Question

A 6.9 kg model rocket is traveling horizontally and due south with a speed of 20.2 m/s when it explodes into two pieces. The velocity of one piece, with a mass of 2.4 kg, is v1 = (-12 m/s)i + (30 m/s)j + (15 m/s)k, with i pointing due east, j pointing due north, and k pointing vertically upward.

(a) What is the velocity of the other piece, in unit-vector notation? Incorrect: Your answer is incorrect. m/s i m/s j m/s k

(b) What is the kinetic energy of the other piece? J (c) How much kinetic energy is produced by the explosion? J

Answer 1

here,

mass of rocket ,m = 6.9 kg

mass of first peice , m1 = 2.4 kg

mass of seccond peice , m2 = 4.5 kg

speed of 1 , v1 = (- 12 i + 30 j + 15 k ) m/s

a)

let the velocity of other peice be v2

using conservation of momentum

m * v = m1 * v1 + m2 * v2

6.9 * (- 20.2 j) = 2.4 * ( - 12 i + 30 j + 15 k ) + 4.5 * ( v2)

solving for v2

v2 = ( 6.4 i - 47 j - 8 k) m/s

b)

the kinetic energy of other peice , KE2 = 0.5 * m2 * ( 6.4^2 + 47^2 + 8^2) J

KE2 = 0.5 * 4.5 * ( 6.4^2 + 47^2 + 8^2) J

KE2 = 5206.4 J

c)

kinetic energy of first peice , KE1 = 0.5 * m1 * ( 12^2 + 30^2 + 15^2)

KE1 = 0.5 * 2.4 * (12^2 + 30^2 + 15^2) J

KE1 = 1522.8 J

initial kinetic energy , KE = 0.5 * m * v^2

KE = 0.5 * 6.9 * 20.2^2 J

KE = 1407.7 J

the kinetic energy produced , dKE = KE2 + KE1 - KE

dKE = 5321.5 J