Here   u = 0,     v = 30 m/s,        t = 10 s
\(\therefore\)    \(a=\frac { u-v }{ t } =\frac { 30-0 }{ 10 } =3m/{ s }^{ 2 }\)
Now  m = 1500 kg,  \(a=3m/{ s }^{ 2 }\)
\(\therefore\)  Required force,   \(F=ma=1500\times 3N=4500N\)

Total mass, \(=3\times { 10 }^{ 4 }+2\times 2\times { 10 }^{ 4 }=7\times { 10 }^{ 4 }\) kg,  \(a=0.2\ { ms }^{ -2 }\)
(i) \(F=ma=7\times { 10 }^{ 4 }\times 0.2=1.4\times { 10 }^{ 4 }N\)
(ii) Force experienced by each wagon \(=1.4\times { 10 }^{ 4 }N\)

Inertia of Motion. The tendency of a body to remain in its state of uniform motion in a straight line, is called 'inertia of motion'.
Examples:
(i) An athlete runs for certain distance before taking a jump so that his inertia of motion may help him to take a longer jump.
(ii) if a horse running fast suddenly stops, the rider is thrown forward if he is not firmly seated.

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.

Force may be defined as a push or a pull which or tends to change the state of rest of uniform motion or direction of motion of a body. The force exerted by the engine makes the train to move from its actual position of rest while the force exerted by the brakes slows down or stops the moving train. The force exerted on the steering wheel of a car changes its direction of motion.

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,   \({ m }_{ 1 }={ m }_{ 2 }=1.5kg,\ { u }_{ 1 }=2.5{ ms }^{ -1 },\ { u }_{ 2 }=-2.5{ ms }^{ -1 }\) 
Let v be the velocity of the combined object after the collision. By conservation of momentum,
Total momenta after collision = Total momenta before collision
 \(\left( { m }_{ 1 }+{ m }_{ 2 } \right) v={ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }\)
\(\left( 1.5+1.5 \right) v=1.5\times 2.5+1.5\times \left( -2.5 \right) \)
 \(3.0\ v=0\)
 \(v=0\ { ms }^{ -1 }.\)

Here, \(m_1=100\) g = 0.1 kg, \(m_2=200\) g = 0.2 kg, \(u_1=2 m {s}^{-1},\) \(u_2=1 m{s}^{-1},\) \(u_1=1.67 m{s}^{-1}\) \(u_2=?\)
accordinf to the law of conservation of momentum,
 \({ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }={ m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 }\)
or  \(0.1\times 2+0.2\times 1=0.1\times 1.67+0.2{ v }_{ 2 }\)
or  \(0.4=0.167+0.2{ v }_{ 2 }\)
or   \({ v }_{ 2 }=\frac { 0.4-0.167 }{ 0.2 } =1.165\quad m{ s }^{ -1 }\)