Answer Happy

I need help with part C
t1​=0.8728 s Correct You can verify your result by computing this time using the kinematic equations that describe the motion of a particle with constant acceleration: t1​=2d/a​ or t1​=v/a Part C What is the velocity of the blocks 4.40 s after the blocks have started moving? Assume that the rope joining the two blocks in long enough so that, at this time, block A is on the frictionless surface while block B is still on the rough surface. (Figure 2). Express your answer numerically in meters per second to three significant figures.
Learning Goal: Integrating the equation of motion, as applied to all particles in a system, yields ∑mi​(vi​)1​+∑∫t1​t2​​Fi​dt=∑mi​(vi​)2​ where mi​ is the th particle's mass, vi​ is the ith particle's velocity, and Fi​ is the external force that acts on the ith particle. This relationship states that the sum of the initial linear momenta, at time t1​, and the impulses of all the external forces acting between times t1​ and t2​ is equal to the sum of the linear momenta of the system, at time t2​. If the system has a mass center, G, the expression becomes Figure
Two blocks, each of mass m=7.60 kg, are connected by a massless rope and start sliding down a slope of incline θ=40.0∘ at t=0.000 s. The slope's top portion is a rough surface whose coefficient of kinetic friction is μk​=0.350. At a distance d=1.40 m from block A's initial position the slope becomes frictionless. (Figure 1) What is the velocity of the blocks when block A reaches this frictional transition point? Assume that the blocks width is negligible Express your answer numerically in meters per second to four significant figures. View Available Hint(s) Correct Part B How long does it take block A to reach the transition point? Express your answer numerically in seconds to four significant figures. View Available Hint(s)