As there is no force after this interval, the body will continue its motion with uniform velocity of 1 m/s.

So after 2 more seconds,** its velocity is 1 m/s.**

**Solution of problem **2

**After the force stops to act, the velocity of the body is = V = distance/time=10/2 m/s = 5 m/s**

Initial velocity = U =0

Time interval for the accelerated motion in presence of the force =t= 2 seconds

Therefore we can find the acceleration a as

a**= (V-U)/t =(5-0)/2 = 2.5 m/s^2 m=Mass of the body is 2 kg. a=2.5 m/s^2 So force F = ma = 2 x 2.5 N= 5 N**

**Solution of problem **3

** Sol: Force = mass x acceleration = 0.5 x 5 N = 2.5 N**

**Solution of problem **4

Initial velocity = U = 50 m/s

Final velocity = V = 70 m/s

Time interval = 2 s

mass = m = 0.01 kg

Initial momentum = mU = 0.01 x 50 =0.5 kg m/s

Final momentum = mV = 0.01 x 70 = 0.7 kg m/s

Acceleration =a= (V-U)/t =(70-50)/2 m/s^2 = 10 m/s^2

Force = m a = 0.01 x 10 N = 0.1 N

**Solution of problem **5

** Initial momentum = mU = 1 x 50 = 50 kg m/s Final momentum = mV = 0 (as V =0) So the change in momentum =mV – mU = 0 – 50 = – 50 kg m/s Force = change of momentum / time = -50/0.05 N = – 1000 N The negative sign denotes a force that opposes the motion, causing retardation. **

**Solution of problem **6

**Mass = 500 kg initial velocity U = 36 kmph= 36 x (5/18) m/s =10 m/s final velocity =0 time interval =10 s**

Therefore, acceleration (retardation here actually) = (0-10)/10 m/s^2 = -1 m/s^2

Therefore, acceleration (retardation here actually) = (0-10)/10 m/s^2 = -1 m/s^2

**The force applied by the brakes = mass x acceleration = 500 x (-1) N = -500 N**

The negative sign of the force denotes that the force is resistive i.e. the force opposes the motion.

The negative sign of the force denotes that the force is resistive i.e. the force opposes the motion.

**Solution of problem **7

** Initial momentum= mass x initial velocity ****= (50/1000)kg x (100) m/s = 5 kg m/s**

final momentum = 0 (as final velocity is zero)

To find Retardation a we will use the following equation:

V^2 = U^2 – 2as

Putting values,

0=(100)^2 – 2a (2/100)

**Retardation** **a= (10000x 100)/4 = 25 x 10^4 m/s^2**

Resistive force = m a = (50/1000) x (25 x 10^4)=12500 N

**Solution of problem** 8

**1**^{st}** case: F = ma = 10 x 5 N = 50 N**

2^{nd}** case F = same as first case = 50 N U = 0 V = 2 m/s t = 1 sec So, Acceleration a =(V-U)/t = 2 m/s^2**

Force = mass X acceleration

Mass of the second block = M = F/a = 50/2 Kg = 25 kg

Force = mass X acceleration

Mass of the second block = M = F/a = 50/2 Kg = 25 kg

### Force and Laws of Motion Class 9 Numericals – Continued

We are going to post more numericals here. As well as we will publish rest of the solutions very soon.**Go to our listing of numericals: here**