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4-1.Newton's Laws of Motion
hard
A block of mass $m$ slides down on a wedge of mass $M$ as shown in figure. Let $\vec a_1$ be the acceleration of the wedge and $\vec a_2$ the acceleration of block w.r.t. ground. $N_1$ is the normal reaction between block and wedge and $N_2$ the normal reaction between wedge and ground. Friction is absent everywhere. Select the incorrect alternative
A$N_2< (M + m)g$
B$N_1 = m ( g\, cos \theta\, -\, |\vec a_1| sin \theta ) $
C$N_1\ sin \theta = M| \vec a_1|$
D$m\ \vec a_2=\,-M\, \vec a_1$
Solution
For entire system, $\overrightarrow{\mathrm{F}_{\mathrm{y}}}=(\mathrm{M}+\mathrm{m}) \mathrm{a}_{\mathrm{cmy}}$
$(\mathrm{M}+\mathrm{m}) \mathrm{g}-\mathrm{N}=(\mathrm{M}+\mathrm{m}) \mathrm{a}_{\mathrm{cmy}}$
$\mathrm{N}=(\mathrm{m}+\mathrm{M})\left(\mathrm{g}-\mathrm{a}_{\mathrm{cmy}}\right)$ $…(1)$
$\mathrm{N}_{1}+\mathrm{ma}_{1} \sin \theta=\mathrm{mg} \cos \theta$
$\mathrm{N}_{1}=\mathrm{m}\left[\mathrm{g} \cos \theta-\mathrm{a}_{1} \sin \theta\right]$ $…(2)$
$For\,M:$
$(\mathrm{M}+\mathrm{m}) \mathrm{g}-\mathrm{N}=(\mathrm{M}+\mathrm{m}) \mathrm{a}_{\mathrm{cmy}}$
$\mathrm{N}=(\mathrm{m}+\mathrm{M})\left(\mathrm{g}-\mathrm{a}_{\mathrm{cmy}}\right)$ $…(1)$
$\mathrm{N}_{1}+\mathrm{ma}_{1} \sin \theta=\mathrm{mg} \cos \theta$
$\mathrm{N}_{1}=\mathrm{m}\left[\mathrm{g} \cos \theta-\mathrm{a}_{1} \sin \theta\right]$ $…(2)$
$For\,M:$
Standard 11
Physics
