Two bodies having the same mass $40\,kg$ are moving in opposite directions, one with a velocity of $10\,m{s^{ - 1}}$ and the other with $7\,m{s^{ - 1}}$. If they collide and move as one body, the velocity of the combination is
A. $10\,m{s^{ - 1}}$
B. $7\,m{s^{ - 1}}$
C. $3\,m{s^{ - 1}}$
D. $1.5\,m{s^{ - 1}}$

Hint: To answer this question, we will use the principle of conservation of linear momentum and then by use of this principle and equations we can calculate the velocity of the combined bodies after contact.

Formula used:
The principle of conservation of linear momentum says that Initial momentum of a system is always equal to final momentum of a system.
${P_i} = {P_f}$
where $P = mv$ denotes the momentum of a body.

Complete step by step solution:
According to the question, we have given that initial conditions of two bodies as $m,{u_1} = 10\,m{s^{ - 1}}$ and $m,{u_2} = - 7\,m{s^{ - 1}}$ as both bodies are moving in opposite directions. so total initial momentum of the system is
${P_i} = m{u_1} + m{u_2} \\
\Rightarrow {P_i} = 3m \to (i) \\ $
Now, let V be the final velocity of combination of two bodies after collision having mass of $m + m = 2m$ then final momentum of the body is,
${P_f} = 2mV \to (ii)$
Now, equation (i) and (ii) we get
$3m = 2mV \\
\therefore V = 1.5m{s^{ - 1}} \\ $
Hence, the correct answer is option D.

Note: The law of conservation of energy, which states that energy is always conserved, and the law of conservation of angular momentum in rotational dynamics, which states that total system angular momentum is always conserved when no external torque is applied to the body, are the other two basic laws of conservation.