Question

Block A has a mass of 5kg
and is placed on the top of
smooth triangular block B
having a mass of 25kg. If the
system is released from rest,
determine the distance B
moves, in metre, when A
reaches the bottom. Neglect
the size of block A.​

Answer 1

Given that,

Mass of block A = 5 kg

Mass of block B = 25 Kg

Suppose the base is 0.5 m

If the system is at rest then the center of mass will be zero.

When block A reaches the bottom then block A moves $x_{A}$ in negative direction and block B moves $x_{B}$ in positive direction.

So, the total distance covered by block A and B is  

$x_{A}+x_{B}=0.5$....(I)

We need to calculate the distance of B

Using formula of center of mass

$X_{cm}=\dfrac{-m_{A}x_{A}+m_{B}x_{B}}{m_{A}+m_{B}}$

$m_{A}x_{A}=m_{B}x_{B}$

Put the value of $x_{A}$ from equation (II) in equation (I)

$\dfrac{m_{B}x_{B}}{m_{A}}+x_{B}=0.5$

$x_{B}(m_{A}+m_{B})=0.5m_{A}$

$x_{B}=\dfrac{0.5m_{A}}{m_{A}+m_{B}}$

Put the value of both masses

$x_{B}=\dfrac{0.5\times5}{5+25}$

$x_{B}=0.083\ m$

Hence, The block of B moves 0.083 m.