Newton's Second Law-Momentum

Newton's Second Law-Momentum

The law refers to the simple case when the net external force on a body is zero.The second law of motion refers to the general situation when there is a net external force acting on the body.It relates the net external force to the acceleration of the body.

Momentum

Momentum of a body is defined to be the product of its mass m and velocity v, and is denoted by p:

​​​​​​​              p = m v               

Momentum is clearly a vector quantity. The following common experiences indicate the importance of this quantity for considering the effect of force on motion.

1. Suppose a light-weight vehicle (say a small car) and a heavy weight vehicle (say a loaded truck) are parked on a horizontal road. We all know that a much greater force is needed to push the truck than the car to bring them to the same speed in same time. Similarly, a greater opposing force is needed to stop a heavy body than a light body in the same time, if they are moving with the same speed.

2. 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.

3. Speed is another important parameter to consider. A bullet fired by a gun can easily pierce human tissue before it stops, resulting in 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.

4. 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 the 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.

                                        ​​​​​​​                                  

                       fig.5.3.Force not only depends on the change in momentum but also on how fast the change is brought about.

                A seasoned cricketer draws in his hands during a catch, allowing greater time for the ball to stop and hence requires a smaller force.

5.​​​​​​​ Observations confirm that the product of mass and velocity (i.e. momentum) is basic to the effect of force on motion. Suppose a fixed force is applied for a certain interval of time on two bodies of different masses, initially at rest, the lighter body picks up a greater speed than the heavier body. However, at the end of the time interval, observations show that each body acquires the same momentum. Thus the same force for the same time causes the same change in momentum for different bodies. This is a crucial clue to the second law of motion.

6. ​​​​​​​In the preceding observations, the vector 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. 5.4). 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.5.4.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 \(\Delta\)t, the velocity of a body of mass m changes from v to v + \(\Delta\)v i.e. its initial momentum p=mv changes by p=mv.According to the Second Law,

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOraiabg6 % HiLoaalaaabaGaeyiLdqKaamiCaaqaaiabgs5aejaadshaaaGaam4B % aiaadkhacaWGgbGaeyypa0Jaam4Aaiabg2da9maalaaabaGaeyiLdq % KaamiCaaqaaiabgs5aejaadshaaaaaaa!477C! F\infty \frac{{\Delta p}}{{\Delta t}}orF = k = \frac{{\Delta p}}{{\Delta t}}\)

where k is a constant of proportionality. Taking the limit \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam % iDaiabgkziUkaaicdaaaa!3AFD! \Delta t \to 0\), the term \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % GHuoarcaWGWbaabaGaeyiLdqKaamiDaaaaaaa!3AC2! \frac{{\Delta p}}{{\Delta t}}\) becomes the derivative or differential co-efficient of p with

respect to t, denoted by \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGKbGaamiCaaqaaiaadsgacaWG0baaaaaa!39C6! \frac{{dp}}{{dt}}\).Thus \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOraiabg2 % da9iaadUgadaWcaaqaaiaadsgacaWGWbaabaGaamizaiaadshaaaaa % aa!3C87! F = k\frac{{dp}}{{dt}}\)

For a body of fixed mass m,

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGKbGaamiCaaqaaiaadsgacaWG0baaaiabg2da9maalaaabaGaamiz % aaqaaiaadsgacaWG0baaamaabmaabaGaamyBaiaadAhaaiaawIcaca % GLPaaacqGH9aqpcaWGTbWaaSaaaeaacaWGKbGaamODaaqaaiaadsga % caWG0baaaiabg2da9iaad2gacaWGHbaaaa!49C9! \frac{{dp}}{{dt}} = \frac{d}{{dt}}\left( {mv} \right) = m\frac{{dv}}{{dt}} = ma\)

i.e the Second Law can also be written as

F = k m a       (5.4)

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. (5.4) to define the unit of force. We, therefore, have the liberty to choose any constant value for k. For simplicity, we choose k = 1. The second law then is \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOramaala % aabaGaamizaiaadchaaeaacaWGKbGaamiDaaaacqGH9aqpcaWGTbGa % amyyaaaa!3D6F! F=\frac{{dp}}{{dt}} = ma\).

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

Let us note at this stage some important points about the second law :

1. In the second law, F = 0 implies 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 :

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGgb % WaaSbaaSqaaiaadIhaaeqaaOGaeyypa0ZaaSaaaeaacaWGKbGaamiC % amaaBaaaleaacaWG4baabeaaaOqaaiaadsgacaWG0baaaiabg2da9i % Gac2gacaGGHbWaaSbaaSqaaiaadIhaaeqaaaGcbaGaamOramaaBaaa % leaacaWG5baabeaakiabg2da9maalaaabaGaamizaiaadchadaWgaa % WcbaGaamyEaaqabaaakeaacaWGKbGaamiDaaaacqGH9aqpciGGTbGa % aiyyamaaBaaaleaacaWG5baabeaaaOqaaiaadAeadaWgaaWcbaGaam % OEaaqabaGccqGH9aqpdaWcaaqaaiaadsgacaWGWbWaaSbaaSqaaiaa % dQhaaeqaaaGcbaGaamizaiaadshaaaGaeyypa0JaciyBaiaacggada % WgaaWcbaGaamOEaaqabaaaaaa!5A45! \begin{array}{l} {F_x} = \frac{{d{p_x}}}{{dt}} = {{\mathop{\rm ma}\nolimits} _x}\\ {F_y} = \frac{{d{p_y}}}{{dt}} = {{\mathop{\rm ma}\nolimits} _y}\\ {F_z} = \frac{{d{p_z}}}{{dt}} = {{\mathop{\rm ma}\nolimits} _z} \end{array}\)

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).

3. The second law of motion given by Eq. (5.5) 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.

the acceleration of the centre of mass of the system about which we shall study in detail in chapter 7. Any internal forces in the system are not to be included in F.

              fig.5.5.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.

4. 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 

ANSWER

The retardation ‘a’ of the bullet (assumed constant) is given by \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabg2 % da9maalaaabaGaeyOeI0IaamyDamaaCaaaleqabaGaaGOmaaaaaOqa % aiaaikdacaWGZbaaaiabg2da9maalaaabaGaeyOeI0IaaGyoaiaaic % dacqGHxdaTcaaI5aGaaGimaaqaaiaaikdacqGHxdaTcaaIWaGaaiOl % aiaaiAdaaaGaamyBaiaadohadaahaaWcbeqaaiabgkHiTiaaikdaaa % GccqGH9aqpcqGHsislcaaI2aGaaG4naiaaiwdacaaIWaGaamyBaiaa % dohadaahaaWcbeqaaiabgkHiTiaaikdaaaaaaa!550A! a = \frac{{ - {u^2}}}{{2s}} = \frac{{ - 90 \times 90}}{{2 \times 0.6}}m{s^{ - 2}} = - 6750m{s^{ - 2}}\) 

The retarding force, by the second law of motion, is= 0.04 kg × 6750 m s^-2 = 270 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 3

The motion of a particle of mass m is decribed by \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadwhacaWG0bGaey4kaSYaaSaaaeaacaaIXaaabaGaaGOmaaaa % caWGNbGaamiDamaaCaaaleqabaGaaGOmaaaaaaa!3F24! y = ut + \frac{1}{2}g{t^2}\).Find the force acting on the particle.

ANSWER

We know,

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG5b % Gaeyypa0JaamyDaiaadshacqGHRaWkdaWcaaqaaiaaigdaaeaacaaI % YaaaaiaadEgacaWG0bWaaWbaaSqabeaacaaIYaaaaaGcbaGaamOtai % aad+gacaWG3baabaGaamODaiabg2da9maalaaabaGaamizaiaadMha % aeaacaWGKbGaamiDaaaacqGH9aqpcaWG1bGaey4kaSIaam4zaiaads % haaeaacaWGHbGaam4yaiaadogacaWGLbGaamiBaiaadwgacaWGYbGa % amyyaiaadshacaWGPbGaam4Baiaad6gacaGGSaGaamyyaiabg2da9m % aalaaabaGaamizaiaadAhaaeaacaWGKbGaamiDaaaacqGH9aqpcaWG % Nbaaaaa!6025! \begin{array}{l} y = ut + \frac{1}{2}g{t^2}\\ Now\\ v = \frac{{dy}}{{dt}} = u + gt\\ acceleration,a = \frac{{dv}}{{dt}} = g \end{array}\)

Then the force is given by Eq. (5.5)

F = ma = mg

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.

Newton's Second Law-Impulse

Newton's Second Law-Impulse

We sometimes encounter examples where a large force acts for a very short duration producing a finite change in momentum of the body. For example, when a ball hits a wall and bounces back, the force on the ball by the wall acts for a very short time when the two are in contact, yet the force is large enough to reverse the momentum of the ball.Often, in these situations, the force and the time duration are difficult to ascertain separately. However, the product of force and time, which is the change in momentum of the body remains a measurable quantity. This product is called impulse:

Impulse = Force × time duration

= Change in momentum      (5.7)

A large force acting for a short time to produce a finite change in momentum is called an impulsive force. In the history of science, impulsive forces were put in a conceptually different category from ordinary forces. Newtonian mechanics has no such distinction. Impulsive force is like any other force – except that it is large and acts for a short time.

EXAMPLE 4

A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m s^–1.If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball. (Assume linear motion of the ball).

ANSWER

Change in momentum

=0.15\(\times\)12-(-0.15\(\times\)12)

=3.6 Ns,

Impulse = 3.6 N s,in the direction from the batsman to the bowler.

This is an example where the force on the ball by the batsman and the time of contact of the ball and the bat are difficult to know, but the impulse is readily calculated.