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A wedge shaped block 'A' of mass 'M' is at rest on a smooth horizontal surface.A small block 'B' of mass 'm' placed at the top edge of inclined plane of length 'L' as shown in the figure.By the time, the block 'B' reaches the bottom end the wedge 'A' will move a distance of

ramya bhaskar , 16 Years ago

Grade 11

2 Answers

Badiuddin askIITians.ismu Expert

Dear ramya

Since there is no external force in horizontal direction in wedge + mass system

so center of mass of wedge + mass syaytem will not move

let x1 is the COM of wedge from origin

and x is the distance moved by wedge

(Mx1 +mLcosΘ)/(m+M) = {M(x1+x) + mx}/(m+M)

x=mLcos Θ/(m+M)

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Badiuddin

Last Activity: 16 Years ago

ankit singh

Since there is no external force in horizontal direction in wedge + mass system

so center of mass of wedge + mass syaytem will not move

let x1 is the COM of wedge from origin

and x is the distance moved by wedge

(Mx1 +mLcosΘ)/(m+M) = {M(x1+x) + mx}/(m+M)

x=mLcos Θ/(m+M)

Last Activity: 5 Years ago

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