# IGNOU MPH-002 Solved Assignment 2024 | MSCPH | IGNOU

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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## IGNOU MPH-002 Assignment Question Paper 2024

1. a) A particle of mass $m$$m$mm$m$ moving to the right with an initial velocity $u$$u$uu$u$ collides elastically with a particle of unknown mass $M$$M$MM$M$ coming from the opposite direction. After the collision $m$$m$mm$m$ has a velocity $u/2$$u/2$u//2u / 2$u/2$ at right angles to the incident direction, and $M$$M$MM$M$ is deflected back making an angle of ${45}^{\circ }$${45}^{\circ }$45^(@)45^{\circ}${45}^{\circ }$ degrees to its incident direction as shown below. Calculate the ratio $M/m$$M/m$M//mM / m$M/m$.
b) Write down the equations of constraint/s for the following systems and classify them as holonomic/nonholonomic and scleronomic/rheonomic:
i) a particle of mass $m$$m$mm$m$ sliding down the surface of a sphere of radius $R$$R$RR$R$ without friction, under the action of gravity.
ii) a simple pendulum of length $L$$L$LL$L$ for which the point of support is oscillating vertically with an angular speed $\omega$$\omega$omega\omega$\omega$ and amplitude $A$$A$AA$A$.
2. Using the D’ Alembert’s Principle determine the equation of motion for an Atwood’s machine with masses $M$$M$MM$M$ and $3M$$3M$3M3 M$3M$.
3. The kinetic and potential energies for a mechanical system with two generalized coordinates ${q}_{1}$${q}_{1}$q_(1)q_1${q}_{1}$ and ${q}_{2}$${q}_{2}$q_(2)q_2${q}_{2}$ are
$T={q}_{2}{}^{2}{\stackrel{˙}{q}}_{1}^{2}+2{\stackrel{˙}{q}}_{2}^{2};V={q}_{1}{}^{2}-{q}_{2}{}^{2}$$T={q}_{2}{}^{2}{\stackrel{˙}{q}}_{1}^{2}+2{\stackrel{˙}{q}}_{2}^{2};V={q}_{1}{}^{2}-{q}_{2}{}^{2}$T=q_(2)^(2)q^(˙)_(1)^(2)+2q^(˙)_(2)^(2);V=q_(1)^(2)-q_(2)^(2)T=q_2{ }^2 \dot{q}_1^2+2 \dot{q}_2^2 ; V=q_1{ }^2-q_2{ }^2$T={q}_{2}{}^{2}{\stackrel{˙}{q}}_{1}^{2}+2{\stackrel{˙}{q}}_{2}^{2};V={q}_{1}{}^{2}-{q}_{2}{}^{2}$
i) Write down the Lagrangian for the system.
ii) Obtain the expressions for the generalized momenta.
iii) Derive the Euler Lagrange equations of motion for the system.
iv) Obtain the energy function for the system.
1. A particle moves under a central force, $f\left(r\right)=-k{r}^{3}\stackrel{^}{r},k>0$$f\left(r\right)=-k{r}^{3}\stackrel{^}{r},k>0$f(r)=-kr^(3) hat(r),k > 0f(r)=-k r^3 \hat{r}, k>0$f\left(r\right)=-k{r}^{3}\stackrel{^}{r},k>0$. Is this an attractive or a repulsive force? Find the radius and energy of the circular orbits.
2. A particle is scattered by a rigid sphere of radius $R$$R$RR$R$, as given in the figure. Determine the differential scattering cross-section and the total scattering cross-section.
1. Consider three identical springs and two masses ${m}_{1}=m$${m}_{1}=m$m_(1)=mm_1=m${m}_{1}=m$ and ${m}_{2}=2m$${m}_{2}=2m$m_(2)=2mm_2=2 m${m}_{2}=2m$, as shown in the figure. The motion of the two masses are constrained along the line joining them. Find the normal mode frequencies and the normal coordinates.
$$a^2=b^2+c^2-2bc\:Cos\left(A\right)$$

## MPH-002 Sample Solution 2024

mph-001-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

# mph-001-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

1. a) A particle of mass $m$$m$mm$m$ moving to the right with an initial velocity $u$$u$uu$u$ collides elastically with a particle of unknown mass $M$$M$MM$M$ coming from the opposite direction. After the collision $m$$m$mm$m$ has a velocity $u/2$$u/2$u//2u / 2$u/2$ at right angles to the incident direction, and $M$$M$MM$M$ is deflected back making an angle of ${45}^{\circ }$${45}^{\circ }$45^(@)45^{\circ}${45}^{\circ }$ degrees to its incident direction as shown below. Calculate the ratio $M/m$$M/m$M//mM / m$M/m$.
In an elastic collision, both momentum and kinetic energy are conserved. Let’s consider the horizontal (x) and vertical (y) components of the momentum separately:
1. Momentum Conservation (Horizontal Direction):
Before the collision, the total horizontal momentum is $mu-Mu$$mu-Mu$mu-Mumu – Mu$mu-Mu$. After the collision, the horizontal momentum of $m$$m$mm$m$ is zero (since it moves at right angles to the incident direction), and the horizontal component of $M$$M$MM$M$‘s momentum is $Mv\mathrm{cos}\left({45}^{\circ }\right)$$Mv\mathrm{cos}\left({45}^{\circ }\right)$Mv cos(45^(@))Mv \cos(45^\circ)$Mv\mathrm{cos}\left({45}^{\circ }\right)$, where $v$$v$vv$v$ is the final velocity of $M$$M$MM$M$. Therefore, we have:
$mu-Mu=Mv\mathrm{cos}\left({45}^{\circ }\right)$$mu-Mu=Mv\mathrm{cos}\left({45}^{\circ }\right)$mu-Mu=Mv cos(45^(@))mu – Mu = Mv \cos(45^\circ)$mu-Mu=Mv\mathrm{cos}\left({45}^{\circ }\right)$
Since $\mathrm{cos}\left({45}^{\circ }\right)=\frac{1}{\sqrt{2}}$$\mathrm{cos}\left({45}^{\circ }\right)=\frac{1}{\sqrt{2}}$cos(45^(@))=(1)/(sqrt2)\cos(45^\circ) = \frac{1}{\sqrt{2}}$\mathrm{cos}\left({45}^{\circ }\right)=\frac{1}{\sqrt{2}}$, this equation simplifies to:
$mu-Mu=\frac{Mv}{\sqrt{2}}$$mu-Mu=\frac{Mv}{\sqrt{2}}$mu-Mu=(Mv)/(sqrt2)mu – Mu = \frac{Mv}{\sqrt{2}}$mu-Mu=\frac{Mv}{\sqrt{2}}$
$u\left(m-M\right)=\frac{Mv}{\sqrt{2}}\phantom{\rule{1em}{0ex}}\text{(1)}$$u\left(m-M\right)=\frac{Mv}{\sqrt{2}}\phantom{\rule{1em}{0ex}}\text{(1)}$u(m-M)=(Mv)/(sqrt2)quad(1)u(m – M) = \frac{Mv}{\sqrt{2}} \quad \text{(1)}$u\left(m-M\right)=\frac{Mv}{\sqrt{2}}\phantom{\rule{1em}{0ex}}\text{(1)}$
2. Momentum Conservation (Vertical Direction):
Before the collision, the total vertical momentum is zero. After the collision, the vertical momentum of $m$$m$mm$m$ is $mu/2$$mu/2$mu//2mu/2$mu/2$, and the vertical component of $M$$M$MM$M$‘s momentum is $Mv\mathrm{sin}\left({45}^{\circ }\right)$$Mv\mathrm{sin}\left({45}^{\circ }\right)$Mv sin(45^(@))Mv \sin(45^\circ)$Mv\mathrm{sin}\left({45}^{\circ }\right)$. Therefore:
$0=mu/2-Mv\mathrm{sin}\left({45}^{\circ }\right)$$0=mu/2-Mv\mathrm{sin}\left({45}^{\circ }\right)$0=mu//2-Mv sin(45^(@))0 = mu/2 – Mv \sin(45^\circ)$0=mu/2-Mv\mathrm{sin}\left({45}^{\circ }\right)$
Using $\mathrm{sin}\left({45}^{\circ }\right)=\frac{1}{\sqrt{2}}$$\mathrm{sin}\left({45}^{\circ }\right)=\frac{1}{\sqrt{2}}$sin(45^(@))=(1)/(sqrt2)\sin(45^\circ) = \frac{1}{\sqrt{2}}$\mathrm{sin}\left({45}^{\circ }\right)=\frac{1}{\sqrt{2}}$, this equation simplifies to:
$\frac{mu}{2}=\frac{Mv}{\sqrt{2}}$$\frac{mu}{2}=\frac{Mv}{\sqrt{2}}$(mu)/(2)=(Mv)/(sqrt2)\frac{mu}{2} = \frac{Mv}{\sqrt{2}}$\frac{mu}{2}=\frac{Mv}{\sqrt{2}}$
$mu=Mv\sqrt{2}\phantom{\rule{1em}{0ex}}\text{(2)}$$mu=Mv\sqrt{2}\phantom{\rule{1em}{0ex}}\text{(2)}$mu=Mvsqrt2quad(2)mu = Mv\sqrt{2} \quad \text{(2)}$mu=Mv\sqrt{2}\phantom{\rule{1em}{0ex}}\text{(2)}$
3. Kinetic Energy Conservation:
The total kinetic energy before and after the collision remains the same. Before the collision, the total kinetic energy is $\frac{1}{2}m{u}^{2}+\frac{1}{2}M{u}^{2}$$\frac{1}{2}m{u}^{2}+\frac{1}{2}M{u}^{2}$(1)/(2)mu^(2)+(1)/(2)Mu^(2)\frac{1}{2}mu^2 + \frac{1}{2}Mu^2$\frac{1}{2}m{u}^{2}+\frac{1}{2}M{u}^{2}$. After the collision, it is $\frac{1}{2}m\left(u/2{\right)}^{2}+\frac{1}{2}M{v}^{2}$$\frac{1}{2}m\left(u/2{\right)}^{2}+\frac{1}{2}M{v}^{2}$(1)/(2)m(u//2)^(2)+(1)/(2)Mv^(2)\frac{1}{2}m(u/2)^2 + \frac{1}{2}Mv^2$\frac{1}{2}m\left(u/2{\right)}^{2}+\frac{1}{2}M{v}^{2}$. Equating these, we get:
$\frac{1}{2}m{u}^{2}+\frac{1}{2}M{u}^{2}=\frac{1}{2}m{\left(\frac{u}{2}\right)}^{2}+\frac{1}{2}M{v}^{2}$$\frac{1}{2}m{u}^{2}+\frac{1}{2}M{u}^{2}=\frac{1}{2}m{\left(\frac{u}{2}\right)}^{2}+\frac{1}{2}M{v}^{2}$(1)/(2)mu^(2)+(1)/(2)Mu^(2)=(1)/(2)m((u)/(2))^(2)+(1)/(2)Mv^(2)\frac{1}{2}mu^2 + \frac{1}{2}Mu^2 = \frac{1}{2}m\left(\frac{u}{2}\right)^2 + \frac{1}{2}Mv^2$\frac{1}{2}m{u}^{2}+\frac{1}{2}M{u}^{2}=\frac{1}{2}m{\left(\frac{u}{2}\right)}^{2}+\frac{1}{2}M{v}^{2}$
Simplifying, we find:
$m{u}^{2}+M{u}^{2}=\frac{1}{4}m{u}^{2}+M{v}^{2}$$m{u}^{2}+M{u}^{2}=\frac{1}{4}m{u}^{2}+M{v}^{2}$mu^(2)+Mu^(2)=(1)/(4)mu^(2)+Mv^(2)mu^2 + Mu^2 = \frac{1}{4}mu^2 + Mv^2$m{u}^{2}+M{u}^{2}=\frac{1}{4}m{u}^{2}+M{v}^{2}$
$\frac{3}{4}m{u}^{2}+M{u}^{2}=M{v}^{2}\phantom{\rule{1em}{0ex}}\text{(3)}$$\frac{3}{4}m{u}^{2}+M{u}^{2}=M{v}^{2}\phantom{\rule{1em}{0ex}}\text{(3)}$(3)/(4)mu^(2)+Mu^(2)=Mv^(2)quad(3)\frac{3}{4}mu^2 + Mu^2 = Mv^2 \quad \text{(3)}$\frac{3}{4}m{u}^{2}+M{u}^{2}=M{v}^{2}\phantom{\rule{1em}{0ex}}\text{(3)}$
To find the ratio $M/m$$M/m$M//mM/m$M/m$, we can use equations (2) and (3). From equation (2), we have $v=\frac{mu}{M\sqrt{2}}$$v=\frac{mu}{M\sqrt{2}}$v=(mu)/(Msqrt2)v = \frac{mu}{M\sqrt{2}}$v=\frac{mu}{M\sqrt{2}}$. Substituting this into equation (3), we get:
$\frac{3}{4}m{u}^{2}+M{u}^{2}=M{\left(\frac{mu}{M\sqrt{2}}\right)}^{2}$$\frac{3}{4}m{u}^{2}+M{u}^{2}=M{\left(\frac{mu}{M\sqrt{2}}\right)}^{2}$(3)/(4)mu^(2)+Mu^(2)=M((mu)/(Msqrt2))^(2)\frac{3}{4}mu^2 + Mu^2 = M\left(\frac{mu}{M\sqrt{2}}\right)^2$\frac{3}{4}m{u}^{2}+M{u}^{2}=M{\left(\frac{mu}{M\sqrt{2}}\right)}^{2}$
$\frac{3}{4}m{u}^{2}+M{u}^{2}=\frac{{m}^{2}{u}^{2}}{2M}$$\frac{3}{4}m{u}^{2}+M{u}^{2}=\frac{{m}^{2}{u}^{2}}{2M}$(3)/(4)mu^(2)+Mu^(2)=(m^(2)u^(2))/(2M)\frac{3}{4}mu^2 + Mu^2 = \frac{m^2u^2}{2M}$\frac{3}{4}m{u}^{2}+M{u}^{2}=\frac{{m}^{2}{u}^{2}}{2M}$
$\frac{3}{4}m+M=\frac{{m}^{2}}{2M}$$\frac{3}{4}m+M=\frac{{m}^{2}}{2M}$(3)/(4)m+M=(m^(2))/(2M)\frac{3}{4}m + M = \frac{m^2}{2M}$\frac{3}{4}m+M=\frac{{m}^{2}}{2M}$
$\frac{3}{4}Mm+{M}^{2}=\frac{{m}^{2}}{2}$$\frac{3}{4}Mm+{M}^{2}=\frac{{m}^{2}}{2}$(3)/(4)Mm+M^(2)=(m^(2))/(2)\frac{3}{4}Mm + M^2 = \frac{m^2}{2}$\frac{3}{4}Mm+{M}^{2}=\frac{{m}^{2}}{2}$
$2{M}^{2}+\frac{3}{2}Mm-\frac{{m}^{2}}{2}=0$$2{M}^{2}+\frac{3}{2}Mm-\frac{{m}^{2}}{2}=0$2M^(2)+(3)/(2)Mm-(m^(2))/(2)=02M^2 + \frac{3}{2}Mm – \frac{m^2}{2} = 0$2{M}^{2}+\frac{3}{2}Mm-\frac{{m}^{2}}{2}=0$
Solving this quadratic equation for $M/m$$M/m$M//mM/m$M/m$, we get:
$\frac{M}{m}=\frac{-\frac{3}{2}±\sqrt{{\left(\frac{3}{2}\right)}^{2}+4×2×\frac{1}{2}}}{4}=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}+4}}{4}=\frac{-\frac{3}{2}±\sqrt{\frac{25}{4}}}{4}=\frac{-\frac{3}{2}±\frac{5}{2}}{4}$$\frac{M}{m}=\frac{-\frac{3}{2}±\sqrt{{\left(\frac{3}{2}\right)}^{2}+4×2×\frac{1}{2}}}{4}=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}+4}}{4}=\frac{-\frac{3}{2}±\sqrt{\frac{25}{4}}}{4}=\frac{-\frac{3}{2}±\frac{5}{2}}{4}$(M)/(m)=(-(3)/(2)+-sqrt(((3)/(2))^(2)+4xx2xx(1)/(2)))/(4)=(-(3)/(2)+-sqrt((9)/(4)+4))/(4)=(-(3)/(2)+-sqrt((25)/(4)))/(4)=(-(3)/(2)+-(5)/(2))/(4)\frac{M}{m} = \frac{-\frac{3}{2} \pm \sqrt{\left(\frac{3}{2}\right)^2 + 4 \times 2 \times \frac{1}{2}}}{4} = \frac{-\frac{3}{2} \pm \sqrt{\frac{9}{4} + 4}}{4} = \frac{-\frac{3}{2} \pm \sqrt{\frac{25}{4}}}{4} = \frac{-\frac{3}{2} \pm \frac{5}{2}}{4}$\frac{M}{m}=\frac{-\frac{3}{2}±\sqrt{{\left(\frac{3}{2}\right)}^{2}+4×2×\frac{1}{2}}}{4}=\frac{-\frac{3}{2}±\sqrt{\frac{9}{4}+4}}{4}=\frac{-\frac{3}{2}±\sqrt{\frac{25}{4}}}{4}=\frac{-\frac{3}{2}±\frac{5}{2}}{4}$
Taking the positive solution (since mass cannot be negative), we have:
$\frac{M}{m}=\frac{\frac{2}{2}}{4}=\frac{1}{2}$$\frac{M}{m}=\frac{\frac{2}{2}}{4}=\frac{1}{2}$(M)/(m)=((2)/(2))/(4)=(1)/(2)\frac{M}{m} = \frac{\frac{2}{2}}{4} = \frac{1}{2}$\frac{M}{m}=\frac{\frac{2}{2}}{4}=\frac{1}{2}$
Therefore, the ratio of the masses is $M/m=1/2$$M/m=1/2$M//m=1//2M/m = 1/2$M/m=1/2$.

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$$c=a\:cos\:B+b\:cos\:A$$

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