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 character of momentum has not been evident. In the examples so far, momentum and change in momentum both have the same direction. But this is not always the case. Suppose a stone is rotated with uniform speed in a horizontal plane by means of a string, the magnitude of momentum is fixed, but its direction changes (Fig.

. ). A force is needed to cause this change in momentum vector. This force is provided by our hand through the string.

Experience suggests that our hand needs to exert a greater force if the stone is rotated at greater speed or in a circle of smaller radius, or both. This corresponds to greater acceleration or equivalently a greater rate of change in momentum vector. This suggests that the greater the rate of change in momentum vector the greater is the force applied. Fig.

. Force is necessary for changing the direction of momentum, even if its magnitude is constant. We can feel this while rotating a stone in a horizontal circle with uniform speed by means of a string. These qualitative observations lead to the second law of motion expressed by Newton as follows : 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 F for time interval ∆ t , the velocity of a body of mass m changes from v to v + ∆ v i.e. its initial momentum p = m v changes by ∆ ∆ v . According to the Second Law, or k ∆ ∆ ∝ ∆ ∆ where k is a constant of proportionality. Taking the limit ∆ t → , the term ∆ ∆ p becomes the derivative or differential co-efficient of p with respect to t , denoted by d d t p .

Thus k ( . ) For a body of fixed mass m , v v a ( . ) i.e the Second Law can also be written as F = k m a ( . ) which shows that force is proportional to the product of mass m and acceleration a .

The unit of force has not been defined so far. In fact, we use Eq. ( . ) to define the unit of force.

We, therefore, have the liberty to choose any constant value for k . For simplicity, we choose k = . The second law then is a ( . ) In SI unit force is one that causes an acceleration of m s - to a mass of kg.

This unit is known as newton : N = kg m s - . Let us note at this stage some important points about the second law : . In the second law, F = implies a = . The second law is obviously consistent with the first law.

. The second law of motion is a vector law. It is equivalent to three equations, one for each component of the vectors : ma x x x ma z z z a = d ( . ) 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. . ).

. The second law of motion given by Eq. ( . ) is applicable to a single point particle.

The force F in the law stands for the net external force on the particle and a stands for acceleration of the particle. It turns out, however, that the law in the same form applies to a rigid body or, even more generally, to a system of particles. In that case, F refers to the total external force on the system and a refers to the acceleration of the system as a whole. More precisely, a is the acceleration of the centre of mass of the system about which we shall study in detail in Chapter .

Any internal forces in the system are not to be included in F . Fig. . Acceleration at an instant is determined by the force at that instant.

The moment after a stone is dropped out of an accelerated train, it has no horizontal acceleration or force, if air resistance is neglected. The stone carries no memory of its acceleration with the train a moment ago. . The second law of motion is a local relation which means that force F at a point in space (location of the particle) at a certain instant of time is related to a at that point at that instant.

Acceleration here and now is determined by the force here and now, not by any history of the motion of the particle (See Fig. . ) . Example .

A bullet of mass . kg moving with a speed of m s – enters a heavy wooden block and is stopped after a distance of 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 – u a – m s – m s .

− − × × The retarding force, by the second law of motion, is = . kg × m s - = N The actual resistive force, and therefore, retardation of the bullet may not be uniform. The answer therefore, only indicates the average resistive force. ⊳ Example .

The motion of a particle of mass m is described by y = + ut gt . Find the force acting on the particle.