Frequency = 5000 Hz
Speed of sound in air = 332 m/s
Speed of sound in water = 1450 m/s
Wavelength of sound in air  \(=\frac{332}{5000}\)
\(=66.4 \times 10^{-3}\)
\(\text { Wavelength of sound water } \lambda_{\text {water }}\)
\(=\frac{1450}{5000} =290 \times 10^{-3} \mathrm{~m} \)
\(\therefore \frac{\lambda_{\text {water }}}{\lambda_{\text {air }}} =\frac{290 \times 10^{3}}{66.4 \times 10^{-3}} =4.367 \)

\(\mathrm{F} =-\mathrm{kx} \)
\(\therefore \mathrm{x} =\left|-\frac{F}{k^{\prime}}\right|=\frac{F}{k^{\prime}} \)
\(\mathrm{F} =\mathrm{Mg} \text { and } k^{\prime}=\frac{k}{2} \)
\(\therefore \mathrm{x} =\frac{M g}{\frac{k}{2}} \)
\(=\frac{2 M g}{k} \)

Pressure is directly proportional to density.

\(\mathrm{COP}=\frac{T_{L}}{T_{H}-T_{L}}\)
\(\mathrm{T}_{\mathrm{L}}=-12+273 =261 \mathrm{~K} \)
\(5=\frac{261}{T_{H}-261} \)
\(\therefore 5\left(T_{H}-261\right) =261 \)
\(5 \mathrm{~T}_{\mathrm{H}}-1305 =261 \)
\(5 \mathrm{~T}_{\mathrm{H}} =261+1305 =1566 \)
\(\therefore T_{H} =\frac{1566}{5} \)
\(=313.2 \mathrm{~K} \)
\(\mathrm{~T}_{\mathrm{H}}=313.2-273 =40.2^{\circ} \mathrm{C} \)

The moments of inertia of two discs are I_a and I_b respectively The angular velocity of the disc A is \(\omega\).
The sum of kinetic energies of two discs before coming in contact is \(k_{1}=\frac{1}{2} I_{a} \omega_{1}^{2}+\frac{1}{2} I_{b} \omega_{2}^{2}\)
\(\text { But angular velocity of the disc be is } \omega_{2}=0 \ \text {(rest)}\)
\(\therefore k_{1}=\frac{1}{2} I_{a} \omega_{1}^{2}\)
The final kinetic energy of the two discs system \(k_{2}=\frac{1}{2} \frac{I_{a}^{2} \omega_{1}^{2}}{I_{a}+I_{b}}\)
The loss of kinetic energy is
\(k_{1}-k_{2} =\frac{1}{2} I_{a} \omega^{2}-\frac{1}{2}\left[\frac{I a^{2} \omega_{1}^{2}}{I_{a}+I_{2 b}}\right] \)
\(=\frac{1}{2} \frac{\left[I_{1}\left(I_{a}+I_{b}\right) \omega^{2}-I_{a}^{2} \omega^{2}\right]}{I_{a}+I_{b}} \)
\(k_{1}-k_{2} =\frac{1}{2} \frac{I_{a} b}{\left(I_{a}+I_{b}\right)} \omega^{2} \)

Using law of conservation of momentum,
\(2 m v =\sqrt{m^{2} v^{2}+m^{2} v^{2}} \)
\(=\sqrt{2 m^{2} v^{2}} \)
\(v =\frac{\sqrt{2} m v}{2 m}=\frac{v}{\sqrt{2}} \)
Energy released in explosion = \(2 \times \frac{1}{2} m v^{2} +\frac{1}{2} \times 2 m \times\left(\frac{v^{2}}{\sqrt{2}}\right)^{2} \)
\(=m v^{2}+m \times \frac{v^{2}}{2} \)
\(=\frac{3}{2} m v^{2} \)

\(v=a t+b t^{2}\)
\(\text { Dimensional equation is } \mathrm{LT}^{-1}\)
\(=a T=b T^{2}\)
\(\therefore \text { The dimension of } b=\frac{\mathrm{LT}^{-1}}{\mathrm{~T}^{2}}=\mathrm{LT}^{-3}\)

The gravitational potential energy U(r) can be calculated by finding the sum of gravitational potential energy of each pair of particles.
\(U=-\frac { { Gm }_{ 1 }m_{ 2 } }{ { r }_{ 12 } } -\frac { { Gm }_{ 1 }m_{ 3 } }{ { r }_{ 13 } } -\frac { { Gm }_{ 1 }m_{ 4 } }{ { r }_{ 14 } } -\frac { { Gm }_{ 2 }m_{ 3 } }{ { r }_{ 23 } } -\frac { { Gm }_{ 2 }m_{ 4 } }{ { r }_{ 24 } } -\frac { { Gm }_{ 3 }m_{ 4 } }{ { r }_{ 34 } } \)
Here r₁₂, r₁₃ ... are distance between pair of particles
\({ r }_{ 14 }^{ 2 }={ R }^{ 2 }+{ R }^{ 2 }=2{ R }^{ 2 }\)
\({ r }_{ 14 }=\sqrt { 2 } R={ r }_{ 12 }={ r }_{ 23 }={ r }_{ 34 }\)
r₁₃ = r =₂₄ = 2R 
\(U=-\frac { { Gm }_{ 1 }m_{ 2 } }{ \sqrt { 2 } R } -\frac { { Gm }_{ 1 }m_{ 3 } }{ { 2R } } -\frac { { Gm }_{ 1 }m_{ 4 } }{ { \sqrt { 2 } R } } -\frac { { Gm }_{ 2 }m_{ 3 } }{ \sqrt { 2 } R } -\frac { { Gm }_{ 2 }m_{ 4 } }{ 2R } -\frac { { Gm }_{ 3 }m_{ 4 } }{ \sqrt { 2 } R } \)
\(U=-\frac { G }{ R } \left[ \frac { { Gm }_{ 1 }m_{ 2 } }{ \sqrt { 2 } } -\frac { { Gm }_{ 1 }m_{ 3 } }{ { 2 } } -\frac { { Gm }_{ 1 }m_{ 4 } }{ { \sqrt { 2 } } } -\frac { { Gm }_{ 2 }m_{ 3 } }{ \sqrt { 2 } } -\frac { { Gm }_{ 2 }m_{ 4 } }{ 2 } -\frac { { Gm }_{ 3 }m_{ 4 } }{ \sqrt { 2 } } \right] \)
If all the masses are equal, then m₁ = m₂ = m₃ = m₄ = M
\(U=-\frac { G{ M }^{ 2 } }{ R } \left[ \frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ { 2 } } +\frac { 1 }{ { \sqrt { 2 } } } +\frac { 1 }{ \sqrt { 2 } } +\frac { 1 }{ 2 } +\frac { 1 }{ \sqrt { 2 } } \right] \)
\(U=-\frac { G{ M }^{ 2 } }{ R } \left[ 1+\frac { 4 }{ \sqrt { 2 } } \right] \)
\(U=-\frac { G{ M }^{ 2 } }{ R } \left[ 1+2\sqrt { 2 } \right] \)
The gravitational potential V(r) at a point O is equal to the sum of the gravitational potentials due to individual mass. Since potential is a scalar, the net potential at point O is the algebraic sum of potentials due to each mass.
\({ V }_{ o }(r)\frac { { Gm }_{ 1 } }{ R } -\frac { { Gm }_{ 2 } }{ { R } } -\frac { { Gm }_{ 3 } }{ { R } } -\frac { { Gm }_{ 4 } }{ R } \)
If m₁ = m₂ = m₃ = m₄ = M 
\({ V }_{ o }(r)\frac { { 4GM } }{ R } \)

(i) Consider two masses m₁ and m₂ separated by a distance r which has gravitational potential energy U(r).
(ii) The gravitational potential due to mass ml at a point P which is at a distance r from ml is obtained by making m₂ equal to unity (m₂=1kg).
(iii) Thus the gravitational potential V(r) due to mass m = 1 at a distance r is,
V(r)=\(\frac { { GM }_{ 1 } }{ { r } } \)

(i) When the gas is suddenly compressed \(\triangle\)Q = 0
\(C=\frac { \triangle Q }{ m\triangle T } \)=0 Specific heat is zero.
(ii) Gas is heated and allowed to expand, DT =0 C = a specific heat is infinite
(ii) Expansion is less than rise in temperature due to supply of heat. \(\triangle\)T is +ve.
Specific heat is +ve.
(iv) Expansion is more than rise in temperature. \(\triangle\)T is -ve
Specific heat is -ve.

Accumulate to kinetic theory:
(i) The molecules of a gas are in a state of continuous random motion. 
(ii) They collide with one another and also with the walls of the vessel. Whenever a molecule collides with the wall. It returns with a changed momentum and an equal momentum is transfused to the wall (conservation of momentum).
Accumulate to Newton's law:
(i) The transfer of momentum to the wall is equal to the force excited on the wall.
(ii) The force excited per unit area of the wall is the pressure of the gas.
Hence a gas excites pres due to the continuous call of its molecules with the walls of the vessel.

From the above picture we observe that y = sin (x−a) for a = 0, a =\(\frac { \pi }{ 4 } ,a=\frac { \pi }{ 2 } ,a=\frac { 3\pi }{ 2 } \)and a = \(\pi\), the function y = sin (x−a) shifts towards right. Further, we can take a = vt and v = \(\frac{\pi}{4}\) , and sketching for different times t = 0s, t = 1s, t = 2s etc., we once again observe that y = sin(x−vt) moves towards the right. Hence, y = sin(x−vt) is a travelling (or progressive) wave moving towards the right. If y = sin(x+vt) then the travelling (or progressive) wave moves towards the left. Thus, any arbitrary function of type y = f(x−vt) characterising the wave must move towards right and similarly, any arbitrary function of type y = f(x+vt) characterizing the wave must move towards left.

A black body is the one which absorbs completely heat radiations of all wavelengths when heated. A black body neither reflects nor transmits any radiation. Its absorptive power is unity.

Reynold's number is a dimensionless number. It is used to find the nature of the flow of fluid, whether it is streamlined or turbulent.
Reynold's number \(R_{e}=\frac{\rho v D}{\eta }\)
ρ - denotes the density of the fluid v - denotes the velocity of the fluid
D - denotes the diameter of the pipe
η - denotes the coefficient of viscosity the fluid

(i) Torque about point D is zero. (as F passes through D).
Torque about point C is zero. (as F passes through C).
Torque about point O is \((\overrightarrow {OC})\times \overrightarrow{F}\) and direction is along GH.
Torque about point B is \((\overrightarrow {BD})\times \overrightarrow{F}\) and direction is along GH.
(The perpendicular distance of \(\overrightarrow {BD}\) with respect to \(\overrightarrow {F}\) is \(\overrightarrow {OC}\).
(ii) Torque about axis CD is zero (as F is parallel to CD).
Torque about axis OC is zero (as F intersects OC).
Torque about axis AB is zero (as F is parallel to AB).
Torque about axis GH is  \((\overrightarrow {OC})\times \overrightarrow{F}\) and direction is along GH.

Initial momentum of the object P_i = 10 \(\times\)15 = 150 kgm s^-1
Final momentum of the object p_f = 0
Δp = 150 - 0 = 150 kg ms^-1
(a) Impulse J = Δp = 150 N s and Average force \({ F }_{ avg }=\frac { \triangle p }{ \triangle t } =\frac { 150 }{ 0.03 } \)= 5000N
b) Impulse J = Δp = 150 N s and Average force F_avg =\(\frac { 150 }{ 10 } \)= 15N

Yes, if body has no linear and angular acceleration then a body in uniform straight line of motion will be in equilibrium.

Total energy at A = Total energy at Q
\({v}^{2}={1\over2}{mv}^{2}+mgh=0+mgR\)
v² = 2g(R-b) or \(v=\sqrt{2g(R-k)}\)

(i) Tangential acceleration = g sin ፀ
(ii) mg cos ፀ acts along OP (outwards)
(iii) tension (T) acts along PO (inwards)
Net force on the body at P acting along
PO = T- mg cos ፀ

This must provide the necessary centripetal force. \(\frac{mv^2}{r}\)
\(\therefore T -mg \ cos \ \theta=\frac{mv^2}{r} \)  
Centripetal acceleration (a_⊥) =  \(\frac { T-mg \ cos\theta }{ m } \)

(i) 1 Mega ohm (M\(\Omega\)) = 10⁶\(\Omega\)
(ii) 1 milliampere (mA) = 10^-3A
(iii) 1 daca gram (da g) = 10g
(iv) 1 nano second (ns) = 10^-9s
(v) 1 microvolt (\(\mu\)Y) = 10^-6y
(vi) 1 centimetre (cm) = 10^-2m

Graphical representation of displacement, velocity and acceleration of a particle vibrating simple harmonically with respect to time t.
(i) Displacement graph is a sine curve. Maximum displacement of the particle is y =+a
(ii) The velocity of the vibrating particle is maximum at the mean position i.e v =+ aw and it is zero at the extreme position.
(iii) The acceleration of the vibrating particle is zero at the mean position and maximum at the extreme position (i.e) ± aw².
The velocity is ahead of displacement by a phase angle of \(\frac { \pi }{ 2 } \) . The acceleration is ahead of the velocity by a phase angle \(\frac { \pi }{ 2 } \) or by aw² phase  \(\pi\) ahead of displacement. (i.e) When the displacement has its greatest positive value, acceleration has its negative maximum value or vice versa.

F ∝ m^a v^b r^c ;
F = k m^a v^b r^c
where k is a dimensionless constant of proportionality. Rewriting above equation in terms of dimensions and taking k= 1, we have
[MLT^-2] = [M]^a [LT^-1]^b [L]^c
= [M^aL^bT^-bL^c]
[MLT^-2] = [M^aL^b+cT^-b]
Comparing the powers of M, L and T on both sides
 a = 1 ; b + c = 1; -b =-2
2 + c = 1 ; b = 2;
a = 1 b = 2 and c = -1
From the above equation we get
F = m^av^br^c
F =m¹v²r^-1
or F = \(\frac { m{ v }^{ 2 } }{ r } \)

Given:
During t₁ = 0.1s, distance covered is s₁ = 0.3m
During next t₂ = 0.1s, distance covered is s₂= 0.398m
We know \(s=ut+\frac{1}{2}at^2\)
In this case u = 0, a = g, \(s=\frac{1}{2}gt^2\)
t = t₁+ t₂ = 0.1 + 0.1 = 0.2
\(\therefore g=\frac{2s}{t^2}\)
For this problem,
\(g=\frac{2(s_2-s_1)}{t^2}\)
\(g=\frac{2(0.398-0.3)}{(0.2)^2}=\frac{2\times(0.098)}{0.04}=\frac{0.196}{0.04}\)
\(g=\frac{19.6}{4}=4.9m/s^2\)
∴ Acceleration due to gravity in that planet = 4.9 m/s²

(i) When a rigid body moves, its center of mass will also move along with the body. For kinematic quantities like velocity (v_CM) and acceleration (a_CM) of the center of mass, we can differentiate the expression for position of center of mass with respect to time once and twice respectively. For, simplicity, let us take the motion along, X direction only.
\(\overset { \rightarrow }{ v } _{ CM }=\frac { d\overset { \rightarrow }{ x } _{ CM } }{ dt } =\frac { \sum { { m }_{ i }\left( \frac { d\overset { \rightarrow }{ x } _{ i } }{ dt } \right) } }{ \sum { { m }_{ i } } } =\frac { \sum { { m }_{ i }\overset { \rightarrow }{ v } _{ i } } }{ \sum { { m }_{ i } } } \)
\(\overset { \rightarrow }{ v } _{ CM }=\frac { \sum { { m }_{ i }\overset { \rightarrow }{ { v }_{ i } } } }{ \sum { { m }_{ i } } } \)
\(\overset { \rightarrow }{ a } _{ CM }=\frac { d }{ dt } \left( \frac { d\overset { \rightarrow }{ x_{ CM } } }{ dt } \right) =\left( \frac { d\overset { \rightarrow }{ v } _{ CM } }{ dt } \right) =\frac { \sum { { m }_{ i }\left( \frac { d\overset { \rightarrow }{ v_{ i } } }{ dt } \right) } }{ \sum { { m }_{ i } } } \)
=\(\frac { \sum { { m }_{ i }\overset { \rightarrow }{ a_{ i } } } }{ \sum { { m }_{ i } } } \)
(ii) In the absence of external force, i.e. \(\overset { \rightarrow }{ F_{ ext } } \)= 0 the individual rigid bodies of a system can move or shift only due to the internal forces. 
(iii) This will not affect the position of the center of mass. This means that the center of mass will be in a state of rest or uniform motion. Hence, \(\overset { \rightarrow }{ { v }_{ CM } } \) will be zero when center of mass is at rest and constant when center of mass has-uniform motion (\(\overset { \rightarrow }{ v_{ CM } } =0 \ or \ \overset { \rightarrow }{ v_{ CM } } \)=constant). There will be no acceleration of center of mass, \(\left( \overset { \rightarrow }{ a_{ CM } } =0 \right) \).
From equation
\(0=\frac { \sum { { m }_{ i }\overset { \rightarrow }{ v_{ i } } } }{ \sum { { m }_{ i } } } \)
\(\overset { \rightarrow }{ v_{ CM } } =\frac { \sum { { m }_{ i }\overset { \rightarrow }{ v_{ i } } } }{ \sum { { m }_{ i } } } ;\overset { \rightarrow }{ a_{ CM } } =0\)
(vi) Here, the individual particles may still move with their respective velocities and accelerations due to internal forces

(i) A body of mass (m) attached to one end of a massless and inextensible string executes circular motion in a vertical plane with the other end of the string fixed. The length of the string becomes the radius \(\vec{r}\) of the circular path.
(ii) The motion of the body by taking the free body diagram (FBD) at a position where the position vector \(\vec{r}\) makes an angle e with the vertically downward direction and the instantaneous velocity is as shown in Figure.
There are two forces acting on the mass.
1. Gravitational force which acts downward
2. Tension along the string.
Applying Newton's second law on the mass, In the tangential direction,

mg sinθ = ma_t
mg sinθ = -m \((\frac{dv}{dt})\)
where, a_t = -\((\frac{dv}{dt})\) is tangential retardation
In the radial direction,
T - mg cosθ = m a_r
T - mg cosθ = \(\frac{mv^{2}}{r}\)
where, a_r = \(\frac{v^{2}}{r}\) is the centripetal acceleration.

Consider an object thrown with initial velocity \(\overrightarrow{u}\) at an angle \(\theta\) with the horizontal.
Then,
\(\overrightarrow{u}=u,\overrightarrow{i}+u,\overrightarrow{j}\)
where u_x =u cos \(\theta\) is the horizontal component and u_y = u sin \(\theta\) the vertical component of velocity.

Since the acceleration due to gravity acts in the direction opposite to the direction of vertical component u_y, this component will gradually reduce to zero at the maximum height of the projectile. At this maximum height, the same gravitational force will push the projectile to move downward and fall to the ground.
But, there is no acceleration along the x direction throughout the motion. So, the horizontal component of velocity (u_x = u cos \(\theta\)) remains the same till the object reaches the ground.
After anytime t, the velocity along horizontal motion
v_x = u_x + a_xt = u_x = u cos\(\theta\)  [∴ a_x = 0]
The horizontal distance travelled by projectile in time t is
s_x = \({u}_{x}t+{{1}\over{2}}a_n{t}^{2}\)
[∴ s_x = x and a_n = 0]
∴ t = \(\frac{x}{u \cos \theta} \) ......(1)
For the vertical motion the velocity after time t is v_y = u_y + a_yt
v_y = u sin \(\theta\) - gt........(2)
(∴ u_g = u sin \(\theta\) and a_y = - g)
The vertical distance travelled by the projectile in the same time r is
s_y = \({u}_{y}t{{1}\over{2}}{a}_{y}{t}^{2}\)
y = sin \(\theta\ t-{{1}\over{2}}{gt}^{2}\)
(Here, (Here, s_y = y, u_y = u sin \(\theta\) and a_y = - g) 
Substitute the value of t from equation (1) in equation (3), we have
\(y=u\ \sin\theta{{x}\over{u\ \cos\ \theta}}-{{1}\over{2}}g{{x^2}\over{u^2{cos}^{2}\theta}}\)
\(y=x\ \tan\theta-{{1}\over{2}}g{{x^2}\over{u^2{cos}^{1}\theta}}\)
Thus the path followed by the projectile is an inverted parabola.
2. Maximum height: The maximum vertical distance travelled by the projectile during its journey is called maximum height. 
For the vertical part of the motion. \({v}_{y}^{2}={u}_{y}^{2}+2{a}_{y}s\)
Here, u_y = u sin \(\theta\),  a_y = -g, S = h_max and at the maximum height v_y = 0.
Hence,
∴ 0 = u² sin² = 2gh_max
\({h}_{max}={{{u}^{2}{sin}^{2}\theta}\over{2g}}\)
3. Horizontal Range (R): The maximum horizontal distance between the point of projection and the point on the horizontal plane where the projectile hits the ground is called horizontal range (R). 
Range R = Horizontal component of velocity x time of flight
= u cos \(\theta\times{T}_{f}\)
\(R=u\ \cos\theta\times{{2u\ \sin\theta}\over{g}}={{2{u}^{2}\sin\theta\cos\theta}\over{}g}\)      \([\therefore T_f=\frac {2u\ sin \theta}{g}]\)
\(\therefore\ R={{u^2\sin 2\theta}\over{g}}\)

The duration for which the sound persists is called reverberation time.

The tension in the string is T = m g = 1.2\(\times\)9.8 = 11.76 N
The mass per unit length is μ = 0.25 kg m^-1 Therefore, velocity of the wave pulse is 
\(v=\sqrt { \frac { T }{ \mu } } =\sqrt { \frac { 11.76 }{ 0.25 } } =6.858 \ ms^{-1}=6.8 \ ms^{-1}\)
The time taken by the pulse to cover the distance of 30 cm is
t =\(\frac { d }{ v } =\frac { 30\times { 10 }^{ -2 } }{ 6.8 } =\)0 044 s = 44 ms Where, ms = milli second.

(a) The velocity
\(\vec { v } =\frac { d\vec { r } }{ dt } =\frac { dx }{ dt } \hat { i } +\frac { dy }{ dt } \hat { j } +\frac { dz }{ dt } \hat { k } \)
We obtain \(\vec { v } (t)=6t\hat { i } +5\hat { j } \)
The velocity has only two components v_x = 6t, depending on time t and v_y = 5 which is independent of time.
The velocity at t = 3 s is \(\vec { v } \)(3) =\(18\hat { i } +5\hat { j } \)
(b) The speed at t = 3 s is
\(v=\sqrt { { 18 }^{ 2 }+5^{ 2 } } =\sqrt { 349 } \approx \)   18.68 m s^-1
(c) The acceleration \(\vec { a } \) is \(\vec { a } =\frac { { d }^{ 2 }\vec { r } }{ { dt }^{ 2 } } =6\hat { i } \)
The acceleration has only the x-component. Note that acceleration here is independent of t, which means \(\vec { a } \) is constant. Even at t = 3 s it has same value a = 6t. The velocity is non uniform, but the acceleration is uniform (constant) in this case.

The first body is very much lighter than the second body  \(\left( { m }_{ 1 }<<{ m }_{ 2 },\frac { { m }_{ 1 } }{ { m }_{ 2 } } <<1 \right) \)then the ratio \(\frac { { m }_{ 1 } }{ { m }_{ 2 } } \approx 0\) and also if the target is at rest (u₂=0)
Dividing numerator and denominator of equation \({ v }_{ 1 }=\left( \frac { { m }_{ 1 }-{ m }_{ 2 } }{ { { m }_{ 1 }+m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { { 2m }_{ 2 } }{ { { m }_{ 1 }+m }_{ 2 } } \right) { u }_{ 2 }\) by m₂ we get
\({ v }_{ 1 }=\left( \frac { \frac { { m }_{ 1 } }{ { m }_{ 2 } } -1 }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) { u }_{ 1 }+\left( \frac { 2 }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) \left( 0 \right) \)
\({ v }_{ 1 }=\left( \frac { 0-1 }{ 0+1 } \right) { u }_{ 1 }\)
v₁=-u₁
Similarly, dividing numerator and denominator of equation 
\({ v }_{ 2 }=\left( \frac { { 2m }_{ 1 } }{ { { m }_{ 1 }+m }_{ 2 } } \right) { u }_{ 1 }+\left( \frac { { m }_{ 2 }-{ m }_{ 1 } }{ { { m }_{ 1 }+m }_{ 2 } } \right) { u }_{ 2 }\), by m_2,  we get
\({ v }_{ 2 }=\left( \frac { 2\frac { { m }_{ 1 } }{ { m }_{ 2 } } }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) { u }_{ 1 }+\left( \frac { 1-\frac { { m }_{ 1 } }{ { m }_{ 2 } } }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) \left( 0 \right) \)
\({ v }_{ 2 }=\left( 0 \right) { u }_{ 1 }+\left( \frac { 1-\frac { { m }_{ 1 } }{ { m }_{ 2 } } }{ \frac { { m }_{ 1 } }{ { m }_{ 2 } } +1 } \right) \left( 0 \right) \)
v₂=0 

For measurement of small masses of atomic/subatomic particles etc., we make use of a mass spectrograph.

Relation between angular momentum and rotational K.E.
Angular momentum, L = IW.
Rotational K.E., EK = \(\frac{1}{2}\)
\(2I={ (I\omega ) }^{ 2 }\)
\({ x }_{ k }=\frac { { L }^{ 2 } }{ 2I } \)
\(\sqrt { { E }_{ k } } =\frac { L }{ \sqrt { 2I } } \)
\(\frac { \sqrt { { E }_{ k } } }{ L } =\frac { I }{ \sqrt { 2I } } =constant\)
If I = 1, graph will be hyperbola.

The centrifugal force appears to act on the particle, only when we analyses the motion from a rotating frame. With respect to an inertial frame there is only centripetal force which is given by the tension in the string. For this reason centrifugal force is called as a 'pseudo force'. A pseudo force has no origin, It arises due to the non-inertial nature of the frame considered.