Conservation of Linear Momentum
The total quantity of motion contained in a body is called linear momentum. It is denoted by $\vec{P}$ and defined as the product of mass and velocity of the body. For a body of mass $m$ and velocity $v$, its linear momentum $\vec{P}$ in vector form is given by:
$$\vec{P} = m\vec{v}$$In terms of magnitude only,
$$P = mv$$SI unit of linear momentum is $\text{kg m s}^{-1}$ or Ns, and dimensional formula is $[MLT^{-1}]$.
Law of Conservation of Linear Momentum
It states that, "the total linear momentum of a system of particles remains constant, provided that no external force acts on the system".
Proof:
From Newton's second law of motion,
$$F = \frac{dP}{dt}$$In absence of force,
$$\frac{dP}{dt} = 0 \;\Rightarrow\; P = \text{constant}$$Proof (for a system of particles):
Let us consider a system of $n$ particles of total mass $M$. Then,
$$M = m_1 + m_2 + \cdots + m_n$$So,
$$P = P_1 + P_2 + \cdots + P_n$$Differentiating the above equation w.r.t. $t$,
$$\frac{dP}{dt} = \frac{dP_1}{dt} + \frac{dP_2}{dt} + \cdots + \frac{dP_n}{dt}$$or,
$$\frac{dP}{dt}\;(=F) = F_1 + F_2 + \cdots + F_n$$In the system of particles there are both external and internal forces.
$$\therefore \frac{dP}{dt} = F_1^{ext} + F_1^{int} + F_2^{ext} + F_2^{int} + \cdots + F_n^{ext} + F_n^{int}$$or,
$$\frac{dP}{dt} = \left(F_1^{ext}+F_2^{ext}+\cdots+F_n^{ext}\right) + \left(F_1^{int}+F_2^{int}+\cdots+F_n^{int}\right)$$Internal forces do not contribute to the total force because they appear in pairs of equal and opposite forces.
$$\Rightarrow F^{int} = 0$$ $$\Rightarrow \frac{dP}{dt} = F^{ext}$$$\therefore$ Total momentum of the system can be changed only by changing the external forces.
Centre of Mass (C.M.)
It is the point at which the whole mass of the body is supposed to be concentrated.
When force is applied to the centre of mass, the object moves in the direction of the force without rotation.
Centre of Mass Frame of Reference (or Inertial Frame of Reference)
The inertial frame or centre-of-mass frame of reference is that frame in which the centre of mass is at rest.
$$\text{i.e.}\quad V_{cm} = 0 \;\Rightarrow\; P_{cm} = M\cdot V_{cm} = 0$$So, it is also called the zero momentum frame of reference.
In absence of external force, the centre of mass moves with constant velocity in the inertial or centre-of-mass frame of reference.
Lab Frame of Reference
The reference frame connected to the lab is called the lab frame of reference.
Motion of Centre of Mass
Let $M$ = constant. Then, the position vector of the centre of mass (C.M.) is:
$$R = \frac{\sum_{i=1}^{n} m_i r_i}{\sum_{i=1}^{n} m_i}$$or,
$$R = \frac{m_1 r_1 + m_2 r_2 + \cdots + m_n r_n}{m_1+m_2+\cdots+m_n}$$or,
$$R = \frac{m_1 r_1 + m_2 r_2 + \cdots + m_n r_n}{M}$$or,
$$MR = m_1 r_1 + m_2 r_2 + \cdots + m_n r_n \qquad \cdots (i)$$Differentiating w.r.t. $t$,
$$M\frac{dR}{dt} = m_1\frac{dr_1}{dt} + m_2\frac{dr_2}{dt} + \cdots + m_n\frac{dr_n}{dt}$$or,
$$MV = m_1v_1 + m_2v_2 + \cdots + m_n v_n \qquad \cdots (ii)$$or,
$$P = \sum m_k v_k \qquad \cdots (iii)$$In absence of external force,
$$P = \sum m_k v_k = \text{constant} \quad\text{and,}$$ $$MV = \text{constant} \qquad (\because P = MV)$$ $$V = \frac{\text{constant}}{M}$$ $$\Rightarrow V = \frac{P}{M} \qquad (\because P = \text{constant})$$$\therefore$ Velocity of the centre of mass remains constant in absence of external force.
Co-ordinate Transformation between Lab Frame and Centre of Mass Frame
Let us consider two frames of reference: the lab frame and the C.M. frame.
Let $R$ = position of C.M. from lab frame, $r$ = position of particle from lab frame, $r'$ = position of particle from C.M. frame.
From the triangle law of vector addition,
$$r = R + r' \qquad \cdots (i)$$Differentiating w.r.t. $t$,
$$\frac{dr}{dt} = \frac{dR}{dt} + \frac{dr'}{dt}$$ $$v = V_{cm} + v'$$Head-on Collision of Two Particles (One Dimensional)
Let $m_1$ and $m_2$ move with initial velocities $u_1$ and $u_2$ ($u_1 > u_2$) and final velocities $v_1$ and $v_2$ respectively.
For elastic collision:
From conservation of linear momentum,
$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$or,
$$m_1(u_1-v_1) = m_2(v_2-u_2) \qquad \cdots (i)$$From conservation of kinetic energy,
$$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$$or,
$$m_1(u_1^2-v_1^2) = m_2(v_2^2-u_2^2) \qquad \cdots (ii)$$Dividing eq.(ii) by eq.(i),
$$u_1+v_1 = v_2+u_2 \qquad \cdots (iii)$$or,
$$u_1-u_2 = v_2-v_1$$or,
$$u_2-u_1 = -(v_2-v_1) \qquad \cdots (iv)$$$\therefore$ Relative velocity before collision $= -$ relative velocity after collision.
From eq.(iii),
$$v_1 = v_2+u_2-u_1 \qquad \cdots (v)$$and,
$$v_2 = u_1+v_1-u_2 \qquad \cdots (vi)$$Replacing the value of $v_1$ from eq.(v) into eq.(i), we get,
$$m_1\{u_1-(v_2+u_2-u_1)\} = m_2(v_2-u_2)$$or,
$$m_1u_1 - m_1v_2 - m_1u_2 + m_1u_1 = m_2v_2 - m_2u_2$$or,
$$2m_1u_1 + (m_2-m_1)u_2 = (m_2+m_1)v_2$$or,
$$v_2 = \frac{2m_1u_1}{(m_1+m_2)} + \frac{(m_2-m_1)u_2}{(m_1+m_2)}$$ $$v_2 = \frac{2m_1u_1}{m_1+m_2} + \frac{(m_1-m_2)u_2}{m_1+m_2}$$Similarly,
$$v_1 = \frac{2m_2u_2}{m_1+m_2} + \frac{(m_1-m_2)u_1}{m_1+m_2}$$Case I: When $m_1 = m_2$
$$v_2 = u_1 \qquad \text{and} \qquad v_1 = u_2$$Therefore, the velocities of the particles exchange.
Case II: When $u_2 = 0$ (second mass is at rest)
Then,
$$v_1 = \frac{(m_1-m_2)u_1}{m_1+m_2} \qquad \text{and,} \qquad v_2 = \frac{2m_1u_1}{m_1+m_2}$$Also, fraction of kinetic energy imparted to the second body is:
$$\frac{\frac{1}{2}m_2v_2^2}{\frac{1}{2}m_1u_1^2} = \frac{m_2\left(\frac{2m_1u_1}{m_1+m_2}\right)^2}{m_1u_1^2} = \frac{4m_1m_2}{(m_1+m_2)^2}$$which can also be written as
$$= \frac{4\frac{m_2}{m_1}}{\left(1+\frac{m_2}{m_1}\right)^2}$$(a) When $m_1 = m_2$:
Fraction of K.E. imparted to second body $= 1$. In this case, after collision, the first particle comes to rest and the second particle moves with the initial kinetic energy of the first particle.
(b) When $m_2 \gg m_1$, then:
$$v_1 \approx -u_1, \qquad v_2 \approx 0$$It indicates that when a lighter particle collides against a much more massive particle at rest, the latter continues to remain at rest and the velocity of the former gets reversed.
(c) When $m_1 \gg m_2$:
$$v_1 \approx u_1, \qquad v_2 \approx 2u_1$$It indicates that the velocity of the massive particle remains practically unaltered on collision with the lighter particle at rest, and the lighter particle acquires nearly twice the initial velocity of the massive particle.
Deflection of a Moving Particle by a Particle at Rest
(Condition: perfectly elastic collision in two dimensions)
(I) In Lab Frame
Let $m_1$ = mass of first body, $m_2$ = mass of second body at rest, $u_1$ = initial velocity of mass $m_1$, $u_2$ = initial velocity of mass $m_2 = 0$, $v_1$ = final velocity of mass $m_1$, $v_2$ = final velocity of mass $m_2$, $\theta_1$ = angle of deflection of $m_1$, $\theta_2$ = angle of deflection of $m_2$.
From conservation of linear momentum along the $x$-axis:
$$m_1u_1 = m_1v_1\cos\theta_1 + m_2v_2\cos\theta_2 \qquad \cdots (1)$$and for the $y$-axis:
$$0 = m_1v_1\sin\theta_1 - m_2v_2\sin\theta_2 \qquad \cdots (2)$$From conservation of kinetic energy,
$$\frac{1}{2}m_1u_1^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 \qquad \cdots (3)$$If we know the initial conditions for $m_1, m_2$ and $u_1$, we have to determine four unknown quantities, i.e. $v_1, v_2, \theta_1, \theta_2$, but we have only three equations connecting them.
So, let us assume $m_1 = m_2$; then eq.(3) becomes:
$$u_1^2 = v_1^2 + v_2^2$$or,
$$v_2^2 = u_1^2 - v_1^2 \qquad \cdots (4)$$And from eq.(2),
$$v_1\sin\theta_1 = v_2\sin\theta_2$$or,
$$\sin\theta_2 = \frac{v_1\sin\theta_1}{v_2} \qquad \cdots (5)$$And from the figure,
$$v_1 = u_1\cos\theta_1 \qquad \cdots (6)$$By solving the above equations, we can determine $v_1, v_2, \theta_1$ and $\theta_2$.
(II) In C.M. Frame
Let $u_1'$ = initial velocity of mass $m_1$ in C.M. frame, $u_2'$ = initial velocity of mass $m_2$ in C.M. frame, $v_1'$ = final velocity of mass $m_1$ in C.M. frame, $v_2'$ = final velocity of mass $m_2$ in C.M. frame.
Since momentum of C.M. ($P_{cm}$) $= 0$. Applying the conservation law before collision:
$$m_1u_1' + m_2u_2' = 0$$or,
$$m_1u_1' = -m_2u_2'$$or,
$$|m_2u_2'| = |m_1u_1'|$$ $$\Rightarrow u_2' = \frac{m_1}{m_2}u_1' \qquad \cdots (1)$$Applying the conservation law after collision,
$$m_1v_1' + m_2v_2' = 0$$or,
$$m_2v_2' = -m_1v_1'$$or,
$$|m_2v_2'| = |m_1v_1'|$$or,
$$m_2v_2' = m_1v_1'$$ $$\Rightarrow v_2' = \frac{m_1}{m_2}v_1' \qquad \cdots (2)$$From conservation of kinetic energy,
$$\frac{1}{2}m_1u_1'^2 + \frac{1}{2}m_2u_2'^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2v_2'^2$$or,
$$\frac{1}{2}m_1u_1'^2 + \frac{1}{2}m_2\left(\frac{m_1}{m_2}u_1'\right)^2 = \frac{1}{2}m_1v_1'^2 + \frac{1}{2}m_2\left(\frac{m_1}{m_2}v_1'\right)^2$$or,
$$m_1u_1'^2\left(1+\frac{m_1}{m_2}\right) = m_1v_1'^2\left(1+\frac{m_1}{m_2}\right)$$or,
$$u_1'^2 = v_1'^2$$or,
$$u_1' = v_1'$$ $$\Rightarrow v_1' = -u_1' \qquad \cdots (3)$$Similarly,
$$\Rightarrow v_2' = -u_2' \qquad \cdots (4)$$Thus, in the centre-of-mass frame of reference, the magnitudes of the velocities of the particles remain unaltered in an elastic collision.
If $u_1$ = velocity of $m_1$ in lab frame, $u_2$ = velocity of $m_2$ in lab frame. Then,
$$u_1' = u_1 - V_{cm} \qquad \cdots (5) \qquad (r = R_{cm}+r',\ v = V_{cm}+v')$$ $$u_2' = u_2 - V_{cm} \qquad \cdots (6)$$Since,
$$V_{cm} = \frac{m_1u_1+m_2u_2}{m_1+m_2} \qquad \left(P_{cm}=MV_{cm} \;\therefore\; V_{cm}=\frac{P}{M}\right)$$Then,
$$u_1' = u_1 - \frac{m_1u_1+m_2u_2}{m_1+m_2}$$or,
$$u_1' = \frac{m_1u_1+m_2u_1-m_1u_1-m_2u_2}{m_1+m_2}$$or,
$$u_1' = \frac{m_2(u_1-u_2)}{m_1+m_2}$$or,
$$u_1' = -\frac{m_2(u_2-u_1)}{m_1+m_2}$$ $$\Rightarrow v_1' = -u_1' = \frac{m_2(u_2-u_1)}{m_1+m_2}$$Again, from eq.(6),
$$u_2' = u_2 - \frac{m_1u_1+m_2u_2}{m_1+m_2}$$or,
$$u_2' = \frac{m_1u_2+m_2u_2-m_1u_1-m_2u_2}{m_1+m_2}$$or,
$$u_2' = \frac{m_1(u_2-u_1)}{m_1+m_2}$$ $$\Rightarrow v_2' = -u_2' = -\frac{m_1(u_2-u_1)}{m_1+m_2}$$Value of Scattering Angle
(i) In C.M. Frame of Reference
In the C.M. frame of reference there is no limitation on the value of scattering angle $\theta$. So, it can have any possible value.
(ii) In Lab Frame of Reference
In this frame of reference, there are some restrictions on the value of the scattering angle $\theta_1$. It will be clear from the following explanation.
From the figure,
$$\tan\theta_1 = \frac{v_1\sin\theta_1}{v_1\cos\theta_1} \qquad \cdots (1)$$In this case, the $y$-component of the first particle in the lab frame and C.M. frame are equal.
$$\Rightarrow v_1\sin\theta_1 = v_1'\sin\theta$$But the $x$-component of this velocity differs by $V_{cm}$ in the two frames of reference.
$$\therefore v_1\cos\theta_1 = v_1'\cos\theta + V$$Replacing this value in equation (1), we get,
$$\tan\theta_1 = \frac{v_1'\sin\theta_1}{v_1'\cos\theta+V} = \frac{\sin\theta_1}{\cos\theta + \frac{V}{v_1'}} \qquad \cdots (2)$$As we know,
$$V = \frac{m_1u_1}{m_1+m_2}$$or,
$$V = \frac{m_1(u_1'+V)}{m_1+m_2}$$or,
$$V = \frac{m_1u_1'}{m_1+m_2} + \frac{m_1V}{m_1+m_2}$$or,
$$V - \frac{m_1V}{m_1+m_2} = \frac{m_1u_1'}{m_1+m_2}$$or,
$$\frac{m_2V}{m_1+m_2} = \frac{m_1u_1'}{m_1+m_2}$$or,
$$m_2V = m_1u_1'$$ $$\therefore V = \frac{m_1}{m_2}u_1' \qquad \cdots (3)$$Since $u_1' = v_1'$ (for magnitude only),
$$\Rightarrow V = \frac{m_1}{m_2}v_1'$$ $$\Rightarrow \frac{V}{v_1'} = \frac{m_1}{m_2}$$Replacing this value in eq.(2), we get,
$$\tan\theta_1 = \frac{\sin\theta}{\cos\theta + \frac{m_1}{m_2}} \qquad \cdots (4)$$Case 1: If $m_1 \gg m_2$
If $m_1$ is greater than $m_2$, then $\frac{m_1}{m_2}>1$. So, the denominator can never be $0$. So, $\theta_1$ must be less than $90^\circ$.
That means, if a massive particle collides with a light particle at rest, then it cannot bounce back.
Case 2: If $m_1 = m_2$
Then $\frac{m_1}{m_2}=1$. The denominator can be zero if $\cos\theta = -1$. In this case, $\tan\theta_1$ can be infinity and $\theta_1$ can have value up to a limiting value $90^\circ$.
That means, collision between two particles of equal mass where the second particle is at rest — then the two particles may move at right angles to each other.
Case 3
The value of $\tan\theta_1$ can also be negative. So, for this case, all values of $\theta_1$ can be possible.
If a light particle collides with a massive one at rest, then it bounces back along the original path.
The collision for which total linear momentum remains conserved but K.E. does not conserve is called inelastic collision.
Perfectly Inelastic Collision in One Dimension
1. In Lab Frame
Let $u_1$ = velocity of mass $m_1$ before collision, $v$ = velocity of both masses after collision (with $u_2 = 0$).
Then, from conservation of momentum,
$$m_1u_1 = (m_1+m_2)v$$ $$v = \frac{m_1u_1}{m_1+m_2}$$Since $\frac{m_1}{m_1+m_2}<1 p="">
$$\Rightarrow u_1>v \quad\text{or}\quad v
Case ii: if $m_1 \gg m_2$,
$$v = u_1$$Case iii: if $m_1 \ll m_2$,
$$v \approx 0$$And,
$$\frac{(K.E.)_{final}}{(K.E.)_{initial}} = \frac{\frac{1}{2}(m_1+m_2)v^2}{\frac{1}{2}m_1u_1^2} = \frac{\frac{1}{2}(m_1+m_2)\cdot\frac{m_1^2u_1^2}{(m_1+m_2)^2}}{\frac{1}{2}m_1u_1^2}$$or,
$$\frac{(K.E.)_{final}}{(K.E.)_{initial}} = \frac{m_1}{m_1+m_2} < 1$$ $$\therefore \frac{(K.E.)_{final}}{(K.E.)_{initial}} < 1$$ $$\Rightarrow (K.E.)_{final} < (K.E.)_{initial}$$2. In Centre of Mass Frame
(use prime for C.M. frame)
Velocity of C.M.,
$$V_{cm} = \frac{m_1u_1+m_2u_2}{m_1+m_2} = \frac{m_1u_1}{m_1+m_2} \qquad \cdots (1) \qquad (\because m_2 \text{ is at rest, so } u_2=0)$$From conservation of linear momentum,
$$m_1u_1' + m_2u_2' = (m_1+m_2)v'$$or,
$$v' = \frac{m_1u_1'+m_2u_2'}{m_1+m_2} \qquad \cdots (2)$$We know,
$$u_1' = u_1 - V_{cm} = u_1 - \frac{m_1u_1}{m_1+m_2} = \frac{m_1u_1+m_2u_1-m_1u_1}{m_1+m_2} = \frac{m_2u_1}{m_1+m_2}$$and,
$$u_2' = u_2 - V_{cm} = 0 - \frac{m_1u_1}{m_1+m_2} = -\frac{m_1u_1}{m_1+m_2}$$Replacing $u_1'$ and $u_2'$ in equation (2), we get,
$$v' = \frac{m_1\cdot\frac{m_2u_1}{m_1+m_2}}{m_1+m_2} - \frac{m_2\cdot\frac{m_1u_1}{m_1+m_2}}{m_1+m_2}$$or,
$$v' = \frac{0}{m_1+m_2} = 0$$$\Rightarrow$ Velocity of the centre of mass is equal to the velocity of the frame itself.
Angular Momentum
The moment of linear momentum is called angular momentum. It is given by:
$$\vec{J} = \vec{r}\times\vec{p}$$Differentiating w.r.t. $t$, we get,
$$\frac{d\vec{J}}{dt} = \frac{d}{dt}(\vec{r}\times\vec{p})$$ $$= \vec{r}\frac{d\vec{p}}{dt} + \vec{p}\frac{d\vec{r}}{dt}$$ $$= \vec{r}\frac{d\vec{p}}{dt} + \vec{p}\times\vec{v}$$ $$= \vec{r}\frac{d\vec{p}}{dt} + m\vec{v}\times\vec{v}$$ $$= \vec{r}\frac{d\vec{p}}{dt} + 0 \qquad (\because \vec{v}\times\vec{v} = v^2\sin 0 = 0)$$ $$\frac{d\vec{J}}{dt} = \vec{r}\times\vec{F} \qquad \left(\because \vec{F}=\frac{d\vec{p}}{dt};\ \text{rate of change of momentum}\right)$$or,
$$\tau = \vec{r}\times\vec{F} \qquad (\because \text{moment of force} = \text{torque})$$If $R$ = position vector of centre of mass, $r_c$ = position vector of particle from C.M., $r$ = position of particle, $V$ = velocity of centre of mass. Then,
$$\vec{r} = \vec{R} + \vec{r_c} \qquad \cdots (1)$$Differentiating w.r.t. $t$, we get,
$$\vec{v} = \vec{V} + \vec{v_c} \qquad \cdots (2)$$Now, angular momentum:
$$\vec{J} = \sum m(\vec{r}\times\vec{v}) \qquad \cdots (3)$$From equations (1), (2), and (3), we get,
$$\vec{J} = \sum m\{(\vec{R}+\vec{r_c})\times(\vec{V}+\vec{v_c})\}$$or,
$$\vec{J} = \sum m(\vec{R}\times\vec{V}) + \sum m(\vec{R}\times\vec{v_c}) + \sum m(\vec{r_c}\times\vec{V}) + \sum m(\vec{r_c}\times\vec{v_c})$$or,
$$\vec{J} = \vec{R}\times\sum m\vec{V} + \vec{R}\times\sum m\vec{v_c} + \sum m\,\vec{r_c}\times\vec{V} + \sum m(\vec{r_c}\times\vec{v_c}) \qquad \cdots (4)$$We have from eq.(1),
$$\vec{r_c} = \vec{r}-\vec{R}$$or,
$$m\vec{r_c} = m\vec{r} - m\vec{R} \qquad \text{(multiplying both sides by }m\text{)}$$or,
$$\sum m\vec{r_c} = \sum m\vec{r} - \sum m\vec{R} \qquad (\sum \text{ on both sides})$$or,
$$\sum m\vec{r_c} = \sum m\vec{r} - M\vec{R} \qquad \cdots (5)$$For centre of mass,
$$\vec{R} = \frac{\sum m\vec{r}}{M}$$or,
$$M\vec{R} = \sum m\vec{r} \qquad \cdots (6)$$From eq.(5) and (6),
$$\sum m\vec{r_c} = 0 \;\Rightarrow\; \sum m\vec{v_c} = 0$$Replacing these values in eq.(4), we get,
$$\vec{J} = \vec{R}\times\sum m\vec{V} + 0 + 0 + \sum m(\vec{r_c}\times\vec{v_c})$$or,
$$\vec{J} = \vec{R}\times\sum m\vec{V} + \sum m(\vec{r_c}\times\vec{v_c})$$or,
$$\vec{J} = \vec{R}\times\vec{P} + \vec{J}_c \qquad \cdots (6)$$where $\vec{J}_c$ = angular momentum of the system about the centre of mass, or spin angular momentum, and $\vec{R}\times\vec{P}$ = orbital angular momentum.
$\therefore$ Total angular momentum = orbital angular momentum + spin angular momentum.
Now, differentiating eq.(6) w.r.t. $t$, we get,
$$\frac{d\vec{J}}{dt} = \frac{d}{dt}(\vec{R}\times\vec{P}) + \frac{d\vec{J}_c}{dt}$$ $$\Rightarrow \tau = \frac{d(\vec{R}\times\vec{P})}{dt} + \frac{d\vec{J}_c}{dt}$$If we take the centre of mass as origin, then, $P = mV = 0$.
$$\Rightarrow \tau = \frac{d\vec{J}_c}{dt}$$Conservation of Angular Momentum
In absence of external torque, the total angular momentum remains constant.
We know,
$$\tau = \frac{d\vec{J}_c}{dt}$$In absence of torque, $\tau = 0$,
$$\Rightarrow \frac{d\vec{J}_c}{dt} = 0$$ $$\Rightarrow \vec{J}_c = 0 = \text{constant}$$This may be called conservation of spin angular momentum.
Again, we know,
$$\tau = \vec{r}\times\vec{F} = rF\sin\theta$$Torque will be absent if:
- Position vector $\vec{r} = 0$
- Force $\vec{F} = 0$
- Angle between position vector and force is $0$ ($r$ and $F$ are in the same direction)
For example, for a planet orbiting the sun, the force is always directed along the radius vector (Sun–Earth line):
$$\tau = Fr\sin 0 = 0$$$\therefore$ Angular momentum of a central force is always constant.
Scattering of a Positive Particle by a Heavy Nucleus
Let us consider a proton ($e$) approaching towards the nucleus of charge $Ze$.
At infinite distance, the repulsive force is zero and the particle moves with velocity $v_0$. When the particle reaches $A$, it experiences a repulsive force and proceeds along the path $AP$.
Let $NA = r_A$ = distance of closest approach, $NB = b$ = impact parameter, $v_A$ = velocity of the particle at $A$.
From conservation of angular momentum,
total initial angular momentum at $\infty$ = total final angular momentum at $A$
Again, from conservation of energy,
$$KE_{initial} + PE_{initial} = KE_{final} + PE_{final}$$or,
$$\frac{1}{2}mv_0^2 + 0 = \frac{1}{2}mv_A^2 + \frac{1}{4\pi\varepsilon_0}\frac{Ze\cdot e}{r_A}$$or,
$$\frac{1}{2}mv_0^2 - \frac{1}{2}mv_A^2 = \frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r_A}$$or,
$$\frac{1}{2}mv_0^2 - \frac{1}{2}m\left(\frac{v_0b}{r_A}\right)^2 = \frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r_A} \qquad [\text{from (1)}]$$ $$\Rightarrow \frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r_A} = \frac{1}{2}mv_0^2\left(1-\frac{b^2}{r_A^2}\right)$$For head-on collision, impact parameter $b = 0$:
$$\therefore \frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r_A} = \frac{1}{2}mv_0^2$$If the colliding particle is an alpha ($\alpha$) particle, then the above equation becomes:
$$\frac{1}{4\pi\varepsilon_0}\frac{Ze\cdot 2e}{r_A} = \frac{1}{2}mv_0^2$$ $$\Rightarrow r_A = \frac{Ze^2}{\pi\varepsilon_0\, mv_0^2}$$System of Variable Mass
Rocket
The rocket is propelled towards the direction opposite to that of the jet (burning fuel). Hence, momentum lost by the jet of fuel must be equal to the momentum gained by the rocket.
Theory
Let $M$ = mass of rocket including fuel and oxidiser, $V$ = velocity of rocket at a given time in the lab frame of reference, $v$ = velocity of jet (exhaust velocity of burnt gas),
$$\frac{dM}{dt} = -\alpha = \text{rate of decrease of mass}$$Assuming forward velocity of the rocket and backward velocity of burnt gas along the same line. Velocity of jet in lab frame $= V-v$.
$\therefore$ Rate of change of momentum of jet $= -\alpha(V-v)$
$$= \frac{dM}{dt}(V-v)$$Neglecting the weight of the rocket and applying Newton's third law of motion, the force acting on the rocket propelling it forward:
$$= \frac{dM}{dt}(V-v)$$Since we have mass of rocket ($M$) and $V$ is its velocity at a given time,
$$\text{Force acting on the rocket} = \frac{d(MV)}{dt}$$ $$\Rightarrow \frac{d(MV)}{dt} = \frac{dM}{dt}(V-v) \qquad \cdots (1)$$ $$\frac{dM}{dt}V + M\frac{dV}{dt} = \frac{dM}{dt}(V) - \frac{dM}{dt}v$$ $$M\frac{dV}{dt} = -\frac{dM}{dt}v \qquad \cdots (2)$$ $$\Rightarrow dV = -\frac{dM}{M}v$$Integrating the above equation, we get,
$$\int dV = \int -\frac{dM}{M}v$$ $$V = -v\log_e M + C \qquad \cdots (3)$$Let $M_0$ = initial mass of rocket (at $t=0$), $V_0$ = initial velocity of rocket (at $t=0$).
Then,
Replacing this value in eq.(3), we get,
$$V = -v\log_e M + V_0 + v\log_e M_0$$ $$V = V_0 + v\log_e\frac{M_0}{M} \qquad \cdots (4)$$Here, $M = M_0-\alpha t$
$$\text{or,}\qquad M = M_0\left(1-\frac{\alpha}{M_0}t\right)$$ $$\text{or,}\qquad M = M_0(1-\beta t) \qquad \text{where } \beta = \frac{\alpha}{M_0}$$ $$\Rightarrow \frac{M_0}{M} = (1-\beta t)^{-1}$$Substituting this value in eq.(4),
$$V = V_0 + v\log_e(1-\beta t)^{-1}$$ $$V = V_0 - v\log_e(1-\beta t) \qquad \cdots (5)$$Numericals
Q1.
A particle of mass 4 gram lies in a potential field given by $U = 200x^2+1500\ \text{erg/gm}$. Obtain the frequency of vibration.
Solution:
Here, $m = 4$ gram, $U = 200x^2+1500\ \text{erg/gm}$, frequency $(f) = ?$
We know,
$$\text{Force} = -\frac{dU}{dx} = -\frac{d}{dx}(200x^2+1500) = -400x$$Now,
$$\text{Acceleration } (a) = \frac{\text{Force}}{\text{Mass}} = \frac{-400x}{4} = -100x$$Where acceleration $= \omega^2 x$,
$$-4\pi^2f^2x = -100x$$ $$f^2 = \frac{100}{4\pi^2} \qquad (\text{for magnitude only})$$ $$f = 2\ \text{Hz}$$Q2.
Find out the magnitude of angular momentum of a bicycle wheel ($M=2$ kg, and $r=40$ cm) when rolling at 30 km/hr. How much torque is required to turn the handlebar through 1 radian in $0.1\ \text{sec}^2$.
Solution:
Given,
$$\text{Mass}(M) = 2\ \text{kg}, \qquad \text{Radius}(r) = 40\ \text{cm} = 40\times10^{-2}\ \text{m}$$ $$\text{Velocity}(v) = 30\ \text{km/hr} = \frac{30\times1000}{60\times60} = \frac{25}{3}\ \text{m/s}$$Now,
$$\text{Angular momentum}(L) = mvr = 2\times\frac{25}{3}\times 40\times10^{-2} = 6.64\ \text{kg m}^2\text{s}^{-1}$$Again, torque $(\tau) = ?$,
We know that,
$$\tau = I\alpha = mr^2\frac{d\omega}{dt}$$ $$= 2\times(40\times10^{-2})^2\times(1/0.1) \qquad \left(\text{since } \frac{d\omega}{dt}=\frac{1}{0.1}\ \text{rad/sec}^2\right)$$ $$\tau = 3.2\ \text{N m}$$Q3.
A rocket of mass 20 kg has 180 kg fuel. The exhaust velocity of fuel is 1.6 km/s. Calculate the minimum rate of consumption of fuel so that the rocket may rise from the ground. Also calculate the ultimate vertical speed gained by the rocket when the rate of consumption of fuel is 2 kg/sec.
Solution:
Given,
Mass of rocket + fuel $(M_0) = 20+180 = 200$ kg
Exhaust velocity of fuel $(v) = 1.6\ \text{km/sec} = 1.6\times10^3\ \text{m/s}$
$\frac{m}{t} = ?$
Case I: For the minimum rate of consumption of fuel,
$$\frac{mv}{t} \geq M_0g$$ $$\Rightarrow \frac{m}{t} \geq \frac{M_0g}{v} = \frac{200\times9.8}{1.6\times10^3} = 1.225\ \text{kg/s}$$Case II:
$$\frac{m}{t} = 2\ \text{kg/s}$$Speed of rocket $V = ?$
$$\text{Time for consumption of fuel} = \frac{\text{mass of fuel}}{m/t} = \frac{180}{2} = 90\ \text{sec}$$Ultimate vertical speed of rocket is given by:
$$V = V_0 + v\log_e\frac{M_0}{M} - gt$$Here, $V_0 = 0$ m/s, $v = 1.6\times10^3$ m/s, $M_0 = 180+20 = 200$ kg, $M = 20$ kg.
Now,
$$V = 0 + 1.6\times10^3\log_e\frac{200}{20} - 9.8\times90$$ $$= 2802.13614\ldots$$ $$V = 2802.14\ \text{m/s}$$