Suppose an object of mass, `m` is moving along a straight line with an initial velocity, `u`. It is uniformly accelerated to velocity, `v` in time, t by the application of a constant force, `F` throughout the time, `t`.
The initial and final momentum of the object will be, `p_1 = m u` and `p_2 = m v` respectively.
The change in momentum `∝ \ \ p_2 – p_1`
`∝ \ \ m v – m u`
`∝ \ \ m xx (v – u)`.
The rate of change of momentum ` ∝ \ \ ( m xx (v − u))/t`
Or, the applied force,
` F \ \ ∝ \ \ ( m xx (v − u))/t`
` F = ( k m ( v - u))/t` .......(9.2)
` = k m a` .........(9.3)
Here `a [ = (v – u)//t ]` is the acceleration, which is the rate of change of velocity. The quantity, `k` is a constant of proportionality. The SI units of mass and acceleration are kg and `m s^(-2)` respectively.
The unit of force is so chosen that the value of the constant, `k` becomes one. For this, one unit of force is defined as the amount that produces an acceleration of `1 m \ \ s^(-2)` in an object of `1` kg mass. That is,
`1` unit of force `= k xx (1 kg) xx (1 m \ \ s^(-2))`.
Thus, the value of `k` becomes `1`. From Eq. (9.3)
`F = m a` .......(9.4)
The unit of force is kg `m \ \ s^(-2)` or newton, which has the symbol `N`. The second law of motion gives us a method to measure the force acting on an object as a product of its mass and acceleration.
The second law of motion is often seen in action in our everyday life. Have you noticed that while catching a fast moving cricket ball, a fielder in the ground gradually pulls his hands backwards with the moving ball?
In doing so, the fielder increases the time during which the high velocity of the moving ball decreases to zero. Thus, the acceleration of the ball is decreased and therefore the impact of catching the fast moving ball (Fig. 9.8) is also reduced.
If the ball is stopped suddenly then its high velocity decreases to zero in a very short interval of time.
Thus, the rate of change of momentum of the ball will be large. Therefore, a large force would have to be applied for holding the catch that may hurt the palm of the fielder.
In a high jump athletic event, the athletes are made to fall either on a cushioned bed or on a sand bed. This is to increase the time of the athlete’s fall to stop after making the jump.
This decreases the rate of change of momentum and hence the force. Try to ponder how a karate player breaks a slab of ice with a single blow.
The first law of motion can be mathematically stated from the mathematical expression for the second law of motion. Eq. (9.4) is
`F = ma`
or ` F = ( m(v-u))/t` ........(9.5)
or `Ft = m v – m u`
That is, when `F = 0, v = u` for whatever time, t is taken. This means that the object will continue moving with uniform velocity, u throughout the time, t. If u is zero then v will also be zero. That is, the object will remain at rest.