From the above figure,
R = mg cos \(\theta\)
F = \(\mu\)R = \(\mu\) mg cos \(\theta\)
Net force on the block down the incline
= mg sin \(\theta\) - F
= mg sin \(\theta\) - \(\mu\) mg cos \(\theta\)
= mg (sin \(\theta\) - \(\mu\) cos \(\theta\))
Here distance moved x = 10 cm = 0.1 m
I equilibrium,
work done = Potential energy of stretched spring
\(mg(sin\ \theta -\mu \ cos\ \theta )x=\frac { 1 }{ 2 } { kx }^{ 2 }\)
or 2mg (sin \(\theta\) - \(\mu\) cos \(\theta\)) = kx
or 2 \(\times\) 1 kg \(\times\) 10 ms^-2(sin 37° - \(\mu\) cos 37°) = 100 \(\times\) 0.1 m
or 20(0.601 - \(\mu\) \(\times\) 0.798) = 10
or  0.601 - 0.798\(\mu\) = \(\frac { 10 }{ 20 } =0.5\)
or -0.798\(\mu\) = 0.5 - 0.601 = -0.101
or \(\mu =\frac { -0.101 }{ -0.798 } =\frac { 101 }{ 798 } =0.126\)
Hence \(\mu\) = 0.126

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 \(m\mu +0=(m+M)v\Rightarrow v=\frac { mu }{ m+M } \) 

Given, m₁ = 9000 kg, u₁ = 36 km/h = =10m/s
m₂ = 9000 kg, u₂ = 0, v₁ = v2 = v
By conservation of momentum
m₁u₁ + m₂u₂ = (m₁ + m₂)v
\(9000\times 10\times 9000\times 0=\left( 9000+9000 \right) v\)   
or     \(v=\frac { 90000 }{ 18000 } =5m/s\)     
Total KE before collision = \(\frac { 1 }{ 2 } { m }_{ 1 }{ u }_{ 1 }^{ 2 }+\frac { 1 }{ 2 } { m }_{ 2 }{ u }_{ 2 }^{ 2 }\) 
 \(=\frac { 1 }{ 2 } \times 9000\times 10\times 10+0\)  
\(\\ =450000J\)
Total KE after collision=\(\frac { 1 }{ 2 } \left( { m }_{ 1 }+{ m }_{ 2 } \right) { v }^{ 2 }\)  
\(=\frac { 1 }{ 2 } \times 2\times 9000\times { \left( 5 \right) }^{ 2 }\) 
\( =225000J\)
Thus, total KE collision < Total KE before collision. Hence, the collision is inelastic.

We assume that the pebble is initially at rest on the cliff.
u = 0, m = 1.00 g = 10³ kg
v = 50 ms^-1 , h = 1.00 km = 10³ m
The change in KE of the pebble is 
\(\Delta K=\frac { 1 }{ 2 } m{ v }^{ 2 }-\frac { 1 }{ 2 } m{ u }^{ 2 }\) = \(\frac { 1 }{ 2 } \) x 10^-3 x (50)² - 0 = 1.25 J
Assuming that g = 10 ms^-2 is constant, the work done by the gravitational force is
W_g = F . h = mgh = 10^-3 x 10 x 10³ = 10.0 J
If W_r is the work done by the resistive force on the pebble, then from the work energy theorem,
 \(\Delta K\) = W_g + W_r
or    W_r = \(\Delta K\) - W_g = 1.25 - 10.0 = - 8.75 J

Here, m = 0.1 kg, u = 0, v = 20 ms^-1, t = 3s
Thrust of the rocket = ma = m \(\frac { v-u }{ t } \) 
= o.i x \(\frac { 20-0 }{ 3 } \) = \(\frac { 2 }{ 3 } \) N    \(\left[ \because v=u+at\quad or\ a=\frac { v-u }{ t } \right] \)
Kinetic energy gained by the rocket
K = \(\frac { 1 }{ 2 } m{ v }^{ 2 }\) = \(\frac { 1 }{ 2 } \) x 0.1 x (20)² = 20 J
Energy content per unit mass of the fuel
= \(\frac { Total\ energy }{ Mass\ of\ the\ fuel } =\frac { 20J }{ 0.02kg } =1000Jk{ g }^{ -1 }\)

(i) Negative In the given case, the direction of force (vertically downward) and displacement (vertically upward) are opposite to each other. Hence, the sign of work done is negative
(ii) Work done is -ve because direction if friction force is opposite to sliding motion.
(iii) Work done by an applied force on  body moving on a rough horizontal plane  is positive because force is being applied in the direction of motion so as to overcome friction .
(iv) Work done is negative because the resistive force of air  acts in a direction opposite to the direction of motion of the vibrating  pendulum.

The nuclear decay may be represented as follows
Fe⁵⁷\(\rightarrow\) Fe⁵⁷ + hv (\(\gamma \)-ray photon)
According to de-Broglie hypothesis, momentum of a photon of energy y E is 
\(p=\frac { E }{ c } =\frac { 14.4\times 1.6\times { 10 }^{ -16 }J }{ 3\times { 1 }0^{ 8 }{ ms }^{ -1 } } \)
\(\\ p=\ 7.68\ \times { 10 }^{ -24 }kg { ms }^{ -1 }\)
By conservation of momentum, the momentum of daughter nucleus, p=momentum of \(\gamma \) -ray photon
\(=d\ 7.68 \ \times { 10 }^{ -24 }kg \ { ms }^{ -1 }\)
The recoil energy of the nucleus will be
\(k= \ \frac { { p }^{ 2 } }{ 2m } =\frac { ({ 7.68 \ \times { 10 }^{ -24 }) }^{ 2 } }{ 2\times \ 56.935 \ \times 1.66 \times { 10 }^{ -27 } }\)
\( \\ =0.32\times { 10 }^{ -21 }\)
\(\\ =\frac { 0.312\times { 10 }^{ -21 } }{ 1.6\times { 10 }^{ -16 } } keV\)
\(K=1.95{ \times 10 }^{ -16 }keV\)

59 hp