Applications of Newton's Laws of Motion

Impulse

A force acts for such a small time interval that we measure the effect of the force, called impulse. Impulse is measured by taking the product of the average force applied and the time interval for which the force is applied.

$$\begin{align*} \therefore \text {Impulse} &= F_{avg} \times \Delta t \\ \text {According to Newton’s law of motion, force} \\ F &= \frac {dp}{dt} \\ \text {or,} Fdt &= dp \\ \text {Integrating both the sides with the limit} \\ \int _0^t fdt &= \int _{p_1}^{p_2} dp = p_1 – p_2 \\ \text {where} p_2 \text {and} p_1 \text {are initial and final momentum.} \\ \therefore \text {Impulse} &= p_2 – p_1 \\ \end{align*}$$

So, impulse of a force is measured by the change in linear momentum produced by the force.

Application of the Impulse

1. While catching a cricket ball, the player lowers his hand to increase the time of impact. As F × t = change in momentum, greater the time of impact smaller is the force and vice-versa. However, the change in momentum or impulse is the same in both cases.
2. China wares, glass wares etc are wrapped in straw or paper before packing them. It is because due to the collision, it will take the longer time to reach the impact through straw or paper and small chances of breaking take place.
3. When a person falls on a cemented floor from a height, the person comes to rest in a short time. So he will get more injury. But if he jumps on a sand floor, he will not get many injuries.

Application of Newton’s Laws of Motion

Problem of Masses and pulley

Let us consider light mass m and heavy mass M are connected to two free ends of an inextensible string which pass over a smooth pulley. Let T be the tension in the string. The mass n and M move upwards and downwards respectively with an acceleration a.

Equation of Motion of Mass M

Resultant downward force acting on mass M is given by

$$\begin{align*} F &= Mg – T \\ \text {Since}\: F &= Ma, \\ Ma &= Mg – T \dots (i) \\ \end{align*}$$

Equation of Motion of Mass m

Resultant upward force acting on mass m is given by

$$\begin{align*} F &= T - Mg \\ \text {Since}, \: F &= Ma, \\ Ma &=T - Mg \dots (ii) \\ \text {adding equations} (i) and (ii), \text {we get} \\ (M + m) a &= (M – m) g \\ \text{or,} a &= \left (\frac {M – m}{M + m} \right )g \dots (iii) \\ \text {Putting the value of ‘a’ in equations (i), we get} M\left (\frac {M – m}{M + m} \right )g &= Mg – T \\ \text {or,} \: T &= \left (\frac {2Mm }{M + m} \right )g\dots(iv) \end{align*}$$

Apparent Weight of a Person in a Lift

When a man stands on a weighing machine in a lift as shown in the figure, the weight of mass acts downwards while the machine offers a reaction R upward given by reading of the machine. But if the man and the weighing machine both are in motion, the machine shows the apparent weight which will be different than the actual weight.

1. When the lift is at rest.
   Suppose a man of mass m standing on a lift at rest. Let R is the reaction of the man given by weighing machine. As acceleration is zero, the net force is zero.
   $$\begin{align*} R – mg &= 0 \\ \text {or,} \: R &= mg \\ \end{align*}$$
   So, apparent weight is equal to the actual weight of the man.
2. When the lift is moving with uniform speed upward or downward.
   As acceleration, a = 0 and net force = 0
   $$\begin{align*} R – mg &= 0 \\ \text {or,} \: R &= mg \\ \end{align*}$$
   So, apparent weight is equal to the actual weight of the man.
3. When the lift is accelerating upward.
   Let the lift is accelerating with uniform acceleration in a in upward direction. The weight of man is acting downward and reaction of the man on the floor is in upward direction. Then,
   $$\begin{align*} F &= R – mg \: \text {or} \: ma = R –mg \\ \text{or,} R &= mg + ma = m(g + a) \\ \text {So normal reaction,} \\ R &= mg + ma \\ \end{align*}$$
   This machine shows more reading than the real weight of the man.
4. When the lifting is accelerating downward.
   Let ‘a’ be the downward acceleration of the lift. As the floor is slipping downward, the net force is given by
   $$\begin{align*} F &= mg – R \\ \text {or,} \: ma &= mg – R \\ \text {or,} \: R &= mg – ma = m(g-a) \\ \end{align*}$$
   The reaction of the floor is smaller than the real weight of the man.
5. Free fall of the lift.
   In free fall, acceleration of the body a = g. So, from the second law of motion, the net force acting on the man is
   $$\begin{align*} F &= mg – R \\ \text {or,} \: ma &= mg – R \\ \text {or,} \: R &= mg – mg = m(g-g) = 0 \\ \end{align*}$$
   So, we feel weightlessness.
6. When downward acceleration is greater than g.
   For acceleration, a>g, from the relation R = mg – ma, R will be negative, and man appears to move in upward direction. The man will stick on the ceiling of the lift.