Two blocks are free to slide along the frictionless wooden track shown below. Th

Question

Two blocks are free to slide along the frictionless wooden track shown below. The block of mass m1 = 5.10 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 the block of mass m2 = 11.0 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

Explanation / Answer

Loss in potential Energy = Gain in Kientic Energy

Therefore

mgh = 0.5*m*v^2

v = sqrt(2gh)

= sqrt(2*9.8*5)

= 9.899 m/sec

Momentum remains Conserved

Therfore

5.10*9.899 = 5.10*v1 + 11*v2

Also Kinetic Energy remains Conserved

Thefore

0.5*5.10*9.899^2 = 0.5*5.10*v1^2 + 0.5*11*v2^2

Therefore we get

v1 = - 3.6275 m/sec , v2 = 6.27 m/sec

Therefore

h = v^2/2g

= (3.6275^2)/(2*9.8)

= 0.671 m

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