Let $M$ be the original mass of the train and $m_d$ be the mass of the detached carriages. The traction force is denoted by $F_T$, and the resistance force is $F_r = 0.1 \times (\text{mass}) \times g$. We apply Newton's second law to the train in two states: before and after the carriages detach. We assume the acceleration due to gravity is $g = 10$ m/s².

[Q1] Ratio of the mass of the detached carriages to the original mass of the train

State 1: Before Detachment The total mass is $M$. The net force equation is:

$$F_T - 0.1Mg = M a_1$$

State 2: After Detachment The remaining mass is $M' = M - m_d$. The traction force $F_T$ remains constant. The net force equation is:

$$F_T - 0.1(M-m_d)g = (M-m_d) a_2$$

Derivation We have a system of two equations with the unknown $F_T$. We can solve for the desired ratio by eliminating $F_T$. From the first equation, we can express the traction force as:

$$F_T = M(a_1 + 0.1g)$$

Substitute this expression for $F_T$ into the second equation:

$$M(a_1 + 0.1g) - 0.1(M-m_d)g = (M-m_d) a_2$$

Expanding and simplifying the terms:

$$Ma_1 + 0.1Mg - 0.1Mg + 0.1m_d g = Ma_2 - m_d a_2$$ $$Ma_1 + 0.1m_d g = Ma_2 - m_d a_2$$

Rearrange the equation to group terms with $M$ and $m_d$:

$$m_d a_2 + 0.1m_d g = Ma_2 - Ma_1$$ $$m_d(a_2 + 0.1g) = M(a_2 - a_1)$$

The ratio of the detached mass to the original mass is:

$$\frac{m_d}{M} = \frac{a_2 - a_1}{a_2 + 0.1g}$$

Calculation Substituting the given values $a_1 = 1$ m/s², $a_2 = 2$ m/s², and using $g=10$ m/s²:

$$\frac{m_d}{M} = \frac{2 - 1}{2 + 0.1(10)} = \frac{1}{2 + 1}$$ $$\frac{m_d}{M} = \frac{1}{3}$$