Question

A spaceship of mass 2.0*106 kg is cruising at a speed of 5.0*106 m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 5.0*105 kg, is blown straight backward with a speed of 2.0*106 m/s. A sec-ond piece, with mass 8.0*105 kg, continues forward at 1.0*106 m/s. What are the direction and speed of the third piece?

Answer 1

Answer: v3 = 1.6*10^7 m/s , and the direction is the same as the initial direction of movement of the spaceship

Explanation:

The initial mass is M = 2.0*10^6kg

the velocity is V = 5.0*10^6m/s

then, the initial moment is P = M*V = 1.0*10^13m*kg/s

now, we must have that the momentum before and afther must be equal, after we have:

one part with:

m1 = 5.0*10^5kg, v1 = -2.0*10^6m/s

then p1 = -1.0*10^12kh*m/s

other part has:

m2 = 8.0*10^5kg, v2 = 1.0*10^6m/s

p2 = 8*10^11kg*m/s

then, the third object must have a momentum:

p3 = P - p1 - p2 = 10.0*10^12 - 0.8*10^12 + 2.0*10^12 = 11.2*10^12

then m3*v3 = 11.2*10^12 kg*m/s so from this we know that the velocity is in the same direction than the initial velocity.

now, we can find m3 by:

M = m1 + m2 + m3

m3 = M - m1 - m2 = 2.0*10^6kg - 0.5*10^6kg - 0.8*10^6kg = 0.7*10^6kg

then v3 = (11.2*10^12)/(07*10^6) m/s = (11.2/0.7)*10^(12 - 6) m/s = 16.0*10^6 m/s

v3 = 1.6*10^7 m/s

Answer 2

Answer:

v3 = 1.6*10^7 m/s

and the directio is straight forward

Explanation:

Given that,

spaceship of mass  M = 2.0*10^6kg

the velocity is V = 5.0*10^6m/s

one section:

mass of m1 = 5.0*10^5kg,

speed  v1 = -2.0*10^6m/s

sec-ond

mass, m2 = 8.0*10^5kg,

speed, v2 = 1.0*10^6m/s

use conservation of momentum

Initial momentum = final momentum

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

(2*10^6)*(5*10^6) = (5.0*10^5)*(-2*10^6) + (8.0*10^5)*(1*10^6) + (2*10^6 -5.0*10^5 -8.0*10^5 )*v3

(10*10^12) = -10*10^11 + 9.0*10^11 + (7.0*10^5 )*v3

v3 = 1.6*10^7 m/s