Question

A body in flight explodes into three equal portions. One portion continues along the original line of flight. The other two go off in directions each inclined at 60° to the original path. The energy released in the explosion is twice as great as the kinetic energy possessed by the body at the time of explosion. Determine the kinetic energy of each fragment immediately after the explosion. Is energy conserved? Why or why not?

Answer 1

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let yo = speed of body before collision, ht Using Momentum conservation, Bytvo = verbo travestó tolv, 30o = vtv - 0 to of (3m) v = mv ² + mv 7 mv ² - 2 13 m2 energy released during explo8 ал. » 4.5 v6 = vt ? fm 0(0, 4.500 = (3x v 1 + V 2 4.54² = 94 ²+V² - 6 vov, to.5v² -4.5V + 6 VV, tegy 51-542

I Dale Page 1.5V, 2 - 6 vo Vy + 4.56 on solving ve ovo tj 36634x1-584562 - 24105 V = 6vot Jove W = 6 Votavo svo - 3 also, vi = 600 - 300 = 1 and, v = 3 -V, = 300 - 3vo 70 (This is possible ce possible relocity of body is V=vo) V =2vo

Ke of two particles which is deflected at an angle Go is - Imv² = 1 m (200) direct K.E. of particle which moves in the of projectile = 1 mv, - I m x V = n No kinetic energy is not conserved.