A wedge $Y$ with mass of $10$ kg and all frictionless surfaces and the inclined surface making $37^\circ$ with horizontal. A block $X$ with mass $2$ kg is placed at the highest point of the wedge as shown in figure is at rest. At $t = 0$ wedge ($Y$) is pulled toward right with constant force ($f$) of $24$ N. Taking the block $X$ at rest at $t = 0$, the time taken by it to slide down $8.8$ m on the slope, while $Y$ is on the move, is ______ s.
(take $\tan(37^\circ) = 3/4$ and $g = 10\ \text{m/s}^2$)
Answer: (A) $2$
Concept. In the wedge's frame the block feels a pseudo-force $mA$; the official answer takes $A = \dfrac{f}{M + m}$.
Step 1. Wedge acceleration. $A = \dfrac{24}{12} = 2\ \text{m/s}^2$.
Step 2. Along the slope. $a = g\sin37° - A\cos37° = 6 - 1.6 = 4.4\ \text{m/s}^2$.
Step 3. Time. $t = \sqrt{\dfrac{2 \times 8.8}{4.4}} = 2$ s.
(An exact treatment of the block's push on the wedge gives $A \approx 1.34\ \text{m/s}^2$ and $t \approx 1.9$ s; the nearest option is still $2$ s.)
Answer. $2$ (Option A).
Solution by Sreeraj P, M.Sc Physics
