Physics - NLM Question with Solution | TestHub

Let A = acceleration of the wedge (horizontal) and a' = acceleration of B relative to the wedge down the incline (45°).

$$\mathrm{A}=\frac{\mathrm{m}\mathrm{g}\sin \theta \cos \theta }{\mathrm{M}+\mathrm{m}{\sin }^2\theta }$$

$$=\frac{0.6\times \mathrm{g}\times \frac{1}{2}}{1.7+0.6\times \frac{1}{2}}$$

$$=\frac{0.3\mathrm{g}}{2}=\frac{3\mathrm{g}}{20}$$

So statement (A) is correct.

$${\mathrm{a}}^{\prime }=\frac{(\mathrm{M}+\mathrm{m})\mathrm{g}\sin \theta }{\mathrm{M}+\mathrm{m}{\sin }^2\theta }$$

$$=\frac{2.3\mathrm{g}(1/\sqrt{2})}{2}=\frac{23\sqrt{2}\mathrm{g}}{40}$$

Absolute acceleration of B - horizontal component (taking wedge's motion into account):

$${\mathrm{a}}^{\prime }\cos \theta -\mathrm{A}=\frac{23\sqrt{2}}{40}\mathrm{g}\times \frac{1}{\sqrt{2}}-\frac{3\mathrm{g}}{20}$$

$$=\frac{23\mathrm{g}}{40}-\frac{6\mathrm{g}}{40}=\frac{17\mathrm{g}}{40}$$

So statement (C) is correct. Vertical component:

$${\mathrm{a}}^{\prime }\sin \theta =\frac{23\sqrt{2}}{40}\mathrm{g}\times \frac{1}{\sqrt{2}}=\frac{23\mathrm{g}}{40}$$

So statement (B) is correct. Hence (A), (B), (C).