Two blocks are connected by a light weight, flexible cord that passes over a frictionless pulley. ifm1=2 kg and m2 = 3 kg, and block 2 is initially at rest 1.96 m above the floor, how long does it take block 2 to reach the floor?

t = 1.41 sec.

Explanation:

If we assume that the acceleration of the blocks is constant, we can apply any of the kinematic equations to get the time since the block 2 was released till it reached the floor.

First, we need to find the value of  acceleration, which is the same for both blocks.

If we take as our system both blocks, and think about the pulley as redirecting the force simply (as tension in the strings behave like internal forces) , we can apply Newton's 2nd Law, as they were moving along the same axis, aiming at opposite directions, as follows:

F = m₂*g - m₁*g = (m₁+m₂)*a (we choose as positive the direction of the acceleration, will be the one defined by the larger mass, in this case m₂)

⇒ a = ( = g/5 m/s²

Once we got the value of a, we can use for instance this kinematic equation, and solve for t:

Δx = 1/2*a*t² ⇒ t² = (2* 1.96m *5)/g = 2 sec² ⇒ t = √2 = 1.41 sec.

0.11 m/s and 0.28 m/s should be correct

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s