Q.A particle of mass $m$ slides on a smooth double inclined plane, as shown in the figure, where the path $AC$ is steeper (makes a larger angle with the horizontal) than the path $AB$.

For a block sliding on a completely frictionless incline of angle $\theta$, resolving the weight $mg$ into components along and perpendicular to the slope gives a perpendicular balance $N=mg\cos\theta$ and an along-slope (accelerating) component $mg\sin\theta=ma$, so $a=g\sin\theta.$

The single most important feature of this result is that the acceleration depends only on the incline angle, never on the object's mass — a heavy block and a light block, released together on the same frictionless slope, accelerate identically and reach the bottom at the same time with the same speed, $v=\sqrt{2gs\sin\theta}$ (from $s$, the incline's length, starting from rest). …