$(a)$ State the law of conservation of momentum.
$(b)$ $A$ body of mass $2 \, kg$,initially moving with a velocity of $10 \, m s^{-1}$,collides with another body of mass $5 \, kg$ at rest. After the collision,the velocity of the first body becomes $1 \, m s^{-1}$. Find the velocity of the second body.

(D) The law of conservation of momentum states that in the absence of an external unbalanced force,the total linear momentum of a system remains conserved (constant).
$(b)$ Given:
Mass of the first body,$m_{1} = 2 \, kg$
Initial velocity of the first body,$u_{1} = 10 \, m s^{-1}$
Mass of the second body,$m_{2} = 5 \, kg$
Initial velocity of the second body,$u_{2} = 0 \, m s^{-1}$ (at rest)
Final velocity of the first body,$v_{1} = 1 \, m s^{-1}$
Let the final velocity of the second body be $v_{2}$.
According to the law of conservation of momentum:
$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$
Substituting the values:
$(2 \times 10) + (5 \times 0) = (2 \times 1) + (5 \times v_{2})$
$20 + 0 = 2 + 5v_{2}$
$20 - 2 = 5v_{2}$
$18 = 5v_{2}$
$v_{2} = \frac{18}{5} = 3.6 \, m s^{-1}$
Thus,the velocity of the second body after the collision is $3.6 \, m s^{-1}$.