A 243 kg projectile, fired with a speed of 145 m/s at a 51.0 degree angle, break

Question

A 243 kg projectile, fired with a speed of 145 m/s at a 51.0 degree angle, breaks into three pieces of equal mass at the highest point of its arc (where its velocity is horizontal). Two of the fragments move with the same speed right after the explosion as the entire projectile had just before the explosion; one of these moves vertically downward and the other horizontally.
Determine the magnitude of the velocity of the third fragment immediately after the explosion.


Determine the direction of the velocity of the third fragment immediately after the explosion.

Determine the energy released in the explosion

Explanation / Answer

at highest point velocity = 145cos(51) = 91.25 initial momentum = 243*91.25 i = 22174.1 i mass of each piece = 243/3 = 81 final momentum = 81*91.25 i - 81*91.25 j + P conservation of momentum 22174.1 i = 81*91.25 i - 81*91.25 j + P 14782.85 i + 7391.25 j = P v3 = P/81 =182.5 i + 91.25 j magnitude = 204.045 m/s direction = tan-1(91.25/182.5) = 26.565 with horizontal initial kinetic energy = .5*243*91.2562 i am leaving these final calculations to you you know final velocities you can get total final KE of three pieces so change in kinetic energy = energy released in the explosion I hope this is clear:)

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