Please help! A spaceship of mass 2.3 Times 10^6 kg is cruising at a speed of 4.2

Question

Please help!

A spaceship of mass 2.3 Times 10^6 kg is cruising at a speed of 4.2 Times 10^6 m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 5.1 Times 10^5 kg , is blown straight backward with a speed of 2.0 Times 10^5 m/s . A second piece, with mass 7.7 Times 10^5 kg , continues forward at 1.0 Times 10^6 m/s. The movement direction of the third piece is What is the speed of the third piece? Assume that the initial speed of the ship is positive. Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Solution:

Initial mass of the spaceship M = 2.3*106 kg

Initial velocity of the spaceship V = 4.2*106 m/s

We assume that thespaceship initial moving in the positive X direction. Thus the initial velocity in unit vector notation is given by,

V = (4.2*106 m/s) i

Mass of the first piece, m1 = 5.1*105 kg

Since this piece is blown straight backwards, its velocity is,

v1 = (-2.0*106 m/s) i

Mass of the second piece, m2 =7.7*105 kg

Since this piece is blown forward, its velocity is,

v2 = (1.0*106 m/s) i

The mass of the third piece is,

m3 = M - m1 - m2

m3 = (2.3*106 kg) – (5.1*105 kg) – (7.7*105 kg)

m3 = 1.02*106 kg

And we have to find its velocity v3

Since the explosion has occurred internally, no external forces are acting and thus the linear momentum is conserved.

M*V = m1 *v1 + m2* v2 + m3 *v3

m3 *v3 = M*V - m1 *v1 - m2* v2

v3 = [M*V - m1 *v1 - m2* v2]/ m3

v3 = [(2.3*106 kg)* (4.2*106 m/s) i – (5.1*105 kg)* (-2.0*106 m/s) i – (7.7*105 kg)* (1.0*106 m/s) i] / [1.02*106 kg]

v3 = [(9.91*1012 kg.m/s) i ]/ [1.02*106 kg]

v3 = (9.7157*106 m/s) i

With two significant figures, the speed of the third piece is in the direction of original spaceship,

v3 = (9.7*106 m/s) i

Hence the answer is 9.7*106 m/s

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