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A block of mass 2kgslides down the face of a smooth 45^o wedge of mass 9 kg as shown in the figure. The wedge is placed on a frictionless horizontal surface. Determine the acceleration of the wedge.

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$2 m / s^{2}$

$\frac{11}{\sqrt{2}} m / s^{2}$

$1 m / s^{2}$

$none of these$

answer is C.

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Detailed Solution

Suppose a is the acceleration of block w.r.t. wedge.

For block:

$N + m A \sin 45^{\circ} = m g \cos 45^{\circ} . . . . . (i)$

For wedge: $N \sin 45^{\circ} = M A . . . . . . . (i i)$

From (i) and (ii) $A = 1 m / s^{2}$

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