A 204-kg projectile, fired with a speed of 130 m/s at a 65.0 degree angle, break

Question

A 204-kg projectile, fired with a speed of 130 m/s at a 65.0 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

M = 204-kg, V = 130 m/s, = 65.0 degrees,

mass of each piece = m = M/3 = 68 kg, M = 3m

velocity before explosion V' = Vcos = 54.9 m/s

after explosion, v1 vertically downward, v2 horizontally, in magnitude v1 = v2 = V', the 3rd velocity = v

horizontal momentum conservation: MV' = m*v2 + m*vx

3m*V' = m*(V' + vx), 3V' = V' + vx, so vx = 2V'

vertical momentum conservation: 0 = -m*v1 + m*vy

0 = m*(-V' + vy), so vy = V'

the magnitude of the velocity of the third fragment immediately after the explosion is v = (vx2 + vy2) = V'*5 = 123 m/s

the direction of the velocity of the third fragment immediately after the explosion is

tan-1(vy/vx) = tan-1(1/2) = 26.7 degrees with horizontal

the energy released in the explosion is

(mV'2/2 + mV'2/2 + mv2/2) - MV'2/2 = m(v2 - V'2)/2 = 2mV'2 = 4.10*105 J

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