question_answer

A block of mass 0.5 kg is placed over top of a wedge of mass 50 kg.

Angle of inclination \[\theta \]is \[37{}^\circ \]with horizontal. Block is released at \[t=0\] and it slide down the incline and reaches bottom with a speed of \[1.2625\,m{{s}^{-1}}.\]

There is no friction. Speed of the wedge with respect to ground, when block reaches at bottom is

A) \[0.01\,m{{s}^{-1}}\]              

B) \[0.02\,m{{s}^{-1}}\]

C) \[0.03\,m{{s}^{-1}}\]              

D) \[0.05\,m{{s}^{-1}}\]

Correct Answer: A

Solution :

Let block reaches bottom with speed u and at that instant block is moving with a velocity v, then by conservation of momentum, we have

\[Mv+m\,(v-u\cos \theta )=0\]

Substituting values in above- equation, we get

\[5.0v+0.5v-0.5\times \frac{4}{5}=0\]

\[\Rightarrow \]\[v(50.5)=0.5\times 1.2625\times \frac{4}{5}\]

\[\Rightarrow \]\[v=\frac{0.5\times 1.2625\times 4}{5\times 50.5}=0.01\,m{{s}^{-\,1}}\]