- #1

agusb1

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- Homework Statement
- Two objects are connected by a light string passing over a light, frictionless pulley as shown in image. The object of mass m1= 5.00 kg is released from rest at a height h= 4.00 m above the table. Using the isolated system model, (a) determine the speed of the object of mass m2 = 3.00 kg just as the 5.00 kg objects leaves hits the table and (b) find the maximum height above the table to which the 3.00 kg object rises

- Relevant Equations
- Change in mechanical energy of the system = 0

I have some conceptual questions about this task. In order to get the correct result (I checked the textbook answer) in part (a) I had to assume that the speed for each block is the same at all instants. And that if one block moves down x meters, the other one will move up that same amount of meters. Is this always the case for objects connected through a a light frictionless pulley?

In part (b), when analyzing the maximum height to which the 3kg objects rises, I had to use the following equation to get the correct answer:

m2 * g* 4 m + 0.5 * m2 * (4.42 m/s)^2 = m2 * g* Hmax

(4.42 m/s is the speed obtained when solving part (a) )

The initial situtation we're analyzing is when m1 hits the table. At that point m2 is at h= 4 m. The final situation is the one in which m2 is at the maximum height, that is, its velocity is zero. Now, my question is how can m2 go further up AFTER m1 has touched the table? Shouldn't it stay at 4 meters? Once the block touches the table, it can't keep going down, so it should stop moving. If one block is moving upwards while the other one is at rest, it means their speeds are different. Doesn't that contradict the idea that both blocks have the same speed and acceleration(in magnitude), and that they move the same amount of meters in opposite directions? m1 would already be at h = 0 when m2 is at h =4 meters, and it seems like it can't have a negative height because there's a table stopping it from moving. So how does one object go from h = 4 m to h= 5 m while the other one doesn't move down one meter? Does this imply that once m1 hits the table, it is no longer a part of the pulley system, so those "rules" no longer apply?

In part (b), when analyzing the maximum height to which the 3kg objects rises, I had to use the following equation to get the correct answer:

m2 * g* 4 m + 0.5 * m2 * (4.42 m/s)^2 = m2 * g* Hmax

(4.42 m/s is the speed obtained when solving part (a) )

The initial situtation we're analyzing is when m1 hits the table. At that point m2 is at h= 4 m. The final situation is the one in which m2 is at the maximum height, that is, its velocity is zero. Now, my question is how can m2 go further up AFTER m1 has touched the table? Shouldn't it stay at 4 meters? Once the block touches the table, it can't keep going down, so it should stop moving. If one block is moving upwards while the other one is at rest, it means their speeds are different. Doesn't that contradict the idea that both blocks have the same speed and acceleration(in magnitude), and that they move the same amount of meters in opposite directions? m1 would already be at h = 0 when m2 is at h =4 meters, and it seems like it can't have a negative height because there's a table stopping it from moving. So how does one object go from h = 4 m to h= 5 m while the other one doesn't move down one meter? Does this imply that once m1 hits the table, it is no longer a part of the pulley system, so those "rules" no longer apply?