JEE MainPhysicsCenter of Mass, Momentum and Collision
A uniform square sheet of side length 12 cm is placed on an xy -plane such that its bottom-left corner is at the origin and its sides lie along the positive x and y axes. A triangular corner of the sheet is folded perfectly over the remaining part. The fold is made along the line joining the midpoints of the top edge (6,12) and the right edge (12,6) , such that the corner originally at (12,12) lands exactly on the po
Correct answer
23
Step-by-step solution
Let the total mass of the square sheet be M . Its area is 144 cm ^2 . The original center of mass of the entire sheet is at (6, 6) . The folded part is a triangle with original vertices (12,6) , (12,12) , and (6,12) . The area of this triangular corner is 1 2 6 6 = 18 cm ^2 . The mass of this folded part is m = M 18 144 = M 8 . The original x -coordinate of the center of mass of this triangular part is: x_i = 12 + 12 + 6 3 = 10 cm After folding, the corner (12,12) lands on (6,6) . The other two vertices of the fold