Physics - NLM Question with Solution | TestHub

Let $a_M$ be the acceleration of the wedge (to the left) and $a_b$ be the acceleration of the block relative to the wedge (down the incline). Equations of motion: For wedge $M$: $N\sin \theta =M{a}_M$ (1) For block $m$ (absolute accelerations $a_x=a_b\cos \theta -a_M$, $a_y=a_b\sin \theta$):$mg-N\cos \theta =m(a_b\sin \theta )$ (2)$N\sin \theta =m(a_b\cos \theta -a_M)$ (3) From (1) and (3): $M{a}_M=m(a_b\cos \theta -a_M)⟹(M+m)a_M=m{a}_b\cos \theta$ (4) From (1), $N=M{a}_M/\sin \theta$. Substitute into (2):$mg-(M{a}_M/\sin \theta )\cos \theta =m{a}_b\sin \theta ⟹mg\sin \theta -M{a}_M\cos \theta =m{a}_b{\sin }^2\theta$ (5) From (4), $a_b=(M+m)a_M/(m\cos \theta )$. Substitute into (5):$mg\sin \theta -M{a}_M\cos \theta =m\frac{(M+m)a_M}{m\cos \theta }{\sin }^2\theta$ $mg\sin \theta \cos \theta -M{a}_M{\cos }^2\theta =(M+m)a_M{\sin }^2\theta$ $mg\sin \theta \cos \theta =a_M(M{\cos }^2\theta +M{\sin }^2\theta +m{\sin }^2\theta )=a_M(M+m{\sin }^2\theta )$. P: $a_M=\frac{mg\sin \theta \cos \theta }{M+m{\sin }^2\theta }$. Given $m=1,M=3,\theta =30^{\circ }$. $\sin \theta =1/2,\cos \theta =\sqrt{3}/2,{\sin }^2\theta =1/4$.$a_M=\frac{1\cdot g\cdot (1/2)\cdot (\sqrt{3}/2)}{3+1\cdot (1/4)}=\frac{g\sqrt{3}/4}{13/4}=\frac{g\sqrt{3}}{13}$. Q: $a_b=\frac{(M+m)a_M}{m\cos \theta }=\frac{(3+1)}{1\cdot (\sqrt{3}/2)}\cdot \frac{g\sqrt{3}}{13}=\frac{4}{\sqrt{3}/2}\cdot \frac{g\sqrt{3}}{13}=\frac{8}{\sqrt{3}}\cdot \frac{g\sqrt{3}}{13}=\frac{8g}{13}$. R: $N=M{a}_M/\sin \theta =3\cdot \frac{g\sqrt{3}}{13}/(1/2)=\frac{6g\sqrt{3}}{13}$. S: Force exerted by wedge on ground is $N_{ground}=(M+m)g-m{a}_y=(M+m)g-m(a_b\sin \theta )$.$N_{ground}=(3+1)g-1\cdot (8g/13)\cdot (1/2)=4g-4g/13=\frac{52g-4g}{13}=\frac{48g}{13}$.