1. **Problem statement:** Two trucks A and B collide on the same track moving in opposite directions. Mass of A is 600 kg, mass of B is $m$ kg. Speeds before collision: A is 4 m/s, B is 2 m/s (opposite direction). After collision, they stick together and move at 0.5 m/s in A's original direction. 2. **Formula and rules:** Use conservation of momentum since no external horizontal forces act during collision. Momentum before collision = Momentum after collision $$m_A v_{A_i} + m_B v_{B_i} = (m_A + m_B) v_f$$ Note: Since B moves opposite to A, take A's direction as positive, so B's velocity is negative. 3. **Calculate $m$:** Given: $m_A = 600$ $v_{A_i} = 4$ $v_{B_i} = -2$ $v_f = 0.5$ Substitute: $$600 \times 4 + m \times (-2) = (600 + m) \times 0.5$$ Simplify: $$2400 - 2m = 300 + 0.5m$$ Bring terms together: $$2400 - 300 = 0.5m + 2m$$ $$2100 = 2.5m$$ Solve for $m$: $$m = \frac{2100}{2.5} = 840$$ So, mass of B is 840 kg. 4. **Calculate impulse on A:** Impulse $J$ is change in momentum: $$J = m_A (v_{A_f} - v_{A_i})$$ Given $v_{A_f} = 0.5$, $v_{A_i} = 4$: $$J = 600 \times (0.5 - 4) = 600 \times (-3.5) = -2100$$ Magnitude of impulse is $2100$ Ns. --- **Final answers:** (a) $m = 840$ kg (b) Impulse magnitude on A = 2100 Ns