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The study of Physics Topics can help us understand and solve real-world problems, from climate change to medical imaging technology.

## Applications Of Newton’s Second Law Of Motion

When two bodies, a heavy one and a light one, are acted upon by the same force for the same time, the light body attains a higher velocity (or higher speed) than the heavy one. But the momentum gained by both the bodies is the same. The link between force and momentum is expressed in Newton’s second law of motion.

According to Newton’s second law of motion : The rate of change of momentum of a body is directly proportional to the applied force, and takes place in the direction in which the force acts. The rate of change of momentum of a body can be obtained by dividing the ‘Change in momentum’ by ‘Time taken’ for change. So, Newton’s second law of motion can be expressed as :

Force ∝ \(\frac{\text { Change in momentum }}{\text { Time taken }}\)

Consider a body of mass m having an initial velocity u. The initial momentum of this body will be mu. Suppose a force F acts on this body for time t and causes the final velocity to become v. The final momentum of this body will be mv. Now, the change in momentum of this body is mv – mu and the time taken for this change is t. So, according to Newton’s second law of motion :

F ∝ \(\frac{m v-m u}{t}\)

or F ∝ \(\frac{m(v-u)}{t}\)

But \(\frac{v-u}{t}\) represents change in velocity with time which is known as acceleration ‘a’. So, by writing ‘a’ in place of -yp- in the above relation, we get:

F ∝ m × a

Thus, the force acting on a body is directly proportional to the product of ‘mass’ of the body and ‘acceleration’ produced in the body by the action of the force, and it acts in the direction of acceleration. This is another definition of Newton’s second law of motion.

The relation F ∝ m × a can be turned into an equation by putting in a constant k.

Thus, F = k × m × a (where k is a constant)

The value of constant k in SI units is 1, so the above equation becomes :

F = m × a

or Force = mass × acceleration

Thus, Newton’s second law of motion gives us a relationship between ‘force’ and ‘acceleration’. When a force acts on a body, it produces acceleration in the body, the acceleration produced may be positive or negative. Newton’s second law of motion also gives us a method of measuring the force in terms of mass and acceleration. The force acting on a body can be calculated by using the formula : F = m × a.

a = \(\frac{F}{m}\)

We can also write the equation F = m × a as :

a = \(\frac{F}{m}\)

It is obvious from the above relation that: The acceleration produced in a body is directly proportional to the force acting on it and inversely proportional to the mass of the body. Since the acceleration produced is inversely proportional to the mass of a body, therefore, if the mass of a body is doubled, its acceleration will be halved. And if the mass is halved then acceleration will get doubled (provided the force remains the same). Moreover, since the acceleration produced is inversely proportional to the mass of the body, it means that it will be easier to move light bodies (having less mass) than heavy bodies (having large mass).

The SI unit of force is newton which is denoted by N. A newton is that force which when acting on a body of mass 1 kg produces an acceleration of 1 m/s^{2} in it. We have just seen that:

F = m × a

Putting m = 1 kg and a = 1 m/s^{2}, F becomes 1 newton.

So, 1 newton = 1 kg × 1 m/s^{2}

In order to get an idea of 1 newton force, we should hold a weight of 100 grams on our outstretched palm. The force exerted by 100 gram weight on our palm is approximately equal to 1 newton.

The first law of motion discussed earlier is, in fact, a special case of the second law, because when the applied force F is zero, then the acceleration ‘a’ is also zero and the body remains in its state of rest or of uniform motion. It is obvious that Newton’s second law gives us a relationship between the force applied to a body and the acceleration produced in the body. The formula F = m × a should be memorized because it will be used to solve numerical problems.

It should be noted that just as a minus sign for acceleration shows that the acceleration is acting in a direction opposite to the motion of the body, in the same way, if a minus sign comes with the force, it will indicate that the force is acting in a direction opposite to that in which the body is moving (just as the force of friction acts in a direction opposite to that of the moving body).

### Applications of Newton’s Second Law of Motion

Some of the observations of our daily life can be explained on the basis of Newton’s second law of motion. In all these cases some technique or arrangement is used to reduce the momentum of a fast moving body more gently (by allowing more time to stop it), so that injury can be prevented or reduced. Here are some examples.

### 1. Catching a Cricket Ball

A cricket player (or fielder) moves his hands backwards on catching a fast cricket ball. This is done to prevent injury to the hands. We can explain it as follows : A fast moving cricket ball has a large momentum. In stopping (or catching) this cricket ball, its momentum has to be reduced to zero. Now, when a cricket player moves back his hands on catching the fast ball, then the time taken to reduce the momentum of ball to zero is increased (see Figure). Due to more time taken to stop the ball, the rate of change of momentum of ball is decreased and hence a small force is exerted on the hands of player. So, the hands of player do not get hurt.

If, however, a cricket player stops a fast moving ball suddenly (keeping his hands stationary), then the large momentum of the ball will be reduced to zero in a very short time. Due to this, the rate of change of momentum of cricket ball will be very large and hence it will exert a large force on player’s hands. The player’s hands will get hurt.

### 2. The Case of a High Jumper

During athletics meet, a high jumping athlete is provided either a cushion or a heap of sand on the ground to fall upon. This is done to prevent injury to the athlete when he falls down after making a high jump. We can explain it as follows : When the high jumper falls on a soft landing site (such as a cushion or a heap of sand), then the jumper takes a longer time to come to a stop. The rate of change of momentum of athlete is less due to which a smaller stopping force acts on the athlete. And the athlete does not get hurt. Thus, the cushion or sand, being soft, reduces the athlete’s momentum more gently. If, however, a high jumping athlete falls from a height on to hard ground, then his momentum will be reduced to zero in a very short time. The rate of change of momentum will be large due to which a large opposing force will act on the athlete. This can cause serious injuries to the athlete.

### 3. The Use of Seat Belts in Cars

These days all the cars are provided with seat belts for passengers to prevent injuries in case of an accident. In a car accident, a fast running car stops suddenly. Due to this the car’s large momentum is reduced to zero in a very short time. The slightly stretchable seat belts worn by the passengers of the car increase the time taken by the passengers to fall forward. Due to longer time, the rate of change of momentum of passengers is reduced and hence less stopping force acts on them. So, the passengers may either not get injured at all or may get less injuries. It is obvious that seat belts reduce the passengers’ momentum more gently and hence prevent injuries.

We will now solve some problems based on Newton’s second law of motion.

**Example Problem 1.**

What force would be needed to produce an acceleration of 4 m/s^{2} in a ball of mass 6 kg?

**Solution.**

The force needed is to be calculated by using the relation :

Force = mass × acceleration or F = m × a

Here, Force, F = ? (To be calculated)

Mass, m = 6 kg

And, Acceleration, a = 4 m/s^{2}

Now, putting these values of m and a in the above equation, we get :

F = 6 × 4

or Force, F = 24 N

Thus, the force needed is of 24 newtons.

**Example Problem 2.**

What is the acceleration produced by a force of 12 newtons exerted on an object of mass 3 kg ?

**Solution.**

Here, Force, F = 12

Mass, m = 3 kg

And, Acceleration, a = ? (To be calculated)

We know that :

F = m × a

So, 12 = 3 × a

3a = 12

a = \(\frac{12}{3}\) m/s^{2}

or Acceleration, a = 4 m/s^{2}

**Example Problem 3.**

Calculate the force required to impart to a car a velocity of 30 m/s in 10 seconds starting from rest. The mass of the car is 1500 kg.

**Solution.**

Here, Mass, m = 1500 kg

Let us calculate the value of acceleration by using the first equation of motion.

Now, Initial velocity, u = 0 (Car starts from rest)

Final velocity, v = 30 m/s

And, Time taken, t = 10 s

Now, putting these values in the equation :

v = u + at

We get, 30 = 0 + a × 10

10a = 30

a = \(\frac{30}{10}\) m/s^{2}

or Acceleration, a = 3 m/s^{2}

Now, putting m = 1500 kg and a = 3 m/s^{2} in equation :

F = m × a

We get, F = 1500 × 3 N

= 4500 N

Thus, the force required in this case is of 4500 newtons.

**Example Problem 4.**

The speed-time graph of a car is given alongside. The car weighs 1000 kg.

(i) What is the distance travelled by the car in the first two seconds?

(ii) What is the braking force applied at the end of 5 seconds to bring the car to a stop within one second?

**Solution:**

(i) We will first calculate the distance travelled in the first two seconds. From the given graph we find that:

Distance travelled in first two seconds = Area under line AB and the time axis

= Area of triangle AB2

= \(\frac{1}{2}\) × Base × Height

= \(\frac{1}{2}\) × 2 × 15

= 15m

Thus, the distance travelled by the car in the first two seconds is 15 metres.

(ii) The braking force is to be calculated by using the formula:

Force = mass × acceleration

or F = m × a

Here, the mass of the car is given as 1000 kg but the acceleration in the last one second (from point C to point D) is to be obtained from the graph shown above. Now, we know that the acceleration is given by the slope of speed-time graph. So,

Acceleration in last one second = Slope of line CD

\(=\frac{\text { Perpendicular }}{\text { Base }}\)

= \(\frac{15}{1}\)

= 15 m/s2

If we look at the above given graph, we will find that the speed of the car is decreasing from point C to point D. That is, the acceleration from C to D is negative and hence it should be written with a minus sign.

Thus,

Acceleration, a = -15 m/s^{2}

Now, putting in = 1000 kg and a = -15 m/s^{2} in formula (1), we get:

Force, F = 1000 × (-15)

= -15000N

Thus, the force applied by the brakes to stop the car is 15000 newtons. The negative sign of force here shows that the force is being applied in a direction opposite to the motion of the car. That is, it is a retarding force.

**Example Problem 5.**

A truck starts from rest and rolls down a hill with constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it il its mass is 7 metric tonnes.

**Solution.**

First of all we will find its acceleration by using the relation:

Distance travelled, s = ut + \(\frac{1}{2}\)at^{2}

Here, Distance travelled, s = 400 m

Initial speed, u = 0 (It starts from rest)

Time, t = 20 s

And, Acceleration, a = ? (To be calculated)

Putting these values in the above formula, we get:

400 = 0 × 20 + \(\frac{1}{2}\) × a × (20)^{2}

400 = 200 a

a = \(\frac{400}{200}\)

Acceleration, a = 2 m/s^{2} …… (1)

We will now calculate the force by using the relation:

F = m × a

Here, Mass, m = 7 metric tonnes

= 7 × 1000 kg

= 7000 kg …… (2)

And, Acceleration, a = 2 m/s^{2} (Calculated above)

So, Force, F = 7000 × 2

F = 14000 N

Thus, the force acting on the truck is 14000 newtons.

**Example Problem 6.**

A force of 5 newtons gives a mass m1 an acceleration of 8 m/s^{2}, and a mass m^{2} an acceleration of 24 m/s^{2}. What acceleration would it give if both the masses are tied together ?

**Solution.**

(i) In the first case :

Force, F = 5 N

Mass, m = m_{1} (To be calculated)

And, Acceleration, a = 8 m/s^{2}

Now, F = m × a

So, 5 = m_{1} × 8

And m_{1} = \(\frac{5}{8}\) kg

m_{1} = 0.625 kg

Thus, the mass m_{1} is 0.625 kg.

(ii) In the second case :

Force, F = 5N

Mass, m = m_{2} (To be calculated)

And, Acceleration, a = 24 m/s^{2}

Now, F = m × a

So, 5 = m_{2} × 24

And, m_{2} = \(\frac{5}{24}\) kg

m_{2} = 0.208 kg

Thus, the mass m_{2} is 0.208 kg

(iii) In the third case :

Force, F = 5 N

Total mass, m = m_{1} + m_{2}

= 0.625 + 0.208

= 0.833 kg

And, Acceleration, a = ? (To be calculated)

Now, putting these values in the relation :

F = m × a

We get: 5 = 0.833 × a

a = \(\frac{5}{0.833}\)

a = 6 m/s^{2}

Thus, if both the masses are tied together, then the acceleration would be 6 m/s^{2}.

**Example Problem 7.**

Which would require a greater force – accelerating a 10 g mass at 5 m/s^{2} or a 20 g mass at 2 m/s^{2}?

**Solution:**

(i) In first case : Force, F = m × a

= \(\frac{10}{1000}\)kg × 5m/s^{2}

= 0.05 N ……. (1)

(ii) In second case : Force, F = \(\frac{20}{1000}\) × 2 m/s^{2}

= 0.04 N …… (2)

Thus, a greater force of 0.05 N is required for accelerating a 10 g mass.