A spaceship of mass 2.1*106 kg is cruising at a speed of 5.8*106 m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 4.6*105 kg, is blown straight backward with a speed of 2.5*106 m/s. A second piece, with mass 7.9*105 kg, continues forward at 1.3*106 m/s.

Part A

What is the speed of the third piece?

Express your answer using two significant figures. m/s

Part B

What is the directionof the third piece?

Straight backward

Given a space ship of mass

M = 2.1 * 10^6 kg

The space ship is moving at

V = 5.8 * 10^6 m/s

This space ship blows into three

The first part is blown backward I. e. in the negative direction

Then, mass of first part and it's velocity is

M1 = 4.6 * 10^5 kg

V1 = - 2.5 * 10^6 m/s

The second part continues in forward motion, I. e positive direction

Then, mass of the second part and it's velocity is

M2 = 7.9 * 10^5 kg

V2 = 1.3 * 10^6 m/s

A. What is the speed of the third piece?

So, we need to know the mass of the third part

M1 + M2 + M3 = M

M3 = M-M1-M2

M3 = 2.1 * 10^6 - 4.6 * 10^5 - 7.9 * 10^5

M3 = 8.5 * 10^5 kg

So, applying conservation of momentum

Momentum before explosion = momentum after explosion

MV = M1•V1 + M2•V2 + M3•V3

2.1 * 10^6 * 5.8 * 10^6 = 4.6 * 10^5 * - 2.5 * 10^6 + 7.9 * 10^5 * 1.3 * 10^6 + 8.5*10^5•V3

1.218 * 10¹³ = - 1.15 * 10¹² + 1.027 * 10¹² + 8.5 * 10^5•V3

1.218 * 10¹³ + 1.15 * 10¹² - 1.027 * 10¹² = 8.5 * 10^5•V3

1.2303 * 10^13 = 8.5 * 10^5 V3

V3 = 1.2303 * 10¹³ / 8.5 * 10^5

V3 = 1.45 * 10^7 m/s

B. The direction is straight forward since the velocity of the part is positive