A 44.0 kg body is moving in the direction of the positive x axis with a speed of

Question

A 44.0 kg body is moving in the direction of the positive x axis with a speed of 308 m/s when, owing to an internal explosion, it breaks into three pieces. One part, whose mass is 8.0 kg moves away from the point of explosion with a speed of 350 m/s along the positive y axis. A second fragment, whose mass is 4.5 kg moves away from the point of explosion with a speed of 300 m/s along the negative x axis.

What is the speed of the third fragment? Ignore effects due to gravity.

How much energy was released in the explosion?

Explanation / Answer

The initial kinetic energy of the mass is:

KEi = 1/2 * m * v² = 0.5 * 44 * 308² = 2087008Joules

m1 = 8 kg
Module V1 = 350 m/s (on the Y axis)
components V1 are:
V1x = 0
V1y = 350

m2 = 4.5 kg
Module V2 = - 300 m/s (on the X axis)
components V2 are:
V2x = - 300
V2y = 0


The m3 mass is:
m3 = m - (m1 + m2) = 44 - 8 - 4.5 = 31.5 kg

The vectorial sum of the speeds of the 3 parts must be zero:
V1(vector) + V2(vector) + V3(vector) = 0
velocity vector components of m3 are:
a) V3x = - (V1x + V2x) = 0 - (- 300) = 300 m/s
b) V3y = - (V1y + V2y) = - 350 - 0 = - 350 m/s

its module is (Pythagorean Theorem):
V3(mod) = (V3x² + V3y²) = (300² + 350²) = 460.9772229 m/s

The final total kinetic energy of the 3 masses is:
KEf = 1/2 * m1 * v1² + 1/2 * m2 * v2² + 1/2 * m3 * v3²
KEf = 0.5 * 8 * 350² + 0,5 * 4.5 * 300² + 0.5 * 31.5 * 460.9772229² = 4039375.001 Joules

The energy released in the explosion is:
E(expl) = KEf - KEi = 4039375.001 - 2087008 = 1952367.001 Joules

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