$(a)$ Two objects of masses $100\, g$ and $200\, g$ are moving along the same line and direction with velocities $2\, m s^{-1}$ and $1\, m s^{-1}$ respectively. They collide and after the collision,the first object moves at a velocity of $1.67\, m s^{-1}$. Determine the velocity of the second object.
$(b)$ If a man jumps out from a boat,the boat moves backwards. Why?

(N/A) Given: $m_1 = 100\, g = 0.1\, kg$,$m_2 = 200\, g = 0.2\, kg$,$u_1 = 2\, m s^{-1}$,$u_2 = 1\, m s^{-1}$,$v_1 = 1.67\, m s^{-1}$.
Using the law of conservation of linear momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$.
Substituting the values: $(0.1 \times 2) + (0.2 \times 1) = (0.1 \times 1.67) + (0.2 \times v_2)$.
$0.2 + 0.2 = 0.167 + 0.2 v_2$.
$0.4 = 0.167 + 0.2 v_2$.
$0.2 v_2 = 0.4 - 0.167 = 0.233$.
$v_2 = 0.233 / 0.2 = 1.165\, m s^{-1}$.
$(b)$ This phenomenon is based on Newton's third law of motion and the law of conservation of momentum. When the man jumps forward,he exerts a force on the boat in the backward direction. Since the boat is floating in water and is not fixed,it moves backwards to conserve the total momentum of the system.