The drawing shows two frictionless inclines that begin atground level ( h = 0 m)

Question

The drawing shows two frictionless inclines that begin atground level (h = 0 m) and slope upward at the same angle. One track is longer than the other, however.Identical blocks are projected up each track with the same initialspeed v0. One the longer track the block slidesupward until it reaches a maximum height H above theground. On the shorter track the block slides upward, flies off theend of the track at a height H1 above theground, and then follows the familiar parabolic trajectory ofprojectile motion. At the highest point of this trajectory, theblock is a height H2 above the end of thetrack. The initial total mechanical energy of each block is thesame and is all kinetic energy. The initial speed of each block isv0 = 6.58m/s, and each incline slopes upward at an angle of = 52.5°. The block on the shorter trackleaves the track at a height of H1 =1.30 m above the ground. Find(a) the height H for the block on the longer track and (b)the total height H1 + H2for the block on the shorter track.

Explanation / Answer

we can conserve energy as the incline is friction less a)from energy coservation final energy=mgh =initial energy =0.5mv*v hence h=v2/2g=6.582/19.6=2.209m b)here the final kinetic energy is not 0 let the velocity at the point where it leaves the contact is u then the horizontal velocity is ucos final kinetic energy =0.5m*u2cos2 by applying the conservation of energy between the point of leavingand ground we have 0.5m*u2+mgh1=0.5mv2 we get u now again applying this principle 0.5m*u2cos2+mg(h1+h2)=o.5mv2 plug in the values

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