A 6.9 kg model rocket is traveling horizontally and due south with a speed of 20
ID: 1582093 • Letter: A
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
Explanation / Answer
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