Hello, PhysicsForum,
I haven't posted a physics question, mainly because I don't have mechanics classes anymore :(
But, studying on my own, I found this problem on the web with the solution:
«61. A 2.0-kg box is attached by a string to a 5.0-kg box. A compressed spring is placed between them. The two boxes are initially at rest on a friction-free track. The string is cut and the spring applies an impulse to both boxes, setting them in motion. The 2.0-kg box is propelled backwards and moves 1.2 meters to the end of the track in 0.50 seconds. Determine the time it takes the 5.0-kg box to move 0.90 meters to the opposite end of the track.
Answer: 0.94 s
For the sake of the discussion, the 2-kg box will be referred to as Box 1 and the 5-kg box will be referred to as box 2.
Given: mbox 1 = 2.0 kg; mbox 2 = 5.0 kg; dbox 1 = 1.2 m; tbox 1 = 0.50 s; dbox 2 = 0.90 m
Find: tbox 2 = ???
The two boxes are initially at rest. The total system momentum is initially 0. After the cutting of the string and the impulse of the spring, the total system momentum must also be 0. Thus, Box 1's backward momentum must be equal to the Box 2's forward momentum. The distance and time for Box 1 must be used to determine its velocity.
v = d/t = (1.2 m) / (0.5 s) = 2.4 m/s
Now the principle of momentum conservation can be used to determine Box 2's velocity.
mbox 1 • vbox 1 = mbox 2 • vbox 2
(2 kg) • (2.4 m/s) = (5 kg) • vbox 2
vbox 2 = (2 kg) • (2.4 m/s) / (5 kg) = 0.96 m/s
The velocity of Box 2 can be used to determine the time it takes it to move a distance of 0.90 meters.
vbox 2 = dbox 2 / time
Time = dbox 2 / vbox 2 = (0.90 m) / (0.96 m/s) = 0.9375 s = ~0.94 s »
http://www.physicsclassroom.com/revi...um/momans4.cfm
Now, what I don't understand from this whole process, is why we consider the speed of the blocks to be uniform. I think that the string, not being specified, could have been long and could therefore provide an acceleration throughout the movement of the blocks. This would make the calculation of the velocity erroneous. Is this correct?
Also, I want to know how the spring separates energy. Let's say the compression of the spring gathered 1000 J of energy. Would each block get 500J? How does the separation of the energy work?
I haven't posted a physics question, mainly because I don't have mechanics classes anymore :(
But, studying on my own, I found this problem on the web with the solution:
«61. A 2.0-kg box is attached by a string to a 5.0-kg box. A compressed spring is placed between them. The two boxes are initially at rest on a friction-free track. The string is cut and the spring applies an impulse to both boxes, setting them in motion. The 2.0-kg box is propelled backwards and moves 1.2 meters to the end of the track in 0.50 seconds. Determine the time it takes the 5.0-kg box to move 0.90 meters to the opposite end of the track.
Answer: 0.94 s
For the sake of the discussion, the 2-kg box will be referred to as Box 1 and the 5-kg box will be referred to as box 2.
Given: mbox 1 = 2.0 kg; mbox 2 = 5.0 kg; dbox 1 = 1.2 m; tbox 1 = 0.50 s; dbox 2 = 0.90 m
Find: tbox 2 = ???
The two boxes are initially at rest. The total system momentum is initially 0. After the cutting of the string and the impulse of the spring, the total system momentum must also be 0. Thus, Box 1's backward momentum must be equal to the Box 2's forward momentum. The distance and time for Box 1 must be used to determine its velocity.
v = d/t = (1.2 m) / (0.5 s) = 2.4 m/s
Now the principle of momentum conservation can be used to determine Box 2's velocity.
mbox 1 • vbox 1 = mbox 2 • vbox 2
(2 kg) • (2.4 m/s) = (5 kg) • vbox 2
vbox 2 = (2 kg) • (2.4 m/s) / (5 kg) = 0.96 m/s
The velocity of Box 2 can be used to determine the time it takes it to move a distance of 0.90 meters.
vbox 2 = dbox 2 / time
Time = dbox 2 / vbox 2 = (0.90 m) / (0.96 m/s) = 0.9375 s = ~0.94 s »
http://www.physicsclassroom.com/revi...um/momans4.cfm
Now, what I don't understand from this whole process, is why we consider the speed of the blocks to be uniform. I think that the string, not being specified, could have been long and could therefore provide an acceleration throughout the movement of the blocks. This would make the calculation of the velocity erroneous. Is this correct?
Also, I want to know how the spring separates energy. Let's say the compression of the spring gathered 1000 J of energy. Would each block get 500J? How does the separation of the energy work?
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