Two blocks are free to slide along the frictionless, wooden track shown in the figure. The block of mass m¡ = 5.00 kg is released from the position shown, at height h = 5.00 m above the flat part of the track. Protruding from its front end is the north pole of a strong magnet, which repels the north pole of an identical magnet embedded in the back end of a block of mass m2 = 10.0 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m; rises after the elastic collision. Ki + Ui = Kg+ m,v," + 0 = 0+ %3D %3D V,= JZ(9.80)(5.0
Two blocks are free to slide along the frictionless, wooden track shown in the figure. The block of mass m¡ = 5.00 kg is released from the position shown, at height h = 5.00 m above the flat part of the track. Protruding from its front end is the north pole of a strong magnet, which repels the north pole of an identical magnet embedded in the back end of a block of mass m2 = 10.0 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m; rises after the elastic collision. Ki + Ui = Kg+ m,v," + 0 = 0+ %3D %3D V,= JZ(9.80)(5.0
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:Two blocks are free to slide along the firictionless, wooden track shown in
the figure. The block of mass m = 5.00 kg is released from the position
shown, at height h = 5.00 m above the flat part of the track. Protruding from
its front end is the north pole of a strong magnet, which repels the north pole
of an identical magnet embedded in the back end of a block of mass m2 =
10.0 kg, initially at rest. The two blocks never touch. Calculate the maximum
height to which m¡ rises after the elastic collision.
K; + Ui = Ks+U;
I m,v,"+ 0 = 0+ m,gh
%3D
V,= SZ(1.80)(5.00)
V,= 1.90 mls
m, - M2
V, =
Mit M2
-(9.10m/s)= - 3. 30 mls
m.ghmax =Ź mi(-3.30)" → hamere
(-330mls)?
homax
219.80 ml.
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