Q 80

Expert-verifiedFound in: Page 291

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

A spaceship of mass 2.0 10^{6} kg is cruising at a speed of 5.0 10^{6} m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 5.0 10^{5} kg, is blown straight backward with a speed of 2.0 10^{6} m/s. A second piece, with mass 8.0 10^{5} kg, continues forward at 1.0 10^{6} m/s. What are the direction and speed of the third piece?

Velocity of the third piece is 14.610^{6} ms^{-1} in forward direction.

We need to find the direction and speed of the third piece .

The spaceship initially has a mass M = 2 x 10^{6} kg. After the explosion, the spaceship splits into three pieces,

Using conservation of momentum,

M = m_{1} + m_{2} + m_{3}

m_{3} = M - m_{1 }- m_{2}

m_{3 }= (210^{6} kg) - (510^{5} kg) - (810^{5} kg)

m_{3} = 710^{5 }kg

= 0.710^{6} kg

As per law of conservation of momentum

And momentum is expressed as

P = mv

Therefore,

M_{s}v_{s} = m_{1 }(-v_{1}) + m_{2}v_{2} + m_{3}v_{3}

v_{1 }is taken as negative because it is moving in backward direction.

Putting values,

v_{3} =

As the value of v_{3} is positive, it will move in forward direction.

94% of StudySmarter users get better grades.

Sign up for free