The two masses shown in the figure are each initially 1.80 m above the ground, a

Question

The two masses shown in the figure are each initially 1.80 m above the ground, and the massless frictionless pulley is fixed 4.8 m above the ground. (mA = 1.5 kg and mB = 2.6 kg.) What maximum height does the lighter object reach after the system is released? [Hint: First determine the acceleration of the lighter mass and then its velocity at the moment the heavier one hits the ground. This is its "launch" speed. Assume the mass doesn't hit the pulley. Ignore the mass of the cord.]
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The two masses shown in the figure are each initially 1.80 m above the ground, and the massless frictionless pulley is fixed 4.8 m above the ground. (mA = 1.5 kg and mB = 2.6 kg.) What maximum height does the lighter object reach after the system is released? [Hint: First determine the acceleration of the lighter mass and then its velocity at the moment the heavier one hits the ground. This is its "launch" speed. Assume the mass doesn't hit the pulley. Ignore the mass of the cord.]

Explanation / Answer

For the 1.5 kg we have T (tension) up which is the positive direction, and the weight of the mass

(1.5x9.8 = 14.7 N) down in the negative direction:
T - 14.7 = (1.5 kg)a

For the 2.6 kg mass the positive direction is down (as we make acceleration positive, and since we guessed that the 2.6 kg mass would accelerate down, then down is the positive direction; the two accelerations must be the same) and we have tension (T) up which is the negative direction in this case, and the weight

(2.6x9.8 = 25.48) which is down (the positive direction in this case):
25.48 - T = (2.6 kg)a

we solve the two equations and two unknowns for the acceleration:

T-14.7 + 25.48 - T = (4.1 kg)a

10.78 N = (4.1 kg)a

a = 2.63 m/s/s

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Next, let's suvat to find the velocity the lighter mass has when it is 3.60 m off the ground. (it is already 1.80 m from the ground, and when the heavier mass hits the ground, it will be 1.80 m above this...)  

We know it starts with u = 0, a = 2.63 m/s/s, and s = 1.80 m

so using v^2 = u^2 + 2as

v1 = 3.08 m/s

Once block 2 hits the ground, FT--->0 and block 1 will have the downward acceleration of g. For this motion of block 1 up to the highest point reached, we have

                  

v^2 = v1^2 + 2a(h - y1)

0 = (3.08m/s)^2 + 2(-9.80m/s2)(h - 3.60m)

h = 4.084 m  

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