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 = 4.90 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_2 = 10.9 kg, initially at rest. The two blocks never touch, but we can treat this as a perfectly elastic collision. Calculate the maximum height to which m_1 rises after the elastic collision.

Explanation / Answer

calculate the maximum height to which m1 rises after the elastic collision

this is elastic collision. So that K.E and momentum are conserved.

The relative speed of the two bodies has the same magnitude and the opposite direction

V1i =(2gh)= (2x9.81x5) = 9.90m/s, V2i = 0

V'i = V1i-V2i = 9.90m/s

V'f =-V'i

VCM =V1i (m1/ (m1+m2)) = 9.90 (4.9/(10.9+4.9))

VCM= 3.07m/s

V1f = VCM+V'f(m1/ (m1+m2))= 3.07+(-9.90 (10.9/4.9+10.9))

V1f = -3.76m/s

h= V1f2/(2g) =0.721m

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