5.5 Newton’s second law of motion

NEWTON’S SECOND LAW OF MOTION (Law of Mass and Acceleration)

Newton’s second law of motion states that the acceleration of an object is dependent upon two variables: the net force \(\vec{F}\) acting upon the object and the mass \(m\) of the object. In other words, the acceleration of an object as measured from an inertial frame is given by the vector sum of all the forces acting on the object divided by its mass.

In terms of equation: \(\vec{a}=\vec{F} / m \quad\) or, \(\vec{F}=m \vec{a}\). \(\quad \ldots\) (5.1)
The inertial frame (A frame of reference in which Newton’s first law is valid is called an inertial frame of reference) is already defined by the first law of motion. A force \(\vec{F}\) acting on a particle of mass \(m\) produces an acceleration \(\vec{F} / m\) in it with respect to an inertial frame. 

What is momentum?

The momentum of a body is defined to be the product of its mass \(m\) and velocity \(\vec{v}\), and is denoted by \(\vec{p}\):

\(\vec{p}=m \vec{v}\) (Momentum is clearly a vector quantity.)

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. Thus, if under the action of a force \(\vec{F}\) for time interval \(\Delta t\), the velocity of a body of mass \(m\) changes from \(\vec{v}\) to \(\vec{v}+\Delta \vec{v}\) i.e. its initial momentum \(\vec{p}=m \vec{v}\) changes by \(\Delta \vec{p}=m \Delta \vec{v}\). According to the Second Law,
\(
\vec{F} \propto \frac{\Delta \vec{p}}{\Delta t} \text { or } \vec{F}=k \frac{\Delta \vec{p}}{\Delta t}
\)
where \(k\) is a constant of proportionality. Taking the limit \(\Delta t \rightarrow 0\), the term \(\frac{\Delta \vec{p}}{\Delta t}\) becomes the derivative or differential co-efficient of \(\vec{p}\) with respect to \(t\), denoted by \(\frac{\mathrm{d} \vec{p}}{\mathrm{d} t}\). Thus

\(\vec{F}=k \frac{\mathrm{d} \vec{p}}{\mathrm{d} t} \dots(5.2)\)
For a body of fixed mass \(m\),
\(\frac{\mathrm{d} \vec{p}}{\mathrm{d} t}=\frac{\mathrm{d}}{\mathrm{d} t}(m \vec{v})=m \frac{\mathrm{d} \vec{v}}{\mathrm{d} t}=m \vec{a} \dots(5.3)\)
Therefore, the Second Law can also be written as
\(
\vec{F}=k ~m \vec{a} \dots (5.4)
\)
which shows that force is proportional to the product of mass \(m\) and acceleration \(\vec{a}\).

The unit of force has not been defined so far. In fact, we use Eq. (5.2) to define the unit of force. If we choose \(k=1\). The second law then is
\(\vec{F}=\frac{\mathrm{d} \vec{p}}{\mathrm{d} t}=m \vec{a} \dots(5.5)\)

In SI unit force is one that causes an acceleration of \(1 \mathrm{~m} \mathrm{~s}^{-2}\) to a mass of \(1 \mathrm{~kg}\). This unit is known as newton: \(1 \mathrm{~N}=1 \mathrm{~kg} \mathrm{~m}  \mathrm{~s}^{-2}\).

Let us note at this stage some important points about the second law :
1. In the second law, \(\vec{F}=0\) implies \(\vec{a}=0\). The second law is obviously consistent with the first law.
2. The second law of motion is a vector law. It is equivalent to three equations, one for each component of the vectors :
\(
\begin{aligned}
& F_x=\frac{\mathrm{d} p_x}{\mathrm{~d} t}=m a_x \\
& F_y=\frac{\mathrm{d} p_y}{\mathrm{~d} t}=m a_y \\
& F_z=\frac{\mathrm{d} p_z}{\mathrm{~d} t}=m a_z \dots(5.6)
\end{aligned}
\)
This means that if a force is not parallel to the velocity of the body, but makes some angle with it, it changes only the component of velocity along the direction of force. The component of velocity normal to the force remains unchanged. For example, in the motion of a projectile under the vertical gravitational force, the horizontal component of velocity remains unchanged (Fig. 5.5).

The following common experiences indicate the importance of this quantity for considering the effect of force on motion.

If two stones, one light, and the other heavy, are dropped from the top of a building, a person on the ground will find it easier to catch the light stone than the heavy stone. The mass of a body is thus an important parameter that determines the effect of force on its motion.

Speed is another important parameter to consider. A bullet fired by a gun can easily pierce human tissue before it stops, resulting in a casualty. The same bullet fired with moderate speed will not cause much damage. Thus for a given mass, the greater the speed, the greater is the opposing force needed to stop the body in a certain time. Taken together, the product of mass and velocity, that is momentum, is evidently a relevant variable of motion. The greater the change in the momentum in a given time, the greater is the force that needs to be applied.

A seasoned cricketer catches a cricket ball coming in with great speed far more easily than a novice, who can hurt his hands in the act. One reason is that the cricketer allows a longer time for his hands to stop the ball. As you may have noticed, he draws in hands backward in the act of catching the ball (Fig. 5.3). The novice, on the other hand, keeps his hands fixed and tries to catch the ball almost instantly. He needs to provide a much greater force to stop the ball instantly, and this hurts. The conclusion is clear: force not only depends on the change in momentum but also on how fast the change is brought about. The same change in momentum brought about in a shorter time needs a greater applied force. In short, the greater the rate of change of momentum, the greater is the force.

Example 5.3:A bullet of mass \(0.04 \mathrm{~kg}\) moving with a speed of \(90 \mathrm{~m} \mathrm{~s}^{-1}\) enters a heavy wooden block and is stopped after a distance of \(60 \mathrm{~cm}\). What is the average resistive force exerted by the block on the bullet?

Answer: The retardation \(a\) of the bullet (assumed constant) is given by
\(
a=\frac{-u^{2}}{2 s}=\frac{-90 \times 90}{2 \times 0.6} \mathrm{~m} \mathrm{~s}^{-2}=-6750 \mathrm{~m} \mathrm{~s}^{-2}
\)

The retarding force, by the second law of motion, is
\(
=0.04 \mathrm{~kg} \times 6750 \mathrm{~m} \mathrm{~s}^{-2}=270 \mathrm{~N}
\)

Example 5.4: The motion of a particle of mass \(m\) is described by \(y=u t+\frac{1}{2} g t^2\). Find the force acting on the particle.

Answer: We know
\(
y=u t+\frac{1}{2} g t^2
\)
Now,
\(
v=\frac{\mathrm{d} y}{\mathrm{~d} t}=u+g t
\)
acceleration, \(a=\frac{\mathrm{d} v}{\mathrm{~d} t}=g\)
Then the force is given by Eq. (5.5)
\(
F=m a=m g
\)
Thus the given equation describes the motion of a particle under acceleration due to gravity and \(y\) is the position coordinate in the direction of \(g\).

What is Impulse Force

A large force acting for a short time to produce a finite change in momentum is called an impulsive force. 

Impulse \(=\) Force \(\times\) time duration = Change in momentum

Example 5.5:A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of \(12 \mathrm{~m} \mathrm{~s}^{-1}\).
If the mass of the ball is \(0.15 \mathrm{~kg}\), determine the impulse imparted to the ball. (Assume linear motion of the ball)

Answer:  Change in momentum
\(
\begin{aligned}
&=0.15 \times 12-(-0.15 \times 12) \\
&=3.6 \mathrm{~N} \mathrm{~s},
\end{aligned}
\)
Impulse \(=3.6 \mathrm{~N} \mathrm{~s}\),  in the direction from the batsman to the bowler.

Working With Newton’s First and Second Law

Newton’s laws refer to a particle and relate the forces acting on the particle with its acceleration and its mass. A systematic algorithm for writing equations from Newton’s laws is as follows:

Step-1: Decide the System

The first step is to decide the system on which the laws of motion are to be applied. The system may be a single particle, a block, a combination of two blocks one kept over the other, two blocks connected by a string, a piece of string, etc. The only restriction is that all parts of the system should have identical acceleration. Consider the situation shown in the figure below. Block B does not slip over A, the disc D slides over the string and all parts of the string are tight. A and B move together. C is not in contact with A or B. But as the length of the string between A and C does not change, the distance moved by C in any time interval is the same as that by A. The same is true for G. The distance moved by G in any time interval is the same as that by A, B or C. The direction of motion is also the same for A, B, C, and G. They have identical accelerations. We can take any of these blocks as a system or any combination of the blocks from these as a system. 

Step 2: Identify the Forces

Once the system is decided, make a list of the forces acting on the system due to all the objects other than the system. Any force applied by the system should not be included in the list of the forces.

Consider the situation shown in the figure below. The boy stands on the floor balancing a heavy load on his head. The load presses the boy, the boy pushes the load upward the boy presses the floor downward, the floor pushes the boy upward, the earth attracts the load downward, the load attracts the earth upward, the boy attracts the earth upward and the earth attracts the boy downward. There are many forces operating in this world. Which of these forces should we include in the list of forces?

The forces are listed in the upper half of the table (5.1). Instead, if we take the load as the system and discuss the equilibrium of the load, the list of forces will be different. These forces appear in the lower half of table (5.1).

Step 3: Make a Free Body Diagram

Now, represent the system by a point in a separate diagram and draw vectors representing the forces acting on the system with this point as the common origin. The forces may lie along a line, may be distributed in a plane (coplanar) or may be distributed in the space (non-planar). We shall rarely encounter situations dealing with non-planar forces. For coplanar forces, the plane of the diagram represents the plane of the forces acting on the system. Indicate the magnitudes and directions of the forces in this diagram. This is called a free-body diagram. The free body diagram for the example discussed above with the boy as the system and with the load as the system are shown in the figure below.

Step 4: Choose Axes and Write Equations

Any three mutually perpendicular directions may be chosen as the \(X-Y-Z\) axes. We give below some suggestions for choosing the axes to solve problems. If the forces are coplanar, only two axes, say \(X\) and \(Y\), taken in the plane of forces are needed. Choose the \(X\)-axis along the direction in which the system is known to have or is likely to have acceleration. A direction perpendicular to it may be chosen as the \(Y\)-axis. If the system is in equilibrium, any mutually perpendicular directions in the plane of the diagram may be chosen as the axes. Write the components of all the forces along the \(X\)-axis and equate their sum to the product of the mass of the system and its acceleration. This gives you one equation. Write the components of the forces along the \(Y\)-axis and equate the sum to zero. This gives you another equation. If the forces are collinear, this second equation is not needed.

Example 5.6: A block of mass \(M\) is pulled on a smooth horizontal table by a string making an angle \(\theta\) with the horizontal as shown in the figure below. If the acceleration of the block is \(a\), find the force applied by the string and by the table on the block.

Answer: Let us consider the block as the system.
The forces on the block are
(a) pull of the earth, \(M g\), vertically downward,
(b) contact force by the table, \(N\), vertically upward,
(c) pull of the string, \(T\), along the string.
The free-body diagram for the block is shown in the figure below.

The acceleration of the block is horizontal and towards the right. Take this direction as the \(X\)-axis and vertically upward direction as the \(Y\)-axis. We have,
component of \(M g\) along the \(X\)-axis \(=0\)
component of \(N\) along the \(X\)-axis \(=0\)
component of \(T\) along the \(X\)-axis \(=T \cos \theta\).
Hence the total force along the \(X\)-axis \(=T \cos \theta\).
Using Newton’s law, \(T \cos \theta=M a \dots(i)\)
Component of \(M g\) along the \(Y\)-axis \(=-M g\)
component of \(N\) along the \(Y\)-axis \(=N\)
component of \(T\) along the \(Y\)-axis \(=T \sin \theta\).
Total force along the \(Y\)-axis \(= N+T \sin \theta-M g\).
Using Newton’s law, \(N+T \sin \theta-M g=0 \dots(ii)\)
From equation (i), \(T=\frac{M a}{\cos \theta} \cdot\) Putting this in equation (ii) \(N=M g-M a \tan \theta\)