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 = 4.96 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 = 9.40 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

Two blocks are free to slide along the frictionless wooden track shown below. The block of mass m1 = 4.96 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 = 9.40 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.

Explanation / Answer

Here

m1gh = 0.5*m1*v^2

Therefore

v = sqrt(2gh)

= sqrt(2*9.8*5)

= 9.899 m/sec

Here Momentum will remain Conserved

Therefore

4.96*9.899 = 4.96*v1 + 9.40*v2 -------------------(1)

Also

Here Kinetic Energy will remain Conserved

Therefore

0.5*4.96*9.899^2 = 0.5*4.96*v1^2 + 0.5*9.40*v2^2 -------------------(2)

Therefore from rquation 1 and 2 , we get

v1 = Final Ve;locity of m1 = - 3.06 m/sec

v2 = Final Veocity of m2 = 6.838 m/sec

So Final height = v1^2/2g

= (3.06^2)/(2*9.8)

= 0.477 m

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