`star` LINEAR MOMENTUM OF A SYSTEM OF PARTICLES

`star` VECTOR PRODUCT OF TWO VECTORS

`star` VECTOR PRODUCT OF TWO VECTORS

`●` Let us recall that the linear momentum of a particle is defined as

`color{blue}{p = m v................(7.12)}`

`●` Let us also recall that Newton’s second law written in symbolic form for a single particle is

`color{blue}{F = (dp)/(dt).................(7.13)}` where `F` is the force on the particle.

`●` Let us consider a system of `n` particles with masses `m_1, m_2 ,...m_n` respectively and velocities `v_1 , v_2 , ,....... v_n` respectively. The particles may be interacting and have external forces acting on them. The linear momentum of the first particle is ` m_1 v_1` , of the second particle is `m_2 v_2` and so on.

`●` For the system of `n` particles, the linear momentum of the system is defined to be the vector sum of all individual particles of the system,

`P = p_1 + p_2 + ..... + p_n`

`color{blue}{= m_1 v_1 + m_2 v_2 .......... + m_n v_n............(7.14)}`

Comparing this with Eq. (7.8)

`color{blue}{P = M V..............(7.15)}`

`●` Thus, the total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its centre of mass. Differentiating Eq. (7.15) with respect to time,

` color{blue}{(dP)/(dt) = M (dV)/(dt) = MA................(7.16)}`

Comparing Eq.(7.16) and Eq. (7.11),

` color{blue}{(dP)/(dt) = F_text(ext)....................(7.17)}`

`●` This is the statement of Newton’s second law extended to a system of particles.

Suppose now, that the sum of external forces acting on a system of particles is zero. Then from Eq.(7.17)

` color{blue}{(dP)/(dt) = 0" or " P = " Constant"...................(7.18)}`

`●` This is the law of conservation of the total linear momentum of a system of particles. Because of Eq. (7.15), this also means that when the total external force on the system is zero the velocity of the centre of mass remains constant. (We assume throughout the discussion on systems of particles in this chapter that the total mass of the system remains constant.)

`●` Note that on account of the internal forces, i.e. the forces exerted by the particles on one another, the individual particles may have complicated trajectories. Yet, if the total external force acting on the system is zero, the centre of mass moves with a constant velocity, i.e., moves uniformly in a straight line like a free particle.

`●` The vector Eq. (7.18a) is equivalent to three scalar equations,

` color{blue}{P_x = c_1, P_y = c_2" and "P_z = c_3............ (7.18 b)}`

`●` Here `P_x, P_y` and `P_z` are the components of the total linear momentum vector `P` along the `x, y` and `z` axes respectively; `c_1, c_2` and `c_3` are constants.

`●` As an example, let us consider the radioactive decay of a moving unstable particle, like the nucleus of radium. A radium nucleus disintegrates into a nucleus of radon and an alpha particle. The forces leading to the decay are internal to the system and the external forces on the system are negligible. So the total linear momentum of the system is the same before and after decay.

`●` The two particles produced in the decay, the radon nucleus and the alpha particle, move in different directions in such a way that their centre of mass moves along the same path along which the original decaying radium nucleus was moving [Fig. 7.13(a)].

`●` If we observe the decay from the frame of reference in which the centre of mass is at rest, the motion of the particles involved in the decay looks particularly simple; the product particles move back to back with their centre of mass remaining at rest as shown in Fig.7.13 (b).

`●` Now In astronomy, binary (double) stars is a common occurrence. If there are no external forces, the centre of mass of a double star moves like a free particle, as shown in Fig.7.14 (a). The trajectories of the two stars of equal mass are also shown in the figure; they look complicated.

`●` If we go to the centre of mass frame, then we find that there the two stars are moving in a circle, about the centre of mass, which is at rest.

`●` Note that the position of the stars have to be diametrically opposite to each other [Fig. 7.14(b)]. Thus in our frame of reference, the trajectories of the stars are a combination of (i) uniform motion in a straight line of the centre of mass and (ii) circular orbits of the stars about the centre of mass.

`●` As can be seen from the two examples, separating the motion of different parts of a system into motion of the centre of mass and motion about the centre of mass is a very useful technique that helps in understanding the motion of the system.

`color{blue}{p = m v................(7.12)}`

`●` Let us also recall that Newton’s second law written in symbolic form for a single particle is

`color{blue}{F = (dp)/(dt).................(7.13)}` where `F` is the force on the particle.

`●` Let us consider a system of `n` particles with masses `m_1, m_2 ,...m_n` respectively and velocities `v_1 , v_2 , ,....... v_n` respectively. The particles may be interacting and have external forces acting on them. The linear momentum of the first particle is ` m_1 v_1` , of the second particle is `m_2 v_2` and so on.

`●` For the system of `n` particles, the linear momentum of the system is defined to be the vector sum of all individual particles of the system,

`P = p_1 + p_2 + ..... + p_n`

`color{blue}{= m_1 v_1 + m_2 v_2 .......... + m_n v_n............(7.14)}`

Comparing this with Eq. (7.8)

`color{blue}{P = M V..............(7.15)}`

`●` Thus, the total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its centre of mass. Differentiating Eq. (7.15) with respect to time,

` color{blue}{(dP)/(dt) = M (dV)/(dt) = MA................(7.16)}`

Comparing Eq.(7.16) and Eq. (7.11),

` color{blue}{(dP)/(dt) = F_text(ext)....................(7.17)}`

`●` This is the statement of Newton’s second law extended to a system of particles.

Suppose now, that the sum of external forces acting on a system of particles is zero. Then from Eq.(7.17)

` color{blue}{(dP)/(dt) = 0" or " P = " Constant"...................(7.18)}`

`color{green}☞` Thus, when the total external force acting on a system of particles is zero, the total linear momentum of the system is constant.

`●` This is the law of conservation of the total linear momentum of a system of particles. Because of Eq. (7.15), this also means that when the total external force on the system is zero the velocity of the centre of mass remains constant. (We assume throughout the discussion on systems of particles in this chapter that the total mass of the system remains constant.)

`●` Note that on account of the internal forces, i.e. the forces exerted by the particles on one another, the individual particles may have complicated trajectories. Yet, if the total external force acting on the system is zero, the centre of mass moves with a constant velocity, i.e., moves uniformly in a straight line like a free particle.

`●` The vector Eq. (7.18a) is equivalent to three scalar equations,

` color{blue}{P_x = c_1, P_y = c_2" and "P_z = c_3............ (7.18 b)}`

`●` Here `P_x, P_y` and `P_z` are the components of the total linear momentum vector `P` along the `x, y` and `z` axes respectively; `c_1, c_2` and `c_3` are constants.

`●` As an example, let us consider the radioactive decay of a moving unstable particle, like the nucleus of radium. A radium nucleus disintegrates into a nucleus of radon and an alpha particle. The forces leading to the decay are internal to the system and the external forces on the system are negligible. So the total linear momentum of the system is the same before and after decay.

`●` The two particles produced in the decay, the radon nucleus and the alpha particle, move in different directions in such a way that their centre of mass moves along the same path along which the original decaying radium nucleus was moving [Fig. 7.13(a)].

`●` If we observe the decay from the frame of reference in which the centre of mass is at rest, the motion of the particles involved in the decay looks particularly simple; the product particles move back to back with their centre of mass remaining at rest as shown in Fig.7.13 (b).

`●` Now In astronomy, binary (double) stars is a common occurrence. If there are no external forces, the centre of mass of a double star moves like a free particle, as shown in Fig.7.14 (a). The trajectories of the two stars of equal mass are also shown in the figure; they look complicated.

`●` If we go to the centre of mass frame, then we find that there the two stars are moving in a circle, about the centre of mass, which is at rest.

`●` Note that the position of the stars have to be diametrically opposite to each other [Fig. 7.14(b)]. Thus in our frame of reference, the trajectories of the stars are a combination of (i) uniform motion in a straight line of the centre of mass and (ii) circular orbits of the stars about the centre of mass.

`●` As can be seen from the two examples, separating the motion of different parts of a system into motion of the centre of mass and motion about the centre of mass is a very useful technique that helps in understanding the motion of the system.

`●` This product is a vector. Two important quantities in the study of rotational motion, namely, moment of a force and angular momentum, are defined as vector products.

`color{purple}bbul{"Definition of Vector Product"}`

`=>` A vector product of two vectors `a` and `b` is a vector `c` such that

(i) magnitude of `c = c = ab sin theta` where `a` and `b` are magnitudes of `a` and `b` and `theta` is the angle between the two vectors.

(ii) `c` is perpendicular to the plane containing `a` and `b`.

(iii) if we take a right handed screw with its head lying in the plane of `a` and `b` and the screw perpendicular to this plane, and if we turn the head in the direction from `a` to `b`, then the tip of the screw advances in the direction of `c`. This right handed screw rule is illustrated in Fig. 7.15a.

`=>` Alternately, if one curls up the fingers of right hand around a line perpendicular to the plane of the vectors `a` and `b` and if the fingers are curled up in the direction from `a` to `b`, then the stretched thumb points in the direction of `c`, as shown in Fig. 7.15b.

`color{green}☞` A simpler version of the right hand rule is the following : Open up your right hand palm and curl the fingers pointing from `a` to `b`. Your stretched thumb points in the direction of `c`.

`color{green}☞` It should be remembered that there are two angles between any two vectors `a` and `b` . In Fig. 7.15 (a) or (b) they correspond to `theta` (as shown) and `(360^0– theta)`. While applying either of the above rules, the rotation should be taken through the smaller angle `(< 180^0)` between `a` and `b`. It is `theta` here.

`●` Because of the cross used to denote the vector product, it is also referred to as cross product.

`►` Note that scalar product of two vectors is commutative as said earlier, `a*b = b*a`

`●` The vector product, however, is not commutative, i.e. `a xx b ne b xx a` The magnitude of both `a xx b` and `b xx a` is the same `( ab sin theta )` ; also, both of them are perpendicular to the plane of `a` and `b`. But the rotation of the right-handed screw in case of `a xx b` is from `a` to `b`, whereas in case of `b xx a` it is from `b` to `a`. This means the two vectors are in opposite directions. We have

`color{blue}{a xx b=-b xx a}`

`►` Another interesting property of a vector product is its behaviour under reflection. Under reflection (i.e. on taking the mirror image) we have `x -> x,y -> y` and `z -> z` . As a result all the components of a vector change sign and thus `a ->-a, b -> −b` . What happens to `a xx b` under reflection?

`a xx b->(−a)xx (−b) = a xx b`

Thus, `a xx b` does not change sign under reflection.

`►` Both scalar and vector products are distributive with respect to vector addition.

Thus,

`a*(b + c) = a*b + a*c`

`color{blue}{a xx (b + c) = a xx b + a xx c}`

`►` We may write `c = a xx b` in the component form. For this we first need to obtain some elementary cross products:

(i) `a xx a = 0` (`0` is a null vector, i.e. a vector with zero magnitude)

This follows since magnitude of `a xx a` is

`a^2 sin 0^o=0`

From this follow the results

`hat i xx hat i =0, hat j xx hat j=0, hat k xx hat k =0`

(ii) `hat i xx hat j = hat k`

`●` Note that the magnitude of `hat i × hat j` is `sin90^0` or `1`, since `hat i` and `hat j` both have unit magnitude and the angle between them is `90^0`. Thus, `hat i × hat j` is a unit vector. A unit vector perpendicular to the plane of `hat i` and `hat j` and related to them by the right hand screw rule is `hat k` . Hence, the above result. You may verify similarly,

`hat j xx hat k = hat i` and `hat k xx hat i = hat j`

From the rule for commutation of the cross product, it follows:

`hat j xx hat i = −hat k, hat k × hat j = −hat i, hat i × hat k = −hat j`

`●` Note if `hat i,hat j, hat k` occur cyclically in the above vector product relation, the vector product is positive. If `hat i , hat j , hat k` do not occur in cyclic order, the vector product is negative.

Now, `a xx b = (a_x hat i + a_y hat j + a_z hat k) xx( b_x hat i + b_y hat j + b_z hat k)`

` = a_x b_y hat k − a_x b_z hat j − a_y b_x hat k + a_y b_z hat i + a_z b_x hat j − a_z b_y hat i`

`=>` We have used the elementary cross products in obtaining the above relation. The expression for `a xx b` can be put in a determinant form which is easy to remember.

`color{purple}bbul{"Definition of Vector Product"}`

`=>` A vector product of two vectors `a` and `b` is a vector `c` such that

(i) magnitude of `c = c = ab sin theta` where `a` and `b` are magnitudes of `a` and `b` and `theta` is the angle between the two vectors.

(ii) `c` is perpendicular to the plane containing `a` and `b`.

(iii) if we take a right handed screw with its head lying in the plane of `a` and `b` and the screw perpendicular to this plane, and if we turn the head in the direction from `a` to `b`, then the tip of the screw advances in the direction of `c`. This right handed screw rule is illustrated in Fig. 7.15a.

`=>` Alternately, if one curls up the fingers of right hand around a line perpendicular to the plane of the vectors `a` and `b` and if the fingers are curled up in the direction from `a` to `b`, then the stretched thumb points in the direction of `c`, as shown in Fig. 7.15b.

`color{green}☞` A simpler version of the right hand rule is the following : Open up your right hand palm and curl the fingers pointing from `a` to `b`. Your stretched thumb points in the direction of `c`.

`color{green}☞` It should be remembered that there are two angles between any two vectors `a` and `b` . In Fig. 7.15 (a) or (b) they correspond to `theta` (as shown) and `(360^0– theta)`. While applying either of the above rules, the rotation should be taken through the smaller angle `(< 180^0)` between `a` and `b`. It is `theta` here.

`●` Because of the cross used to denote the vector product, it is also referred to as cross product.

`►` Note that scalar product of two vectors is commutative as said earlier, `a*b = b*a`

`●` The vector product, however, is not commutative, i.e. `a xx b ne b xx a` The magnitude of both `a xx b` and `b xx a` is the same `( ab sin theta )` ; also, both of them are perpendicular to the plane of `a` and `b`. But the rotation of the right-handed screw in case of `a xx b` is from `a` to `b`, whereas in case of `b xx a` it is from `b` to `a`. This means the two vectors are in opposite directions. We have

`color{blue}{a xx b=-b xx a}`

`►` Another interesting property of a vector product is its behaviour under reflection. Under reflection (i.e. on taking the mirror image) we have `x -> x,y -> y` and `z -> z` . As a result all the components of a vector change sign and thus `a ->-a, b -> −b` . What happens to `a xx b` under reflection?

`a xx b->(−a)xx (−b) = a xx b`

Thus, `a xx b` does not change sign under reflection.

`►` Both scalar and vector products are distributive with respect to vector addition.

Thus,

`a*(b + c) = a*b + a*c`

`color{blue}{a xx (b + c) = a xx b + a xx c}`

`►` We may write `c = a xx b` in the component form. For this we first need to obtain some elementary cross products:

(i) `a xx a = 0` (`0` is a null vector, i.e. a vector with zero magnitude)

This follows since magnitude of `a xx a` is

`a^2 sin 0^o=0`

From this follow the results

`hat i xx hat i =0, hat j xx hat j=0, hat k xx hat k =0`

(ii) `hat i xx hat j = hat k`

`●` Note that the magnitude of `hat i × hat j` is `sin90^0` or `1`, since `hat i` and `hat j` both have unit magnitude and the angle between them is `90^0`. Thus, `hat i × hat j` is a unit vector. A unit vector perpendicular to the plane of `hat i` and `hat j` and related to them by the right hand screw rule is `hat k` . Hence, the above result. You may verify similarly,

`hat j xx hat k = hat i` and `hat k xx hat i = hat j`

From the rule for commutation of the cross product, it follows:

`hat j xx hat i = −hat k, hat k × hat j = −hat i, hat i × hat k = −hat j`

`●` Note if `hat i,hat j, hat k` occur cyclically in the above vector product relation, the vector product is positive. If `hat i , hat j , hat k` do not occur in cyclic order, the vector product is negative.

Now, `a xx b = (a_x hat i + a_y hat j + a_z hat k) xx( b_x hat i + b_y hat j + b_z hat k)`

` = a_x b_y hat k − a_x b_z hat j − a_y b_x hat k + a_y b_z hat i + a_z b_x hat j − a_z b_y hat i`

`color{red}{axxb=( a_y b_z − a_z b_ i)hati + (a_z b_x − a_x b_z ) hat j + (a_x b_y − a_y b_x)hat k}`

`=>` We have used the elementary cross products in obtaining the above relation. The expression for `a xx b` can be put in a determinant form which is easy to remember.

`color{orange}{a xx b = |(hat i , hat j , hat k),(a_x , a_y , a_z),(b_x, b_y , b_z)|}`

Q 3210734619

Find the scalar and vector products of two vectors. `a = (3 hat i – 4hat j + 5 hat k )` and `b = – (2 hat i + hat j – 3 hat k )`

`color{blue}{a * b} =(3 hat i - 4 hat j + 5 hat k) * (-2 hat i + hat j - 3 hat k)`

`=-6 -4 -15`

`-25`

`a xx b = | (hat i , hat j , hat k),(3,-4,5),(-2, 1, -3)| = 7 hat i - hat j - 5 hat k`

Note `b xx a = −7 hat i + hat j + 5 hat k`