Problem Analysis

A train of length $l$ moves over a hill with dimensions $a, b, c$.
Given: $l = 168$ m, $a = 60$ m, $b = 80$ m, $c = 100$ m.
Note that $a^2 + b^2 = 60^2 + 80^2 = 3600 + 6400 = 10000 = c^2$. The hill forms a right-angled triangle.
Since $l (168) > a+b (140)$, the train completely covers the hill during the maximum potential energy configuration.

Step 1: Calculate Maximum Potential Energy

Let $\lambda = M/l$ be the mass per unit length. The maximum PE occurs when the entire hill is covered by the train. The remaining part of the train lies on the flat ground ($PE=0$).
The PE of a rod of mass $m$ and length $L$ inclined at angle $\theta$ is $mg (H_{cm}) = mg (h/2)$.

• Part on side a: Length $a$. Mass $m_a = \lambda a$. Height of COM is $h/2$. $$ U_a = (\lambda a) g \frac{h}{2} $$
• Part on side b: Length $b$. Mass $m_b = \lambda b$. Height of COM is $h/2$. $$ U_b = (\lambda b) g \frac{h}{2} $$

Total Max PE: $$ U_{max} = \frac{\lambda g h}{2} (a + b) $$ Altitude $h$ of the triangle with sides $a, b, c$: Area = $\frac{1}{2}ab = \frac{1}{2}ch$. $$ h = \frac{ab}{c} $$ Substituting $h$: $$ U_{max} = \frac{\lambda g}{2} \left( \frac{ab}{c} \right) (a+b) $$

Step 2: Conservation of Energy

The initial kinetic energy of the train on the flat track must be sufficient to provide this potential energy. $$ \text{Initial KE} = U_{max} $$ $$ \frac{1}{2} M v^2 = \frac{1}{2} (\lambda l) v^2 = \frac{\lambda g ab (a+b)}{2c} $$ Canceling $\frac{\lambda}{2}$: $$ l v^2 = \frac{g ab (a+b)}{c} $$ $$ v = \sqrt{ \frac{g ab (a+b)}{cl} } $$

Step 3: Numerical Calculation

Given: $g=10$, $a=60$, $b=80$, $c=100$, $l=168$. $$ v = \sqrt{ \frac{10 \times 60 \times 80 \times (60+80)}{100 \times 168} } $$ $$ v = \sqrt{ \frac{48000 \times 140}{16800} } $$ $$ v = \sqrt{ \frac{48000}{120} } \quad (\text{Since } 16800/140 = 120) $$ $$ v = \sqrt{ 400 } = 20 \text{ m/s} $$