(a) Newton third law of motion states that action and reaction are equal and opposite and they act on different bodies.
(b) Action and reaction act on different bodies.
(c) This is because the action and reaction forces always act on different bodies.
(d) When a shot is fired from the gun with some force in the forward direction, the shot in turn exerts an equal and opposite force on the gun, causing it to recoil or kick backwards.

Here,  m = 1 kg,   \(u = 20 {m}{s}^{-1}\), v = 0, s = 50 m
As  \({v}^{2}-{u}^{2}=2as\)
\(\therefore\)   \({0}^{2}-{20}^{2}=2a \times 50\)
or   \(a = -\frac{400}{100}=-4 {m}{s}^{-2}\)
Force of friction, \(F = ma = 1 \times (-4)=-4 N.\)

Here, u = 0,  s = 400 m, t = 20 s
\(\therefore\)  \(s = ut + \frac {1}{2} at^{2}\)
Or \(400 = 0+ \frac {1}{2} a (20)^{2}\)    or 
\(a = \frac {400 \times 2} {400}= 2 m/s^2\)
\(\therefore\)  Force, \(F = ma = 7000 \times 2 = 14,000 N.\)

1. The inertia of a body is proportional to its mass.
2. If we kick a football, it moves a large distance. But if we kick a ball of stone of the same size, it hardly moves. The stone oppose the change in motion to a larger exten due to its larger mass and henece larger inertia.

Inertia. Inertia is the natural tendency of a body to resist any change in its state of rest or uniform motion in a straight line. For example, a book lying on a table will remain there until an external force is applied on to remove or displace it from that position. Inertia os of three types:

(i) inertia of rest, (ii) inertia of motion and (iii) inertia of direction.

Newton's Laws of motion. Sir Issac Newton further studied the ideas of Galileo's on force and motion. He arrived at three laws of motion which are called Newton's laws of motion. These laws may be stated as follows:
First Law. A body at rest or in uniform motion will remain at rest in uniform motion unlesss an unbalanced force acts upon it.
Second Law. The rule of change of momentum of a body is directly proportional to the applied unbalanced force and the change takes place in the direction of the force.
Third Law. Action and reaction are equal and opposite and they act on different bodies.

(a)  Measurement of force from Newton's second law. Suppose a force F acts on a body of mass m and changes its velocity from u to v in t seconds. Then
Initial momentum of the body, \({p}_{1}=mu\)
Final momentum of the body,  \({p}_{2}=mu\)
Change of momentum  \(= {p}_{2}-{p}_{1}=mv-mu=m\left(v-u\right)\)
Time taken \(= t\)
\(\therefore\) Rate of change of momentum  \(= \frac {Change \ of\ momentum} {time \ taken} = \frac {m\left(v-u\right)}{t}=ma\) 
Where a  is the acceleration of the body.
According to Newton's second law, the rate of change of momentum is directly proportional to the applied force, so
 \(F \infty \ ma.\)             or F = kma
Where k is constant. The unit of force is so chosen that k is equal to one. If m = 1, a = 1 and F = 1, then
1 = k. 1. 1    or  k = 1
F = ma     or   \(Force = Mass \times Acceleration.\)  
So a unit force is that force which produces a unit acceleration in a body of unit mass.
(b) The S.I. unit of force is Newton. One Newton is that force which produces an acceleration of \(1 m/ {s}^{2}\) in a body of mass 1
1 newton \(= 1 kg \times 1 m/s^2\) or \(1N = 1 kg m/s^2\)  
Thus, the second law of motion gives us a method to measure force. If the mass and acceleration of a body are known, we can determine the force acting on it.

Newton's first law of motion. According to this law, a body at rest or in uniform motion will remain at rest or in uniform motion unless an unbalanced force acts upon it. This law consists of three parts:
(i) First part, says that a body at rest continues in its state of rest. For example, a person standing in a bus falls backward when the bus suddenly starts moving forward. When the bus moves, the lower part of his body begins to move along with the bus while the upper part of his body continues to remain at rest due to inertia. That is why a person falls backward when the bus starts.
(ii) Second part, says that a body uniform motion continues moving in straight line path will a uniform speed. for example, when a moving bus suddenly stops, a person sittng in it falls forward. As the bus stops, the lower part of his body comes to rest along with the bus while upper part of his body continues to remain in motion due to inertia and so he fall forward.
(iii) Third part, says that a body moving with a uniform speed in a straight line cannot change itself its direction of motion. For example, when a bus takes a sharp turn, a person sitting in the bus gets a force acting away from the center of the curved path due to his tendency to move in the original direction. He has to hold on a support to prevent himself from swaying away in the turning bus.

(a) Both vehicles experience equal forces of action and reaction.
(b) Momentum change for both vehicles is same.
(c) Car experiences greater acceleration due to its smaller mass.
(d) Due to its smaller inertia or opposite to the force exerted on it.

Here     m = 1,200 kg
Initial velocity,  \(v = 90 \ km/h = 90 \times \frac {5}{18} \ m \ s^{-1}=25 \ m s ^{-1}\)
Final velocity,    \(u = 18 km/h = 18 \times \frac {5}{18} = 5m\quad s^{-1}\)
Time,    t  =  4 s
Accleration,   \(a = \frac {v-u}{t}=\frac {5-25}{4}=-5 m \ s^{-2}\)
magnitude of accleration  \(= 5 m \ s^{-2}.\)
Change in momentum \(= m \left( v- u \right)=1,200 \left(5-25\right)\)
 \(= - 24,000 \ kq \ m \ s^{-1}.\)
Magnitude of change in momentum \(= 24,000 \ kq \ m \ s^{-1}.\)
Magnitude of force \(= \frac {Change \ in \ momentum}{time \ taken}=\frac {24,000}{4}=6,000 \ N.\)

Here,  m = 10 kg,   u = 0,  s = 80 cm = 0.80 m,  \(a = 10 m / s^{-2}\)
Let v be the velocity gained by the dumb-bell as it reaches the floor.
Aa      \(v_2-u_2=2as\)
\(\therefore\)  \(v_2-0_2= 2 \times 10 \times 0.80 = 16\)
or    \(v = 4 \quad m {s}^{-1}\)
Momentum transferred by the dumb-ball to the floor
 \(p = mv = 10 \times 4 = 40 \ kg \ m \ s^{-1}.\)

For boy: \({ m }_{ 1 }=60\ kg,\)   \({ u }_{ 1 }=3m\ { s }^{ -1 }\) \({ v }_{ 1 }=v\) (say)
For trolley: \(\\ { m }_{ 2 }=140\ kg,{ u }_{ 2 }=1.5\ { ms }^{ -1 },{ v }_{ 2 }=v \)
By conservation of momentum
\({ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }=\left( { m }_{ 1 }+{ m }_{ 2 } \right) v\)
\(60\times 3+140\times 1.5=\left( 60+140 \right) v\)
or   \(v=\frac { 180+210 }{ 200 } =\frac { 390 }{ 200 } =1.95{ ms }^{ -1 }.\)

Here,  \(m_1 = 2\) kg,  \(u_1 = 10 m{s}^{-1}\)   \(v_1 = 1 {m}{s}^{-1}\)
and \(m_2 = 5kg,\)   \(v_2 = 0,\)  \(v_2 =?\)
By conservation of momentum,
 \({ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }={ m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 }\)
\(2\times 10+5\times 0=5\times 1+5{ v }_{ 2 }\)
\(\therefore\)   \({ u }_{ 2 }=\frac { 15 }{ 5 } =3{ ms }^{ -1 }.\)

Let the final velocity of the bag along with the bullet embedded in it be u.
\(\therefore\)  For bullet,   \(m_1=20\) g = 0.02 kg,  \(u_1 = 500 m/s,\)   \(v_1=v\)
 For block,    \(m_2= 1\) kg,    \(u_2=0,\)    \(v_2=v\)
According to the law of conservation of momentum,
Total momentum before collision  = Total momentum after collision
\({ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }={ m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 }\)
or \({ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }=\left( { m }_{ 1 }+{ m }_{ 2 } \right) v\)    \([\because v_1=v_2=v]\)
or   \(v=\frac { { m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)
\(=\frac { 0.02\times 500+1\times 0 }{ 0.02+1 } =\frac { 10 }{ 1.02 } =\frac { 1000 }{ 102 } =9.87\quad m/s\)