1) A 206kg projectile, fired with a speed of 148m/s at a 67.0 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.
A) Determine the magnitude of the velocity of the third fragment immediately after the explosion. ANSWER IN M/S
B) Determine the direction of the velocity of the third fragment immediately after the explosion. PLEASE ANSWER DEGREE ABOVE THE HORIZONTAL.
C) Determine the energy released in the explosion. ANSWER IN J.
CwCc2012-12-03T09:57:24Z
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First find the velocity of the projectile at the top: v = v0 cos(67) = 57.8 m/s
Use the conservation of momentum: mv i = (m/3) v i - (m/3) v j + (m/3) v3 v3 = 2v i + v j |v3| = sqrt((2v)^2 + v^2) = sqrt(5) v = 2.24 * 57.8 m/s = 129 m/s dir(v3) = atan(v/2v) = 26.6 degrees, above horizontal
To find the energy of the explosion, calculate the difference between the final and initial energies: E = 1/2 (m/3) v^2 + 1/2 (m/3) v^2 + 1/2 (m/3) (sqrt(5) v)^2 - 1/2 mv^2 = (2/3) mv^2 = (2/3) (206 kg) (57.8 m/s)^2 = 459 kJ
1) it will have to equal the magnitude of the others in the other way so like 3 fragments 1 explodes -y direction 1 explodes x direction so 4th quadrant
so the magnitude of the other fragment will be positive y, negative x direction
so it will be x^2 + y^2 =r^2 third fragment explodes at exactly 135 degrees (90+45, since sides are even) and then the magnitude is 209.3036 m/s
A)209 m/s B)135 degrees above horizontal... or 45 north of west
C) 1/2mv^2 is kinetics energy so 2*K + 1times a different K mv1^2 + 1/2mv2^2 206*148^2 + 1/2*206*209^2 K = 9,011,367 J