Question

A spaceship of mass 2.30 x 10^6 kg is cruising at a speed of 6.00 x 10^6 m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 5.00 x 10^5 kg , is blown straight backward with a speed of 1.80 x 10^6 m/s . A second piece, with mass 8.40 x 10^5 kg , continues forward at 8.00 x 10^5 m/s .

Part A) What is the speed of the third piece? Express your answer with the appropriate units.

Answer 1

Answer:
`v_3=14.61* 10^6 m/s`

Step-by-step explanation:

Given

mass of spaceship
`M=2.30* 10^6 kg`

speed of spaceship
`u=6* 10^6 m/s`

After blowing in to three pieces

`m_1=5* 10^5 kg`

`v_1=-1.80* 10^6 m/s`

`m_2=8.40* 10^5 kg`

`v_2=8* 10^5 m/s`

`m_3=M-m_1-m_2`

`m_3=2.30* 10^6-5* 10^5-8.40* 10^5`

`m_3=9.6* 10^5 kg`

consider forward motion to be Positive

Let
`v_3` be the velocity of third particle

conserving momentum as no external Force is applied

`Mu=m_1v_1+m_2v_2+m_3v_3`

`2.30* 10^6* 6* 10^6=5* 10^5\cdot (-1.80* 10^6)+8.40* 10^5* 8* 10^5+9.6* 10^5\cdot v_3`

`13.8* 10^(12)=-0.9* 10^(12)+0.672* 10^(12)+9.6* 10^5\cdot v_3`

`v_3=(14.028* 10^(12))/(9.6* 10^5)`

`v_3=14.61* 10^6 m/s`