Problem $1(15$ points$)$: At time t $=0,$ the velocity of cylinder A is vA$,1=0\mathrm{.}3$ m$/$s down.

Using the linear impulse$-$momentum principle, determine the velocity vB$,2$ of cylinder B

$($positive if down, negative if up$)$ at time t $=2$ s$.$ Assume no mechanical interference and

neglect all friction.

Hint: You may follow the steps below to construct the solution.

Step $1$: Study the pulley system based on what we have learned in the chapter of particles

kinematics. Find the relations of the velocities of A and B$.$ Clearly indicate the

positive directions.

Step $2$: For each particle, i$.$e$.,$ A and B$,$ draw a set of impulse$-$momentum diagram. Re$-$

member to include all external forces applied. Let the tension force of the cable

be T $.$ What are the forces applied to A and B$,$ respectively, by the pulley system?

$($Use free$-$body diagrams to determine.$)$

Step $3$: Based on the impulse$-$momentum diagrams, derive two equations $($one for A and

one for B$)$ using the impulse$-$momentum principle.

Step $4$: Substitute some of the velocity terms following the relations of the velocities of A

and B from Step $1.$

Step $5$: Now, we should have two equations with two unknowns: R t$2$

t$1$ T dt and vB$,2.$ Solve

for vB$,2$ by cancelling out R t$2$

t$1$ T dt$.$

Step $6$: Check that the final answer is vB$,2=1\mathrm{.}652$ m$/$s