The recoil speed of rifle V = \(mv\over M\) is inversely proportional to its mass. So for a heavy rifle the kick is less stronger.

When the monkey is climbing down with an acceleration, then

mg - T=ma
\(\Rightarrow\)T=mg - ma = m (g - a)
or T=40 kg x (10 - 4) ms^-2 = 240 N
The rope will not break.

(i) When a horse is trying to pull a cart, he pushes the ground backward at an angle from the horizontal. According to Newton's third law of motion, the ground also apply equal reaction force on the feet of the horse in opposite direction. The vertical component of reaction balances the weight of the horse and horizontal component is responsible for motion of the cart. In empty space, there is no reaction force, therefore, a horse cannot pull a cart.
(ii) When bus is moving, the passengers sitting in it are also in motion and moving with the speed of the bus. When bus stops suddenly then lower part of the body of passengers which is in contact with the bus slow down with the bus but upper part of the bodies of the passengers continue to remain in motion in initial direction due to its inertia of motion and hence are thrown forward.
(iii) In pulling a lawn mower, a force F is applied in upward direction, making an angle e with the horizontal [Fig.(a)].Its vertical component in upward direction decreasing the effective weight of the mower.

In pushing a lawn mower, a force F is applied in downward direction, making an angle \(\theta\) with the horizontal. Its vertical component is in downward direction increasing the effective weight of the mower. Therefore, it is easier to pull a lawn mower than to push it.
(iv) In holding a catch, the impulse imparted to the hands = F x \(\triangle\)t = change in momentum of the ball when a cricketer lowers his hands to take a catch, he increases the time taken to stop the ball. As time t increases the force F applied on the hands of the cricketer by the ball decreases and his hands feel less hurt.

25 cm

(i) The part correctly describes the trajectory of the stone after the spring breaks because when a stone tied to one end of a string is whirled round in a circle then velocity of the stone at any point is along the tangent at that point. If the string breaks suddenly, then stone flies off tangentially, along the direction of its velocity.
(ii) The part correctly describes the trajectory of the stone after the spring breaks because when a stone tied to one end of a string is whirled round in a circle then velocity of the stone at any point is along the tangent at that point. If the string breaks suddenly, then stone flies off tangentially, along the direction of its velocity.
(iii) The part correctly describes the trajectory of the stone after the spring breaks because when a stone tied to one end of a string is whirled round in a circle then velocity of the stone at any point is along the tangent at that point. If the string breaks suddenly, then stone flies off tangentially, along the direction of its velocity.

The carts with rubber tyres are easier to ply than those with iron types because the coefficient of friction between rubber and concrete is less than between iron and the road.

As the applied force increases, the static friction also increases and becomes equal to the applied force to make the object stationery. That is why static friction is called a self-adjusting force.

Given, mass, m = 72.2 kg
Gravity acceleration, g = 9.8 m/s² 
Scale reading = apparent weight = R = ?
(i)  While descending with constant velocity, a = 0
\(\therefore \)    R =  mg
R = 72.2 x 9.8
\(\Longrightarrow \)  R = 707.56 N
(ii)  While ascending with a = 3.2 m/s² 
R = m ( g + a )
R = 72.2 ( 9.8 + 3.2 ) = 938.6 N

Given, F = 36 dyne at an angle of 6000 .
∴∴   Component of force along x-direction
F_x = F cos 6000 = 36 x 1/2 = 18 dyne
 But  F_x = ma_x ,
a_x = \(\frac{F_x}{m} =\frac {18}{18} = 1\) cm/s²

We know that, F = rate of change of momentum
F = 1 kg-m/s² = 1 N

Let us analyse the motion of particle with respect to table which is moving with cabin with an angular speed of \(\omega\) . Along AB centrifugal force of magnitude m\(\omega^2\) R will act at A on the particle which can be treated as constant from A to B as L < < R.
\(\therefore\)acceleration of particle along AB with respect to cabin a = \(\omega^2\) R (constant) Required time' t' is given by
\(S=ut+{1\over 2}at^2\)
\(\Rightarrow L=0+{1\over2}\times \omega^2Rt^2\)
\(\Rightarrow t=\sqrt{2L\over \omega^2R}\)

In a closed glass cage, air inside is bound with the cage. Therefore,
(i) there would be no change in weight of the cage if the bird flies with a constant velocity.
(ii) the cage becomes heavier, when bird flies upwards with an acceleration.
(iii) the cage appears lighter, when bird flies downwards with an acceleration.

The situation is shown in figure below
    
(i) When the floor is removed, the forces on the person are
(a) weight mg downward.
(b) normal force N due to the wall towards the centre.
(c) frictional force Is parallel to the wall, upwards.
The person is moving in a circle with a uniform speed, so its acceleration is v2 /r towards the centre. Newton's law for the horizontal direction (second law) and for the vertical direction (first law) give
 \(N=m{ v }^{ 2 }/r\quad \quad ...(i)\)    
\(\\ { f }_{ s }=mg\quad \quad \quad ...(ii)\)
For the minimum speed, when the floor may be removed, the friction is limiting one and so equals \({ \mu }_{ s }N\) 
This gives 
 \({ \mu }_{ s }N=mg\) 
\(\\ \frac { { \mu }_{ s }m{ v }^{ 2 } }{ r } =mg\)
\(\\ v=\sqrt { \frac { rg }{ { \mu }_{ s } } } =\sqrt { \frac { 2m\times 10m/{ s }^{ 2 } }{ 0.2 } } =10m/s\)

\(\because \)  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)

Given, r =10⁴ km = 10⁴ \(\times \) 1000 m = 10⁷ m,
m = 2500 kg
v = 4 kms^-1 = 4\(\times \)10³ ms^-1
Now,centripeta force is F = \(\frac { m{ v }^{ 2 } }{ r } \) 
F = \(\frac { 2500\times (4\times { 10 }^{ 3 }{ ) }^{ 2 } }{ { 10 }^{ 7 } } =\frac { 2500\times 16\times { 10 }^{ 6 } }{ { 10 }^{ 7 } }\)
\( \\ F=\quad \frac { 250\times 16\times { 10 }^{ 7 } }{ { 10 }^{ 7 } }\)
\(= 4000N\)

Figure shows the various forces acting on the rod AB when it is leaning between the wall and the ground without slipping.
Since the rod is in equilibrium, the net force as well as the net torque on it must each be zero. Considering the forces acting on the rod: we have
\(R_1+(-F')=0 or \ R_1 =\mu_2 R_2\) and \(R_2+F+(-W)=0 or \ R_2 +\mu_1 R_1=W\)

Considering next the moments of all forces about A, we have
R₂ x 0 B = W x 0 N + \(\mu\)₂R₂ x OA  or R₂ x A B cos\(\theta =W \times {AB cos \theta \over 2}+\mu_2 R_2 \times AB sin \theta\) 
\(or \ (R_2 -{W\over 2})cos \theta =(\mu_2 R_2)sin \theta\)
\(or \ tan \theta ={({R_2-{W\over 2}})\over \mu_2 R_2}\)
Now W= R₂ + \(\mu\)₁R₁ = R₂ + \(\mu\)₁ ( \(\mu\)₂ R₂)
=R₂ (1 +\(\mu\)₁\(\mu\)₂)
\(\therefore R_2-W/2 =R_2-{R_2\over 2}(1+\mu_1 \mu_2)\)
\(={R_2 \over 2}-{R_2\over 2}\mu_1 \mu_2={R_2 \over 2}(1-\mu_1 \mu_2)\)
\(\therefore tan \theta ={R_2\over 2}{(1-\mu_1 \mu_2)\over \mu_2 R_2}={(1-\mu_1\mu_2)\over 2\mu_2}\)
or \(\theta= tan^{-1} \{ {(1-\mu_1\mu_2)\over 2\mu_2} \}\)

The free-body diagram of the given problem is shown in the figure (a).

mg = 15 x 9.8 = 147 N
Since block is at rest, so Tc = mg = 15 x 9.8 = 147 N

Resolve T_A and T_B into x and y components. x-component of T_A is given by
T_xA = - T_A cos 28° = - T_A (0.8830)
= - 0.8830 T_A
y-component of T_A is given by
T_yA = T_A sin 28° = 0.4690 T_A
Similarly, x-component of T_B is given by
T_xB = T_B cos 47° = 0.6820 T_B
and y-component of T_B is given by
T_yB = T_B sin 47° = 0.7310 T_B.
These components are represented
Since the knot is in equilibrium, so
(i)\(\sum\)x-components of tensions = 0
i.e., T_xA + T_xB = 0
or - 0.8830 T_A + 0.6820 T_B = 0
or TA =\({0.6820\over0.8830}\) T_B = 0.772 TB ...(i)
(ii)\(\sum\)y-components of tensions = 0
i.e., T_YA + T_yB - T_c = 0
or 0.4690 T_A + 0.7310 T_B - 147 = 0
or 0.4690 T_A + 0.7310 T_B = 147
Substituting the value of equation (i), we get
0.4690 x 0.772 T_B + 0.7310 T_B = 147
or 1.093 T_B =147
or T_B = \({147\over1.093}\) = 134.5 N
Now substituting the values of T_B in eqn. (i), we get
T_A = 0.772 x 134.5 N = 103.8 N
Thus, T_A = 103.8 N; T_B = 134.5 Nand T_c = 147 N.

Let the direction in which the molecules rebound after striking the surface be taken as positive.
=Momentum of each molecule after striking the surface
= mv₂ = 5 x 10^-26 kg x 500 ms^-1
Momentum of each molecule before striking the surface
= mv₁ = 5 x 10^-26 kg x (- 500)
28 x 10²³ molecules strike the surface per second.
∴ Change in momentum of the molecules due to striking the surface in 1second
28 x 10²³ [(5 x 10^-26 x 500) - 5 x 10^-26 (- 500)] kg ms^-1
= 28 x 10²³ x 5 x 10^-26 x 1000 kg ms^-1
= 140 kg ms^-1
∴ Rate of change of momentum
=\(\frac { 140kg\ { ms }^{ -1 } }{ 1s } \) =140 kg ms^-2
But rate of change of momentum is equal to the impressed force.
∴ Force exerted by the surface on the molecule
= 140 kg ms^-2 = 140
By Newton's third law of motion, this must also be the magnitude of the force exerted by the molecules on the surface.
∴ Force exerted by the molecules on the surface
=140 N
Area of the surface = 14 cm²
= 14 x 10^-4 m²
∴ Pressure on the surface
\(\frac { Force }{ Area } =\frac { 140N }{ 14\times { 10 }^{ -4 }{ m }^{ 2 } } \)
= 10⁵ N m^-2.

Given mass of the block = m
Coefficient of friction between the block and the wall = \(\mu \)
Let a force F be applied on the block 10 hold the block against the wall. The normal reaction of mass be Nand force offriction acting upward be f· In equilibrium, vertical and horizontal forces should be balanced separately.

f = mg          ...(i)
F = N             ..(ii)
But force of friction (f) = \(\mu N\) 
=\(\mu N\) [using Eq(ii)]        .....(iii)
From Eqs. (i) and (iii),we get
\(\mu F=mg\ or\ F=\frac { mg }{ \mu } \) 

When the nail fixed on the table is removed, system of two bodies moves with an acceleration a in the forward direction. The acceleration can be found by using
Newton's second law. Draw the FBD for each body as shown in the figure
 
For mass m₁, T = m₁a               ....(i)
For mass m₂, m₂g - T = m₂a      .....(ii)
Adding Eqs. (i) and (ii), we get
m₁a + m₂a = m₂g
a = \(\frac { { m }_{ 2 }g }{ { m }_{ 1 }+{ m }_{ 2 } } =\frac { 10\times 10 }{ 20+10 } =\frac { 100 }{ 30 } =3.33m/{ s }^{ 2 }\)
Tension, T = m₁a = 10 x 3.33 = 33.3 N
When the bodies stop, acceleration, a will be zero. Suppose, the tension becomes T. As the net force on each body is zero, so for body m2
T = m₂g = 10 x 10 = 100N

Mass of the body,m = 0.04 kg
The position time graph OA from t = 0 to t = 2s is a straight line therefore body is moving with a constant velocity.
Velocity of the body,v=Slope of x-t graph
= \(\frac { 2-0 }{ 2-0 } =1cm/s={ 10 }^{ -2 }m/s\quad [\because 1\quad cm={ 10 }^{ -2 }m]\)
Part AB of position time graph is also a straight line.Therefore velocity of the body
 \({ v }^{ \prime }=\frac { 0-2 }{ 0-2 } =-1\quad cm/s=-{ 10 }^{ -2 }cm/s\) 
Negative sign shows that the direction of velocity is reversed after 2 s and it is being repeated.
A suitable physical context for this motion is a ball moving with a constant velocity of 10^-2 m/s between two walls located at x=0 and at x=2m and rebounded repeadedtly on striking each wall.
magnitude of the impulse imparted to the ball after every two seconds
= chage in momentum of teh ball
= mv - mv^' = m(v - v^')
= \(0.04[{ 10 }^{ -2 }-(-{ 10 }^{ -2 })]=8\times { 10 }^{ -4 }kg-m/s\) 

When a man is standing on a weighing scale, it will read the normal reaction R as apparent weight.
Given , mass of man (m) = 70 kg
In each case the weighing scale will read the reaction R, i.e. the apparent weight.
(i) As lift is moving upward with a uniform speed, therefore, its acceleration a = 0
\(\therefore \)  Normal reaction w = R = mg = 70 x 10 N = 700 N
w acts vertically downwards and R acts vertically upwards.
\(\therefore \)  Reading on weighing scale = \(\frac { 700 }{ 10 } \) = 70 kg
(ii) Acceleration of the lift, a = 5 m/ s² (\(\downarrow \) )
\(\therefore \)  Normal reaction, R = m (g - a) = 70 ( 10 - 5) N
= 70 x 5N = 350 N
\(\therefore \)  Reading on weighing scale = \(\frac { 350\quad N }{ 10\quad m/{ s }^{ 2 } } \) = 35 kg
(iii) Acceleration of the lift, a = 5 m/s² (\(\uparrow \) ) 
\(\therefore\)  Normal reaction R = m ( g + a )
= 70 ( 10 + 5 ) = 1050 N
\(\therefore\)   Reading on weighing scale = \(\frac { 1050N }{ 10m/{ s }^{ 2 } } \) = 105 kg
(iv) Acceleration of the lift when it is falling freely under gravity
a = g (\(\downarrow\) )
\(\therefore\)  Normal reaction, R = m ( g - g ) = 0
\(\therefore\)  Reading on weighing scale = 0

(i) Boy A fell on the cemented floor and come to rest abruptly. So, the change in momentum of boy A took place in small interval of time. Hence, a large force was exerted on the boy A by the floor. Therefore, he got injured.
On the other hand, the sand under the weight of boy B yielded and the change in momentum of boy B took place in large interval of time. Therefore, less force was exerted on boy B by the floor and he was not injured.
(ii) Satish is concerned about the welfare of others.He understands his duty to help the needy people.