m = 0.02 kg, M = 100 kg, v = 80 m s^-1, V = ?
\(V=-{mv\over M}=-{0.020 kg \times 80 ms^{-1}\over 100kg}\)
= 0.016 m / s^-1=- 1.6 cm s^-1
Negative sign indicates that the gun moves in a direction opposite to the direction of motion of the bullet.

25 cm

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.

(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.

Here, radius r =20 cm
Centripetal acceleration, = 980 cms^-2
We know that centripetal acceleration, a = rω²
\(\omega =\sqrt { \frac { a }{ r } } =\sqrt { \frac { 980 }{ 20 } }\)
\(\omega =\sqrt { 49 } =7rad/s\)

When a body is moving along a circular path, speed always remains constant and a centripetal force is acting on the body.

Step 1: Given that:
The mass(m) of the car= 1500kg
Initial velocity of the car(u) = 90kmh⁻¹=90 × 518 ms⁻¹ =25ms⁻¹
Final velocity of the car(v) = 0
Step 2: Calculation of impulse:
We have,
Impulse= Change in momentum of the body = mv−mu
Thus, Impulse on the car will be;
= 1500kg × 0 − 1500kg × 25ms⁻¹
=−37500kgms⁻¹
=−3.75×10⁴kgms⁻¹
Thus,Impulse on the car will be 3.75 × 10⁴kgms⁻¹

Given, Mass of the object (m) = 500 g = 0.5 kg
Speed of the object (v) = 25 m/s
(i) Impulse imparted to the object
= change in the momentum = mv - mu
= m ( v - u ) = 0.5 ( 25 - 0) = 12.5 N-s
(ii) Velocity of the object after rebounding = -\(\frac { 25 }{ 2 } \) m/s
 v' = -12.5 m/s
\(\therefore \)  Change in momentum = m ( v' - v )
= 0.5 ( - 12.5 - 25 ) = -18.75 N-s

Porcelain objects are wrapped in paper or straw before packing to reduce  the chances during transportation. During transportation  sudden jerks or even fall can take place. Forces are created at the point of collision and the force takes longer time to reach the porcelain objects through paper or straw for same change in momentum as  F = \({ \Delta p }/{ \Delta t }\) and therefore a lesser force acts on object.

Impulse and change in momentum are along the same direction. Therefore, angle between these two vectors is zero degree.

Yes, if the external forces acting on the body caan be represented in magnitude and direction by the sides of a closed polygon taken in the same order.

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

According to the Newton's second law of motion. F = ma, for given acceleration a if m is large. F should be more i.e. greater force will be required to put a larger mass in motion.

Mass of body A(ml) = 5 kg
Mass of body B(m2) =10 kg
Coefficient of friction between the bodies and the table (\(\mu \)) = 015
Force applied horizontally at A, F = 200 N
(i) Reaction of partition
Limiting friction acting to the left
f =\(\mu \)R =\(\mu \)(m₁ + m₂)g
= 0.15(5+ 10) x 9.8 = 22.05N
\(\therefore\) Net force acting on the partition towards the right
F' =F - f
= 200 - 22.05 = 177.95 N
According to the Newton's third law of motion,
Reaction of partition = Net force acting on the partition
= 177.95 N [towards the left]
(ii) Action-reaction forces between A and B Let t. be the force of limiting friction acting on body A and FI be the net force applied by body A on body B.
 
f₁ = \(\mu \) R₁
= \(\mu \)m₁g [.: R₁ = m₁g]
= 0.15 x  5 x 9.8
= 7.35 N (towards the left).
Net force applied by body A on body B,
F₁ = F - f₁ = 200 -7.35
= 192.65 N [towards the right]
According to Newton's third law of motion, reaction force applied by body B on body, i.e.
A = F₁ = 192.65 N (towards left)

Given, Speed of the stream of water, \(\nu \) = 15 m/s
Area of cross-section of the  tube, A = 10^-2 m²
Volume of water coming out per second from the tube
V = A\(\nu \) = 10^-2 x 15 = 15 x 10^-2 m³
Density of water = 10³ kg/m³
\(\therefore \)  Mass of the water coming out of the tube per second
m = V\(\rho \)     \(\left[ \because Density=\frac { mass }{ volume } \right] \) 
=  15 x 10^-2 x 10³ kg
= 150 kg/s
Force exerted on the wall by the impact of water = change in momentum per second
= m\(\nu \)  = 150 x 15 N
= 2250 N

When an object is thrown vertically upwards or it falls vertically downward under gravity, then an acceleration g = 10 m/s² acts downward due to the earth's gravitational pull.
Mass of pebble (m) = 0.05 kg
(i) During upward motion
Net force acting on pebble (F) = ma = 0.05 x 10 N
= 0.50 N (vertically downwards)
(ii) During downward motion
Net force acting on pebble (F) = ma = 0.05 x 10 N
= 0.50 N (vertically downward)
(iii) At the highest point
Net force acting on pebble
(F) = ma = 0.05 x 10 N
= 0.50 N  (vertically downward)
If pebble was thrown at an angle of 45\(^0\)  with the horizontal direction, then acceleration acting on it and therefore force acting on it will remain unchanged, i.e. 0.50 N (vertically downward).

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
\({ T }_{ max }=200\quad N=\frac { mv^{ 2 }max }{ r } \)
\(\\ { v }_{ max }=\sqrt { \frac { { T }_{ max }\times r }{ m } } =\sqrt { \frac { 200\times 1.5 }{ 0.25 } } =34.6{ m }/{ s }\)

(i) 5.2 N backward
(ii) 450 N on the boat
(iii) 180 N on the hockey stick

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^')
= \(0.04[{ 10 }^{ -2 }-(-{ 10 }^{ -2 })]=8\times { 10 }^{ -4 }kg-m/s\) 

Step 1: Given
Mass =80 kg, circle of radius =10 m, speed =6.1 m/s
Step 2: Determining the concept
This problem is based on the concept of uniform circular motion. Uniform circular motion is a motion in which an object moves in a circular path with constant velocity. Also, it involves Newton's second law of motion.
Formula:
The velocity in uniform circular motion is given by,
\(\mathrm{v}=\frac{2 \pi \mathrm{R}}{\mathrm{T}}\)
where, v is the velocity, R is the radius and T is the time period
According to Newton's second law of motion
\(\mathrm{F}_{\mathrm{N}}-\mathrm{mg}=\mathrm{ma}_{\mathrm{c}}\)
where \({ }^{a_c}\) is an acceleration, g is an acceleration due to gravity, m is mass, \({ }^{F_N}\) is the normal force and R is the radius.
Step 3: (a) Determining the period of the motion
Using equation (i) the time can be written as,
\(\mathrm{T}=2 \pi \mathrm{R} / \mathrm{V}-2 \pi(10 \mathrm{M}) /(6.1 \mathrm{~m} / \mathrm{s})=10 \mathrm{~s}\)
Hence, the period of the motion is $10 \mathrm{~s}$.
Step 4: (b)Determining the magnitude of the normal force on the addict from the seat when both go through the highest point of the circular path
In this case, Normal force \({ }^{F_N}\)is directed upward, gravitational force mg is to downward and acceleration is also directed to the down. Thus, by using Newton's 2nd law,
\(\mathrm{F}_{\mathrm{N}}-\mathrm{mg}=\mathrm{ma}_{\mathrm{c}}\) (Centripetal acceleration is due to the circular motion)
\(\mathrm{F}_{\mathrm{N}}=\mathrm{m}\left(\mathrm{g}-\mathrm{v}^2 / \mathrm{R}\right)=486 \mathrm{~N} \approx 4.9 \times 10^2 \mathrm{~N}\)
Hence, the magnitude of the normal force on the addict from the seat when both go through the highest point of the circular path is \({ }^{4.9 \times 10^2 \mathrm{~N}}\).
Step 5: (c) Determining the magnitude of the normal force on the addict from the seat when both go through the lowest point
Now, reverse both the normal force direction and the acceleration direction
Thus, \(\mathrm{F}=\mathrm{m}\left(\mathrm{g}+\mathrm{v}^2 / \mathrm{R}\right)-1081 \mathrm{~N} \approx 1.1 \mathrm{kN}\)
Hence, the magnitude of the normal force on the addict from the seat when both go through the lowest point is \({ }^{1.1 \mathrm{kN}}.\)

Force F = ma, therefore force acting on a particle in unaccelerated (a = 0) motion is zero.
(i) As drop of rain is falling downward with a constant speed, therefore its acceleration is zero.
According to Newton's second law of motion, net force acting on drop F = ma = 0.
(ii) In floating condition, the weight of the body is balanced by the upthrust. Therefore, net force acting on a cork floating on water = 0
(iii) As kite is held stationary in the sky, therefore acceleration of the kite is zero. Therefore, net force acting on the car F = ma = 0.
(iv) As car is moving with a constant velocity, therefore, its acceleration is zero. i.e., a = 0 therefore, net force acting on the car F = ma = 0.
(v) As electron is in a space where there is no electric field, magnetic field and gravitational (material) objects, therefore, no electric, magnetic and gravitational force is acting on it. Hence, net force acting on electron is zero.