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.

Let the masses of the two blocks be \(m_1\) and \(m_2\)  respectively, and a and 5a be their respective accelerations.
For first block   \(F_1=m_1.a\)
For second block    \(F_2=m_2.5a\)
\(\because\)   \(F_1=F_2\)
\(\therefore\)    \({ m }_{ 1 }a={ m }_{ 2 }.5a\)  or    \(\frac { { m }_{ 1 } }{ { m }_{ 2 } } =\frac { 5 }{ 1 } =5:1.\)

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.

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, \(m_1=100\) 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.