(3) two blocks are free to slide along the frictionless wooden track. a block of mass m1=5.00kg is released from A. Protruding from its front end is the north pole of a strong magnet, repelling the north pole of an identical magnet embedded in the back end of the block of mass m2=10.0kg, initially at rest. the two blocks never touch. Calculate the maximum height to which m1 rises after the elastic collision.(adsbygoogle = window.adsbygoogle || []).push({});

Do you use conservation of kinetic energy and momentum to figure out the final velocity of the block M1 after the collision. Then let all that kinetic energy be converted to gravitational potential energy to determine how high it rises???

So GPE = KE , mgh= 1/2mv^2

if (5)(9.81)(5) = 1/2 (10)v^2

u find initial velocity?...

do u sub that into

vfinal = (m1-m2)/(m1 + m2) *vinitial?

to find vfinal...?

and then sub that back into...

mgh = 1/2 mv^2

to solve for h=1/2mv^2/mg?

OR iS THis completely wrong? somebody let me know if im on the right track.. or where i veered off the track...

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# Physics -momentum/collision

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