Answer
$m_{bowl}=4m_{mouse}$
Work Step by Step
Please see the attached image below.
Here we use the principle of conservation of momentum.
$\rightarrow m_{1}u_{1}+m_{2}u_{2}=m_{1}V_{1}+m_{2}V_{2}$
Let's plug known values into this equation.
$0+0=m_{1}\frac{(\frac{D}{2}-\frac{D}{10})}{t}+m_{2}\frac{(\frac{-D}{10})}{t}$
$\frac{m_{2}D}{10}= m_{1}(\frac{5D-D}{10})=\gt m_{2}=4m_{1}$
$$m_{bowl}=4m_{mouse}$$
