Question $1$

Atword Machine, left side $4$ kg $,$ right side $5$ kg $,$ negligible pulley mass and negligible string mass.

The constant acceleration $=9\mathrm{.}8*($difference of the two mass values$)/($sum of the two mass values$)=9\mathrm{.}8*\frac{5-4}{5+4}=,$ meter $\frac{?}{sec}\frac{?}{sec}.$

After a certain duration, the velocity magnitude became $1$mete$\frac{r}{sec},$ and the $4$ kg accelerated upward while the $5$ kg downward. The time duration to achieve $1\frac{m}{s}$ must be second.

Question $2$

Atwood Marchine left side $4$ kg and $1$ kg $,$ right side $5$ kg $,$ negligible pulley mass and negligible string mase

The $4$ kg collided with $1$ kg and both masses moved up together. Using momentum conservation, the new velocity magnitude of the $4$ kg $-1$ kg composite after collision became meter per second. ${?}_{.}$ The collision generated an exchange impulsive magnitude of Newton$-$sec$.$

The relatively faster $5$ kg pulled onto the slower $4$ kg $-1$ kg composter such that all objects moved together in a collision model. The new velocity magnitude became meter per second. The pullingcollision generated an exchange impulse magnitude of Newton$-$sec$.$ The final Atwood Machine configuration of $4kg-1kg$ composite on the left and $5$ kg on the right must have an acceleration of $\text{'}$s$/$s$.$

The alternative model of $1$ kg @ $0\frac{m}{s}$ collided with $(4+5)$ kg @ $1\frac{m}{s}$ would give a new velocity magnitude of $\frac{m}{s}.$ The collision generated an exchange impulse magnitude of Newtonsec.