Laws of Motion Part 2

Concept:In a non-inertial frame (the accelerating wedge), a pseudo‑force acts on the block opposite to the wedge’s acceleration.The net acceleration along the incline is found by resolving real and pseudo forces parallel to the incline.Explanation:Let the length of the incline (hypotenuse) be $s$.From the geometry: $\cos \theta = \frac{L}{s}$ → $s = \frac{L}{\cos \theta}$.Work in the frame of the wedge (accelerating left with $a_0$).A pseudo‑force $ma_0$ acts on the block to the right (opposite to $a_0$).Resolve forces along the incline:• Component of gravity down the incline: $mg \sin \theta$.• Component of pseudo‑force up the incline (since it acts to the right and the incline slopes down to the right? check: pseudo‑force is horizontal to right; its component along the incline is $ma_0 \cos \theta$ up the incline because the incline angle $\theta$ measured from horizontal? Actually, for wedge moving left, pseudo‑force is to right. The incline goes down to the right (as typical). The component of pseudo‑force along incline is $ma_0 \cos \theta$ pointing up the incline (opposing motion). So net force along incline down = $mg\sin\theta - ma_0\cos\theta$.Thus relative acceleration $a_{\text{rel}} = g \sin \theta - a_0 \cos \theta$.Block starts from rest, so using $s = \frac{1}{2} a_{\text{rel}} t^2$:$\frac{L}{\cos \theta} = \frac{1}{2} (g\sin\theta - a_0\cos\theta) t^2$.Solve for $t^2$:$t^2 = \frac{2L}{\cos\theta (g\sin\theta - a_0\cos\theta)} = \frac{2L}{g\sin\theta\cos\theta - a_0\cos^2\theta}$.Use identities: $2\sin\theta\cos\theta = \sin2\theta$ and $\cos^2\theta = \frac{1+\cos2\theta}{2}$.Multiply numerator and denominator by 2:$t^2 = \frac{4L}{2g\sin\theta\cos\theta - 2a_0\cos^2\theta} = \frac{4L}{g\sin2\theta - a_0(1+\cos2\theta)}$.Therefore $t = \sqrt{ \frac{4L}{g\sin2\theta - a_0(1+\cos2\theta)} }$.Answer:$t = \sqrt{ \frac{4L}{g\sin2\theta - a_0(1+\cos2\theta)} }$ → Option B.