(i) The values displayed by Vipul were intelligent, helping nature, awareness and sympathy.
(ii) Momentum of a body is defined as the product of its mass and the velocity with which it is moving.
Momentum = Mass x Velocity Momentum of a body depends upon it mass and the velocity.

(i) Let a be the acceleration of the system to right. All the three frictional forces f₁=μm₁g, f₂ = μm₂g, and f₃ =μm₃g will be directed to the left as the motion of bodies is to the right. Hence, for the whole system

F-μm₁g-μm₂g-μm₃g=(m₁+m₂+m₃)a
F= (m₁+m₂+m₃)(a+μg)
=(1.5 + 2 + 1)(3 + 0.2 x 9.8) = 22.3 N
(ii) The force exerted by the 1.5 kg block on the 2 kg block = F - m₁ (a + μg)
= 22.3 - 1.5 (3 + 0.2 x 9.8)
= 22.3 - 7.44 = 14.86 N

Here M = 20 kg, m = 5 kg and \(\mu_k\) = 0.1
Here net pulling force F = mg - f_k = mg -\(\mu_k\)N

= mg -\(\mu_k\).Mg = 5 x 10 - 0.1 x 20 x 10
= 50 - 20 = 30 N
\(\therefore\)Acceleration of the system a = \({F\over (m+M)}\)
\(={30N\over (5+20)kg}\)
= 1.2 m s^-2
\(\therefore\)Tension in string T = mg - ma
= 5 x 10 - 5 x 1.2
= 50 - 6 = 44 N.

The impulse of the force exerted by the bat is given by the change in the momentum of the ball. Now
Initial momentum of the ball = \({150\over 1000}\times 12Kg \ ms^{-1}=1.8Kg \ ms^{-1}\)
Final momentum of the ball =\(-{150\over 1000}\times 20Kg \ ms^{-1}=-3.0Kg \ ms^{-1}\)
Change in the momentum of the ball = [1.8 - (- 3.0)] kg ms^-1 = 4.8 kg ms^-1
This equals the impulse of the force exerted by the bat. Since
Impulse = force x time
we have
Average force exerted =\({Impulse \over time}\)
\(={4.8Kg \ ms^{-1}\over 0.01s}=480Kg \ ms^{-1}=480N\)

The direction of the frictional force is opposite to the direction of motion. When a body is moved, say from A to B and then back to A, work is required to be done both during forward and backward motion. So, the net work done in a round trip is not zero. Hence, the frictional force is a non-conservative force.

(a) When the lift is moving upwards with a uniform velocity 5 ms^-1 (acceleration is zero), the reaction R or the pressure on the base is
R = mg = 70 x 9.8 N = 686 N
(b) When the lift is moving upwards with a uniform acceleration of 2.2ms^-2 , the reaction R' or the pressure on the base increases and is given by
R' = m (g + a) = 70 (9.8 + 2.2) N = 840 N
(c) When the lift descends with a uniform acceleration of 2.8 ms^-2, the reaction R" is given by
R" = m (g - a) = 70 (9.8 - 2.8) N = 490 N
(d) When the lift falls freely under gravity, the reaction R'" is given by
R'" = m (g - g) = 0
i.e., the object appears to have become weightless.

Case (i):Let a₁  be the acceleration of the system when 9 kg mass hangs freely and T the tension in the string.
M₁ g- T = m₁ a₁
T -m₂g sin 30° = m₂a₁
\(\Rightarrow g(m_1-m_2 sin 30^o)=\)a₁(m₁+m₂)
\(\Rightarrow a_1={g({9-6\times{1\over2}\over 15})}={6g\over 15}={2g\over 5}\)
Case (ii): Let a₂ be the acceleration of the system when 6 kg mass hangs freely and T'the tension in the string.
m₂g-T' = m₂a₂
T'-m₁gsin 30^o = m₁a₂
\(\Rightarrow\) g(m₂-m₁ sin 30^o) = a₂(m₂ + m₁)
\(\Rightarrow\) \(g(6-9\times{1\over 2})=a_2(6+9)\)
\(\Rightarrow a_2={3g\over 30}={g\over 10}\)
If S is the length of the plane,
In case (i), \(s={1\over 2}a_1t^2_1\)
In case (ii), \(s={1\over 2}a_2t^2_2\)
\(\Rightarrow a_1t^2_1=a_2t^2_1\)
\(\Rightarrow {t_1\over t_2}=\sqrt{a_2\over a_1}=\sqrt{g/10\over 2g/ 5}=\sqrt{1\over4}\Rightarrow t_1:t_2=1:2\)​​​​

In applying brakes, suppose F_B is the force required to stop the truck in distance (d)
\(\therefore F_B \times d ={1\over 2}mv^2 or F_B={1\over2}Fr\)
In taking a turn of radius d, the force required is
\(F_r ={mv^2\over d}=2F_B \ or \ F_B={1\over2}F_r\)
Therefore, it is better to apply brakes.

The horizontally directed force 400 N and weight 20 kg of the block are resolved into two mutually perpendicular components, parallel and perpendicular to the plane as shown.
N =20 g cos 45° + 400 sin 45° = 421.4 N
The frictional force experienced by the block F = \(\mu\)N =\(\mu\) x 421.4 = 421.4 \(\mu\)N.

As the accelerated motion is taking place up the plane.
400 cos 45° - 20g sin 45° - f = 20a
\({400\over \sqrt{2}}-{20\times 9.8\over \sqrt{2}}-421.4\mu=20a=20\)
\(\mu=({400\over \sqrt{2}}-{196\over \sqrt{2}}-12)\times {1\over 421.4}\)
\(={282.8-138.6-12\over 421.4}=0.3137\)
The coefficient of sliding friction between the block and the incline = 0.3137

A single isolated force is not possible. This follows from Newton's third law of motion, according to which to every action, there is an equal and opposite reaction. So, the forces must always exist in pairs. When we talk of a single force, we are considering only one aspect of mutual interaction.

Here as the string is inextensible, acceleration of two blocks will be same. Also, string is massless so tension throughout the string will be same. Contact force will be normal force only.
Let acceleration of each block is a, tension in string is T and contact force between M₁, and surface is N₂ and contact force between M₂ and surface is N₂.
Applying Newton's second law for the blocks;

For M₁, F - T M_l a ..........(i)
M₁ g - N₁= 0 ..........(ii)
For M₂             T=M₂ a ...........(iii)
M₂ g- N₂ = 0 .............(iv)
Solving equations (i) and (iii)
\(a={F\over M_1+M_2} and \ T={M_2F\over M_1+M_2}\).

\(\because \)  Mass of each coin = m
Number of total coins = 10
(iii) Reaction of the 6th coin on the 7th coin
= -(force exerted on 6th coin)
= -(weight of 4 coins)
= -4 mg N (vertically upward)

Speed of the aircraft, v = 720 km/h
\(=720\times \frac { 5 }{ 18 } { m }/{ s }\quad \quad \quad \left[ \because 1km/h=\frac { 5 }{ 18 } { m }/{ s } \right] \)
= 200 m/s
Angel of banking, \(\theta \) = 15^o
Acceleration due to gravity, g = 9.8m/s²
At turn, \(\tan { \theta } \) = \(\frac { { v }^{ 2 } }{ rg } \)
or     \(r=\frac { { v }^{ 2 } }{ g\tan { \theta } } =\frac { (200)^{ 2 } }{ 9.8\times \tan { 15^{ 0 } } } =\frac { 40000 }{ 9.8\times 0.2679 } \) 
\(\\ r=15420m=15.24\times 10^{ 3 }m=15.24km\)

Mass of stone, m = 0.25 kg, Radius of the string, r = 1.5 m
Frequency, v = 40rev/min = \(\frac { 40 }{ 60 } \) rev/s = \(\frac { 2 }{ 3 } \) rev/s
Centripetal force required for circular motion is obtained from the tension in the string.
\(\therefore \) Tension in the string = Centripetal force
T = \(mr{ \omega }^{ 2 }\)
\(= mr\left( 2\pi n \right) ^{ 2 } \ \left[ \therefore \omega =2\pi v \right] \)
\(= mr4\pi ^{ 2 }{ v }^{ 2 }\)
\(T=0.25\times 1.5\times 4\times \left( \frac { 22 }{ 7 } \right) ^{ 2 }\times \left( \frac { 2 }{ 3 } \right) ^{ 2 }= 6.6N\)
Maximum tension which can be withstand by the string
\({ T }_{ max }=200\quad N=\frac { mv^{ 2 }max }{ r } \)
\(\\ { v }_{ max }=\sqrt { \frac { { T }_{ max }\times r }{ m } } =\sqrt { \frac { 200\times 1.5 }{ 0.25 } } =34.6{ m }/{ s }\)

\(\because \)   Mass of the helicopter, m₁ = 1000 kg
Mass of the crew and the passengers, m₂ = 300 kg
Acceleration of the helicopter, a = 15 m/ s²
Acceleration  due to gravity, g = 10 m/s²
According to Newton's third law of motion, for every action there is an equal and opposite reaction.
\(\therefore \)  Force on the helicopter due to the surrounding air 
= 32500 N ( upward direction)