\(sin\ 30°=\frac { h }{ AB } \)
\(\frac { 1 }{ 2 } =\frac { 0.5 }{ AB } \)
\(\therefore\)   AB = 1 m

Force of friction between the particle and the inclined plane
= \(\mu\) N
=  \(\mu\) \(\times\) mg cos \(\theta\)
\(=\frac { 1 }{ \sqrt { 3 } } \times 0.2\times 10\times cos30°\)
\(=\frac { 1 }{ \sqrt { 3 } } \times 0.2\times 10\times \frac { \sqrt { 3 } }{ 2 } \)
= 1 N
\(\therefore\) The work done by the particle in moving from A to B against the force of friction
=\(\mu\) mg cos \(\theta\) \(\times\) AB
= 1 N \(\times\) 1 m
= 1J
Let \(\overrightarrow { v } \) be the velocity of the particle of B
\(\therefore\) w = change in energy from A to B
\(1J=\left[ \frac { 1 }{ 2 } \times 0.2\times { (5) }^{ 2 }+0 \right] -\left[ \frac { 1 }{ 2 } \times 0.2\times { V }^{ 2 }+0.2\times 10\times 0.5 \right] \)
\(\Rightarrow \quad 1=2.5-[0.1\quad { V }^{ 2 }+1]\)
\(\Rightarrow \quad 1=2.5-0.1\quad { V }^{ 2 }-1\)
\(\Rightarrow\) 0.1 V² = 1.5 - 1
\(\Rightarrow\) 0.1 V² = 0.5
\(\Rightarrow\) V² = 5
\(\Rightarrow\) V = \(\sqrt { 5 } \) = 2.24 ms^-1

The balls A and B are identical
\(\therefore\) masses are same, \(\theta\) = 30°
\(\therefore \quad \phi =90°-30°=60°\)
Initiallly v₁ = 10 ms^-1 and v₂ = 0
\(\therefore\) According to law of conservation of momentum for x-component.
u = v₁ cos 30° + v₂ cos 60°
\(10={ v }_{ 1 }\times \frac { \sqrt { 3 } }{ 2 } +{ v }_{ 2 }\times \frac { 1 }{ 2 } \)
\(\therefore \quad \sqrt { 3 } { v }_{ 1 }+{ v }_{ 2 }=20\)               (i)
For y-component,
0 = v₁ sin 30° - v₂ sin 60°
\(0={ v }_{ 1 }\times \frac { 1 }{ 2 } -{ v }_{ 2 }\times \frac { \sqrt { 3 } }{ 2 } \)
\(\therefore { v }_{ 1 }=\sqrt { 3 } { v }_{ 2 }\)                          (ii)
From eq. (i) and (ii) we get
\(\sqrt { 3 } \times \sqrt { 3 } { v }_{ 2 }+{ v }_{ 2 }=20\)
\(\Rightarrow { 3v }_{ 2 }+{ v }_{ 2 }=20\)
\(\Rightarrow \quad 4{ v }_{ 2 }=20\)
\(\therefore \quad { v }_{ 2 }=5\quad { ms }^{ -1 }\)
From eq. (ii)
\({ v }_{ 1 }=\sqrt { 3 } \times 5=5\sqrt { 3 } \)ms^-1
Here, the velocities of A and B are \(\sqrt [ 5 ]{ 3 } { ms }^{ -1 }\) and 5 ms^-1 respectively.

Let a particle A of mass m and having velocity u collides with particle B of equal mass but at rest. Let the collision be oblique elastic collision and after collision the balls A and B move with velocities v₁ and v₂ respectively inclined at an angle e from each other.

Applying principle of conservation of linear momentum, we get
mu = mv₁ + mv₂ or u = v₁+v₂
or        u² = ( v₁+v₂).( v₁+v₂)
= v₁² + v₂² + 2v₁v₂ \(\cos { \theta } \)    ....  (i)
Again as total KE before collision = total KE after collision
\(\therefore \ \quad \frac { 1 }{ 2 } \)mu² = \( \frac { 1 }{ 2 } \)mv₁² + \( \frac { 1 }{ 2 } \)mv₂²
\(\Rightarrow \)     u² = v₁² + v₂²                    ...(ii)
Comparing Eqs, (i) and (ii), we get 2v₁v₂ \(\cos { \theta } \) =0
As in an oblique collision both v₁ and v₂ are finite, hence \(\cos { \theta } \) = 0
\(\Rightarrow \)      \(\theta =\cos ^{ -1 }{ (0) } =\frac { \pi }{ 2 } \)
Thus, particles A and B are moving in mutually perpendicular directions after the collision.

(i) Let a particle of mass m moving with an initial velocity u collides inelastically with another particle of mass M initially at rest. Let after collision, the combined entity moves with a velocity v.Then, from the conservation of linear momentum, we have
mu + 0 = (m + M) v
\(\Rightarrow \quad v=\frac{m u}{m+M}\)
(ii) Initial kinetic energy of system before collision
\(K=\frac { 1 }{ 2 } m{ \mu }^{ 2 }\)
and final kinetic energy of system after collision
\({ k }^{ ' }=\frac { 1 }{ 2 } (m+M){ v }^{ 2 }=\frac { 1 }{ 2 } (m+M).\left( \frac { { mu }^{ 2 } }{ m+M } \right) \)
Therefore Loss in kinetic energy during collision
\(=\frac { 1 }{ 2 } m{ \mu }^{ 2 }-\frac { 1 }{ 2 } \frac { m{ \mu }^{ 2 } }{ 2(m+M) } \)
\(=\frac { 1 }{ 2 } m{ \mu }^{ 2 }\left[ 1-\frac { m }{ M+m } \right] =\frac { 1 }{ 2 } m{ \mu }^{ 2 }.\left( \frac { M }{ M+m } \right) \)
and fractional loss in kinetic energy during collision
\(\frac { \Delta K }{ K } =\frac { M }{ M+m } \)

As the stone weight acts downwards  and upthrust acts upwards
Thus, net weight of stone can be calculated.
specific gravity = Relative density =  \(\frac{Density\ of \ stone}{Density\ of \ water}\)
thus, density of stone = 3 \(\times \)10 8kg/m³
Volume of stone  = \( \frac{mass \ of\ stone}{Density \ of \ stone}\) 
= \( \frac{5}{3\times10^3}\)
= \(\frac{5}{3}\) \(\times \)10³m³
Upthrust on stone  = Weight of liquid displaced
= \(\frac{5}{3}\)\(\times \) 10^-3\(\times \) \(\rho\) _water\(\times \)g
= \(\frac{5}{3}\) \(\times \)10^-3 \(\times \)10³\(\times \)10
= \(\frac{50}{3}\) N
Thus, net weight of stone,
mg' = mg - upthrust
mg' = (5 \(\times \)10-\(\frac{50}{3}\))
=\( \frac{100}{3}\) N
Now force required to raise the stone , F= \( \frac{100}{3}\) N
Work required to raise the stone
W = Fd = \(\frac{100}{3} \) \(\times \) 5
= \(\frac{500}{3}\) J
= 166.6 J

(a) No, the total kinetic energy, does not remin conserved during the short time whwn two billiard balls are in contact. At that time, balls are at rest and their KE has been transfor,med into elastic potential energy of balls.
(b) Yes, In an elastic collision, the total linear momentum of the system always remains conserved.
(c) No, In an inelastic collision, there is always a loss of kinetic energy, i.e., the total kinetic energy of the billiard balls before collision will always be greater than that after collision.
Yes, The total linear momentum of the system of billiards balls will remain conserved even in the case of an inelastic collision.
(d) Elastic,In the given case, the forces involved are conservation. This is because they depend on the separation between the centres of the billiard balls. Hence, the collision is elastic.

(i) False, internal as well as external forces can change the kinetic energy.Again forces of conservative may change the potential energy of a system.
(ii) False, for non-conservative force the work done over a closed loop is not zero.
(iii) It is usually true but not always true.As an exmple in the exlposion of a cracker final kinetic energy is greater than the initial kinetic energy. Again final kinetic energy of gun-bullet system after firing is more than initial kinetic energy before collision.

Let M= mass of man, m= mass of boy V= speed of m a n, v= speed of boy
Given \(\frac{1}{2} M V^2=\frac{1}{2}\left(\frac{1}{2} m v^2\right)\), As \(m=\frac{M}{2}\)
So \(\frac{1}{2} M V^2=\frac{1}{2}\left(\frac{1}{2} \times \frac{M}{2} m v^2\right)\)
Hence \(v^2=4 V^2\) or v=2 V
When the man speeds up by 1 m/s, then we get
\(\frac{1}{2} M(V+1)^2=\frac{1}{2} m v^2=\frac{1}{2} \frac{M}{2}\left(4 V^2\right)\)
or \((V+1)^2=2 V^2\) or \(V^2-2 V-1=0\)
Solving we get \(V=2.4 \mathrm{~ms}^{-1}\) and \(v=2 V=4.8 \mathrm{~ms}^{-1}\)

Here, mass, m=0.4 kg, speed, u = 9.8 ms^-1,
acceleration, \(\mathrm{a}=-\mathrm{g}=-9.8 \mathrm{~m} / \mathrm{s}^2\) and time, \(t=\frac{1}{2} \mathrm{~s}\)
From equation of motion
\(s=u t+\frac{1}{2} a t^2=\left(9.8 \times \frac{1}{2}\right)+\left(-\frac{1}{2} \times 9.8 \times \frac{1}{2} \times \frac{1}{2}\right)=3.67 \mathrm{~m}\)
therefore Potential energy \(=m g h=(0.4)(9.8)(3.67)=14.386 J\)