A man of $$60$$ kg is running on the road and suddenly jumps into a stationary trolly car of mass $$120$$ kg. Then the trolly car starts moving with velocity $$2$$ m s$$^{-1}$$. The velocity of the running man was ______ m s$$^{-1}$$, when he jumps into the car.

We apply the principle of conservation of linear momentum to solve this problem. The mass of the man is $$m_1 = 60$$ kg, the mass of the trolley is $$m_2 = 120$$ kg, the initial velocity of the trolley is $$0$$ m/s (stationary), and the final velocity of the (man + trolley) system is $$v_f = 2$$ m/s. Let the velocity of the running man be $$v$$ m/s.

Total momentum before equals total momentum after. Substituting into this gives:

$$m_1 \times v + m_2 \times 0 = (m_1 + m_2) \times v_f$$

$$60 \times v + 120 \times 0 = (60 + 120) \times 2$$

$$60v = 180 \times 2$$

$$60v = 360$$

$$v = \frac{360}{60}$$

$$v = 6 \text{ m/s}$$

Hence, the velocity of the running man when he jumps into the car is 6 m/s.