91Ó°ÊÓ

Given Data:

The speed of massive block after descends is: v =3m/s

The speed of massive block at rest is: u =0 m/s

The total mass of blocks is: m+M=22 kg

The distance descended by massive block is:h=1.20 m

The mass of each block in problem can be found by calculating the acceleration of each block and then apply the equilibrium equations for both blocks.

The acceleration of each block is given as:

$v^2=u^2+2ah$

a is the acceleration of each block.

Substitute all the values in the above equation.

$$\begin{gathered} {\left(3\mathrm{m}/\mathrm{s}\right)}^2={\left(0\mathrm{m}/\mathrm{s}\right)}^2+2\mathrm{a}\left(1.20\mathrm{m}\right) \\ \mathrm{a}=3.75\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Apply the equilibrium for small block mass.

$$\begin{gathered} ma=T-mg \\ T=m\left(g+a\right)......\left(2\right) \end{gathered}$$

Here, T is the tension in the light rope, gis the gravitational acceleration.

Apply the equilibrium for massive block mass.

$$\begin{gathered} -Ma=T-Mg \\ T=M\left(g-a\right)......\left(2\right) \end{gathered}$$

Equate the above equations to find the mass of each block.

$$\begin{gathered} m\left(g+a\right)=M\left(g-a\right) \\ g\left(M-m\right)=a\left(M+m\right) \end{gathered}$$

Substitute all the values in the above equation.

$$\begin{gathered} \left(9.8m/s^2\right)\left(M-m\right)=\left(3.75m/s^2\right)\left(22kg\right) \\ M-m=8.42kg......\left(3\right) \end{gathered}$$

The total mass of the blocks is given as:

$m+M=22kg......\left(4\right)$

On solving equation (3) and equation (4), we get

M=15.21kg

m=6.79 kg

Therefore, the mass of both blocks are 6.79 kg and 15.21 kg.