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
 \(p = mv = 10 \times 4 = 40 \ kg \ m \ s^{-1}.\)

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,  \(m_1 = 2\) 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,
 \({ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }={ m }_{ 1 }{ v }_{ 1 }+{ m }_{ 2 }{ v }_{ 2 }\)
\(2\times 10+5\times 0=5\times 1+5{ v }_{ 2 }\)
\(\therefore\)   \({ u }_{ 2 }=\frac { 15 }{ 5 } =3{ ms }^{ -1 }.\)

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   \({ v }_{ 2 }=\frac { 0.4-0.167 }{ 0.2 } =1.165\quad m{ s }^{ -1 }\)

Law of conservation of momentum. This law states that if a number of bodies are interacting with each other (i.e., exerting forces on each other), their total momentum remains conserved before and after the interaction, provided there is no external force acting on them.
Derivation from Newton's second law of motion. Let \(p_1\) and \(p_2\)  represent the sum of momentum of a group of objects before and after the collision, respectively. Let t be the time elapsed during the collision.
According to Newton's Second law of motion,
External force = Rate of change of momentum
or \(F = \frac {p_2-p_1}{t}\)
If there is no external force, that is F = 0, then
 \(\frac {p_2-{p}_{1}}{t}=0\) or \(p_2 = p_1\)   
Hence in the absence of an external force, the total momentum of a group of objects remains unchanged or conserved during the collision. This is the law of conservation of momentum.

Mass and inertia. The mass of a body is a measure of its inertia. The larger the mass of a body, the larger is inertia or opposite offered by a body to change its state of motion. This can be understood from the following examples:
(i) If we kick a football, it flies a long way. if we kick a stone of the same size, it hardly moves instead, we may get an injury in our leg while hitting the stone. The stone has more inertia than the football.
(ii) we may cause a bicycle to pick up a large velocity by applying a certain force. But the same force will produce a negligible change in the motion of a train. This is because, in comparison to the bicycle, the train has more to oppose any change in its state of motion because of its larger mass. in order words, the train has more inertia than the bicycle.
(iii) The SI unit of mass and inertia is kilogram (kg).

Unbalanced forces. If the resultant of the several forces acting on a body is not zero, the forces are said to be unbalanced forces. Unbalanced forces produce a change in the state of rest or uniform motion of a body.
Examples:
(i) In a tug-of-war, when one of the two teams pulls the rope with a larger force, it is able to pull the weaker team towards it. Here teh two forces are not balanced. Therefore, is results in the motion of the weaker team towards the larger force along the rope.
(ii) When we stop pedaling a bicycle, it begins to slow down the road has small cracks and bumps. The bicycle has to overcome these imperfections, which slow down it. Forces which slow down the moving objects in this way recalled forces of friction.
so we can say that objects continue to move, with the same velocity unless acted upon by unbalanced forces.