91Ó°ÊÓ

i) Mass of block 1, $m_{1}is6.6kg$.

ii) The speed of block 2 is $v_{2i}$.

You use the concept of conservation of momentum to find the mass. You use the equations given in the book 9-75 and 9-76 and can solve for the mass of block 2.

$$\begin{gathered} V_{1f}=\frac{2m_2}{m_1+m_2}{v}_{2i} \\ {V}_{2f}=\frac{\left(m_2-m_1\right)}{m_1+m_2}{v}_{2i} \end{gathered}$$

We can use the below equations to find mass as,

$$\begin{gathered} V_{1f}=\frac{2m_2}{m_1+m_2}{v}_{2i} \\ {V}_{2f}=\frac{\left(m_2-m_1\right)}{m_1+m_2}{v}_{2i} \end{gathered}$$

Here, $v_{2i}$is the initial velocity of block -2.

The velocity of block 2, after bouncing off the wall, will be the same as before but with a negative sign; we can write it as,

$V_{1f}=-V_{2f}$

Substitute the values in the above expression, and we get,

$$\begin{gathered} \frac{2m_2}{m_1+m_2}{v}_{2i}=\frac{\left(m_2-m_1\right)}{m_1+m_2}{v}_{2i} \\ 2m_2=-m_2-m_1 \\ 3m_2=m_1 \\ {m}_2=\frac{m_1}{3} \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} m_2=\frac{6.6}{3} \\ =2.2kg \end{gathered}$$

Thus, the mass of block 2 is $m_2=2.2kg$.