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 m_1 = 4.92 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 m_z = 9.40 kg initially at rest. The two blocks never touch. Calculate the maximum height to which m_1 rises after the elastic collision.

Explanation / Answer

speed of m1 when it hits m2

v = sqrt(2gh) = sqrt(2*9.8*5)

= 9.9 m/s

let the speed after collision be v1 and v2

by the conservation of momentum we have

4.92*9.9 + 0 = 4.92*v1 + 9.4v2 -------(1)

velocity of approach = velocity of recess

v2 - v1 = u1 - u2

v2 - v1 = 9.9 ---------(2)

multiply (2) by 4.92 and adding we have

14.32 v2 = 97.416

v2 = 6.8 m/s

v1 = 6.8-9.9 = -3.097 m

so the height reached by m1 after collision = v1^2 /2g

= (3.097^2) / 2*9.8 = 0.49 m

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