Question

A block A(mass=4m)is placed in the top of a wedge B of base length l (mass=20M) as shown in figure.When the system is released from rest.Find the distance moved by the wedge B till the block A reaches at lowest end of wedge assume all surfaces are frictionless.​

Answer 1

$\huge{\underline{\sf{\red{Answer\leadsto \dfrac{L}{6}}}}}$

Given:

Mass of Block A = 4M

Mass of Block B = 20M

Length of Base = L

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Formula

Position of Center of mass = $\dfrac{X_BM_B+X_AM_A}{M_A+M_B}$

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Solution:

Using the Formula above and given information

$\bigcirc$ Initial Position of center of Mass

$\implies\dfrac{X_B.20 M+4M}{24M} \\ \\ = \dfrac{\cancel{4}\cancel{M}(5X_B + L)}{\cancel{24}\cancel{M}} \\ \\ \leadsto \dfrac{5X_B +l}{6}\huge{\star2}$

$\bigcirc$ Final Position of center of mass =

$\implies\dfrac{(X_B+x)\cancel{20}\cancel{M}+\cancel{4}\cancel{M}x}{\cancel{24}\cancel{M}} \\ \\ = \dfrac{5(X_B+x)+x}{6}\huge{\star1}$

$\because$ there is no horizontal force on system

so centre of mass initially= centre of mass finally

$\star1 = \star2 \\ \\ \therefore \dfrac{\cancel{5X_B}+l}{\cancel6} = \dfrac{\cancel{5X_B}+5x+x}{\cancel6} \\ \\ L=6x$

$\large{\boxed{x=\dfrac{L}{6}}}}$

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Answer 2

$\mathfrak{\huge{\orange{\underline{\underline{Answer :}}}}}$

$\huge{\underline{\sf{\red{Answer\leadsto \dfrac{L}{6}}}}}$

Given:

Mass of Block A = 4M

Mass of Block B = 20M

Length of Base = L

Formula

Position of Center of mass = $\dfrac{X_BM_B+X_AM_A}{M_A+M_B}$

_____________________________

Solution:

Using the Formula above and given information

$\bigcirc$ Initial Position of center of Mass

$\implies\dfrac{X_B.20 M+4M}{24M} \\ \\ = \dfrac{\cancel{4}\cancel{M}(5X_B + L)}{\cancel{24}\cancel{M}} \\ \\ \leadsto \dfrac{5X_B +l}{6}\huge{\star2}$

$\bigcirc$ Final Position of center of mass =

$\implies\dfrac{(X_B+x)\cancel{20}\cancel{M}+\cancel{4}\cancel{M}x}{\cancel{24}\cancel{M}} \\ \\ = \dfrac{5(X_B+x)+x}{6}\huge{\star1}$

$\because$ there is no horizontal force on system

so centre of mass initially= centre of mass finally

$\star1 = \star2 \\ \\ \therefore \dfrac{\cancel{5X_B}+l}{\cancel6} = \dfrac{\cancel{5X_B}+5x+x}{\cancel6} \\ \\ L=6x$

$\large{\boxed{x=\dfrac{L}{6}}}}$

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