p=mv=2.3•10⁶•6•10⁶=13.8•10¹² kg•m/s.
p₁= m₁v₁= 5.2•10⁵•2•10⁶=10.4•10¹¹kg•m/s.
p₂=m₂v₂=8.4•10⁵•9•10 ⁵=75.6•10¹ºkg•m/s.
p₃=m₃v₃.
m₃=m-m₁-m₂=2.3•10⁶ -5.2•10⁵-8.4•10⁵=(23-5.2-8.4)•10⁵=9.4•10⁵ kg
Law of conservation of linear momentum
p=- p₁+p₂+p₃
p₃=p- p₁+p₂.
v₃=(p+ p₁-p₂)/m₃=
=( 13.8•10¹²+10.4•10¹¹-75.6•10¹º)/ 9.4•10⁵=
=(1380+104-75.6) 10¹º/9.4•10⁵=149.8•10⁵=
=1.498•10⁷ m/s
A spaceship of mass 2.3×10^6 kg is cruising at a speed of 6.0×10^6 m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 5.2×10^5 kg, is blown straight backward with a speed of 2.0×10^6 m/s. A second piece, with mass 8.4×10^5 kg, continues forward at 9.0×10^5 m/s. What is the speed of the third piece?
1 answer